Introduction Main Theorem Proof Corollaries
Jones polynomial and cosmetic surgery
Kazuhiro Ichihara
(Nihon Univ., College of Humanities & Sciences)
Joint work with
Zhongtao Wu
(Chinese University of Hong Kong) KKOOK 2016
Osaka Electro-Comm. Univ., Aug. 22, 2016.
Introduction Main Theorem Proof Corollaries
Dehn surgery
E(K): the exterior of a
knot K in the 3-sphere S
3(i.e., S
3−(open tubular neighborhood of K))
Dehn surgery:
Gluing a solid torus to E(K)
along a slope γ ;
γ mf
We denote the obtained mfd. by
K(r)when [γ] = r ∈
Q. Recall:
For a knot K in S
3, by using a
meridian-longitudesystem, { slope on ∂N (K) } ←→
1:1
p q
∪
1
0
Introduction Main Theorem Proof Corollaries
Cosmetic surgery
Cosmetic Surgery Conjecture [Kirby’s list Problem 1.81(A)]
Dehn surgeries on inequivalent slopes are never purely cosmetic.
•
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
2are purely cosmetic if there exists an
ori.-pres.homeo. between K(r
1) & K(r
2).
Remark: (Mathieu, 1992)
There exist examples for “orientation reversing” case.
In fact, (18k + 9)/(3k + 1)- and (18k + 9)/(3k + 2)-surgeries
on the trefoil yield ori.-reversingly homeomorphic pairs for k ≥ 0.
Introduction Main Theorem Proof Corollaries
Cosmetic surgery
Cosmetic Surgery Conjecture [Kirby’s list Problem 1.81(A)]
Dehn surgeries on inequivalent slopes are never purely cosmetic.
•
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
2are purely cosmetic if there exists an
ori.-pres.homeo. between K(r
1) & K(r
2).
Remark: (Mathieu, 1992)
There exist examples for “orientation reversing” case.
In fact, (18k + 9)/(3k + 1)- and (18k + 9)/(3k + 2)-surgeries
on the trefoil yield ori.-reversingly homeomorphic pairs for k ≥ 0.
Introduction Main Theorem Proof Corollaries
Main Theorem
Our main result gives a severe restriction for a knot in S
3to admit purely cosmetic surgery in terms of its Jones polynomial.
Main Theorem
Let V
K(t) be the Jones polynomial of a knot K in S
3. If a knot K satisfies either V
K00(1) 6= 0 or V
K000(1) 6= 0 ,
then K(r) K(r
0) as oriented mfds. for distinct slopes r and r
0.
Remark
Boyer and Lines obtained a similar result for a knot K with ∆
00K(1) 6= 0 by using the Casson invariant.
(∆
K(t): the normalized Alexander polynomial) Since V
K00(1) = −3∆
00K(1), our result can be viewed as
an extension of [Proposition 5.1, Boyer-Lines (1990)].
Introduction Main Theorem Proof Corollaries
Main Theorem
Our main result gives a severe restriction for a knot in S
3to admit purely cosmetic surgery in terms of its Jones polynomial.
Main Theorem
Let V
K(t) be the Jones polynomial of a knot K in S
3. If a knot K satisfies either V
K00(1) 6= 0 or V
K000(1) 6= 0 ,
then K(r) K(r
0) as oriented mfds. for distinct slopes r and r
0. Remark
Boyer and Lines obtained a similar result for a knot K with ∆
00K(1) 6= 0 by using the Casson invariant.
(∆
K(t): the normalized Alexander polynomial) Since V
K00(1) = −3∆
00K(1), our result can be viewed as
an extension of [Proposition 5.1, Boyer-Lines (1990)].
Introduction Main Theorem Proof Corollaries
Lescop’s λ
2invariant
The essential new ingredient for our result is the following.
Lescop’s λ
2invariant (2009)
The invariant
λ2:= W
2◦ Z
2, where W
2is a linear form on A
nwith W
2( ) = 1 and W
2( ) = 0.
A
n: the vector space generated by degree n Jacobi diagrams subject to AS and IHX relations
Zn
: the degree n part of the
Kontsevich-Kuperberg-Thurston invariantof rational homology spheres taking its value in A
nIntroduction Main Theorem Proof Corollaries
From λ
2to w
3Fact [Theorem 7.1, Lescop (2009)]
The invariant λ
2satisfies the surgery formula
λ2(K(pq)) = (q
p)2λ002(K) + (q
p)w3(K) +c(q
p)a2(K) +λ2(L(p, q))
for all knots K ⊂ S
3.
a
2(K): the z
2-coefficient of the Conway polynomial ∇
K(z) L(p, q): the lens space (obtained by p/q surgery on the unknot) λ
002(K) & c(
qp): explicit constants defined in [Lescop, 2009]
w3(K)
is a knot invariant, which is shown as follows.
Introduction Main Theorem Proof Corollaries
The invariant w
3Lemma
For all knots K ⊂ S
3, w
3(K) = 1
72 V
K000(1) + 1
24 V
K00(1).
This can be shown in the same line as [Prop. 4.2, Nikkuni (2005)]
by using the skein relation for w
3given by Lescop.
Remark:
For all knots K ⊂ S
3, V
K00(1) = −6a
2(K ) = −3∆
00K(1).
Introduction Main Theorem Proof Corollaries
Outline of Proof of Main Theorem
Suppose that a knot K has either V
K00(1) 6= 0 or V
K000(1) 6= 0 .
Case: V
K00(1) 6= 0
Since V
K00(1) = −3∆
00K(1), by [Proposition 5.1, Boyer-Lines (1990)],
K(r) K(r
0) as oriented mfds. for distinct slopes r and r
0.
Case: V
K000(1) 6= 0 with V
K00(1) = 0 ⇒ w
3(K) 6= 0 by Lemma.
Now assume that K(r) ∼ = K(r
0) for r, r
0∈
Qas oriented mfds.
Then, by [Theorem 1.2, Ni-Wu (2015)],
r=−r0must hold.
Introduction Main Theorem Proof Corollaries
Outline of Proof of Main Theorem
Suppose that a knot K has either V
K00(1) 6= 0 or V
K000(1) 6= 0 . Case: V
K00(1) 6= 0
Since V
K00(1) = −3∆
00K(1), by [Proposition 5.1, Boyer-Lines (1990)], K(r) K(r
0) as oriented mfds. for distinct slopes r and r
0.
Case: V
K000(1) 6= 0 with V
K00(1) = 0 ⇒ w
3(K) 6= 0 by Lemma.
Now assume that K(r) ∼ = K(r
0) for r, r
0∈
Qas oriented mfds.
Then, by [Theorem 1.2, Ni-Wu (2015)],
r=−r0must hold.
Introduction Main Theorem Proof Corollaries
Outline of Proof of Main Theorem
Suppose that a knot K has either V
K00(1) 6= 0 or V
K000(1) 6= 0 . Case: V
K00(1) 6= 0
Since V
K00(1) = −3∆
00K(1), by [Proposition 5.1, Boyer-Lines (1990)], K(r) K(r
0) as oriented mfds. for distinct slopes r and r
0. Case: V
K000(1) 6= 0 with V
K00(1) = 0 ⇒ w
3(K) 6= 0 by Lemma.
Now assume that K(r) ∼ = K(r
0) for r, r
0∈
Qas oriented mfds.
Then, by [Theorem 1.2, Ni-Wu (2015)],
r=−r0must hold.
Introduction Main Theorem Proof Corollaries
Outline of Proof of Main Theorem
Suppose that a knot K has either V
K00(1) 6= 0 or V
K000(1) 6= 0 . Case: V
K00(1) 6= 0
Since V
K00(1) = −3∆
00K(1), by [Proposition 5.1, Boyer-Lines (1990)], K(r) K(r
0) as oriented mfds. for distinct slopes r and r
0. Case: V
K000(1) 6= 0 with V
K00(1) = 0 ⇒ w
3(K) 6= 0 by Lemma.
Now assume that K(r) ∼ = K(r
0) for r, r
0∈
Qas oriented mfds.
Then, by [Theorem 1.2, Ni-Wu (2015)],
r=−r0must hold.
Introduction Main Theorem Proof Corollaries
Let us consider λ
2(K(p/q)) & λ
2(K(−p/q)).
Applying the surgery formula for λ
2, with V
K00(1) = a
2(K) = 0,
λ2(K(pq))−λ2(K(−pq)) = 2(qp)w3(K)+λ2(L(p, q))−λ2(L(p,−q))Here we see that L(p, q) ∼ = L(p, −q) as oriented manifolds by; [Theorem 1.2(b), Ni-Wu (2015)]
If K(p/q) ∼ = K (−p/q) as oriented mfds, q
2≡ −1 (mod p) . Recall:
L(p, q
1) ∼ = L(p, q
2) as oriented mfds iff q
1≡ q
±12(mod p).
By w
3(K) 6= 0, λ
2(K(
pq)) 6= λ
2(K(−
pq)) . A contradiction.
Introduction Main Theorem Proof Corollaries
Let us consider λ
2(K(p/q)) & λ
2(K(−p/q)).
Applying the surgery formula for λ
2, with V
K00(1) = a
2(K) = 0,
λ2(K(pq))−λ2(K(−pq)) = 2(qp)w3(K)+λ2(L(p, q))−λ2(L(p,−q))Here we see that L(p, q) ∼ = L(p, −q) as oriented manifolds by; [Theorem 1.2(b), Ni-Wu (2015)]
If K(p/q) ∼ = K (−p/q) as oriented mfds, q
2≡ −1 (mod p) . Recall:
L(p, q
1) ∼ = L(p, q
2) as oriented mfds iff q
1≡ q
±12(mod p).
By w
3(K) 6= 0, λ
2(K(
pq)) 6= λ
2(K(−
pq)) . A contradiction.
Introduction Main Theorem Proof Corollaries
Let us consider λ
2(K(p/q)) & λ
2(K(−p/q)).
Applying the surgery formula for λ
2, with V
K00(1) = a
2(K) = 0,
λ2(K(pq))−λ2(K(−pq)) = 2(qp)w3(K)+λ2(L(p, q))−λ2(L(p,−q))Here we see that L(p, q) ∼ = L(p, −q) as oriented manifolds by;
[Theorem 1.2(b), Ni-Wu (2015)]
If K(p/q) ∼ = K (−p/q) as oriented mfds, q
2≡ −1 (mod p) . Recall:
L(p, q
1) ∼ = L(p, q
2) as oriented mfds iff q
1≡ q
±12(mod p).
By w
3(K) 6= 0, λ
2(K(
pq)) 6= λ
2(K(−
pq)) . A contradiction.
Introduction Main Theorem Proof Corollaries
Finite type invariants of degrees ≤ 3
Corollary
If a knot K has the finite type invariants v
2(K) 6= 0 or v
3(K) 6= 0, then K(r) K(r
0) for any two distinct slopes r and r
0.
v
2& v
3: the finite type invariants of order 2 and 3 respectively normalized by the conditions that
· v
2(m(K)) = v
2(K) and v
3(m(K)) = −v
3(K)
for any knot K and its mirror image m(K),
· v
2(3
1) = v
3(3
1) = 1 for the right hand trefoil 3
1.
Then we see that: v
2(K) = a
2(K) = −
16V
K00(1)
v
3(K) = −2w
3(K ) = −
361(V
K000(1) + 3V
K00(1)).
Introduction Main Theorem Proof Corollaries
vs Heegaard Floer Homology
Corollary
The cosmetic surgery conjecture is true for all knots with no more than 11 crossings, except possibly
10
33, 10
118, 10
146,
11a
91, 11a
138, 11a
285, 11n
86, 11n
157.
Remark
Ozsv´ ath and Szab´ o gave the example of K = 9
44, which is a genus
two knot such that K(1) and K (−1) have the same Heegaard
Floer homology.
Introduction Main Theorem Proof Corollaries
2-bridge knots
Kb1,c1,···,bm,cm: 2-bridge knot of Conway formC(2b1,2c1,· · ·,2bm,2cm)
. Corollary
The knotKx,1,−x,x,1,−x admits no purely cosmetic surgeries forx≥1.
Remark: Known methods cannot obtain this (cf. I-Saito, 2016).
To show that, we have the next, which is of interest independently.
Proposition
v
3(K
b1,c1,···,bm,cm) = −2w
3(K
b1,c1,···,bm,cm)
= 1
2
m
X
k=1
c
k(
k
X
i=1
b
i)
2−
m
X
i=1
b
i(
m
X
k=i
c
k)
2!
Introduction Main Theorem Proof Corollaries