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交代絡み目の二重分岐被覆に関する
Greeneの予想への アプローチ
畠山 えりか
Hiroshima Univ.
December 22, 2016
Backgrounds
.Conjecture. (Greene 2011) ..
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If a pair of links have homeomorphic double branched coverings, then either both are alternating or both are non-alternating.
I would like to study this conjecture in a special case, by using an idea suggested by M. Boileau.
Alternating knot
.Definition. (alternating link) ..
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An alternating link is a link which possesses a link diagram in which crossings alternate between under- and overpasses.
Example of alternating links
Rational tangle
A tangle is a pair ofB3 and two prop- erly embedded arcs in the ball.
In particular, if the arcs are “slope qp arcs on the pillowcase” then the tan-
gle is called a rational tangle. (B3, t(13)) 2-bridge link
Double branched covering
.Definition. (double branched covering) ..
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p: Σ2(L)→S3 is the double branched covering ofS3 branched over Lif p satisfies the following.
Example of double branched coverings
For the 2-bridge link K(p, q),
Σ2(K(p, q))∼=L(p, q) : lens space of type(p, q)
.Theorem. (Hodgson 1985) ..
K : link inS3
∼ ∼
Greene conjecture
.Conjecture. (Greene 2011) ..
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If a pair of links have homeomorphic double branched coverings, then either both are alternating or both are non-alternating.
I would like to study this conjecture in a special case, by using an idea suggested by M. Boileau.
.Definition. (π-hyperbolic) ..
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K is said to be π-hyperbolicif Σ2(K) is a hyperbolic 3-manifold.
.Theorem. (Boileau-Flapan 1995) ..
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K,K′ :π-hyperbolic knots Σ2(K)∼= Σ2(K′)
⇔K′ is obtained from K by applying Constructions (S) or (P), described below.
Construction (S)
.Definition. (strongly invertible) ..
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A knot K⊂S3 is said to be strongly invertible
⇔ ∃h:S3 →S3 orientation preserving involution s.t.
Fix(h) is a circle.
h(K) =K
Fix(h)∩ K = {2 points}
Construction (S)
K : strongly invertible knot
Construction (S)
The graph consists of 2 vertices of va- lence 3 and 3 arcsk1,k2 andk0. The unknotted circle O is k1∪k2. O′ := k2∪k0 is also the trivial knot.
So the graph is O′∪k1.
O∪k0 = k1∪k2∪k0
= k1∪O′
Construction (S)
K : non-alternating strongly invertible knot
By Construction (S), Σ2(K)∼= Σ2(K′).
Construction (P)
.Definition. (periodic knot of period 2) ..
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A knot K⊂S3 is said to be aperiodic knot of period 2
⇔ ∃f :S3→S3 orientation preserving involution s.t.
Fix(f) is a circle.
f(K) =K Fix(f) ∩K =∅
Construction (P)
K : periodic knot of period 2
Remark to Boileau-Flapan’s theorem again
.Theorem. (Boileau-Flapan 1995) ..
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K,K′ :π-hyperbolic knots Σ2(K)∼= Σ2(K′)
⇔K′ is obtained from K by applying Constructions (S) or (P).
Let K,K′ be 6⋆ knots s.t. Σ2(K)∼= Σ2(K′).
Case (S) K′ is obtained from K by applying Construction (S).
Case (P) K′ is obtained from K by applying Construction (P).
.Result.
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The conjecture holds for
{ Case(P).
a part of Case(S).
A family of π-hyperbolic alternating knots
6⋆ knot
ri : rational number This diagram is alternating.
⇔ r1, ..., r6 : all positive or all negative
r1 =r4 =r6= 12 r2 =r3 =r5= 13
.Proposition.
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The alternating knots K=K(r1, r2, r3,−r4,−r5,−r6) are π-hyperbolic if its diagram is not of the following shape.
(Proof) This follows from the following facts.
(1) By orbifold theorem, the π-orbifold associated with(S3, K) is geometric.
(2) By Menasco’s result,(S3, K) is prime and has no essential Conway sphere.
(3) (S3, K) is neither a torus knot or an arborescent knot.
Plan
Plan K : alternating 6⋆ knot
(s-i) Classify the involutions of(S3, K) which make K to be strongly invertible.
(s-ii) Check whether we can apply Construction (S) to each involutionh.
i.e. Check whetherk1∪k0 ork2∪k0 is a trivial knot.
(s-iii) Decide whether the new knotK′, obtained from the involutionh by applying Construction (S), is alternating.
(p-i) Classify the involutions of(S3, K) which make K to be periodic of period 2.
(p-ii) Check whether we can apply Construction (P) to each involutionf.
i.e. Check whetherKˇ is a trivial knot.
(p-iii) Decide whether the new knotK′, obtained from the involutionf by applying Construction (P), is alternating.
Taite flype conjecture
Though, in general, (s-i), (p-i) are not easy, it is in principle possible to complete these tasks by the following theorem.
.Taite flype conjecture (Menasco-Thistlethwaite 1991) ..
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Two reduced prime oriented alternating link diagrams represent the same link type if and only if they are related by the flype operation.
.Consequence A. (see [Kawauchi, A survey of knot theory, Theorem 10.7.7])
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Involutions of 6⋆ knots
By Consequence A, there are at most three kinds of involutions :
⃝ Type 1 → strongly invertible
⃝ Type 2→
{ strongly invertible periodic of period2
× Type 3→ non-alternating
Example
Example
Sketch of proof (Type 1 : Construction (S))
ri := pqi
i,i= 1,4
Fix(h) intersecting K in 2 points outside tangles.
K : knot
⇔
r2≡ 10 or 11, r3≡ 10 r2≡ 01 or 11, r3≡ 01 r2≡ 10 or 01, r3≡ 11
(mod 2) .
Sketch of proof (Type 1 : Construction (S))
.Lemma. (Morimoto-Sakuma-Yokota 1996) ..
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(B3, t(pq)): rational tangle with slope pq The following hold.
(1) q : odd ⇒ (B3, t(qp)∪ Fix(g))/g ∼=(B3, t(2pq )).
(2) q : even ⇒ (B3, t(qp)∪S)/g ∼=(B3, t(
q 2
p)), where S=cl(Fix(g)\v1v2).
Sketch of proof (Type 1 : Construction (S))
Sketch of proof (Type 1 : Construction (S))
There are exactly 2 cases as follows.
Hence, we can’t apply Construction (S) of B-F of this involutionh.
Remark
Remark If we drop the condition that K is alternating, then there are infinitely many knots such that O′ orO′′ is trivial.
Sketch of proof (Type 2 : Construction (P))
ri := qpi
i,i= 1,4
Fix(h) is disjoint fromK outside tangles.
K : knot
⇔
{ r1 ≡r4 (mod 2) r2 ̸≡r3 (mod 2)
.Observation.
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K : periodic knot of period 2 ⇔ pq11, pq4
4 mod 2∈ {10,01}.
Sketch of proof (Type 2 : Construction (P))
.Lemma.
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(B3, t(pq)): rational tangle with slope pq The following hold.
(1) pq mod 2∈ {11} ⇒(B3, t(qp)∪S)/g∼= (B3, t(p−qq
2
)), where S=cl(Fix(g)\v1v2).
(2) pq mod 2∈ {10,01} ⇒(B3, t(pq)∪Fix(g))/g ∼=(B3, t(p2q−q)).
Sketch of proof (Type 2 : Construction (P))
Kˇ is reduced alternating. ...Kˇ is non-trivial.
Hence, we can’t apply construction (P) of B-F of this involution h.
Example (Type 2 : Construction (S))
ri := qpi
i,i= 1,4
Fix(h) is disjoint fromK outside tangles.
K : knot
⇔r2 ̸≡r3 (mod 2)
.Observation.
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Example (Type 2 : Construction (S))
Example (Type 2 : Construction (S))
.Question.
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...Is the knot O′ non-trivial?
O′′ is reduced alternating.
...O′′ is non-trivial.
.Conclusion.
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