Link lengths and their growth powers
Sungjong No*, Youngsik Huh† , Seungsang Oh‡
Ewha Womans University*, Hanyang University† , Korea University‡
2014. 8. 26.
Introduction
Some results
Contents
1
Introduction
2
Some results
3
Sketch of proof
4
Comments
Sketch of proof
Comments
Introduction
Some results
Contents
1
Introduction
2
Some results
3
Sketch of proof
4
Comments
Sketch of proof
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Introduction
Some results
Sketch of proof
Comments
Introduction
Definitions
A polygonal knot is a simple closed curve in R3 obtained by
joining finitely many points with line segments called sticks.
A lattice knot is a polygonal knot in the cubic lattice
Z3 = (R × Z × Z) ∪ (Z × R × Z) ∪ (Z × Z × R).
stick number (s(K)) : minimum number of sticks required
to construct a polygonal representation of a knot or link K
in R3 .
Introduction
Some results
Sketch of proof
Comments
Introduction
Definitions
lattice stick number (sL (K)) : minimum number of sticks
required to construct a polygonal representation of K in Z3 .
minimum lattice length (Len(K)): minimum number of
unit length edges required to construct a polygonal
representation of K in Z3 .
minimum ropelength : For a smooth simple closed curve c
in R3 , take a uniform tube whose core is c. And consider
the quotient of the arc length of c by the radius of
cross-sectional disk of the tube. Then Rop(K) is an
infimum of such quotient over all smooth curves c realizing
K.
Introduction
Some results
Sketch of proof
Introduction
Example
sL (31 ) = 12
Len(31 ) = 24
Rop(31 ) ≈ 32.74
Comments
Introduction
Some results
Contents
1
Introduction
2
Some results
3
Sketch of proof
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Comments
Sketch of proof
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Introduction
Some results
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Some results
Negami (1991)
√
5+ 8c(K)+9
≤ s(K) for a link K except for the trivial knot, and
2
s(K) ≤ 2c(K) for a link K which has neither the Hopf link as a
connected sum factor nor a splittable trivial component.
Huh-Oh (2011)
s(K) ≤ 32 (c(K) + 1) for any nontrivial knot K. In particular,
s(K) ≤ 32 c(K) for any non-alternating prime knot.
Introduction
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Some results
Diao-Ernst (2007)
p
3 c(K) + 1 + 3 ≤ sL (K) for any link.
Hong-No-Oh (2013)
sL (K) ≤ 3c(K) + 2 for any nontrivial non-trefoil knot K. In
particular, sL (K) ≤ 3c(K) − 4 for any non-alternating prime
knot.
Comments
Introduction
Some results
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Some results
Rop(K) ≤ 2Len(K).
3
4
( 4π
11 c(K)) ≤ Rop(K) and the power
(four-thirds power law)
3
4
is optimal.
Len(K) ≤ O(c(K)[ln(c(K))]5 ) (Diao, Ernst, Por and
Ziegler)
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Introduction
Some results
Sketch of proof
Some results
Question
Is the power of lower and upper bounds are optimal?
If then, how about between power of upper bound and
lower bound?
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Introduction
Some results
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Some results
Main Theorems
Theorem 1
For any real number r between 12 and 1, there is an infinite
family Fr of non-splittable prime links such that s(K) and
sL (K) have the order O(c(K)r ) for every K ∈ Fr .
Theorem 2
For any real number r between 34 and 1, there is an infinite
family Fr of non-splittable prime links such that Len(K) and
Rop(K) have the order O(c(K)r ) for every K ∈ Fr .
Comments
Introduction
Some results
Contents
1
Introduction
2
Some results
3
Sketch of proof
4
Comments
Sketch of proof
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Introduction
Some results
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Sketch of proof
(p, q)-Hopf link (= Hp,q )
q
p
Comments
Introduction
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Sketch of proof
Proof of Theorem 1
Let m and n be integers with 0 ≤ m ≤ n. We consider an
infinite family {Hk = Hkm ,kn | k = 1, 2, 3, . . .}. Then,
s(Hk ) = 3(k m + k n ) , sL (Hk ) = 4(k m + k n ) and c(Hk ) = 2k m+n
Then we have
n
s(Hk ) ≈ sL (Hk ) ≈ O(c(Hk ) m+n ).
Introduction
Some results
Sketch of proof
Comments
Sketch of proof
For any rational number r between 12 and 1, there exist integers
0
m0 and n0 such that r = m0n+n
and 0 ≤ m0 ≤ n0 . Therefore
0
m
n
the family {Hk 0 ,k 0 | k = 1, 2, 3, . . .} satisfies the statement of
the theorem.
For any real number r between 21 and 1, take a sequence {ri } of
rational numbers in the interval converging to r.
Introduction
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Sketch of proof
Proof of Theorem 2
Firstly we derive a lower bound of Rop(Hp2 ,q2 ).
We can get a lower bound
2π(q + 1)p2 + 2π(p + 1)q 2 ≤ Rop(Hp2 ,q2 ).
Comments
Introduction
Some results
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Sketch of proof
Now we derive an upper bound of Len(Hp2 ,q2 ). Like Figure,
2
stack p parallel rectangles with size (d qp e + 1) × (p + 1). And
2
stack other p parallel rectangles with size (d qp e + 3) × (p + 3) in
the same way. Repeat this procedure. Then their total length is
p
q2
X
q2
2p d e + p + 4n − 2 = 2p2 d e + 6p3 .
p
p
n=1
Introduction
Some results
Sketch of proof
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Sketch of proof
Now we link the remaining q 2 parallel rectangles into the
previous part.
Then the length of this lattice link is
p
q2
q2 X
q2
2p d e + 6p3 + 4d e
(p + 2n − 1) = 10p2 d e + 6p3 .
p
p
p
2
n=1
Introduction
Some results
Sketch of proof
Comments
Sketch of proof
2π(q + 1)p2 + 2π(p + 1)q 2 ≤ Rop(Hp2 ,q2 )
≤ 2 Len(Hp2 ,q2 ) ≤ 20p2 d
q2
e + 12p3 .
p
We consider an infinite family {Hk = Hk2m ,k2n | k = 1, 2, 3, . . .}.
Then we have
m+2n
Len(Hk ) ≈ Rop(Hk ) ≈ O(c(Hk ) 2m+2n ).
For any rational number r between 43 and 1, there exist integers
m0 +2n0
m0 and n0 such that r = 2m
and 0 ≤ m0 ≤ n0 .
0 +2n0
The proof is completed.
Introduction
Some results
Contents
1
Introduction
2
Some results
3
Sketch of proof
4
Comments
Sketch of proof
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Introduction
Some results
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Problem
How about a knot? In other words, can we find an
examples for knots?
What is the optimal upper bound of ropelength?
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Introduction
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Comments
For 2-bridge knots,
Len(K) ≤ 8c(K) + 2
Rop(K) ≤ 11.39c(K) + 12.37
Sketch of proof
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Introduction
Some results
Sketch of proof
Thank you
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