Introduction to Geometric Knot Theory · Introduction to Geometric Knot Theory Elizabeth Denne...
Transcript of Introduction to Geometric Knot Theory · Introduction to Geometric Knot Theory Elizabeth Denne...
Introduction to Geometric Knot Theory
Elizabeth Denne
Smith College
CCSU Math Colloquium, October 9, 2009
What is a knot?
DefinitionA knot K is an embedding of a circle in R3. (Intuition: a smooth orpolygonal closed curve without self-intersections.)A link is an embedding of a disjoint union of circles in R3.
Trefoil knotFigure 8 knot
Elizabeth Denne (Smith College) Geometric Knot Theory 9th October 2009 2 / 24
When are two knots the same?
K1 and K2 are equivalent if K1 can be continuously moved to K2.(Technically, they are ambient isotopic.)A knot is trivial or unknotted if it is equivalent to a circle.A knot is tame if it is equivalent to a smooth knot.
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When are two knots the same?
K1 and K2 are equivalent if K1 can be continuously moved to K2.(Technically, they are ambient isotopic.)A knot is trivial or unknotted if it is equivalent to a circle.A knot is tame if it is equivalent to a smooth knot.
Elizabeth Denne (Smith College) Geometric Knot Theory 9th October 2009 3 / 24
When are two knots the same?
K1 and K2 are equivalent if K1 can be continuously moved to K2.(Technically, they are ambient isotopic.)A knot is trivial or unknotted if it is equivalent to a circle.A knot is tame if it is equivalent to a smooth knot.
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Tame knots
A wild knot
p . .
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How can you tell if two knots are equivalent?
IdeaUse topological invariants denoted by I. If K1 is equivalent to K2 thenI(K1) = I(K2),
Crossing number (knot tables)
cr(K ) = minK∈[K ]
( mindirections
(# crossings of K ))
Alexander or Jones polynomialKnot group π1(R3 \ K )
PrincipleIf I(K1) 6= I(K2) then K1 is not equivalent to K2. Knots are distinguishedusing invariants.
Elizabeth Denne (Smith College) Geometric Knot Theory 9th October 2009 5 / 24
How can you tell if two knots are equivalent?
IdeaUse topological invariants denoted by I. If K1 is equivalent to K2 thenI(K1) = I(K2),
Crossing number (knot tables)
cr(K ) = minK∈[K ]
( mindirections
(# crossings of K ))
Alexander or Jones polynomialKnot group π1(R3 \ K )
PrincipleIf I(K1) 6= I(K2) then K1 is not equivalent to K2. Knots are distinguishedusing invariants.
Elizabeth Denne (Smith College) Geometric Knot Theory 9th October 2009 5 / 24
How can you tell if two knots are equivalent?
IdeaUse topological invariants denoted by I. If K1 is equivalent to K2 thenI(K1) = I(K2),
Crossing number (knot tables)
cr(K ) = minK∈[K ]
( mindirections
(# crossings of K ))
Alexander or Jones polynomialKnot group π1(R3 \ K )
PrincipleIf I(K1) 6= I(K2) then K1 is not equivalent to K2. Knots are distinguishedusing invariants.
Elizabeth Denne (Smith College) Geometric Knot Theory 9th October 2009 5 / 24
How can you tell if two knots are equivalent?
IdeaUse topological invariants denoted by I. If K1 is equivalent to K2 thenI(K1) = I(K2),
Crossing number (knot tables)
cr(K ) = minK∈[K ]
( mindirections
(# crossings of K ))
Alexander or Jones polynomialKnot group π1(R3 \ K )
PrincipleIf I(K1) 6= I(K2) then K1 is not equivalent to K2. Knots are distinguishedusing invariants.
Elizabeth Denne (Smith College) Geometric Knot Theory 9th October 2009 5 / 24
Knot and Link Table
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Elizabeth Denne (Smith College) Geometric Knot Theory 9th October 2009 7 / 24
Total curvature κ(γ) of a curve γ
For a polygon P, κ(P) is the sum of exterior angles.For a curve γ, take polygons P with vertices on γ, then
κ(γ) = maxP
κ(P)
Equivalently for a smooth knot K , κ(K ) is the total angle throughwhich the unit tangent vector turns through.
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Total curvature κ(γ) of a curve γ
For a polygon P, κ(P) is the sum of exterior angles.For a curve γ, take polygons P with vertices on γ, then
κ(γ) = maxP
κ(P)
Equivalently for a smooth knot K , κ(K ) is the total angle throughwhich the unit tangent vector turns through.
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Some history ...
1929 Fenchel proved that for a closed curve γ in R3, the total curvatureκ(γ) ≥ 2π with equality iff γ is a convex planar curve.
1947 Borsuk proved same for γ in Rn and conjectured that the totalcurvature of a nontrivial knot ≥ 4π.
Theorem (1950 Fáry-Milnor)For any nontrivial tame knot K , the total curvature κ(K ) ≥ 4π.
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Bridge number
DefinitionThe bridge number br(K ) of a knot K is
br(K ) = minK∈[K ]
( mindirections
(# maxima of K ))
br (Unknot)=1, br (Trefoil)= 2 = br (Figure 8)br(810) = 3
Theorem (1949 Milnor)Given a tame knot K , κ(K ) > 2πbr(K ) and br(K ) ≥ 2 for nontrivialknots.
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The ropelength problem
GoalMinimize ropelength — the length of a knot or link which has a tube offixed diameter around it.
Tight knots are those knots which minimize ropelength.
QuestionWhat do tight knots look like? What is the ropelength of a tight knot?
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DefinitionThe ropelength R(K ) of a knot K is the quotient of its length overthickness R(K ) = Len(K )/τ(K ).
DefinitionThe thickness τ(K ) is the diameter of the largest embedded normaltube around the knot K .Consider local and global conditions here.
K
K}12
}12
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Theorem (2002 CKS and GMSvdM)There is a ropelength minimizer in any (tame) link type; any minimizeris C1,1, with bounded curvature.
CKS also described a family of tight links.Alas, there are many links for which the minimum ropelength isnot known. For example a keychain with ≥ 7 keys.The minimum ropelength is not known for any knot type.Numerical simulations have shown that the trefoil is the shortestpossible knot with minimum ropelength about 16.372.
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Theorem (2002 CKS and GMSvdM)There is a ropelength minimizer in any (tame) link type; any minimizeris C1,1, with bounded curvature.
CKS also described a family of tight links.Alas, there are many links for which the minimum ropelength isnot known. For example a keychain with ≥ 7 keys.The minimum ropelength is not known for any knot type.Numerical simulations have shown that the trefoil is the shortestpossible knot with minimum ropelength about 16.372.
Elizabeth Denne (Smith College) Geometric Knot Theory 9th October 2009 13 / 24
Theorem (2002 CKS and GMSvdM)There is a ropelength minimizer in any (tame) link type; any minimizeris C1,1, with bounded curvature.
CKS also described a family of tight links.Alas, there are many links for which the minimum ropelength isnot known. For example a keychain with ≥ 7 keys.The minimum ropelength is not known for any knot type.Numerical simulations have shown that the trefoil is the shortestpossible knot with minimum ropelength about 16.372.
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Difficult ProblemRelate the ropelength to link type using topological invariants.
Example: Let L be a link type with minimum crossing number n andminimum ropelength R(L). Find constants c1, c2,p,q such that
c1n p ≤ R(L) ≤ c2nq
There are families of torus links for which c1n3/4 ≤ R(L).Diao, Ernst and Yu (2004) proved R(L) ≤ c2n3/2.It is conjectured that R(L) ≤ c3n1.
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Difficult ProblemRelate the ropelength to link type using topological invariants.
Example: Let L be a link type with minimum crossing number n andminimum ropelength R(L). Find constants c1, c2,p,q such that
c1n p ≤ R(L) ≤ c2nq
There are families of torus links for which c1n3/4 ≤ R(L).Diao, Ernst and Yu (2004) proved R(L) ≤ c2n3/2.It is conjectured that R(L) ≤ c3n1.
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Quadrisecants
quadrisecant(alternating)
quadrisecant(simple)
trisecant
secant
Secant = 2-secant
Trisecant = 3-secant
Quadrisecant = 4-secant
DefinitionAn n-secant line is an oriented line in R3 which intersects K in at leastn places. An n-secant is an ordered n-tuple of points in K which lie onan n-secant line.
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Quadrisecants
Three types of quadrisecants: simple, flipped and alternating.Determined by comparing the order of abcd along the line andalong the unoriented knot — there are |S4/D4| = 3 dihedralorderings.
a cca
acbdsimple
abdcflipped alternating
abcd
bb
dbd
dc
Order:
a
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Theorem (D-, 2004)
Every nontrivial tame knot in R3 has at least one alternatingquadrisecant.
Quadrisecant history:1933 Pannwitz: Every nontrivial generic polygonal knot has atleast 2u2 quadrisecants.1994 Kuperberg: Every nontrivial tame knot has at least onequadrisecant.2004 Budney, Conant, Scannell, Sinha: The coefficient of z2 in theConway polynomial may be computed by counting alternatingquadrisecants with an appropriate multiplicity.2004 D- : Every nontrivial C1,1 knot in R3 has at least oneessential alternating quadrisecant.
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Theorem (D-, 2004)
Every nontrivial tame knot in R3 has at least one alternatingquadrisecant.
Quadrisecant history:1933 Pannwitz: Every nontrivial generic polygonal knot has atleast 2u2 quadrisecants.1994 Kuperberg: Every nontrivial tame knot has at least onequadrisecant.2004 Budney, Conant, Scannell, Sinha: The coefficient of z2 in theConway polynomial may be computed by counting alternatingquadrisecants with an appropriate multiplicity.2004 D- : Every nontrivial C1,1 knot in R3 has at least oneessential alternating quadrisecant.
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StrategyAssume the knot K has unit thickness, then ropelength is justthe length of K (the core curve).Use quadrisecants and extra geometry to get a lower bound forropelength.Extra geometry due to unit thickness of K :
I Curvature bounded above by 2.I Given p ∈ K , let B(p) be the unit ball centered at p. Then B(p)
contains a single unknotted arc of K with length at most π.I Other arcs of the knot stay outside this unit ball.
a rα
β
p
a′1
s
b′ b
∠apb = θ
Length arc(ab) is√
r2 − 1 +(θ − (α + β)
)+√
s2 − 1
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StrategyAssume the knot K has unit thickness, then ropelength is justthe length of K (the core curve).Use quadrisecants and extra geometry to get a lower bound forropelength.Extra geometry due to unit thickness of K :
I Curvature bounded above by 2.I Given p ∈ K , let B(p) be the unit ball centered at p. Then B(p)
contains a single unknotted arc of K with length at most π.I Other arcs of the knot stay outside this unit ball.
a rα
β
p
a′1
s
b′ b
∠apb = θ
Length arc(ab) is√
r2 − 1 +(θ − (α + β)
)+√
s2 − 1
Elizabeth Denne (Smith College) Geometric Knot Theory 9th October 2009 18 / 24
StrategyAssume the knot K has unit thickness, then ropelength is justthe length of K (the core curve).Use quadrisecants and extra geometry to get a lower bound forropelength.Extra geometry due to unit thickness of K :
I Curvature bounded above by 2.I Given p ∈ K , let B(p) be the unit ball centered at p. Then B(p)
contains a single unknotted arc of K with length at most π.I Other arcs of the knot stay outside this unit ball.
a rα
β
p
a′1
s
b′ b
∠apb = θ
Length arc(ab) is√
r2 − 1 +(θ − (α + β)
)+√
s2 − 1
Elizabeth Denne (Smith College) Geometric Knot Theory 9th October 2009 18 / 24
StrategyAssume the knot K has unit thickness, then ropelength is justthe length of K (the core curve).Use quadrisecants and extra geometry to get a lower bound forropelength.Extra geometry due to unit thickness of K :
I Curvature bounded above by 2.I Given p ∈ K , let B(p) be the unit ball centered at p. Then B(p)
contains a single unknotted arc of K with length at most π.I Other arcs of the knot stay outside this unit ball.
a rα
β
p
a′1
s
b′ b
∠apb = θ
Length arc(ab) is√
r2 − 1 +(θ − (α + β)
)+√
s2 − 1
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ComputationsGiven quadrisecant abcd , define
r := |b − a|, s := |c − b|, t := |d − c|.Find Len(K ) in terms of r , s and t .Let f (r) :=
√r2 − 1 + arcsin(1
r ).Len(γac) ≥ f (r) + f (s), Len(γda) ≥ f (r) + s + f (t).
dc
tsr
a b r s
f(r)f(s)
2
a br s
ct
d
f (r)
r1
s
f (s)
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ComputationsGiven quadrisecant abcd , define
r := |b − a|, s := |c − b|, t := |d − c|.Find Len(K ) in terms of r , s and t .Let f (r) :=
√r2 − 1 + arcsin(1
r ).Len(γac) ≥ f (r) + f (s), Len(γda) ≥ f (r) + s + f (t).
dc
tsr
a b r s
f(r)f(s)
2
a br s
ct
d
f (r)
r1
s
f (s)
Elizabeth Denne (Smith College) Geometric Knot Theory 9th October 2009 19 / 24
ComputationsGiven quadrisecant abcd , define
r := |b − a|, s := |c − b|, t := |d − c|.Find Len(K ) in terms of r , s and t .Let f (r) :=
√r2 − 1 + arcsin(1
r ).Len(γac) ≥ f (r) + f (s), Len(γda) ≥ f (r) + s + f (t).
dc
tsr
a b r s
f(r)f(s)
2
a br s
ct
d
f (r)
r1
s
f (s)
Elizabeth Denne (Smith College) Geometric Knot Theory 9th October 2009 19 / 24
The results
Theorem (D-, Diao, Sullivan 2006)Any nontrivial knot has ropelength at least 15.66.
Recall: the tight trefoil knot has ropelength ≈ 16.372, so we werepretty close.
Note the ropelength of a knot withsimple essential quadrisecants is at least 15.94,flipped essential quadrisecants is at least 13.936.
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Thank you!
Acknowledgements:My fantastic coauthors: Jason Cantarella, Yuanan Diao, JohnSullivan, Nancy Wrinkle.
Further reading:J. Cantarella, R.B. Kusner, J.M. Sullivan, On the minimumropelength of knots and links. Invent. Math 150:2 (2002)pp 257–286.E. Denne, Y. Diao, J.M. Sullivan, Quadrisecants give new boundsfor ropelength. Geom. Topol. 10 (2006) pp 1–26.Y. Diao, C. Ernst, X. Yu, Hamiltonian knot projections and lengthsof thick knots. Topology Appl. 136 (2004), no. 1-3, pp 7–36.
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DefinitionLet α, β, γ be disjoint arcs from a to b. Let X := R3 r (α ∪ γ) and leth(α) be a parallel curve to α ∪ β, chosen so that α ∪ γ has linkingnumber zero with δ. Then (α, β) is inessential if δ is also nullhomotopic in X . We say (α, β) is essential if it is not inessential.
α
a bβ
γ
h(α)
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Lemma (DDS 2006)If γab is an essential arc in a unit-thickness knot then
Len(γab) ≥
{2π − 2 arcsin(|a− b|/2) if |a− b| < 2,π if |a− b| ≥ 2.
a b
p
Πa Πb
Πq
q
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NotesIf K is the unknot then arc γab is inessential for a,b ∈ K .If both arcs γab and γba for a,b ∈ K are inessential then K is theunknot.
DefinitionSecant ab is essential if both γab and γba are essential.An essential alternating quadrisecant abcd is an alternatingquadrisecant which is essential in the second segment bc
a cca
acbdsimple
abdcflipped alternating
abcd
bb
dbd
dc
Order:
a
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