Invariants of Legendrian Knots and the Legendrian...

Invariants of Legendrian Knotsand the Legendrian Mirror Problem

Joshua M. Sabloff

Joint with G. Civan, J. Etnyre, P. Koprowski, A. Walker

Wesleyan University, April 2009

• Introduce an intriguing problem in Legendrian knot theory• Show you how to use combinatorial and algebraic techniques to

produce a series of “non-classical” invariants of Legendrian knots• Give infinitely many solutions to the problem

Outline

1 Geometric Notions and Questions

2 The First Invariant: Augmentations

3 The Second Invariant: Linearized Contact Homology

4 Products

Outline

4 Products

Outline

4 Products

Outline

4 Products

Where Are We?

4 Products

Geometric Notions and Questions

Knots in R3

A knot is a smooth embedding γ : S1 → R3.

Two knots are equivalent if one can be deformed into the other throughother knots.

Joshua M. Sabloff (et al.) Legendrian Invariants and Mirrors Wesleyan Colloquium 5 / 33

Knots in R3

A knot is a smooth embedding γ : S1 → R3.

Two knots are equivalent if one can be deformed into the other throughother knots.

Mirrors

QuestionWhen is a knot equivalent to its mirror image (under(x , y , z) → (x , y ,−z))?

Sometimes . . . Sometimes not!

Mirrors

Sometimes . . .

Sometimes not!

Mirrors

Sometimes . . . Sometimes not!

Legendrian Knots

Legendrian knots arise in the study of contact topology.

• A contact structure ξ on R3 (or

any smooth 3-manifold) is acompletely non-integrable2-plane field, such as the onespanned by:

{j, i + yk}.

• A knot γ is Legendrian if it isalways tangent to ξ:

z ′(t) = y(t) x ′(t)

Legendrian Knots

{j, i + yk}.

z ′(t) = y(t) x ′(t)

Legendrian Knots

{j, i + yk}.

z ′(t) = y(t) x ′(t)

VisualizationFront Projections

Project the knot to the xz plane.

• Since y(t) = z′(t)x ′(t) , the y coordinate

is the slope of the curve• . . . so no vertical tangents, only

cusps• . . . so crossings are always like:

is the slope of the curve

• . . . so no vertical tangents, onlycusps

• . . . so crossings are always like:

cusps• . . . so crossings are always like:

VisualizationLagrangian Projections

Project the knot to the xy plane.

• Since z(t) =∫ t

0 y(t)x ′(t) dt , the zcoordinate can be recovered

• . . . and by Green’s theorem, theprojection must bound zero signedarea

Classical Invariants

• The Thurston-Bennequin number tbmeasures the linking between theknot and a push-off transverse to ξ

• tb may be computed from thewrithe of the xy diagram

• . . . or writhe−#right cusps in thexz diagram

• The rotation number r measurestwisting of γ′(t) inside a trivializationof ξ

• r may be computed from therotation number of the xy diagram

• . . . or #down cusps−#up cuspsin the xz diagram

• tb = −2• r = ±1

• tb = −2

• r = ±1

• tb = −2

• r = ±1

• tb = −2

• r = ±1

• tb = −2• r = ±1

Classification?

The classical invariants completely classify Legendrian unknots[Eliashberg-Fraser], torus knots and the figure eight knot [Etnyre-Honda], . . . .Call these knot types simple.

Unknot [Eliashberg-Fraser]

0 1 2 3–1–2–3

Classification?

The classical invariants completely classify Legendrian unknots[Eliashberg-Fraser], torus knots and the figure eight knot [Etnyre-Honda], . . . .Call these knot types simple.

Unknot [Eliashberg-Fraser]

0 1 2 3–1–2–3

The Plot Thickens

But the classical invariants do not classify all Legendrian knots!

Legendrian Mirrors

QuestionWhen is a Legendrian knot with r = 0 equivalent to its “Legendrianmirror” under (x , y , z) → (x ,−y ,−z)?

Note that this is a rotation in R3, not a reflection, but it reverses theco-orientation of ξ

Legendrian Mirrors

Note that this is a rotation in R3, not a reflection, but it reverses theco-orientation of ξ

Legendrian Mirrors

Note that this is a rotation in R3, not a reflection, but it reverses theco-orientation of ξThis reflects the front diagram:

Legendrian Mirrors

Note that this is a rotation in R3, not a reflection, but it reverses theco-orientation of ξAnd also the Lagrangian diagram (but also switches the crossings):

Distinct Legendrian Mirrors?

Ng found examples of 62 knots . . .

TheoremThere exists an infinite family of Legendrian knots that are notLegendrian isotopic to their Legendrian mirrors.

In fact, I’ll show you an infinite family of infinite families, distinguishedby ever-deeper invariants.

Where Are We?

4 Products

The First Invariant: Augmentations

Special Subsets of Crossings

The goal is to pick out special subsets of the crossings in a Lagrangiandiagram.

• First, decorate the crossings.• Look for smoothly immersed

disks with convex cornerswhose boundary lies in theknot diagram. There is one +corner, any # of − corners.

• Select a subset of crossings iffor each fixed corner c, thenumber of disks with a positivecorner at c and all negativecorners in the chosen subsetis even.

• First, decorate the crossings.

• Look for smoothly immerseddisks with convex cornerswhose boundary lies in theknot diagram. There is one +corner, any # of − corners.

+ +––

+ +––+ +

––

+ +––

+ +––+ +

––

+ +––

+ +––+ +

––

+ +––

+ +––+ +

––

An Example

This trefoil knot has 5 augmentations.

Gradings

The crossings are graded by a “Conley-Zehnder index” in Z/2r(K )Z.

Choose a capping path γi for the double point i , beginning at theovercrossing:

|qi | ≡ 2r(γi)−12

mod 2r(K )

|qi | = 0 here. . .

Gradings

The crossings are graded by a “Conley-Zehnder index” in Z/2r(K )Z.Choose a capping path γi for the double point i , beginning at theovercrossing:

|qi | ≡ 2r(γi)−12

mod 2r(K )

|qi | = 0 here. . .

Gradings

|qi | ≡ 2r(γi)−12

mod 2r(K )

|qi | = 0 here. . .

Gradings

|qi | ≡ 2r(γi)−12

mod 2r(K )

|qi | = 0 here. . .

The Augmentation Invariant

We can form an invariant Aug(D) by forming a (normalized) count of“graded” augmentations of a diagram D.

TheoremAug(D) is a Legendrian invariant (now denoted Aug(K ))

Example

The augmentation number can distinguish the 52 knots, but not a knotfrom its Legendrian mirror.

Example

Where Are We?

4 Products

The Second Invariant: Linearized Contact Homology

Linearized CohomologyThe Vector Space

Label the crossings in an Lagrangian diagram with {1, . . . , n}. Define avector space A over Z/2Z generated by labels {q1, . . . , qn}.

We can split A as a direct sum according to the gradings of thegenerators.

Example

For the trefoil, A0 = 〈q3, q4, q5〉 and A1 = 〈q1, q2〉.

Example

Linearized CohomologyThe Differential

Given an augmentation ε, let δε : A → A be the linear map defined asfollows:

The contribution of qkto δεqi is the number ofdisks with − at qi , + atqk , and maybe other −corners in ε.

+ +––

+ +––+ +

––

δεq5 = q1 + q2

+ +––

+ +––+ +

––

δεq5 = 2q1 + 2q2 = 0

The Algebra in the Linearized Invariant

Theorem (Chekanov)

δε ◦ δε = 0

Thus im δε ⊂ ker δε, and we can form the linearized homologyLCH∗

ε = ker δε/im δε.

Example

For the trefoil above, the only nonzero differential is

δεq4 = q1 + q2

Thus, we have LCH∗ε = 〈[q1], [q3], [q5]〉.

Theorem (Chekanov)

δε ◦ δε = 0

Example

δεq4 = q1 + q2

Theorem (Chekanov)

δε ◦ δε = 0

Example

δεq4 = q1 + q2

The Linearized Invariant

Theorem (Chekanov)

The set{

LCH∗ε

is an invariant of Legendrian isotopy.

Example

The linearized invariant can also distinguish the 52 knots, but still not aknot from its Legendrian mirror.

The Linearized Invariant

Theorem (Chekanov)

The set{

LCH∗ε

is an invariant of Legendrian isotopy.

Example

The linearized invariant can also distinguish the 52 knots, but still not aknot from its Legendrian mirror.

Where Are We?

4 Products

Products

The Cup Product

To capture the mirror example, we need a potentially non-symmetricoperation on the linearized cohomology.

. . . so define a linear map mε : A⊗ A → A as follows:

The contribution of qkto mε(qi , qj) is thenumber of disks with −at qi and qj (in thatcounterclockwiseorder!), + at qk , andmaybe other − cornersin ε.

+ +––

+ +––+ +

––

Products

The Cup Product

+ +––

+ +––+ +

––

Products

The Cup Product

+ +––

+ +––+ +

––

mε(q5 ⊗ q3) = q1

Products

What is Invariant?

The product map mε descends to a map

µε : LCH jε ⊗ LCHk

ε → LCH j+k+1ε

on cohomology, and . . .

TheoremThe set of linearized homology algebras {(LCH∗

ε , µε)}ε is invariantunder Legendrian isotopy.

Products

What is Invariant?

The product map mε descends to a map

ε → LCH j+k+1ε

on cohomology, and . . .

TheoremThe set of linearized homology algebras {(LCH∗

ε , µε)}ε is invariantunder Legendrian isotopy.

Products

Detecting Mirrors

Want to find a knot (say, with only one ε) with nontrivial

ε → LCH j+k+1ε

but trivialµε : LCHk

ε ⊗ LCH jε → LCH j+k+1

Products

Detecting Mirrors

Want to find a knot (say, with only one ε) with nontrivial

ε → LCH j+k+1ε

but trivialµε : LCHk

ε ⊗ LCH jε → LCH j+k+1

Products

An Infinite Family of Mirrors

Nontrivial cup product in the following degrees:

LCH l−m−1 ⊗ LCH1−k+m → LCH1−k+l

. . . but none when order is reversed. Hence it’s not isotopic to itsLegendrian mirror!

Products

An Infinite Family of Mirrors

Nontrivial cup product in the following degrees:

LCH l−m−1 ⊗ LCH1−k+m → LCH1−k+l

. . . but none when order is reversed. Hence it’s not isotopic to itsLegendrian mirror!

Products

An Infinite Family for the Massey Product

Using disks with three negative corners and a little algebra, we candefine a three-fold “Massey product” that can also be used todistinguish knots from their mirrors:

Products

An Infinite Family for the Massey Product

Using disks with three negative corners and a little algebra, we candefine a three-fold “Massey product” that can also be used todistinguish knots from their mirrors:

Products

An Infinite Family of Infinite Families

In fact, there is an A∞ structure {mεk} on A∗ that induces higher-order

Massey products on LCH∗ε .

Products

An Infinite Family of Infinite Families

In fact, there is an A∞ structure {mεk} on A∗ that induces higher-order

Massey products on LCH∗ε .

Summary

Where Do We Come From? What Are We? WhereAre We Going?

What else can we do / say here?

• Cup product? “Poincaré Duality”!• Larger context and motivation: Morse-Witten-Floer theory• Geometric meaning

Summary

• Cup product? “Poincaré Duality”!

• Larger context and motivation: Morse-Witten-Floer theory• Geometric meaning

Summary

• Cup product? “Poincaré Duality”!• Larger context and motivation: Morse-Witten-Floer theory

• Geometric meaning

Summary

• Cup product? “Poincaré Duality”!• Larger context and motivation: Morse-Witten-Floer theory• Geometric meaning

Summary

Hopefully, you have seen that the Legendrian mirror problem canmotivate some interesting structure in invariants of Legendrian knotsusing an interplay of:

• Combinatorics (counting disks!)• Algebra (linearized contact homology and its product structure!)• Geometric / analytic motivation (err . . . actually I hid this today, but

you might now be motivated to find out!)

Summary

• Combinatorics (counting disks!)

• Algebra (linearized contact homology and its product structure!)• Geometric / analytic motivation (err . . . actually I hid this today, but

Summary

• Combinatorics (counting disks!)• Algebra (linearized contact homology and its product structure!)

• Geometric / analytic motivation (err . . . actually I hid this today, butyou might now be motivated to find out!)

Summary

• Combinatorics (counting disks!)• Algebra (linearized contact homology and its product structure!)• Geometric / analytic motivation (err . . . actually I hid this today, but

Invariants of Legendrian Knots and the Legendrian...

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