Odd homology of tangles and cobordisms

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Odd homology of tangles and cobordisms Krzysztof Putyra Jagiellonian University, Kraków XXVII Knots in Washington 10 th January 2009

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Odd homology of tangles and cobordisms. Krzysztof Putyra Jagiellonian University , Kraków XXVII Knots in Washington 10 th January 2009. 0 -smoothing. 1 -smoothing. Mikhail Khovanov. Cube of resolutions. 110. 100. 000. –. 111. 101. 010. –. –. 001. –. d. d. d. C -3. - PowerPoint PPT Presentation

Transcript of Odd homology of tangles and cobordisms

Page 1: Odd homology of  tangles  and  cobordisms

Odd homologyof tangles and cobordisms

Krzysztof PutyraJagiellonian University, Kraków

XXVII Knots in Washington10th January 2009

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Cube of resolutions

0-smoothing 1-smoothing

Mikhail Khovano

v

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Cube of resolutions1

2

3

C-3 C-2 C-1 C0d d d

edges are cobordis

ms

direct sums create the complex

000

100

010

001

110

101

011

111

vertices are smoothed diagrams –

Mikhail Khovano

v

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Cube of resolutions1

2

3

C-3 C-2 C-1 C0d d d

edges are cobordism

s with arrows

direct sums create the complex(applying

some edgeassignment

)

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100

010

001

110

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011

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vertices are smoothed diagrams

Peter Ozsvath

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Khovanov functorsee Khovanov: arXiv:math/9908171

FKh: Cob → ℤ-Mod

symmetric:

Edge assignment is given explicite.

Category of cobordisms is symmetric:

ORS ‘projective’ functor see Ozsvath, Rasmussen, Szabo:

arXiv:0710.4300

FORS: ArCob → ℤ-Mod

not symmetric:

Edge assignment is given by homological properties.

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Motivation Invariance of the odd Khovanov complex may be proved at the level of topology and new theories may arise.

Fact (Bar-Natan) Invariance of the Khovanov complex can be proved at the level of topology.Question Can Cob be changed to make FORS a functor?

Anwser Yes: cobordisms with chronology

Main question

Dror Bar-Natan

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ChCob: cobordisms with chronology & arrowsChronology τ is a Morse function with exactly one critical point over each critical value.Critical points of index 1 have arrows:- τ defines a flow φ on M- critical point of τ are fix points φ- arrows choose one of the

in/outcoming trajectory for a critical point.

Chronology isotopy is a smooth homotopy H satisfying:- H0 = τ0

- H1 = τ1

- Ht is a chronology

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ChCob: cobordisms with chronology & arrows

Critical points cannot be permuted:

Critical points do not vanish:

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ChCob: cobordisms with chronology & arrowsTheorem The category 2ChCob is generated by the following:

with the full set of relations given by:

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ChCob: cobordisms with chronology & arrows

Theorem 2ChCob with changes of chronologies is a 2-category.

Change of chronology is a smooth homotopy H s.th.- H0 = τ0, H1 = τ1

- Ht is a chronology except t1,…,tn, where one of the following occurs:

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ChCob(B): cobordisms with cornersFor tangles we need cobordisms with corners:

• input and output has same endpoints

• projection is a chronology• choose orientation for each

critical point• all up to isotopies preserving π

being a chronology

ChCob(B)‘s form a planar algebra with planar operators:

1 3 2 (M1,M2,M3)M1

M3

M2

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Which conditions should a functor

F: ChCob ℤ-Mod

satisfies to produce homologies?

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Chronology change condition

This square needs to be anti-commutative after multiplying some egdes with invertible elements (edge assignment proccess).

These two compositions could differ by an invertible element only!

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Chronology change conditionExtend cobordisms to formal sums over a commutative ring R.Find a representation of changes of chronology in U(R) s.th.

α M1 … Ms = β M1 … Ms => α = β

Fact WLOG creation and removing critical points can be represented by 1.Hint Consider the functor given by:

α β Id on others gen’s

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Chronology change conditionExtend cobordisms to formal sums over a commutative ring R.Find a representation of changes of chronology in U(R) s.th.

α M1 … Ms = β M1 … Ms => α = β

Fact WLOG creation and removing critical points can be represented by 1.

Proposition The representation is given by

where X2 = Y2 = 1 and Z is a unit.

X Y Z

XY 1

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Chronology change condition

This square needs to be anti-commutative after multiplying some egdes with invertible elements (edge assignment proccess).

These two compositions could differ by an invertible element only!

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Edge assignmentProposition For any cube of resolutions C(D) there exists an edge assignment e → φ(e)e making the cube anticommutative.Sketch of proof Each square S corresponds to a change of chronology with some coefficient λ. The cochain

ψ(S) = -λis a cocycle:

P i = 1

6

By the ch. ch. condition:

dψ(C) = Π -λi = 1

and by the contractibility of a 3-cube:

ψ = dφ

6

i = 1

P = λrPP = λrP = λrλf PP = λrP = λrλf P = ... = Π λiP

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Edge assignmentProposition For any cube of resolutions C(D) there exists an edge assignment e → φ(e)e making the cube anticommutative.Proposition For any cube of resolutions C(D) different egde assign-ments produce isomorphic complexes.

Sketch of proof Let φ1 and φ2 be edge assignments for a cube C(D). Then

d(φ1φ2-1) = dφ1dφ2

-1 = ψψ-1 = 1

Thus φ1φ2-1 is a cocycle, hence a coboundary. Putting

φ1 = dηφ2

we obtain an isomorphism of complexes ηid: Kh(D,φ1) →Kh(D,φ2).

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Edge assignment

Proposition Denote by D1 and D2 a tangle diagram D with different choices of arrows. Then there exist edge assignments φ1 and φ2 s.th. complexes C(D1, φ1) and C(D2 , φ2) are isomorphic.Corollary Upto isomophisms the complex Kh(D) depends only on the tangle diagram D.

Proposition For any cube of resolutions C(D) there exists an edge assignment e → φ(e)e making the cube anticommutative.Proposition For any cube of resolutions C(D) different egde assign-ments produce isomorphic complexes.

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S / T / 4Tu relationscompare with Bar-Natan: arXiv:math/0410495

Theorem The complex Kh(D) is invariant under chain homotopies and the following relations:

where X, Y and Z are given by the ch.ch.c.Dror Bar-Natan

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HomologiesM FXYZ (M)

Rv+⊕Rv–

1 v+

v+ + ZY v–

v+ 0 v– 1

v– Y v–

v+

v+

v+

v+

v+ Z-1v–

v–

v+

v– Z v+

v+

v–

v–

v–

v+ v+v+

v+ v–v–

v– ZX v–v+

v– 0v–

v–

v+

v+

v–

v–

v–

X

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HomologiesObservation The most general ring is ℤ[X, Y, Z±1]/(X2 = Y2 = 1).

I Equivalence: (X, Y, Z) (-X, -Y, -Z)

-

and Id on others generators.

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HomologiesObservation The most general ring is ℤ[X, Y, Z±1]/(X2 = Y2 = 1).

I Equivalence: (X, Y, Z) (-X, -Y, -Z)

II Equivalence: (X, Y, Z) (X, Y, 1)(X, Y, Z) (V, m, Δ, η, ε, P)(X, Y, 1) (V, m’, Δ’, η’, ε’, P’)Take φ: V V as follows: φ(v+) = v+

φ(v-) = Zv-

Define Φn: Vn Vn:Φn = φn-1 … φ id.

ThenΦ: (V, m, Δ, η, ε, P) (V, m’, ZΔ’, η’, ε’, P’)

Use now the functor given by

Z and Id on others generators

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HomologiesObservation The most general ring is ℤ[X, Y, Z±1]/(X2 = Y2 = 1).

I Equivalence: (X, Y, Z) (-X, -Y, -Z)

II Equivalence: (X, Y, Z) (X, Y, 1)

Corollary There exist only two theories over an integral domain. Observation Homologies KhXYZ are dual to KhYXZ :

KhXYZ (T*) = KhYXZ(T)*

Corollary Odd link homologies are self-dual.

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Tangle cobordismsTheorem For any cobordism M between tangles T1 ans T2 there exists a map

Kh(M): Kh(T1) Kh(T2)defined upto a unit.Sketch of proof (local part like in Bar-Natan’s)Need to define chain maps for the following elementary cobordisms and its inverses:

first row: chain maps from the prove of invariance theoremsecond row: the cobordisms themselves.

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Tangle cobordisms

Satisfied due to the invariance theorem.

I type of moves: Reidemeister moves with inverses („do nothing”)

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Tangle cobordismsII type of moves: circular moves („do nothing”)

- flat tangle is Kh-simple (any automorphism of Kh(T) is a multi-

plication by a unit)- appending a crossing preserves Kh-simplicity

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Tangle cobordismsIII type of moves: non-reversible moves

Need to construct maps explicite.

Problem No planar algebra in the category of complexes: having planar operator D and chain maps f: A A’, g: B B’, the induced map

D(f, g): D(A, B) D(A’, B’)may not be a chain map!

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000 100

110

111011

001

010

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*001*0

11*

*11

00*01* 10*

0*1

0*0

1*1*01

F0

Local to global: partial complexes

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000 100

110

111011

001

010

101

*001*0

11*

*11

00*01* 10*

0*1

0*0

1*1*01

F0

F1

Local to global: partial complexes

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000 100

110

111011

001

010

101

*001*0

11*

*11

00*01* 10*

0*1

0*0

1*1*01

F0

F*

F1

Local to global: partial complexes

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Summing a cube of complexes000 100

110

111011

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*001*0

11*

*11

00*01* 10*

0*1

0*0

1*1*01

F0

F*

F1

KomnF – cube of partial complexes

example: Kom2F(0) = KomF0

Proposition Komn Komm = Komm+n

Local to global: partial complexes

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Tangle cobordismsBack to proof Take two tangles

T = D(T1, T2) T’ = D(T1’, T2)and an elementary cobordisms M: T1 T1’. For each smoothed diagram ST2 of T2 we have a morphism

D(Kh(M), Id): D(Kh(T1), ST2) D(Kh(T1), ST2)

- show it always has an edge assignment- any map given by one of the relation movies induced a

chain map equal Id (D is a functor of one variable)

These give a cube map of partial complexesf: KomnC(T) KomnC(T ’)

where n is the number of crossings of T2.

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References1. D. Bar-Natan, Khovanov's homology for tangles and

cobordisms, Geometry and Topology 9 (2005), 1443-1499

2. J S Carter, M Saito, Knotted surfaces and their diagrams, Mathematical Surveys and Monographs 55, AMS, Providence, RI(1998)

3. V. F. R. Jones, Planar Algebras I, arXiv:math/9909027v1

4. M. Khovanov, A categorication of the Jones polynomial, Duke Mathematical Journal 101 (2000), 359-426

5. P. Osvath, J. Rasmussen, Z. Szabo, Odd Khovanov homology, arXiv:0710.4300v1

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Thank youfor your attention