Odd homologyof tangles and cobordisms

Krzysztof PutyraJagiellonian University, Kraków

XXVII Knots in Washington10th January 2009

Cube of resolutions

0-smoothing 1-smoothing

Mikhail Khovano

v

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

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

)

000

100

010

001

110

101

011

111

vertices are smoothed diagrams

Peter Ozsvath

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.

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

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

ChCob: cobordisms with chronology & arrows

Critical points cannot be permuted:

Critical points do not vanish:

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

with the full set of relations given by:

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:

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

Which conditions should a functor

F: ChCob ℤ-Mod

satisfies to produce homologies?

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!

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

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

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!

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

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).

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.

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

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

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.

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

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.

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.

Tangle cobordisms

Satisfied due to the invariance theorem.

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

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

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!

000 100

110

111011

001

010

101

*001*0

11*

*11

00*01* 10*

0*1

0*0

1*1*01

F0

Local to global: partial complexes

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

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

Summing a cube of complexes000 100

110

111011

001

010

101

*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

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.

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

Thank youfor your attention