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

Cube of resolutions 0 -smoothing 1 -smoothing Mikhail Khovanov

Cube of resolutions C -3 C -2 C -1 C0C0 d dd edges are cobordisms direct sums create the complex vertices are smoothed diagrams – – – – Mikhail Khovanov

Cube of resolutions C -3 C -2 C -1 C0C0 d dd edges are cobordisms with arrows direct sums create the complex (applying some edge assignment) vertices are smoothed diagrams Peter Ozsvath

Khovanov functor see Khovanov: arXiv:math/ F Kh : Cob → ℤ -Mod symmetric: Edge assignment is given explicite. Category of cobordisms is symmetric: ORS ‘projective’ functor see Ozsvath, Rasmussen, Szabo: arXiv: F ORS : 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 F ORS a functor? Anwser Yes: cobordisms with chronology Main question Dror Bar-Natan

ChCob: cobordisms with chronology & arrows Chronology 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: - H 0 = τ 0 - H 1 = τ 1 - H t is a chronology

ChCob: cobordisms with chronology & arrows Critical points cannot be permuted: Critical points do not vanish:

ChCob: cobordisms with chronology & arrows Theorem 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. - H 0 = τ 0, H 1 = τ 1 - H t is a chronology except t 1,…,t n, where one of the following occurs:

ChCob(B): cobordisms with corners For 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: 132 (M 1,M 2,M 3 ) M1M1 M3M3 M2M2

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 condition Extend cobordisms to formal sums over a commutative ring R. Find a representation of changes of chronology in U(R) s.th. α M 1 … M s = β M 1 … M s => 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 condition Extend cobordisms to formal sums over a commutative ring R. Find a representation of changes of chronology in U(R) s.th. α M 1 … M s = β M 1 … M s => α = β Fact WLOG creation and removing critical points can be represented by 1. Proposition The representation is given by where X 2 = Y 2 = 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 assignment Proposition 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) = -λ is a cocycle: P i = 1 6 By the ch. ch. condition: dψ(C) = Π -λ i = 1 and by the contractibility of a 3 -cube: ψ = dφ 6 i = 1 P = λ r PP = λ r P = λ r λ f P P = λ r P = λ r λ f P =... = Π λ i P

Edge assignment 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. Sketch of proof Let φ 1 and φ 2 be edge assignments for a cube C(D). Then d(φ 1 φ 2 -1 ) = dφ 1 dφ 2 -1 = ψψ -1 = 1 Thus φ 1 φ 2 -1 is a cocycle, hence a coboundary. Putting φ 1 = d ηφ 2 we obtain an isomorphism of complexes ηid: Kh(D,φ 1 ) → Kh(D,φ 2 ).

Edge assignment Proposition Denote by D 1 and D 2 a tangle diagram D with different choices of arrows. Then there exist edge assignments φ 1 and φ 2 s.th. complexes C(D 1, φ 1 ) and C(D 2, φ 2 ) 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 relations compare with Bar-Natan: arXiv:math/ 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

Homologies M F XYZ (M) Rv+⊕Rv–Rv+⊕Rv– 1  v + v +   + ZY  v –   v +  0 v –  1 v –   Y  v – v+v+v+v+ v+v+v+v+ v +   Z -1  v – v–v+v–v+   Z  v + v+v–v+v– v–v–v–v– v+ v+v+v+ v+v+ v+ v–v–v+ v–v– v –   ZX v – v + v– 0v–v– 0v– v–v+v–v+ v+v–v+v– v–v–v–v–   X   X 

Homologies Observation The most general ring is ℤ [X, Y, Z ±1 ]/(X 2 = Y 2 = 1). I Equivalence: (X, Y, Z)  (-X, -Y, -Z) - and Id on others generators.

Homologies Observation The most general ring is ℤ [X, Y, Z ±1 ]/(X 2 = Y 2 = 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 Φ n : V  n  V  n : Φ n = φ n-1  …  φ  id. Then Φ: (V, m, Δ, η, ε, P)  (V, m’, ZΔ’, η’, ε’, P’) Use now the functor given by Z and Id on others generators

Homologies Observation The most general ring is ℤ [X, Y, Z ±1 ]/(X 2 = Y 2 = 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 Kh XYZ are dual to Kh YXZ : Kh XYZ (T*) = Kh YXZ (T)* Corollary Odd link homologies are self-dual.

Tangle cobordisms Theorem For any cobordism M between tangles T 1 ans T 2 there exists a map Kh(M): Kh(T 1 )  Kh(T 2 ) 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 theorem second row: the cobordisms themselves.

Tangle cobordisms Satisfied due to the invariance theorem. I type of moves: Reidemeister moves with inverses („do nothing”)

Tangle cobordisms II 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 cobordisms III 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!

*00 1*0 11* *11 00* 01* 10* 0*1 0*0 1*1 *01 F0F0 Local to global: partial complexes

*00 1*0 11* *11 00* 01* 10* 0*1 0*0 1*1 *01 F0F0 F1F1 Local to global: partial complexes

*00 1*0 11* *11 00* 01* 10* 0*1 0*0 1*1 *01 F0F0 F*F* F1F1 Local to global: partial complexes

Summing a cube of complexes *00 1*0 11* *11 00* 01* 10* 0*1 0*0 1*1 *01 F0F0 F*F* F1F1 Kom n F – cube of partial complexes example: Kom 2 F( 0) = KomF 0 Proposition Kom n  Kom m = Kom m+n Local to global: partial complexes

Tangle cobordisms Back to proof Take two tangles T = D(T 1, T 2 ) T’ = D(T 1 ’, T 2 ) and an elementary cobordisms M: T 1  T 1 ’. For each smoothed diagram ST 2 of T 2 we have a morphism D(Kh(M), Id): D(Kh(T 1 ), ST 2 )  D(Kh(T 1 ), ST 2 ) - 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 complexes f: Kom n C(T)  Kom n C(T ’) where n is the number of crossings of T 2.

References 1.D. Bar-Natan, Khovanov's homology for tangles and cobordisms, Geometry and Topology 9 (2005), 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/ v1 4.M. Khovanov, A categorication of the Jones polynomial, Duke Mathematical Journal 101 (2000), P. Osvath, J. Rasmussen, Z. Szabo, Odd Khovanov homology, arXiv: v1

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