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Odd homology of tangles and cobordisms

This paper discusses the relationship between cobordisms and the odd homology of tangles, utilizing Mikhail Khovanov's cube of resolutions. It explores various smoothing techniques and the significance of edges as cobordisms in constructing complexes. Key contributions include examining the symmetry of the category of cobordisms and conditions necessary for a functor from cobordisms to homological structures. The investigation aims to enhance the understanding of invariance in the Khovanov complex, potentially resulting in new theories rooted in topology.

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Odd homology of tangles and cobordisms

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  1. Oddhomologyof tangles and cobordisms Krzysztof Putyra JagiellonianUniversity, Kraków XXVII KnotsinWashington 10th January 2009

  2. 0-smoothing 1-smoothing Mikhail Khovanov Cube of resolutions

  3. 110 100 000 – 111 101 010 – – 001 – d d d C-3 C-2 C-1 C0 Mikhail Khovanov Cube of resolutions 1 3 2 edgesarecobordisms verticesaresmootheddiagrams 011 direct sums create the complex

  4. 110 100 000 111 101 010 001 d d d C-3 C-2 C-1 C0 Cube of resolutions 1 3 2 edgesarecobordismswitharrows verticesaresmootheddiagrams 011 direct sums create the complex (applying some edge assignment) Peter Ozsvath

  5. Khovanov functor seeKhovanov: 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 notsymmetric: Edge assignmentisgiven by homologicalproperties.

  6. Main question Fact (Bar-Natan) Invariance of the Khovanov complex can be proved at the level of topology. QuestionCan Cob be changed to make FORS a functor? Motivation Invariance of the odd Khovanov complex may be proved at the level of topology and new theories may arise. Dror Bar-Natan Anwser Yes: cobordisms with chronology

  7. ChCob: cobordisms with chronology & arrows Chronologyτis a Morse function with exactly one critical point over each critical value. • Critical points of index 1 have arrows: • τdefines a flowφ on M • critical point of τarefixpointsφ • arrowschoose one of thein/outcoming • trajectory for a critical point. Chronology isotopyis a smooth homotopyH satisfying: - H0 = τ0 - H1= τ1 - Ht is a chronology

  8. Criticalpointscannot be permuted: Critical pointsdo not vanish: ChCob: cobordisms with chronology & arrows

  9. with the full set of relations given by: ChCob: cobordisms with chronology & arrows TheoremThe category 2ChCob is generated by the following:

  10. ChCob: cobordisms with chronology & arrows • Change of chronologyis a smoothhomotopyHs.th. • H0 = τ0, H1 = τ1 • Htis a chronologyexceptt1,…,tn, where one of thefollowingoccurs: Theorem2ChCobwithchanges of chronologiesis a 2-category.

  11. ChCob(B): cobordisms with corners For tangles we needcobordismswithcorners: • inputand outputhassame endpoints • projectionis a chronology • chooseorientation for eachcriticalpoint • allup to isotopiespreservingπbeing • a chronology ChCob(B)‘s form a planar algebra withplanaroperators: M1 1 3 2 M2 (M1,M2,M3) M3

  12. Which conditions should a functor F: ChCob ℤ-Mod satisfiesto produce homologies?

  13. 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!

  14. Chronology change condition Extendcobordisms to formalsumsover a commutative ring R. Find a representation of changes of chronologyinU(R) s.th. α M1 … Ms = β M1 … Ms => α = β Fact WLOG creation and removingcriticalpointscan be represented by 1. HintConsiderthefunctorgiven by: Id on othersgen’s α β

  15. Chronology change condition Extendcobordisms to formalsumsover a commutative ring R. Find a representation of changes of chronologyinU(R) s.th. α M1 … Ms = β M1 … Ms => α = β Fact WLOG creation and removingcriticalpointscan be represented by 1. PropositionTherepresentationisgiven by Z X Y 1 XY whereX2 = Y2 = 1 and Zis a unit.

  16. 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!

  17. By the ch. ch. condition: dψ(C) = Π-λi = 1 and by the contractibility of a 3-cube: ψ = dφ 6 i = 1 Edge assignment PropositionFor any cube of resolutions C(D) there exists an edge assignment e → φ(e)e making the cube anticommutative. Sketch of proofEach square S corresponds to a change of chronology with somecoefficient λ. The cochain ψ(S) = -λ is a cocycle: 6 P = λrP = λrλf P = ... = ΠλiP P = λrP P = λrP = λrλf P P i = 1

  18. Sketch of proof Let φ1 and φ2 be edge assignments for a cube C(D). Then d(φ1φ2-1) = dφ1dφ2-1 = ψψ-1 = 1 Edge assignment PropositionFor any cube of resolutions C(D) there exists an edge assignment e → φ(e)e making the cube anticommutative. PropositionFor any cube of resolutions C(D) different egde assign-ments produce isomorphic complexes. Thusφ1φ2-1is a cocycle, hence a coboundary. Putting φ1 = dηφ2 we obtain an isomorphism of complexesηid: Kh(D,φ1) → Kh(D,φ2).

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

  20. S / T / 4Tu relations compare with Bar-Natan: arXiv:math/0410495 TheoremThecomplexKh(D) isinvariant under chainhomotopies and thefollowingrelations: Dror Bar-Natan where X, Y and Z are given by the ch.ch.c.

  21. Homologies v+ v+ v– v– v+ v+ v+ Z-1 v– v– v+ v– Z  v+ v+ v– v– Y  v–  X  v+ v+ v+ v+ v– v– v– 0 v– v– ZX v– v+ v– v+ v+ v– v– v–

  22. Homologies ObservationThe most general ring isℤ[X, Y, Z±1]/(X2 = Y2 = 1). I Equivalence: (X, Y, Z)  (-X, -Y, -Z) - and Id on othersgenerators.

  23. Homologies ObservationThe most general ring isℤ[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- DefineΦn: VnVn: Φn = φn-1 … φ id. Then Φ: (V, m, Δ, η, ε, P)  (V, m’, ZΔ’, η’, ε’, P’) Usenowthefunctorgiven by Z and Id on othersgenerators

  24. Homologies ObservationThe most general ring isℤ[X, Y, Z±1]/(X2 = Y2 = 1). I Equivalence: (X, Y, Z)  (-X, -Y, -Z) II Equivalence: (X, Y, Z)  (X, Y, 1) CorollaryThereexistonlytwotheoriesover an integraldomain. ObservationHomologiesKhXYZare dual to KhYXZ: KhXYZ (T*) = KhYXZ(T)* CorollaryOdd link homologiesareself-dual.

  25. Tanglecobordisms TheoremFor anycobordism M betweentangles T1 ans T2thereexists a map Kh(M): Kh(T1)  Kh(T2) definedupto a unit. Sketch of proof (local part likeinBar-Natan’s) Need to definechainmaps for thefollowingelementarycobordisms and itsinverses: first row: chainmapsfromtheprove of invariancetheorem secondrow:thecobordismsthemselves.

  26. Tanglecobordisms I type of moves: Reidemeistermoveswithinverses („do nothing”) Satisfieddue to theinvariancetheorem.

  27. Tanglecobordisms II type of moves: circularmoves („do nothing”) • flattangleisKh-simple(any automorphism of Kh(T) is a multi- • plication by a unit) • appending a crossingpreservesKh-simplicity

  28. Tanglecobordisms III type of moves: non-reversiblemoves Need to constructmaps explicite. Problem No planar algebra inthecategory of complexes: havingplanar operator D and chainmapsf: AA’, g: BB’, theinduced map D(f, g): D(A, B) D(A’, B’) may not be a chain map!

  29. Local to global: partialcomplexes 000 100 *00 F0 1*0 0*0 110 010 00* 10* 01* *01 11* 001 1*1 101 0*1 *11 011 111

  30. Local to global: partialcomplexes 000 100 *00 F0 1*0 0*0 110 010 00* 10* 01* *01 11* 001 1*1 101 0*1 F1 *11 011 111

  31. Local to global: partialcomplexes 000 100 *00 F0 1*0 0*0 110 010 00* 10* 01* F* *01 11* 001 1*1 101 0*1 F1 *11 011 111

  32. Local to global: partialcomplexes Summing a cube of complexes 000 100 *00 F0 1*0 0*0 110 010 00* 10* 01* F* *01 11* 001 1*1 101 0*1 F1 *11 011 111 KomnF – cube of partialcomplexes example: Kom2F(0) = KomF0 PropositionKomnKomm = Komm+n

  33. Tanglecobordisms Back to proof Take twotangles T = D(T1, T2) T’ = D(T1’, T2) and an elementarycobordismsM: T1T1’. For eachsmoothed diagram ST2 of T2 we have a morphism D(Kh(M), Id): D(Kh(T1), ST2)  D(Kh(T1), ST2) Thesegive a cube map of partialcomplexes f: KomnC(T) KomnC(T ’) wherenisthenumber of crossings of T2. • show italwayshas an edgeassignment • any map given by one of therelationmoviesinduced a chain map • equalId (Dis a functor of one variable)

  34. References D. Bar-Natan, Khovanov's homology for tangles and cobordisms, Geometry and Topology 9 (2005), 1443-1499 J S Carter, M Saito, Knotted surfaces and their diagrams, MathematicalSurveys and Monographs 55, AMS, Providence, RI(1998) V. F. R. Jones, Planar Algebras I, arXiv:math/9909027v1 M. Khovanov, A categorication of the Jones polynomial, Duke Mathematical Journal 101 (2000), 359-426 P. Osvath, J. Rasmussen, Z. Szabo, OddKhovanovhomology, arXiv:0710.4300v1

  35. Thankyoufor yourattention

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