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Odd homology of tangles and cobordisms

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.

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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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