odd homology of tangles and cobordisms
DESCRIPTION
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 PresentationTRANSCRIPT
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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
)
000
100
010
001
110
101
011
111
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
101
*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
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
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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