knot for knovicestc4117/assets/knot_theory.pdf · 2019. 8. 30. · knot for knovices tudor ciurca ,...

Knot for Knovices

Tudor Ciurca∗, Kenny Lau†, Alex Cahill‡, Derek Leung§

Imperial College London

17 June 2019

Abstract

This is a survey of some key results and constructions in knot theory,and is completed as part of the M2R group project. After developingsome basic notions about knots and links, we will step into basic knotarithmetic and operations, exploring fundamental theorems such as Rei-demeister Theorem and Schubert’s Theorem. Next, it would be a naturalmanner to step into the world of invariants. We will introduce variousinteger-valued and polynomial invariants, before wrapping up this chap-ter with Vassiliev’s masterpiece of family of invariants. After that, wewill explore Khovanov homology, which is the categorification of Jones’polynomial. It is indeed a powerful invariant, as we will see it solves theunknot recognition problem. Finally, we will switch our focus to the knotgroup and its Wirtinger presentation to end.

∗Email: [email protected]†Email: [email protected], kc [email protected]‡Email: [email protected]§Email: [email protected]

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Contents

1 Introduction 31.1 What is a knot? . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 When are two knots the same? . . . . . . . . . . . . . . . . . . . 41.3 Making finite and discrete models . . . . . . . . . . . . . . . . . . 51.4 Projections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.5 Links . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.6 Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Reidemeister’s Theorem and Seifert surfaces 132.1 Reidemeister’s Theorem . . . . . . . . . . . . . . . . . . . . . . . 132.2 Basic knot arithmetic . . . . . . . . . . . . . . . . . . . . . . . . 142.3 Seifert surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.4 Schubert’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 18

3 Integer-valued and polynomial invariants 213.1 An overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.2 3-colouring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.3 p-colouring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.4 Seifert matrix and Alexander polynomial . . . . . . . . . . . . . . 323.5 Jones polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.6 Vassiliev invariants . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4 From Jones’ polynomial to Khovanov homology 484.1 Alternative version of Jones polynomial . . . . . . . . . . . . . . 484.2 Khovanov homology . . . . . . . . . . . . . . . . . . . . . . . . . 494.3 Khovanov homology is an oriented link invariant . . . . . . . . . 554.4 Properties of Khovanov homology . . . . . . . . . . . . . . . . . . 60

5 The knot group 655.1 Definition and basic properties . . . . . . . . . . . . . . . . . . . 655.2 Wirtinger presentation . . . . . . . . . . . . . . . . . . . . . . . . 675.3 Meridian and longitude . . . . . . . . . . . . . . . . . . . . . . . 73

6 Bibliography 76

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1 Introduction

1.1 What is a knot?

Definition 1.1. An oriented knot is an injective continuous map f : S1 →R3, where S1 := (x, y) ∈ R2 | x2 + y2 = 1 ⊂ R2 is the unit circle.

Definition 1.2. A knot is a subset of R3 that is homeomorphic to S1.

Examples include the unknot:

01 : (x, y) 7→ (x, y, 0)

and the trefoil knot:

31 : (cos t, sin t) 7→ (r = 2 + cos(3t), θ = 2t, z = sin(3t))

which in Cartesian coordinates becomes:

31 : (x, y) 7→ ((4x3 − 3x)(2x2 − 1), (4x3 − 3x+ 2)(2xy), 3y − 4y3)

Their projections are depicted below.

(a) 01: unknot (b) 31: trefoil

Figure 1: Projection of two knots

Definition 1.3. If f is an oriented knot, then t 7→ f(−t) is the orientation-reversal of the oriented knot.

Definition 1.4. If f is an oriented knot, then the image of f is a knot, calledthe underlying knot.

Remark 1.5. The underlying knot “forgets the orientation”, since the orientation-reversal has the same underlying knot.

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Since S3 := (x, y, z, w) ∈ R4 | x2 + y2 + z2 + w2 = 1, the unit 3-sphere inR4, is the one-point compactification of R3, hence by noting that the image ofa knot is already compact , one can instead define an oriented knot to be aninjective continuous map f : S1 → S3 without loss of generality, and similarlyfor a knot. No such maps can be surjective, for such a map would be bijective;but S1 is compact, S3 is Hausdorff, which implies that such a map would bea homeomorphism; but S1 and S3 are not homeomorphic, as one can removetwo points from S3 and still get a connected space, yet the removal of any twopoints from S1 results in a disconnected space. Therefore, from an orientedknot S1 → S3 we can pick a point not in the image, and then use stereographicprojection on the complement of the point to make an oriented knot S1 → R3;similarly for a knot.In the following we would identify S3 with R3 ∪ ∞ freely, as well as switchthe definition of an (oriented) knot freely between an injective continuous mapS1 → R3 and an injective continuous map S1 → S3.

1.2 When are two knots the same?

One would count “deformations” (to be defined) of a knot as the same knot,so we would need a notion of equivalence of knots that includes at least allthe translations, non-zero dilations, and rotations. Deformations should becontinuous, which leads to this attempted definition:

Attempted definition: Two knots f, g : S1 → R3 are equivalent ifthere is a family of knots Fs : S1 → R3 indexed by s ∈ [0, 1] suchthat F0 = f , F1 = g, and [0, 1]× S1 : (s, t) 7→ Fs(t) is continuous.

This definition is similar to the definition of a homotopy, but here we requireeach intermediate function Fs : S1 → R3 to be injective as well. However, onecan check similarly that this equivalence is indeed an equivalence relation:

• That f ∼ f is witnessed by Fs(t) := f(t).

• Suppose f ∼ g is witnessed by Fs. Then g ∼ f is witnessed by F1−s.

• Suppose f ∼ g is witnessed by Fs and g ∼ h is witnessed by Gs. Thenf ∼ h is witnessed by Hs where Hs := F2s for s ∈

[0, 1

2

]and Hs := G2s−1

for s ∈[

12 , 1]. It is continuous because F1 = G0 = g.

The problem with this definition of equivalence, however, is that every knotis equivalent to the unknot under this definition. To see this, take any knotf : S1 → R3. Since S1 is compact, there is t such that the x-coordinate off(t) is maximum. Using that point as reference, break open the knot from thatpoint while connecting the two resulting points with an arc growing in size tothe right, then shrink everything from the left to the reference point, graduallyre-parametrising the function so that the limit is still injective, and then finallyuse a translation to turn it into the standard unknot.The problem can be fixed by considering the whole space instead of just theknot:

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Definition 1.6. Two oriented knots f, g : S1 → R3 are equivalent if theyare ambient isotopic, i.e. there is a family of homeomorphisms Fs : R3 → R3

for s ∈ [0, 1] with F0(x) = x for all x and F1(f(t)) = g(t) for all t, such that[0, 1]× R3 → R3 : (s, x) 7→ Fs(x) is continuous.

Definition 1.7. Two knots K1,K2 ⊂ R3 are equivalent if there is a familyof homeomorphisms Fs : R3 → R3 for s ∈ [0, 1] with F0(x) = x for all x andF1(K1) = K2, such that [0, 1]× R3 → R3 : (s, x) 7→ Fs(x) is continuous.

Remark 1.8. • These two notions of equivalence are both equivalence re-lations, using a similar argument.

• If two oriented knots are equivalent, then clearly their underlying knotsare also equivalent.

• A version of the converse is true for nice enough knots, called “tame”knots, defined below in Definition 1.13. If two oriented knots are niceenough and their underlying knots are equivalent, then the two orientedknots are either equivalent, or one is equivalent to the orientation-reversalof the other (Theorem 1.16 below).

Given a continuous family of homeomorphisms Fs : R3 → R3 where F0 isthe identity, we say that F1 is isotopic to the identity. A homeomorphismR3 → R3 is isotopic to the identity if and only if it is orientation-preserving(Theorem 3, Sanderson 1960), where a homeomorphism h : R3 → R3 is said tobe orientation-preserving iff for every x ∈ R3 the induced map h∗ : H3(R3,R3−x)→ H3(R3,R3−f(x)) is the same map as the map induced by translationhx(y) := y − x + h(x) for all x ∈ R3, where H•(A,B) is the relative singularhomology. To facilitate computations, one notes that for any x ∈ R3, if his differentiable at x and the derivative at that point is non-singular, then hpreserves orientation at x iff det(Dhx) > 0. This gives us a criterion that issimpler at times:

Theorem 1.9. Two oriented knots f, g : S1 → R3 are equivalent iff there is anorientation-preserving homeomorphism h : R3 → R3 such that h f = g.Similarly, two knots K1,K2 ⊂ R3 are equivalent iff there is an orientation-preserving homeomorphism h : R3 → R3 such that h(K1) = K2.

1.3 Making finite and discrete models

To facilitate theory as well as actual computations, we need a finite and discreteversion of knots. Unfortunately our current definition of knots is too broadwhich allows for some pathological examples called wild knots, which we do notprefer to compute with.

Definition 1.10. A piecewise-linear knot is an oriented knot f : S1 → R3

such that there are t1, · · · , tn ∈ S1 with tn = t1 such that f restricted to [ti, ti+1]is linear for every i = 1→ n− 1.

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Its underlying knot is called a polygonal knot, and is characterized by theexistence of p1, · · · , pn ∈ R3 with pn = p1 such that the knot is [p1, p2]∪ [p2, p3]∪· · · ∪ [pn−1, pn] and that the line segments do not intersect except the fact thatthe end of [pi, pi+1] is the beginning of [pi+1, pi+2].

Here is a polygonal version of the trefoil:

Figure 2: Polygonal realization of trefoil knot

Definition 1.11. A knot K is said to be locally flat if for every p ∈ K thereis an open neighbourhood N 3 p with a homeomorphism h : N → B3 takingN ∩K to the straight line connecting (0, 0,−1) to (0, 0, 1).

Theorem 1.12. The following are equivalent for an oriented knot f : S1 → R3:

1. f is equivalent to a piecewise-linear knot.

2. f is equivalent to a smooth knot, i.e. a smooth embedding S1 → R3.

3. f : S1 → R3 extends to an injective continuous map f : S1 ×B2 → R3.

4. The underlying knot of f is locally flat.

Here Bn is the standard open ball in n dimensions, i.e. ~v ∈ Rn : ‖~v‖ < 1.

Proof. 1 =⇒ 2: Suppose f is equivalent to a piecewise-linear knot p. Aroundeach corner of p one can draw small neighbourhoods such that the intersectionof the knot and each small neighbourhood is the two line segments touchingthe corner. Then one can use bump functions (Lemma 2.14, Lee 2003) toperturb p inside those small neighbourhoods and deform the knot into a smooth

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embedding using a deformation that satisfies the hypothesis of the topologicalisotopy extension theorem (Corollary 1.4, Edwards and Kirby 1971), so thisdeformation extends to an ambient isotopy, showing that p is equivalent to asmooth knot, so f is also equivalent to a smooth knot.2 =⇒ 3: Suppose f is equivalent to a smooth knot s, with F being theorientation-preserving homeomorphism. The image of s is a 1-dimensional sub-manifold of R3, and has a tubular neighbourhood by (Theorem 5.1, Chapter4, Hirsch 1976), i.e. a smooth embedding s : S1 × B2 → R3 extending s.Composing with F−1 gives the desired result.3 =⇒ 4: For t ∈ S1 ⊂ R2 let U = S1−−t ⊂ S1, and U = U×B2 ⊂ S1×B2.U is homeomorphic to B3 and there is a homeomorphism that takes U ×0 tothe straight line connecting (0, 0,−1) to (0, 0, 1). B3 is an open subset of R3,

so by invariance of domain (Theorem 2B.3, Hatcher 2001), f(U)

is an open

subset of R3 as well, and f restricted to U is a homeomorphism onto f(U)

,

which is the desired neighbourhood.4 =⇒ 1: R3 is a triangulated 3-manifold, so it follows from (Theorem 8.1,Moise 1954).

Definition 1.13. An oriented knot satisfying the above equivalent conditions iscalled a tame knot.

Remark 1.14. From the fourth equivalent definition we see that an orientedknot being tame depends only on its underlying knot.

From now on we will only work with tame knots unless otherwise stated.The first benefit is that for tame knots there are only two orientations:

Theorem 1.15. Suppose f and g are tame knots with the same underlyingknot. Then f is either equivalent to g or to the orientation-reversal (Definition1.3) of g.

Proof. Consider s : S1 → S1 such that f(t) = g(s(t)). It can be shown thats is a well-defined homeomorphism. Label the circle [0, 2π]. Lift s to a con-tinuous function s : [0, 2π] → R. Then s is either strictly increasing or strictlydecreasing.If s is strictly decreasing, then replacing g(t) with g(−t) to begin with wouldmake s strictly increasing, so we now assume s is strictly increasing and showthat f is equivalent to g.Consider H : [0, 1]×S1 → R, H(r, t) := (1− r)s(t) + rt. For each r, t 7→ H(r, t)is strictly increasing, since both s(t) and t are strictly increasing and (1 − r)and r are both non-negative and at least one positive. We can now note that(r, t) 7→ g(H(r, t)) is a deformation of f to g with locally flat knots, since theintermediate knots have the same underlying knot. The deformation can bemade to satisfy the hypothesis of the topological isotopy extension theorem(Corollary 1.4, Edwards and Kirby 1971), so this deformation extends to anambient isotopy, showing that f is equivalent to g.

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Corollary 1.16. Suppose f and g are tame knots with equivalent underlyingknots. Then f is either equivalent to g or to the orientation-reversal (Definition1.3) of g.

We also need a finite and discrete version of equivalence. A slight deformationcan be approximated by replacing one side of a given external triangle with theremaining two sides, as defined below:

Definition 1.17. An elementary isotopy applied to a polygonal knot p withvertices p1, · · · , pn, specified by an index i and a triangle [pi, pi+1, q] which onlyintersects the knot at [pi, pi+1], is a polygonal knot formed by replacing the linesegment [pi, pi+1] with the two line segments [pi, q] and [q, pi+1].

We let the elementary isotopies generate an equivalence between polygonalknots:

Definition 1.18. Two polygonal knots p and q are said to be isotopic, alsoknown as combinatorially equivalent, if there is a finite sequence of polygonalknots p = p0, p1, · · · , pn = q, such that each pi and pi+1 are related by anelementary isotopy or the inverse of an elementary isotopy.

One wonders whether this finite and discrete version of equivalence is the sameas the notion of equivalence defined earlier:

Theorem 1.19. Two polygonal knots p and q are equivalent iff they are isotopic.

Proof. See (Proposition 1.10, Burde, Zieschang, and Heusener, 2013).

1.4 Projections

Since most media only allow 2-dimensional diagrams, including this text, wemust find a way to project our knot to R2 that preserves the data of the knot.This motivates the following definition:

Definition 1.20. Given a 2-dimensional subspace E and a polygon knot p, wesay that p has regular projection to E if finitely many points on E are doublepoints, i.e. mapped by two distinct points on p, and the other points on theimage are single points, i.e. mapped by only one point on p, and no vertex isprojected to a double point. The double points are called the crossings.

In fact for a given polygon knot p, such subspaces are very abundant:

Theorem 1.21. For a given polygon knot p, the set of subspaces E such thatp has regular projection to E is an open dense subset of all the 2-dimensionalsubspaces of R3, i.e. the Grassmannian Gr(2,R3). That is to say, for any E,there are subspaces arbitrary close to E admitting a regular projection of p; andif E admits regular projection of p, then any subspace sufficiently close to Eadmits a regular projection of p.

Proof. See (Proposition 1.12, Burde, Zieschang, and Heusener, 2013).

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Remark 1.22. On a double point, we use a solid line to denote the point ontop and a broken line to denote the point below, after choosing a direction ofheight. It is with this principle that Figure 1 is produced.

1.5 Links

The basic definitions and theorems related to knots generalise easily to links,disjoint unions of knots. We will thus state the theorems in this subsectionwithout proof.

Definition 1.23. An oriented link with n components is an injective contin-uous map f :

∐ni=1 S

1 → R3 or f :∐ni=1 S

1 → S3. An oriented link decomposesinto n oriented knots.

Definition 1.24. A link is a subset of R3 or S3 that is homeomorphic to∐ni=1 S

1. A link decomposes into n path-connected components, each of whichis a knot.

Examples include the unlink and the Hopf link:

(a) Unlink (b) Hopf link

Figure 3: Projection of two links

Definition 1.25. If f is an oriented link, then the image of f is a link, calledthe underlying link.

Definition 1.26. If f is an oriented link, then one can perform orientation-reversal (Definition 1.3) on each component to get 2n links, where n is thenumber of components.

Remark 1.27. The underlying link “forgets the orientation”, since each orientation-reversal has the same underlying link.

Definition 1.28. Two oriented links f, g :∐ni=1 S

1 → R3 are equivalent ifthey are ambient isotopic.

Definition 1.29. Two link L1, L2 ⊂ R3 are equivalent if there is a familyof homeomorphisms Fs : R3 → R3 for s ∈ [0, 1] with F0(x) = x for all x andF1(L1) = L2, such that [0, 1]× R3 → R3 : (s, x) 7→ Fs(x) is continuous.

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Remark 1.30. • These two notions of equivalence are both equivalence re-lations.

• If two oriented links are equivalent, then their underlying knots are alsoequivalent.

• If two tame oriented links (Definition 1.35) have equivalent underlyinglinks, then one is equivalent to an oriental-reversal (Definition 1.26) ofthe other (Theorem 1.38).

Theorem 1.31. Two oriented link f, g :∐ni=1 S

1 → R3 are equivalent iff thereis an orientation-preserving homeomorphism h : R3 → R3 such that h f = g.Similarly, two links L1, L2 ⊂ R3 are equivalent iff there is an orientation-preserving homeomorphism h : R3 → R3 such that h(L1) = L2.

Definition 1.32. A piecewise-linear link is an oriented link with each com-ponent piecewise-linear.Its underlying link is called a polygonal link, and is characterized by that eachcomponent is polygonal.

Definition 1.33. A link L is said to be locally flat if every component islocally flat.

Theorem 1.34. The following are equivalent for an oriented link f :∐ni=1 S

1 →R3:

1. f is equivalent to a piecewise-linear link.

2. f is equivalent to a smooth link, i.e. a smooth embedding S1 → R3.

3. f :∐ni=1 S

1 → R3 extends to an injective continuous map f :∐ni=1 S

1 ×B2 → R3.

4. The underlying link of f is locally flat.

Definition 1.35. An oriented link satisfying the above equivalent conditions iscalled a tame link.

From now on we will only work with tame links, unless otherwise specified.

Remark 1.36. From the fourth equivalent definition we see that an orientedlink being tame depends only on its underlying link.

Theorem 1.37. Suppose f and g are tame links with the same underlying link.Then f is either equivalent to g or to an orientation-reversal (Definition 1.26)of g.

Corollary 1.38. Suppose f and g are tame link with equivalent underlyinglinks. Then f is either equivalent to g or to the orientation-reversal (Definition1.26) of g.

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Definition 1.39. An elementary isotopy applied to a polygonal link p isan elementary isotopy applied to one of its component, except that the trianglecannot touch other components.

Definition 1.40. Two polygonal links p and q are said to be isotopic, alsoknown as combinatorially equivalent, if there is a finite sequence of polyg-onal links p = p0, p1, · · · , pn = q, such that each pi and pi+1 are related by anelementary isotopy or the inverse of an elementary isotopy.

Theorem 1.41. Two polygonal knots p and q are equivalent iff they are isotopic.

Definition 1.42. Given a 2-dimensional subspace E and a polygon link p, wesay that p has regular projection to E if finitely many points on E are doublepoints, i.e. mapped by two distinct points on p, and the other points on theimage are single points, i.e. mapped by only one point on p, and no vertex isprojected to a double point. The double points are called the crossings.

Theorem 1.43. For a given polygon link p, the set of subspaces E such thatp has regular projection to E is an open dense subset of all the 2-dimensionalsubspaces of R3, i.e. the Grassmannian Gr(2,R3). That is to say, for any E,there are subspaces arbitrary close to E admitting a regular projection of p; andif E admits regular projection of p, then any subspace sufficiently close to Eadmits a regular projection of p.

1.6 Diagrams

Theorem 1.44. Up to planar isotopy, there are only finitely many diagramswith a given number of crossings.

Proof. This can be shown using a simple combinatorial argument.

Remark 1.45. People have compiled tables of knots and links up to a certainnumber of crossings and assigned names to the knots according to the numberof crossings. For example, the knot 31 is the 1st minimal knot with 3 crossingsup to mirroring, and is the trefoil knot (refer to Figure 1). In fact it is the onlyminimal knot with 3 crossings up to mirroring, where minimal means that anydiagram of any equivalent knot has at least 3 crossings.

In a diagram, at a crossing one can switch the two lines and produce a diagramof another link. For example, the diagrams of the unlink and of the Hopf linkin Figure 2 are related by such a switching.

Definition 1.46. A switching of a crossing in a diagram of a link is anotherdiagram formed by replacing the solid line by a broken line and the broken lineby a solid line at the crossing in the diagram of the link.

Theorem 1.47. Switching all the crossings corresponds to reflecting the knotalong the plane.

Proof. Geometrically clear.

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Definition 1.48. The mirror image of a knot or link is the knot or linkproduced by reflecting across a plane.

Definition 1.49. An arc in a diagram is a connected component of the solid-broken diagram, i.e. a maximal segment of the knot consisting of only over-passings as opposed to under-passings.

Example 1.50. The diagram of the trefoil knot above has 3 arcs.

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2 Reidemeister’s Theorem and Seifert surfaces

2.1 Reidemeister’s Theorem

Recall we can project any link L onto any generic surface to obtain a diagramof the knot. Unfortunately, any knot can be represented by infinitely manydifferent diagrams, which makes it unclear what we can read from a diagram.The following theorem gives an elegant solution to this issue.

Theorem 2.1. Let D1, D2 be the diagrams of two links. Then the two linksare isotopic (Definition 1.40) if and only if D1 = M1M2...MnD2, where Mi iseither a planar isotopy or one of the Reidemeister moves illustrated below.

Ω1: →

Ω2: →

Ω3: →

Figure 4: The three Reidemeister moves: Ω1, Ω2, Ω3

Proof. Note it suffices to prove only the “only if” direction. Now, let D1, D2

be two planar diagrams of two isotopic polygonal links K1,K2. Without loss ofgenerality, K2 can be obtained from K1 by performing one elementary isotopy.Using Definition 1.17, we notate the original segment to be [pi,1, pi,2], and thenew vertex to be q. Without loss of generality, we can assume all projectionsinvolved in this proof are regular, and that the edges stemming from the ends of[pi,1, pi,2] is exterior to the triangle [pi,1, pi,2, q]. This is valid by applying zeroor one Ω1 move.Note the interior of this triangle would overlap various branches in D2. Tosimplify the case, in the diagram, we can triangulate [pi,1, pi,2, q] into elementarypieces of four different types. (The explicit cases are left to the reader as anexercise.) After drafting them out, we can observe that each case is in fact oneof Ω2, Ω3 and planar isotopy. Hence, we showed that each elementary isotopy

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can be decomposed into a finite sequence of planar isotopy and Reidemeistermoves.

Remark 2.2. We note that Ω1,Ω2,Ω3 is not the full list of Reidemeister movesin the most formal sense. Each of them has their own variations, where thenumber of variations depends on if we count orientations or not. However, theproof above applies to all variations, because these extra cases can be discardedby applying similar moves to them.

Remark 2.3. We can regard the three Reidemeister moves Ω1,Ω2,Ω3 as un-twisting, pulling apart and sliding respectively in an informal sense.

2.2 Basic knot arithmetic

Let K1,K2 be two oriented knots.

Definition 2.4. The connected sum, K1#K2 is defined as follows: Take anysmall band in R3 which meets the knots only at its ends. “Connect” the twoknots by cutting the two ends of K1,K2, and joining the other two sides of theband. The orientation of the connected sum is then defined naturally.

Figure 5: Connected sum of left and right trefoil: the square knot

Remark 2.5. Note this operation is commutative and associative, and is well-defined if we consider isotopy classes of knots.

Remark 2.6. Given a connected sum K, it is possible to find a sphere S thatcuts the knot at exactly two distinct points. In this case, we say S factors theknot.

Remark 2.7. Note that the connected sum has to be defined under orientedknots as if we do not give an orientation, as it is not guaranteed that the oper-ation is well-defined. Take the two pretzel knots P (3, 5, 7) and P (3, 5, 9) as anexample, the four possibilities fall into four different isotopy classes. However,

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we can omit the orientation if we define the connected sum under the set ofinvertible knots, knots with both orientations equivalent to each other. This isthe reason why Figure 5 has no orientation shown, since all knots with crossingnumber ≤ 7 are invertible. Wikipedia has the number of invertible knots up tocrossing number 16.

As an additional sidenote, we can also now define the disconnected sum of twoknots as the disjoint union of them. We notate this by K1 tK2. We will alsoconstruct some more notions related to divisibility of links.

Definition 2.8. Given knots L, K. We say L divides K iff there exists a knotM such that K = L#M .

Definition 2.9. For a non-trivial knot K, we call it prime if there exists nonon-trivial L and M such that K = L#M ; otherwise we call K a compositeknot. Note that the unknot is defined to be neither prime nor composite.

Fact. Given knots K, L. Then K is ambient isotopic to L if K = L#U , whereU is the unknot.

Till now, we can already prove knots form a monoid under connected sum. So,a next natural question is whether inverses exist, i.e. for some knot K, can wefind some knot L such that K#L = U , where U is the unknot? Unfortunately,this is not the case, and we shall prove this in the following subsection, after weestablished some more theory. In fact, no non-trivial knots have inverses.

2.3 Seifert surfaces

Seifert surfaces are integral to the investigation of knots, simply because itprovides an alternative way of dealing with knot theory in a more topologicalmanner.

Theorem 2.10. For every oriented knot (or link) K, there exists an orientableand connected surface F ⊂ R3 such that K is the boundary of F .

Proof. Part 1: The Seifert algorithm

1. Choose an oriented diagram D of K.

2. Choose an arbitrary point on the knot.

3. Walk along the orientation until we return to our starting position, and atevery crossing jump to the other arc, still walking along the orientation.

4. Repeat this process for other starting points, and obtain a disjoint unionof circles, called the Seifert circles.

5. We note that the above process can be seen as taking the natural smoothat every crossing of the knot.

6. For concentric circles, raise the inner circles upward.

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7. Fill in the circles to obtain a disjoint union of discs in R3.

8. At every crossing of the original diagram, connect the two correspondingdiscs with a half-twisted band, so that the boundary of the band gives theoriginal crossing.

9. The resulting surface is the surface required.

Part 2: Showing that the resulting surface is connectedAs one moves along the original knot, one connects all the circles in the process.

Part 3: Showing that the resulting surface is orientableFor each disc, as we travel along the boundary using the prescribed orientation,the disc is either always to the left or always to the right. In the first case, colourthe top of the disc with red and the bottom of the disc with blue (pretend thatthe disc is two atoms thick). In the second case, colour the top of the disc withblue and the bottom with red. Note that travelling along the band flips thecolouring for adjacent circles, and preserves the colouring for concentric circles.By a geometrical argument we can see that, similarly: for adjacent circles,travelling along the band changes whether the disc is to the left or to the rightof the boundary; for concentric circles, this property is preserved. Therefore, thecolouring is consistent as we travel along the bands from discs to discs, whichshows that the surface is orientable, as it has two distinct sides.

Part 4: Showing that the boundary of the resulting surface is theoriginal knotAt each crossing, we carefully used a half-twisted band to connect the two circlesso that the boundary of the band gives the original crossing. Filling in the circlesthen taking the boundary gives us the original circle. Therefore, this is clear.

We now define a Seifert surface as follows:

Definition 2.11. A Seifert surface for a knot K is a connected orientabletwo-dimensional manifold with boundary embedded in R3 whose boundary is K.

Corollary 2.12. Any knot has a Seifert surface.

Remark 2.13. Every Seifert surface has only one boundary component, hencethere is a unique knot assigned to each Seifert surface. The converse is false,since the surface generated by the Seifert algorithm depends on the oriented linkdiagram.

Definition 2.14. The genus g(K) of a knot K is the minimal genus acrossall the possible Seifert surfaces for K.

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Fact. A knot has genus 0 iff it is the unknot. Assume K is the unknot. Then Kbounds a disc, which obviously has genus 0. For the other direction, suppose aknot L has genus 0. Then we know it has a Seifert surface of genus 0, i.e. a disc.It is ambient isotopic to a polygonal disc, and we can triangulate such polygonaldisc such that they give a sequence of elementary isotopies (and inverses thereof)to deform the polygonal knot back to a triangle, which is the unknot.

Fact. Every knot with genus 1 is prime. This is a corollary of the additivity ofgenus (Theorem 2.17 below).

To any Seifert surface we can assign a unique knot simply by considering theboundary. However, given a knot K, there may exist more than one Seifertsurface associated with the knot. In fact, given a knot K, and one of its Seifertsurface S which is genus g, we can undergo topological surgery to obtain a newSeifert surface S′ with genus g+1. This is the reason why in particular we haveto set genus to be minimal in the definition above.

Genus turns out to be a integer-valued knot invariant, meaning if K and L aretwo equivalent knots, we have g(K) = g(L).

Theorem 2.15. Let K, L be two equivalent knots. Then, the genus of K isequal to that of L.

Proof. It suffices to prove g(K) ≥ g(L). Let S be a Seifert surface for K suchthat g(S) = g(K). Let Fs be the ambient isotopy bringing K to L. Now wehave F1(S) is a Seifert surface for L. S = F0(S) is homeomorphic to F1(S), sothey have the same genus.

We now attempt to prove additivity of genus. Before that, we will need a lemma.

Lemma 2.16. Suppose H is a Seifert surface of minimal genus for knot K =K1#K2. Let sphere S factor knot K into a disjoint union K1 tK2. If H andS meet transversely, then there exists an isotopy of H that the intersection of itand S is a simple arc.

Proof. Note H ∩ S is a set consisting a simple arc and possibly finitely manydisjoint circles. Take any circle σ. σ must bound a disc on H ∩ S, since if thisis not the case, we can construct a new Seifert surface with one less genus. Ifσ splits S, then we can fill in a disc inside σ such that genus is decreased by 1,contradicting minimality of genus. Or else, we can cut S along σ, and attachtwo discs along their boundaries to the cuts. This surface also contradictsminimality of genus. Finally, an isotopy of H is given by shrinking H insideinterior of S.

Theorem 2.17. g(K1#K2) = g(K1) + g(K2).

Proof. We show that g(K1#K2) ≤ g(K1) + g(K2): For knots K1,K2, considertheir Seifert surfaces with minimal genera. Assuming without loss of generalitythat they have empty intersection, we can connect the two surfaces by adding a

17

small copy of the unit disc, similar to what we did in the definition of connectedsum. In that way we obtain a Seifert surface of genus g(K1)+g(K2) for K1#K2.The other direction is much more challenging. Let us consider a Seifert surfaceH of K1#K2 of minimal genus. By Remark 2.6, we have a sphere S thatfactors K1#K2, i.e. splitting K1#K2 into K1 and K2. We now consider theintersection of H and S, where we assume transversality. Now H ∩ S consistsof an arc that connects the two points of factoring and a set of pairwise disjointcircles (note it is possible that this set is empty). However, all the disjoint circlescan be discarded by the lemma above. By cutting H along the remaining arc,we obtain a disjoint union of Seifert surfaces of K1 and K2 respectively.

Corollary 2.18. Any non-trivial knot has no inverse.

Proof. Assume a non-trivial knot K. Then g(K) 6= 0 by a fact above. Aninverse of knot K (name it by L) would satisfy g(K) + g(L) = g(U) = 0, whichis a contradiction since genus is non-negative.

Corollary 2.19. Let K be a non-trivial knot, L be a knot. Then K#L isnon-trivial.

Proof. We note g(−) is a non-negative function. Suppose K#L is the trivialknot. Then g(L) = g(K#L)− g(K) < 0.

2.4 Schubert’s Theorem

In this section, we will give out results related to prime factorization of knots, aswell as unfolding the Schubert’s Theorem, stating the existence and uniqueness(up to reordering) of a prime factorization of any non-trivial knot.

Theorem 2.20. For any non-trivial knot K, we can write K as:

K = K1#K2#K3...#Kq

where each Ki is a prime knot. We call such expansion a prime factorizationof K.

Proof. If K is prime then it is obvious. So we assume K is composite, by itsdefinition there are non-trivial knots L, M such that K = L#M . We can againcheck if L, M are prime knots and repeat the process above. Note this procedurecannot repeat more than g(K) times, since the only knot with genus 0 is thetrivial knot. Hence, we will always get a finite sequence of knots with connectedsum K.

To prove uniqueness, We would need some additional notions and lemmas. Notethe following definition is very uninsightful, i.e. it has no use other than provingthis theorem, but nevertheless we put it here.

Definition 2.21. Let K be a knot. Let S be a set of pairwise disjoint spheresS1, S2, ..., Sr, such that:

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• Each sphere cuts the knot at exactly two points.

• These spheres factor K into any number of prime knots.

Then S is called a decomposing sphere system (dss) for K. Note in the remainderof this section we will abuse notation of S both referring to the finite amount ofballs and the (infinite amount of) points that made up the spheres for the sakeof simplicity.

Lemma 2.22. Let S be a dss for K. Then the factoring of K consists of |S|+ 1prime knots.

Proof. We conduct strong induction on |S|. Suppose this is true for all |S| < n.Now let Bl be an innermost ball, i.e. it contains no other ball. Such ball mustexist since all spheres are disjoint. Now we define a new knot K by the unionof γ and K complement Bl, where γ is the arc on Sl joining the two pointsof factoring. It is easy to see that K#Kl is again K. Noting a dss for K has|S| − 1 elements (hence factoring of K consists of |S| prime knots by inductionhypothesis) ends the proof.

Equivalence of dss is defined below with well-definedness proved above:

Definition 2.23. S1 and S2 are equivalent if they represent the same (m+ 1)prime factor knots for a knot K.

Lemma 2.24. Let S be a dss for K. Let T ⊂ S3 be a 2-sphere pairwise disjointto all elements in S, and notate its associated balls T1, T2. Define S′ := (S \Sj) ∪ T , where Sj and T determines the same prime knot (w.r.t. S and S′

resp.). If Bj is a ball in T1 such that no ball in T1 contains Bj, then S isequivalent to S′.

Proof. We notate the two (associated) interior and exterior balls of St as Bt, Ctrespectively, for all t from 1 to r. Suppose j 6= k. Without loss of generalityBk ⊂ T1, and we can split into two cases, i.e. whether there exists l such thatBk ⊂ Bl ⊂ T1 or not. If the answer is positive, then balls Bk, Ck represent theknot Kk and “K \ Kk” w.r.t. both systems. Otherwise, Bk, Ck represent thesame knots in S, but in S′, they correspond to Ki and “K \Kj” respectively.But “K \ Kj” is equal to “K \ Ki” by observation (keep in mind that Bk is“maximal”). Hence, S and S′ share the same representation of factor knots.

Remark 2.25. We should be aware that in any non-empty dss S, we have atleast one innermost ball, i.e. a ball containing no other ball, since |S| is finite.Similarly, given two systems S1 and S2 of a knot, if A is an element of S1

with empty intersection with S2, A must be outermost within some ball B ofS2. This is because for this to not happen, all balls of S2 must be contained inA, i.e. A contains some prime knots which is not contained in S2, arriving toa contradiction since S2 should be sufficient to reveal the knot.

Theorem 2.26. Suppose S1 and S2 are dss for the same knot K. Then S1 isequivalent to S2.

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Proof. We will do induction on |S1 ∩ S2|.Base case: Induction on |S1|+|S2|. (Note: Base cases of this induction is trivialto check, we would not state it here.) Assume true for |S1| + |S2| < n. Noteany sphere in S1 has no intersection with S2. Hence by remark 2.25, we havesome sphere S1p maximal within S2q. Define S3 := (S2 \ S2q) ∪ S1p. HenceS2 is equivalent to S3 by construction and Lemma 2.24. Then, it suffices toprove S1 is equivalent to S3. Now, construct K w.r.t. Kj as in the proof of

Lemma 2.22. K = K#Kj . Now note S1 \ Sj and S3 \ Sj are equivalent and

use induction hypothesis on K.

Induction step: Assume true for |S1 ∩ S2| < n. We can assume S1 ∩ S2 isnon-empty, hence by Remark 2.25, we have a innermost ball Q with respect toboth S1 and S2. We name its respective sphere R. Note we can assume R andS2 has non-empty intersection, or else we can proceed with a manner similar tothe base step, using remark 2.25. Hence there exists an “innermost” curve µ ofR ∩ S2. This induces a disk D with µ as its boundary, and its interior havingempty intersection with S2. Considering the definition of a dss, we have µ ⊂ Ffor some F ∈ S2. Hence D ⊂ E, where E is the ball with respect to F . Note Erepresents a prime knot, hence if we slice F into two balls by D, only one ballwould contain the prime knot (the other will represent the unknot). We let V bethat ball, W be the respective sphere. Now we “ε -shrink” W to W ′ such thatW ′ still contains the prime knot, but no longer has µ as boundary. We assumeW ′ is transverse to all other spheres in S2. Now we define S4 := (S2 \F )∪W ′.By Lemma 2.24, S4 is equivalent to S2. Since W ′ ∩R does not contain µ, butF ∩R contains µ, we reduced number of elements in S1 ∩ S2 by 1.

We finish this section by noting that Schubert’s Theorem is the combination ofTheorem 2.20 and an easy corollary of Theorem 2.26. Note a similar theoremholds for links without reservation. The proof is much longer, and it is due to[Hashizume 1958].

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3 Integer-valued and polynomial invariants

3.1 An overview

One of the major issues mathematicians are trying to solve in knot theory is therecognition problem, i.e. to determine whether two given knots are related by afinite sequence of elementary isotopies. Reidemeister Theorem in the previouschapter did give us a insightful pathway. In this chapter, we will introduce an-other approach: invariants. We first define the notion of a link (knot) invariant:

Definition 3.1. Given a set S, an S-valued knot invariant is a functionF from knots to S such that if K1 and K2 are equivalent knots then F (K1) =F (K2).

Definition 3.2. Given a set S, an S-valued link invariant is a function Ffrom links to S such that if L1 and L2 are equivalent links then F (L1) = F (L2).

We give a simple criterion for a function to be a knot (resp. link) invariant:

Theorem 3.3. A function f : Knots → S (resp. f : Links → S) is a knot(resp. link) invariant iff it is invariant under the three Reidemeister moves.

Proof. Trivial by Theorem 2.1.

Knot invariants allow us to distinguish knots. Given two knots K1 and K2 andan invariant F , if we can show that F (K1) 6= F (K2), then we can conclude thatK1 and K2 are not equivalent knots. A similar remark can be made for links.However, most of the more useful invariants involve advanced algebra, and wewill cover some of those in the following subsections. We would first introducethree elementary link invariants and some properties about them.

Definition 3.4. All crossings of any oriented link diagrams can be distinguishedinto two types: positive (Figure left) and negative (Figure right).

Definition 3.5. Let D be a diagram of an oriented link N . We give eachcrossing of this link a number:

• 0 if this is a self-intersection;

• −1 if this is not a self-intersection and the crossing is positive;

• +1 if this is not a self-intersection and the crossing is negative.

Then, the linking number of D, L(D), is defined by adding all these numbersup, then dividing the sum by 2. Corollary 3.12 below explains why it is indeedan integer-valued invariant.

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Theorem 3.6. Suppose D, E are diagrams of an oriented link N . Then L(D) =L(E). Hence, the notion L(N) is well-defined as a link invariant.

Proof. By Theorem 3.3, it suffices to prove L is invariant under all Reidemeistermoves. First, a crossing resolvable by an Ω1 move is a self-intersection. If weassume the two strings in an Ω2 move comes from different components of thelink N , then it takes away a positive and negative crossing from the diagram,hence it differs from L(−) by 1− 1 = 0. Ω3 does not change the linking numberas well, as the type of the central crossing is unchanged, and the types of theother two crossings would not change either, since Ω3 simply involves shiftingthe line not involved in the central crossing. (refer to Remark 2.3)

Corollary 3.7. Any oriented knot K has linking number 0.

Fact. Let N be a two-component link, M be its mirror image, and P be N withall orientations reversed. Then L(M) = L(P ) = −L(N). Similar results stillhold for n-component links with n > 2.

Definition 3.8. The unknotting number u(K) of a knot K is defined as theminimum natural number n such that there is a diagram D of K and a collectionof n crossings of D which, when switched, transforms D into a diagram of theunknot.

Note we can expand this definition to the notion of unlinking number, whichdescribes a similar invariant except our aim is to transform a link into an unlink.

Definition 3.9. The crossing number c(K) of a knot K is defined as the min-imum number of crossings across all diagrams of K.

Fact. Note two knots need not be equivalent if some of their invariants areequal. The map that sends all knots to zero, with unknot and trefoil, serves asan counterexample. They are inequivalent by Theorem 3.23 below.

To prove u(K) is indeed a well-defined invariant, we need the following theorem:

Theorem 3.10. All knots can be transformed into the unknot by switching afinite amount of crossings.

Proof. Take any knot K and some diagram of knot K, D. Draw a line parallel toy-axis and tangent to the knot such that all points on the knot has x-coordinatelarger than that of any point on the line except points of tangency. Now choosea point of tangency, and parametrize the knot t 7→ (x(t), y(t), z(t)), where 0 ≤t ≤ 1. Note such parametrisation is injective except at our chosen point. Weconstruct a new knot by t 7→ (x(t), y(t), t), and then connect the two endpointsby drawing a line parallel to z-axis instead. This is obviously the unknot, asit is contained in the xz-plane after some stretching and pulling. We end theproof by noting the new and old knots are related by switching all crossingswith its two corresponding points ((x(u), y(u), z(u)), (x(v), y(v), z(v))) havingthe following relationship: (u− v)(z(u)− z(v)) < 0.

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Corollary 3.11. Let K be a knot. Then, u(K) ≤ c(K)2 .

Proof. From the proof above, u(K) ≤ c(K). If we switch more than c(K)2

crossings to obtain the unknot by the mechanism above, we use the alternativemap below instead: t 7→ (x(t), y(t), 1− t), then connecting the two endpoints bydrawing a line parallel to z-axis. By similar steps to the proof above, we knowthis alternative map is also the unknot, and number of switchings required is

less than c(K)− c(K)2 .

Corollary 3.12. Despite the factor of 12 , the linking number is always an in-

teger.

Proof. We would assume Theorem 3.10 is true for links. (The philosophy of theproof is the same.) Let K be a link. Note that for every change of crossing, thelinking number is either increased by 2 (from -1 to +1), decreased by 2 (from+1 to -1), or unchanged, which is the case of a self-intersection. Hence, 2∗L(K)always has the same parity with 2 ∗ L(U), which is equal to 0.

We will give a few examples of all invariants mentioned above to end this section.

1. The unknot. Trivially by definition it has crossing number and unknottingnumber 0. Its linking number is 0.

2. The left trefoil. It has crossing number 3: since 31 is a diagram of thetrefoil, the crossing number is bounded above by 3; all diagrams withcrossing number less than 3 must be diagrams of the unknot. This can beverified as there are only 1+2+4×3×3×2 = 75 possibilities to consider.Its unknotting number is 1: first of all, it is not equivalent to the unknotby Theorem 3.23 below, so it is non-zero. To see it is actually 1, switchany crossing and label the nearer crossing (in terms of an anti-clockwisepath) C1. Then reflect the upper part along a tangent at C1, and twistaccordingly to get an unknot. Its linking number is 0 since all crossingsare self-intersections.

3. Whitehead link. Its crossing number is 5, one self-intersection of the “8”,and four intersections of the “8” and the loop. Its unlinking number is 2,trivial by choosing any two oppposite non-self-intersections. (Note if youjust choose one the resultant is the Hopf link.) Its linking number is 0regardless the orientation of the two components.

4. Hopf link. Its crossing number is 2 and its unlinking number is 1. Itslinking number is 1 or −1 depending on the orientation.

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(a) Unknot (b) Trefoil

(c) Whitehead link (d) Hopf link

3.2 3-colouring

The first invariant of links we will study can be computed using combinatorics.Even though it remains limited, the 3-colouring invariant and its connectionwith linear algebra enables us to distinguish links with a high number of cross-ings. To learn this material, [25] was used but ultimately we presented our ownformulation of the proofs and computed our own examples.

Definition 3.13. Let D be a diagram of an unoriented link L. Given 3 colours,define T (D), the set of 3-colourings of D, to be the ways of colouring eacharc of the diagram with one of the 3 colours such that at every crossing of D,the three incidents arcs are either all of the same colour or are all coloureddifferently.

Definition 3.14. Define τ(D) to be the number of possible 3-colourings of D,i.e. |T (D)|.

Definition 3.15. D is said to be tricolourable or (3-colourable) if there existsa non-trivial assignment of colours A ∈ T (D), i.e. an assignment that uses atleast 2 colours.

Example 3.16. Example of a possible 3-coloured trefoil:

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Figure 8: 3-coloured trefoil

Theorem 3.17. Tricolourability and the number of 3-colourings τ(D) is a linkinvariant.

Proof. By Theorem 3.3, it suffices to show that the number of 3-colourings ispreserved under each Reidemeister move. To do this, we establish a bijectionbetween the 3-colourings of a diagram before applying Reidemeister move andthe 3-colourings of a diagram after applying a Reidemeister move.For Ω1, let a be the arc starting on the left. Then two of the three arcs at thecrossing are the same colour a. So for this crossing to satisfy the tricolourabilityproperty, the third arc must also be colour a. Hence Ω1 preserves our property.

Figure 9: Invariance under R1

For R2, let the bottom and top arcs of the left hand side be colour a and b.Then for our knot diagram to satisfy our given condition , a = b implies thatthe 3 distinct arcs on the right must be the same colour (i.e case 1 below). Ifa 6= b, then the third arc on the right hand side must be colour c distinct from

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a or b to satisfy our property (i.e. case 2 below). So we can create a bijectionsuch that R2 preserves our property.

(a) Case 1

(b) Case 2


For R3, we define a colouring on the left hand side and look for a correspondingbijection to the right hand side by enumerating all the possible cases. Assum-ing that we are working locally and that we are keeping all other colouringsconstant, then we must enumerate the possible colourings for the 3 crossings.Let’s consider x to be the number of crossings where the 3 incidents arcs are thesame colour. Then x can take the values 0,1,2,3. However, suppose x = 2 thenthe third crossing has at least two incidents arcs with the same colour and sox must be 3 to satisfy the tricolourability property. Therefore x can only takethe values 0 (case a), 1 (case b,c,d : depending on which crossing x is appliedto) and 3 (case e). In all 5 cases we have a well-defined bijection satisfying the3-colouring property.

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(a) Case a

(b) Case b

(c) Case c

(d) Case d

(e) Case e


However, the strength of this invariant lies in its relation with linear algebra.

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Indeed, if we define our 3 colours to be 0, 1 and 2 in the field F3 then wecan observe that a crossing satisfying the 3-colouring property is equivalent toxi + xj + xk = 0 where xi, xj , xk ∈ 0, 1, 2 is defined as the assigned colour tothe incident arc Ai, Aj , Ak at the crossing Ca.

Theorem 3.18. T (D) = (x1, x2, ..., xk) ∈ Fm3 : xi + xj + xk = 0 at eachcrossing involving arcs Ai, Aj , AkTheorem 3.19. Let D be the diagram of a link. Then, T (D) is a vector spaceand τ(D) = 3dimT (D).

Proof. T (D) is the solution space of homogeneous linear equations. Hence,it is a vector space. In addition our field has only 3 elements. Let D havem arcs and n crossings. Then define A to be an m × n matrix encoding them linear equations of T (D) such that T (D) = x ∈ Fm3 : Ax = 0. Thenτ(D) = 3dimT (D) = 3dim(Ker(A)).

Example 3.20. The first example we shall study is the trefoil where we havelabelled the arcs and the crossings in the following way:

Figure 12: 31 knot with labelled arcs and crossings

The corresponding matrix A is: 1 1 11 1 11 1 1

Noting that Ax = 0 ⇐⇒

1 1 11 1 11 1 1

x = 0 ⇐⇒

1 1 10 0 00 0 0

x = 0, we see

that dim Ker(A) = 2 and τ(31) = 3dim Ker(A) = 9.

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Example 3.21. For this example, we draw the knot 51 and label it as follows:


This gives rise to the following system of equations:

Ax = 0

⇐⇒

0 1 1 0 10 1 0 1 11 1 0 1 01 0 1 1 01 0 1 0 1

x = 0

⇐⇒

1 1 0 1 00 −1 1 0 00 0 0 −1 10 0 0 −1 10 0 1 0 −1

x = 0

⇐⇒

1 1 0 1 00 −1 1 0 00 0 −1 0 10 0 0 −1 10 0 0 0 0

x = 0

So dim Ker(A) = 1 and τ(51) = 3.

Example 3.22. The final example is the figure 8 knot labelled as follows :

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0 1 1 11 1 0 11 1 1 01 0 1 1

∼

1 1 0 10 1 1 10 −1 1 00 0 1 −1

∼

1 1 0 10 1 1 10 0 −1 10 0 0 0

so dim Ker(41) = 1 and τ(41) = 3.

Therefore, we have computed that τ(31) = 9, τ(41) = 3, τ(51) = 3. Additionally,τ(01) = 3, since one can only assign the 3 trivial colourings to the unknot.Therefore we can distinguish the trefoil 31 from the 3 other knots. However, weare unable to draw any conclusions concerning the others.

Theorem 3.23. The trefoil 31 is distinct from the unknot 01, the figure-8 knot41, and the cinquefoil knot 51.

3.3 p-colouring

An important generalisation of the 3-colouring invariant is the p-colouring in-variant where p is a prime. This invariant uses the fact that if p is a prime thenFp := Z/pZ is a field. Therefore, similarly to the 3-invariant, the set of solutionsto a certain collection of equations specified below form a vector space, and solinear algebra can be employed.

Definition 3.24. Let D be a diagram of a link L, then Tp(D) is the set ofcolourings of the arcs of D using the p colours from Fp such that at each crossing,letting Ai be the over-passing arc and Aj, Ak the under-passing arcs, we have2xi − xj − xk ≡ 0 (mod p), where each xi ∈ Fp is the colour assigned to Ai.

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Theorem 3.25. Tp(D) is a vector space and the number of p-colourings τp(D)is a power of p. In addition , τp(D) is a link invariant.

Proof. This proof is similar to the 3-colouring proof.

Example 3.26. We will use the 5-colouring invariant to distinguish the knots01, 41 and 51 which we failed to do with 3-colouring. We can already assumethat τ(01) = 5 as the unknot only has the 5 trivial colourings.


This gives rise to the matrix A:0 −1 −1 0 20 2 0 −1 −1−1 −1 0 2 02 0 −1 −1 0−1 0 2 0 −1

∼

1 0 0 0 −10 1 0 0 −10 0 1 0 −10 0 0 1 −10 0 0 0 0

so dim ker(A) = 2 and τ5(51) = 25

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Example 3.27.


This gives rise to the matrix :0 −1 −1 22 −1 0 −1−1 2 −1 0−1 0 2 −1

∼−1 2 −1 00 −2 3 −10 −1 −1 20 3 −2 −1

∼−1 2 −1 00 −2 3 −10 −1 −1 20 0 0 0

so dim ker(A)=1 and τ5(41) = 5

We can therefore conclude that the 5-colouring invariant has helped us to distin-guish the 51 and the 41 knot as τ5(51) 6= τ5(41). However, we were still unableto distinguish the 41 knot and the unknot and shall try to explore this in thenext invariants.

3.4 Seifert matrix and Alexander polynomial

The Alexander polynomial is the first polynomial invariant, and it is the onlypolynomial invariant since its discovery for over sixty years. Not only does itprovide much more information as an invariant, but the fact that a polynomialcorresponds to an infinite sequence of integers also allows for easier recognition.There are multiple ways to derive the Alexander polynomial, and we wouldchoose a way such that minimal definitions are to be introduced for convenience.A lot of technicalities are also omitted for this reason, so at some point we willonly state important facts without proving them. Also note from this pointonwards till the end of the document, we will denote the unknot as U .So, for a (minimal) Seifert surface T for a link L, we will first find 2g(L)+µ(L)−1loops in the following way (µ(L) denotes the number of components of the link,which is obviously a link invariant): For each genus (i.e. handle), we identifythe inner horizontal equator and any vertical loop around it (i.e. meridianand longitude). For each additional component, we identify the boundary ofthose components as a loop (note knots are simple closed polygonal curves).(Equivalently, we choose any basis of the first homology group of T .) We namethese curves as li, i = 1, 2, ..., 2g(L) + µ(L)− 1(= c).

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Next, thicken the Seifert surface by S× [0, 1], where by convention we denote Sto be S × 0. Project li onto S × 1, and call these loops l′i. We can finallydefine the Seifert matrix:

Definition 3.28. The Seifert matrix M of a Seifert surface (for a link) isthe c× c matrix with entries defined by Mij := L(li ∪ l′j). The Seifert matrix ofthe trivial knot is defined as the empty matrix by convention.

Definition 3.29. The Alexander polynomial ∆A(L) of link L is a polynomialin indeterminate t, defined by ∆A(L)(t) := det(M − tMT ). Note it is only well-defined as a link invariant up to normalization by multiplying some Laurentmonomial, such that the constant term is positive.

Proving the Alexander polynomial is actually a link invariant is out of the scopeof this document, hence we omit it as well. [22] pp.4-5 gives a clear constructionof this proof.

Note the Seifert matrix itself is an interesting topic to discuss, as it containsa lot of information regarding Seifert surfaces. It is a link invariant only upto (S-)equivalence, and has much more interesting properties as well. We willomit those since our sole purpose of introducing Seifert matrices is to derive theAlexander polynomial. Interested readers should refer to [10] Ch.5.

The Alexander polynomial is very powerful (at its time), but it is difficult tocompute without other techniques. It is also difficult to actually locate the loopsmentioned above when the knots become more complicated. This changed whenJohn Conway discovered that a variation of the Alexander polynomial (nowcalled the Alexander-Conway polynomial) satisfies a skein relation:

1. ∆C(O) = 1, where O is any diagram of U .

2. ∆C(L+) − ∆C(L−) = t∆C(L0), where L+, L−, L0 are identical link dia-grams except for one crossing, with differences explained by the notation.

Remark 3.30. The property linking the classical Alexander polynomial andAlexander-Conway polynomial is ∆A(L)(t) = ∆C(L)(t

12 − t− 1

2 ). (Proof of thisrelation can be found in Kunio, pp.110-112) This relation is well-defined sinceAlexander polynomial is symmetric.

Alexander polynomial is not a “good” invariant compared to Jones polynomial,which we will soon see in the next section. Nevertheless, it carries a lot ofinteresting properties. We will prove some of them below.

Theorem 3.31. ∆A(L) satisfies:

1. ∆A(L)(t) = ∆A(L)(t−1), i.e. ∆A(L) is symmetric.

2. ∆A(K)(1) = 1 if K is an oriented knot.

3. ∆A(Us) = 0, where Us is the trivial link with t ≥ 2 components.

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4. ∆A(L)(1) = 0 if L is an oriented link with more than 1 component.

Proof. 1. Let k be the order of M . ∆A(L)(t−1) =det(M − t−1MT ) =t−kdet(tM −MT ) = (−t)−kdet(M − tMT ) = ∆A(L)(t).

2. Note by the remark above, we have ∆A(L)(1) = ∆C(L)(0). Remains toprove that ∆C(L)(0) = ∆C(O)(0). This is affirmed by repeated applica-tion of the second property of the skein relation in the special case t = 0.Recall for all knots we can switch some of the crossings to obtain the un-knot, and the property assures us that the Alexander polynomial whent = 0 is conserved when crossings are switched.

3. Let Os0 be the trivial diagram of Us, and pick any two circles in Os0.Let Os+ and Os− be identical to Os0 except the two circles picked are“connected-summed” by a positive and negative crossing respectively.Note Os+ and Os− is equivalent to O(s−1)0

, since they are both dia-grams for Us−1. Hence by the second property of skein relation, 0 =∆C(Us−1)−∆C(Us−1) = t∆C(Us), yielding the result.

4. A corollary of 3, the proof is similar to that of 2.

Remark 3.32. There is an elegant theorem as follows: For every symmetricLaurent polynomial p with p(1) = 1, there exists a knot with Alexander poly-nomial p. Equivalently, for all polynomials q in t2 with constant term 1, thereexists a knot with Alexander-Conway polynomial q. The proof requires too muchmachinery to be put here, one can refer to [Rolfsen pp.171] for a proof. Basicknowledge on homology theory and infinite cyclic covers should be assumed.

One of the limitations of Alexander polynomial is that it is not able to differ-entiate mirror-images and reverse orientations.

Theorem 3.33. Let K be an oriented knot.

1. If K is the orientation-reversal of K, then ∆A(K) = ∆A(K).

2. If K is the mirror image of K, then ∆A(K) = ∆A(K).

Proof. Note: This is a sketchy sketch instead of a rigorous proof, since we donot intend to introduce the notion of equivalence of Seifert matrices. Interestedreaders should go through Murasugi’s Knot Theory and its Applications Ch.5-6for a better understanding.

1. This is proved by the claim that the Seifert matrix of K is transpose tothat of K. To see why the claim is true, take regular diagrams of K, Krespectively. Name them D, D respectively. D and D are related by allorientations reversed as well. Consider the c loops we have constructed atthe beginning of this section. Since all orientations are reversed, we haveL(li ∪ l′j) = L(lj ∪ l′i).

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2. Similarly, this is proved by the claim that the Seifert matrix of K is thenegative of transpose to that of K. Proof of such claim is similar as well.

(Note: Oriented links have the same properties, and the methodology of provingit is similar.)

The last theorem of this section relates connected-sum to Alexander polynomial.This theorem, in co-ordination of the Schubert’s Theorem, allows us to focusour study of Seifert matrices and Alexander polynomials onto prime knots only.

Theorem 3.34. Given two oriented links K1,K2. Suppose they have Seifertmatrices MK1 ,MK2 , Alexander polynomials ∆A(K1),∆A(K2). Then, MK1#K2 =MK1

⊕MK2, and ∆A(K1#K2) = ∆A(K1)∆A(K2).

Proof. We first obtain Seifert surfaces of K1,K2 by Seifert’s algorithm. Callthem F1, F2. We now construct a Seifert surface of K1#K2 by adding a smallband. (Refer to Definition 2.4) The first part is now trivial by considering thatany loop in F1 and F2 are disjoint if we assume F1 and F2 are disjoint. Thesecond part is a corollary of the first part, which is trivial by considering thedefinition of Alexander polynomial and properties of determinant.

3.5 Jones polynomial

This section begins with [24], although the proofs are done by us. All links andlink diagrams are unoriented unless stated otherwise. Let L be an link diagram.We denote by L+

x the smoothing

7→about some crossing x in L. This smoothing is done locally, producing a newlink diagram. Denote by L−x similarly the smoothing

7→The well-definedness of these smoothing processes is clear. Let h(L) denote thenumber of crossings in L. Since smoothing removes a crossing from the linkdiagram, it follows that

h(L+x ) = h(L)− 1

h(L−x ) = h(L)− 1

for any crossing x in L. Therefore applying the smoothing process successivelymust terminate at some link L0 satisfying h(L0) = 0. From previous results thismeans that L0 is a disjoint union of unknots.

35

The Jones polynomial is constructed recursively, keeping track of these smooth-ings as they are made. In order to define it, we need to first define the Kauff-man bracket. Let LinkDiagrams be the class of link diagrams. The Kauffmanbracket is defined as a function

〈·〉 : LinkDiagrams→ Z[A]

satisfying

1. 〈U〉 = −(A2 +A−2)

2. 〈L〉 = A〈L+x 〉+A−1〈L−x 〉

3. 〈L t U〉 = 〈U〉 · 〈L〉

for any link diagram L and any crossing x in L. To show that this definition iscomplete and well-defined, we will prove that it is equivalent to a closed formof the Kauffman bracket.

For any link diagram L, define a state s as a choice of smoothing (positive ornegative) at every crossing in s. We then denote by w(s) the number of positivesmoothings minus the number of negative smoothings in the state s. Denotealso by |s| the number of disjoint unknots upon the smoothing of L at s. LetΣ(L) be the set of all states of L.

Theorem 3.35. For any link diagram L we have the explicit formula

〈L〉 =∑

s∈Σ(L)

Aw(s)(−A2 −A−2)|s|

and this is an equivalent definition for the Kauffman bracket.

Proof. 1. Assume first the recursive definition. We prove by induction on thecrossing number that the explicit formula holds. For h(L) = 0 we havethat L is a disjoint union of unknots so w(s) = 0 for the only state s of Lgiven by L itself, with no smoothings whatsoever. As a result

〈L〉 = 〈U〉|s| = (−A2 −A−2)|s|

by applying (1) and (3) from the recursive definition.

Let L be a link diagram with at least one crossing and assume that theexplicit formula holds for all link diagrams with crossing numbers less thath(L).

36

〈L〉= A〈L+

x 〉+A−1〈L−x 〉= A

∑s∈Σ(L+

x )

Aw(s)(−A2 −A−2)|s| +A−1∑

s∈Σ(L−x )

Aw(s)(−A2 −A−2)|s|

=∑

s∈Σ(L+x )

Aw(s)+1(−A2 −A−2)|s| +∑

s∈Σ(L−x )

Aw(s)−1(−A2 −A−2)|s|

for some arbitrary crossing x in L. Let s be a state in L. Then s is eitherpositive or negative at x, so it belongs to exactly one of the sums above.The coefficients A and A−1 in front of the sums will contribute to the signat x in w(s). As a result

〈L〉 =∑

s∈Σ(L)

Aw(s)(−A2 −A−2)|s|

2. Assume the explicit definition. The unknot U has a single state s givenby U with no smoothings whatsoever. Hence w(s) = 0 and |S| = 1 so

〈U〉 = (−A2 −A−2)

Let x be a crossing in L. Then any state s that is positive at x restrictsto a state s∗ on L+

x with w(s∗) = w(s) − 1. Likewise, if s were negativeat x, we would have w(s∗) = w(s) + 1. It follows that

〈L〉 =∑

s∈Σ(L)

Aw(s)(−A2 −A−2)|s| = A∑

s∈Σ(L+x )

Aw(s)(−A2 −A−2)|s|+

+A−1∑

s∈Σ(L−x )

Aw(s)(−A2 −A−2)|s| = A〈L+x 〉+A−1〈L−x 〉

To prove the third step, Let L = L∗ t U . Then states on L are in one toone correspondence with states on L∗, since U has no crossings and doesnot interact with L∗. However, every complete smoothing of L will haveone more unknot than the corresponding smoothing on L∗. It follows that

〈L〉 = (−A2 −A−2)〈L∗〉

This completes the equivalence.

37

Definition 3.36. Let L be an oriented link diagram. A crossing will henceforthbe called positive if it looks like

and negative if it looks like

The writhe wr(L) of L is defined as the number of positive crossings n+ minusthe number of negative crossings n−. It is easy to see that the orientation chosendoes not matter if L were a knot. Hence the writhe of an (unoriented) knot iswell defined.

Every oriented link (diagram) has an underlying unoriented link (diagram).We define the Kauffman bracket 〈L〉 of an oriented link diagram L to be theKauffman bracket of the underlying unoriented link diagram.

Definition 3.37. Let L be an oriented link. Let L∗ be an oriented link diagramfor L. Define

fL(A) = (−A3)−wr(L∗) 〈L∗〉〈U〉

The Jones polynomial of L is given by

JL(t) = fL(t−1/4)

The Jones polynomial is in fact a Laurent series in the variable t1/4. This isbecause fL(A) is a Laurent series in A by theorem 3.35, which demonstratesthat 〈L∗〉 is divisible by −A2 −A−2 since |s| ≥ 1 for any state s.

Theorem 3.38. The Jones polynomial is an oriented link invariant.

Proof. We must prove invariance under the three Reidemeister moves.

1.

〈〉 = A〈〉+A−1〈〉 =

= A(−A2 −A−2)〈〉+A−1〈〉 =

= −A3〈〉

All these smoothings are carried out locally. The polynomial fL is pre-served under this move because the factor −A3 will cancel out, since thekink contributes 1 to the writhe, irrespective of the orientation chosen forthe above link component. Hence the Jones polynomial is preserved.

38

2.

〈〉 = A〈〉+A−1〈〉 =

= A2〈〉+ 〈〉+ 〈〉+A−2〈〉 =

〈〉+ (A2 +A−2)〈〉+ (−A2 −A−2)〈〉 =

= 〈〉

So the Kauffman bracket itself is invariant under this move. It followsthat the Jones polynomial is invariant under this move, since the writhe isunchanged by the second Reidemeister move for any chosen orientation.

3.

〈〉 = A〈〉+A−1〈〉

However

〈〉 = 〈〉 = 〈〉

by two applications of the second Reidemeister move. Finally

〈〉 = A〈〉+A−1〈〉

so the required equality follows, since the writhe is unchanged by the thirdReidemeister move for any chosen orientation.

Since the writhe of a knot is well defined, we also get the following.

Corollary 3.39. The Jones polynomial is a knot invariant.

Example 3.40. The first Jones polynomial we shall compute is the Hopf Link.We choose to define the orientation of the Hopf Link as shown in the diagrambelow. The first step is to compute the Kauffman Bracket. We choose to firstcompute the smoothing of the top crossing.

39

(a) Horizontal Smoothing L−x(b) Vertical Smoothing L+

x

(c) Oriented Hopf Link L

〈L〉 = A〈L+x 〉+A−1〈L−x 〉

= A +A−1

Now by smoothing out the bottom crossing of both L+x and L−1

x we get 4 newknots :

40

(a) Vertical Smoothing L+x (1) (b) Horizontal Smoothing L+

x (2)

(c) Vertical Smoothing L−x (1) (d) Horizontal Smoothing L−x (2)

Applying recursively the Kauffman Bracket property, we have:

〈L+x 〉 = A〈L+

x (1)〉+A−1〈L+x (2)〉

〈L−x 〉 = A〈L−x (1)〉+A−1〈L−x (2)〉

= A +A−1

= A(−A2 −A−2)2 +A−1(−A2 −A−2)= A5 +A

= A +A−1

= A(−A2 −A−2) +A−1(−A2 −A−2)2

= A−1 +A−5

But we showed previously that:

〈L〉 = A〈L+x 〉+A−1〈L−x 〉

41

Hence ,

= A(A+A5) +A−1(A−1 +A−5) = A6 +A2 +A−2 +A−6

We can compute that writhe of our oriented Hopf Link and as both crossingshave a negative orientation we have : wr(L) = −2

fL(A) = (−A3)2A6 +A2 +A−2 +A−6

−A2 −A−2= A6(−A4 −A−4) = −A10 −A2

So by taking A = t−14 , the Jones polynomial of our oriented Hopf Link is

JL(t) = −t−5/2 − t−1/2

.

Theorem 3.41. An alternative way to compute the Jones Polynomial is to usethe skein relation:

t−1JL+(t)− tJL−(t) = (t1/2 − t−1/2)JL0

(t)

where JL+(t), JL−(t), JL0

(t) are the Jones polynomial of the link L which onlydiffer locally at a crossing according to the following 3 diagrams:

(a) L+(b) L−

(c) L0

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Proof. The proof is omitted, however, it can be found in ([25],pp. 55).

Example 3.42. We will apply this method by computing the Jones polynomialof the 41 knot. We consider the central crossing highlighted below :

)

(a) L+

(b) L−

(c) L0

We quickly observe that L− is the figure eight knot 41 we are trying to compute.In addition L+ is simply the unknot and L0 is the Hopf Link we computed inExample 3.38 with the reverse orientation. By a similar argument to Example3.38 we find that the Hopf Link with the orientation of L− has Jones polynomialJL−(t) = −t−5/2 − t1/2.Applying the skein relation:

t−1JL+(t)− tJL−(t) = (t1/2 − t−1/2)JL0

(t)

t−1(t)− tJL−(t) = (t1/2 − t−1/2)(−t−5/2 − t1/2)

−tJL−(t) = −t3 + t2 − t+ 1− t−1

JL−(t) = t2 − t1 + 1− t−1 + t−2

So J41(t) = t2 − t+ 1− t−1 + t−2.

43

To conclude, we could for instance use the Jones polynomial to distinguish theunknot 01 and the 51 knot which was impossible with the previous invariantswe had and in particular the p-colouring invariant. 51 has Jones polynomialJ51

(t) = t2 + t4 − t5 + t6 − t7 whereas J51(t) = 1 so we can indeed distinguish

these two knots.The Jones polynomial nevertheless isn’t flawless as 51 and 10132 mirror bothhave the Jones polynomial J01(t) = t2 + t4− t5 + t6− t7 so we cannot distinguishthem with this invariant.

3.6 Vassiliev invariants

Vassiliev invariants are one of many major advancements in low-dimensionaltopology, as well as knot theory. Rather than one single invariant, it representsa whole class of invariants. In this final subsection, we would first define whatis a Vassiliev invariant, and our final aim is to investigate into FundamentalTheorem of Vassiliev Invariants, which marks an elegant end to this section.

We will first introduce the notion of a singular knot. Note in this chapter, if wemention about a knot, it must not be a singular knot.

Definition 3.43. A singular knot is the image of a continuous map f : S1 → R3

which is injective except at a finite set of transversal self-intersections.

Definition 3.44. Let K be a singular knot, f be the respective continuous map.A point p ∈ K is an α-point of K if the preimage of p contains two points, andthe tangent vectors with respect to these two points are linearly independent.

Now, we will name all self-intersections as α-points, and other crossings as β-points. We also need the notion of equivalence of singular knots. Our geometricintuition we can say two singular knots are equivalent if one can continuously de-form into another without affecting any new self-intersections and the transver-sality of the self-intersections. It remains to formalize this concept. We wouldneed the definition of a flat disk.

Definition 3.45. Given an α-point A of a singular knot K. Let B be a smallclosed neighbourhood around A. Let γ ⊂ B be a curve such that its interioris “flat” (i.e. is subset of some plane) and contains B ∩K. Then we say theinterior of γ is a flat disk. If there exists a flat disk for every α-point of K,we take one flat disk from each α-point and denote the set containing these flatdisks as F (K).

Definition 3.46. Given two singular knots K, L. They are said to be equivalentif there exists a orientation-preserving homeomorphism φ : S3 → S3 such that

1. φ(K) = L,

2. There exists F (K), F (L) such that φ(F (K)) = F (L).

Reidemeister’s Theorem applies on singular knots as well, with a small twist:

44

Theorem 3.47. Given two singular knots K, L, and their respective singulardiagrams D, E. K and L are equivalent iff D = M1M2...MnE, where Mi iseither planar isotopy or Reidemeister moves not involving α-points, or Ω3 withan α-point as central crossing. We notate this special case Ω.

Proof. Omitted.

Previous workings have shown that skein relations are crucial to calculationof invariants. For singular knots, we introduce the Vassiliev skein relation asfollows:

Proposition 3.48. Let V be any S-valued knot invariant, where S is an abelian

group. We have V ( ) = V ( )− V ( ).

Through this relation, we can generalize all knot invariants to singular knotswith exactly n α-points by applying the Vassiliev skein relation n times. Notewe can also view the skein relation as resolving an α-point to a β-point. Suchgeneralization is independent of the order of α-points chosen, since one can ex-press all such generalizations into summation of the same 2n terms, and additionis commutative. Note we can denote the generalization of the knot invariant tosingular knots with exactly k α-points the nth derivative of the knot invariant.

Definition 3.49. A knot invariant V is a Vassiliev invariant of order ≤ n ifV (n+1) ≡ 0. (Notate V (n) as the nth derivative of V .)

It turns out a lot of the invariants we have discussed before, including thecoefficients of Alexander-Conway polynomial and that of Jones polynomial, areVassiliev invariants. We would include the former as an example here.

Theorem 3.50. The nth coefficient of the Alexander-Conway polynomial is aVassiliev invariant of order ≤ n.

Proof. Using the skein relation of Alexander-Conway polynomial and the Vas-

siliev skein relation, we have ∆C( ) = ∆C( )−∆C( ) = t∆C( ).(We abuse notation here: ∆C means Alexander-Conway polynomial and all of itsgeneralizations as well) Suppose we have a singular knot Q with k α-points. LetR be the complete smoothing of Q. Then ∆C(Q) = tk∆C(R). So if k ≥ (n+1),∆C(Q) would be multiple of tn+1, hence coefficient of tn becomes 0.

Definition 3.51. Note singular knots are functions from S1 to R3, so for eachsingular knot there exists a parametrizing circle. A chord diagram of an singularknot K, σ(K), is defined as marking n pairs of points on the parametrizingcircle, then drawing n chords to connect those points two by two such that theendpoints of the chord map to the same point in R3.

Theorem 3.52. Let V be a Vassiliev invariant of order ≤ n. For any two sin-gular knots with n α-points K1, K2, we have σ(K1) = σ(K2) implies V (K1) =V (K2).

45

Proof. Assume σ(K1) = σ(K2). Then by the definition above, we can deducethat there is a one-to-one correspondence of α-points of K1 and K2. Now weplace K1 and K2 in R3 such that their α-points coincide and a small neigh-bourhood of each α-point coincide. Now we can construct a deformation ofK1 to K2 preserving the position of all α-points and their respective neigh-bourhoods. If there are (a finite number of) α-points produced in the processof deformation, we can assume generic conditions on them, i.e. they occur atdifferent times during the process of deformation. Vassiliev skein relation tells

us 0 = V ( ) = V ( ) − V ( ) (first equality by definition of Vas-siliev invariants of order n), hence we know in such occasion the value of V isunchanged.

We now denote Vn as the R-module of all Vassiliev invariants of order ≤ n, andAn be the set of all chord diagrams of order n. Let R be a commutative ring.From Theorem 3.52, we obtain a linear transformation δn : Vn −→ RAn, RAn

being the R-module of all functions from An to R, such that δn(V )(σ(K)) =V (K) for all Vassiliev invariants (of order ≤ n) V and singular knots with nα-points K. By definition, we have Ker δn = Vn−1. Hence δn descends to aninjective linear transformation δ′n : Vn/Vn−1 −→ RAn.

We now try to compute the image of δn. We need the following notions.

Definition 3.53. A function f ∈ RAn satisfies the 4-term relation (4T) if

f( ) - f( ) + f( ) - f( ) = 0.

Definition 3.54. Given a singular knot K and its chord diagram σ(K). Let τbe a chord in the diagram that does not intersect with any other chord. We callτ as an isolated chord.

Definition 3.55. A function f ∈ RAn satisfies the 1-term relation (1T) iff(σ) = 0 whenever σ contains an isolated chord.

Definition 3.56. A weight system of order n is a function f ∈ RAn thatsatisfies (4T). An unframed weight system of order n is a weight system oforder n that satisfies (1T).

Remark 3.57. From Definition 3.56, we denote the subspace of weight systemsof order n Wf

n , and the subspace of unframed weight systems of order n Wn.Immediately we have Wn ⊆ Wf

n ⊆ RAn.

Lemma 3.58. Any Vassiliev invariant V satisfies the following relation:

V( ) - V( ) + V( - V( ) = 0.

46

This lemma is proved by applying the Vassiliev skein relation onto all of thefour components, then note that under the alternating sum, these componentscancel each other out.

Theorem 3.59. Let R = C. Then δn sends any Vassiliev invariant to anunframed weight system.

Proof. Suffices to prove any f ∈ RAn that can be written as f = δn(V ) forsome V ∈ Vn satisfies both (1T) and (4T).

Part 1: 1TLet K be a singular knot where its chord diagram σ(K) has an isolated chord τ .Let the α-point corresponding to τ be ε. We can hence divide the knot into twoparts such that they do not have α-points in common, since τ is isolated. Byswitching crossings, we obtain an alternative form of K, K such that the twohalves lie on two sides of some plane respectively. Performing Vassiliev skeinrelation on ε, we have V (K) = V (K) = V (K2)− V (K1) = 0, where K1 and K2

represent ε being transformed into a positive and negative β-point respectively,and we note that they differ just by one twist.

Part 2: 4T(Sketch) Denote the diagrams of a 4T relation as D1, D2, D3, D4. The gist ofthis proof is to allocate a singular knot Ki with partial knot diagrams Mi (as inLemma 3.58), satisfying σ(Ki) = Di. This is a trivial step since Mi are designedto suit this purpose. We finish the proof by directly applying the lemma andnoting Theorem 3.52.

Note for a total proof of the image, we have to prove the converse statement aswell, but we will skip this since this is out of scope of this document. Interestedreaders should refer to [19] Ch.8. Basic knowledge on Morse knots and Kont-sevich integral have to be assumed. We will end this subsection by proving thetwo special cases of the Vassiliev invariants, i.e. k = 0 and k = 1.

Proposition 3.60. V0 is the set of all constant functions. Hence, dim(V0) = 1.Moreover, V1 = V0.

Proof. Let K be any knot. Pick f ∈ V0. By definition, we have the first deriva-tive of f to be identically zero. Note by Theorem 3.10, K can be transformedto U by a finite number of switching crossings.

Since 0 = V ( ) = V ( ) − V ( ) (first equality is valid since f hasfirst derivative zero), we know f(.) remains constant by switching crossings.f(K) = f(U). Using a similar approach, one can show that if g ∈ V1, g(.) =g(Z) = 0, where Z is the singular knot with one α-point and no β-points.

47

4 From Jones’ polynomial to Khovanov homology

4.1 Alternative version of Jones polynomial

In this chapter, let U denote the unknot. Every link (diagram) is unorientedunless stated otherwise. We continue to follow [24, p. 4-5]. We give an alternatedefinition of Jones’ polynomial. Define the modified Kauffman bracket as afunction

〈·〉2 : LinkDiagrams→ Z[q]

satisfying the following

1. 〈U〉2 = q + q−1

2. 〈L〉2 = 〈L+x 〉2 − q〈L−x 〉2

3. 〈L t U〉2 = 〈U〉2 · 〈L〉2

Carry out the substitution A2 = −q−1 in the original Kauffman bracket. Thenit is clear that (1) and (3) match with those in the original Kauffman bracket.However

〈L+x 〉2 − q〈L−x 〉2 = 〈L+

x 〉2 +A−2〈L−x 〉2 = A−1(A〈L+x 〉2 +A−1〈L−x 〉2)

so (2) differs by a factor of A−1. By induction the two versions of the Kauffmanbracket differ by A−1 raised to the power of the crossing number of the link. Asa result

〈·〉2 = A−h(·)〈·〉

Now let L be an oriented link diagram. Since wr(L) = n+ − n− and h(L) =n+ + n− we have

JL(q) = (−A3)−wr(L) 〈L〉〈U〉

= (−A3)−n++n− ·An++n−〈L〉2

q + q−1=

= (−1)n−−n+A4n−−2n+〈L〉2

q + q−1= (−1)n−−n+(−q)n+−2n−

〈L〉2q + q−1

=

= (−1)n−qn+−2n−〈L〉2

q + q−1

We will leave the Jones polynomial in this form. There is a correspondingexplicit formula for the modified Kauffman bracket. Let L be a link diagram.An enhanced state for L is a choice of smoothing for every crossing in L togetherwith a label ±1 for every unknot in the complete smoothing of L induced bythe enhanced state. Denote by Σ+(L) the set of all enhanced states of L.

For an enhanced state s let i(s) be the number of negative smoothings in s andlet λ(s) be the sum of the labels across all unknots in the complete smoothing.

48

Let j(s) = λ(s)+i(s). Keep in mind that every enhanced state has an underlyingstate.

Theorem 4.1. Let L be a link diagram. Then

〈L〉2 =∑

s∈Σ+(L)

(−1)i(s)qj(s)

Proof. Let k(s) be the number of positive smoothings in an enhanced state s.Then h(L) = k(s) + i(s) and w(s) = k(s)− i(s) and so

〈L〉2 = A−h(L)∑

s∈Σ(L)

Aw(s)(−A2 −A−2)|s| =∑

s∈Σ(L)

A−2i(s)(q + q−1)|s| =

=∑

s∈Σ(L)

qi(s)(−1)i(s)(q + q−1)|s|

Now in the above expression (q+q−1)|s| is a generating function for all enhancedstates with underlying state s, where the coefficient to qt is the number of suchenhanced states where λ = t. All such states will have the same value of i(s)and so it follows combinatorially that

〈L〉2 =∑

s∈Σ+(L)

(−1)i(s)qj(s)

This theorem can also be proven inductively from the recursive definition of themodified Kauffman bracket. It is a nicer explicit definition since it is a sum ofmonomials.

4.2 Khovanov homology

The Jones polynomial misses some information about the enhanced states ofa link diagram. Khovanov developed a homology theory inspired by the Jonespolynomial which captures more data about these enhanced states whilst re-maining an oriented link invariant. In fact the coefficients of the Jones polyno-mial are Euler characteristics of the graded pieces of Khovanov homology.

To learn the material for this section, a combination of [24] and [13] were used,but ultimately we present our own formulation.

Let k be a fixed ring. For every link diagram L we will construct a correspond-ing chain complex of graded k-modules called the Khovanov complex. Fix anordering for the crossings in L. Lets say there are n crossings. A state can nowbe represented by a positively ordered sequence (x1 . . . xn) of elements ±1 thatrepresent the type of smoothing at each crossing.

49

Let s be a state of L. We will attach to every such state a k-module Cs. Wewill have a map δst from Cs into Ct for two states t, s if t is given by switchinga positive smoothing in s to a negative one at some crossing x in L. As a resulti(t) = i(s) + 1. We refer to i(·) as the homological grading.

The k-modules are Cs = V ⊗|s| where V = k[x]/(x2). This gives a factor of V forevery unknot in the complete smoothing of L induced by s. The elements 1, xwill correspond to positive and negative labels respectively for the unknot at-tached, and span V over k. One can therefore assign elements of Cs to enhancedstates with underlying state s, and these elements will span Cs.

The maps δst will be determined by what occurs locally at the crossing inquestion. For a state s and a crossing x for which s is positively smoothed letsx denote the number of negatively smoothed crossings in s less than x underthe fixed ordering. There are two cases to consider for the map δst.

Case 1 : Switching the smoothings at a crossing x looks like

In which case δst is given by the local map (−1)sxm where m sends

1⊗ 1 7→ 1

1⊗ x 7→ x

x⊗ 1 7→ x

x⊗ x 7→ 0

Case 2 : Switching the smoothings at a crossing x looks like

In which case δst is given by the local map (−1)sx∆ where ∆ sends

1 7→ 1⊗ x+ x⊗ 1

x 7→ x⊗ x

50

In both these cases the rest of the map corresponds to unknots not involved inthe smoothing change and is given by identity. It is clear that δst is a k-linearmorphism in both cases.

We will now construct the Khovanov complex. We define

Ci =⊕

s∈Σ(L)|i(s)=i

Cs

as the k-module with homological grading i. We have the maps

δi : Ci → Ci+1

given by

δi =∑

i(s)=i,i(t)=i+1

δst

where δst = 0 when it is not defined. We will check later that these are differ-ential maps. There is also a quantum grading j(·) on each Ci induced by

λ(1) = 1, λ(x) = −1

then setting j = i+ λ and making it additive across tensors, so that

j(a⊗ b) = j(a) + j(b) ∀a, b ∈ V

The connection to j from the previous section should be clear. Applying j andλ to enhanced states realized as elements of Ci gives the original result.

In order to check that the δi preserve the quantum grading, it suffices to checkthat δst does so for every pair of states s, t. For this it suffices to check that mand ∆ preserve the quantum grading.

Proposition 4.2. m and ∆ preserve the quantum grading

Proof. For m we have

j(m(1⊗ 1)) = λ(1) + i(1⊗ 1) + 1 = λ(1⊗ 1) + i(1⊗ 1) = j(1⊗ 1)

j(m(1⊗ x)) = λ(x) + i(1⊗ x) + 1 = λ(1⊗ x) + i(1⊗ x) = j(1⊗ x)

j(m(x⊗ 1)) = λ(x) + i(x⊗ 1) + 1 = λ(x⊗ 1) + i(x⊗ 1) = j(x⊗ 1)

Finally m(x ⊗ x) = 0 so the quantum grading is also preserved here, since 0belongs to each quantum graded piece. For ∆ we have

j(∆(1)) = λ(1⊗ x+ x⊗ 1) + i(1) + 1 = λ(1) + i(1) = j(1)

j(∆(x)) = λ(x⊗ x) + i(x) + 1 = λ(x) + i(x) = j(x)

51

We refer to Ci,j as the submodule of Ci consisting of elements with quantumgrading j (as well as the element 0). The maps δi descend down to maps

δi,j : Ci,j → Ci+1,j

by the previous proposition. This will give a graded chain complex of L denotedby [[L]], when the ring k and the order on the crossings of L is implicit. Let[[L]][i] denote the shifting of the homological grading of [[L]] by +i. Let [[L]]jdenote the shifting of the quantum grading of [[L]] by +j.

We now turn to oriented links. The chain complex [[·]] for an oriented linkdiagram is defined for the underlying unoriented link diagram.

Definition 4.3. Let L be an oriented link diagram together with an order onthe crossings of L. The Khovanov complex of L with coefficient ring k is definedas

CHK•,•k (L) = [[L]][−n−]n+ − 2n−

where n+ and n− are defined as in definition 3.36.

Proposition 4.4. With the setup as above, δi is a differential map for every i.

Proof. Let s = (. . . xi . . . xj . . . ) be a state with xi = xj = 1. Let r be thenegative smoothing of s at xi, t the negative smoothing of s at xj and u thenegative smoothing of s at both xi and xj . It is sufficient to show that δtuδst =−δruδsr. We will do so for all possible local cases.

1. The map m is identical to the multiplication map for the k-algebra V , andso it satisfies the associative identity. Moreover, txi

= sxiand 1+sxj

= rxj

and so

(−1)rxjm ((−1)sxim⊗ id) = (−1)rxj+sxim (m⊗ id) =

= −(−1)sxj+txim (id⊗m) = −(−1)txim ((−1)sxj id⊗m)

2. One can check that the map ∆ is a comultiplication for some coalgebraon V . As a result it is coassociative and so

((−1)rxj ∆⊗ id) (−1)sxi ∆ = (−1)rxj+sxi (∆⊗ id) ∆ =

= −(−1)sxj+txi (id⊗∆) ∆ = −((−1)txi id⊗∆) (−1)sxj ∆

3. We consider the maps

∆ m :1⊗ 1 7→ 1⊗ x+ x⊗ 1

1⊗ x 7→ x⊗ xx⊗ 1 7→ x⊗ xx⊗ x 7→ 0

(1)

52

and

(id⊗m) (∆⊗ id) :1⊗ 1 7→ 1⊗ x+ x⊗ 1

1⊗ x 7→ x⊗ xx⊗ 1 7→ x⊗ xx⊗ x 7→ 0

(2)

These are the same maps. Now adding the signs yields

(−1)rxj ∆ (−1)sxim = −((−1)txi id⊗m) ((−1)sxj ∆⊗ id)

and

(−1)txi ∆ (−1)sxjm = −((−1)rxj id⊗m) ((−1)sxi ∆⊗ id)

It can be shown completely symmetrically that

(−1)rxj ∆ (−1)sxim = −((−1)txim⊗ id) ((−1)sxj id⊗∆)

and

(−1)txi ∆ (−1)sxjm = −((−1)rxjm⊗ id) ((−1)sxi id⊗∆)

These are all the nontrivial local cases and so we are done.

Definition 4.5. The Khovanov homology groups of an oriented link diagram Lwith coefficient ring k are defined as the homology groups of the chain complexCHK•,•k (L) for any choice of order on the crossings of L. That is,

Hi,j(L, k) =ker(δi+n−,j−n++2n−)

im(δi+n−−1,j−n++2n−)

One can check that the ordering chosen for the crossings in the link diagramdo not affect the Khovanov homology groups. This is because −1 will alwaysbe a unit in k, and choosing a different ordering will simply alter the partialdifferential maps δst by ±1. This has no effect on the homology of the chaincomplex.

Definition 4.6. Let M be a module over an integral domain k. We define therank of M over k as

Rankk(M) = dimfrac(k)(frac(k)⊗M)

53

Theorem 4.7. We have the following relation between the Jones polynomialand the graded Euler characteristic of Khovanov homology with coefficients inan integral domain.

(q + q−1)JL(q) = χ(L) =∑i,j∈Z

(−1)iRankk(Hi,j(L, k))qj

Proof. The graded Euler characteristic of Khovanov homology is given by thepolynomial

χ(L) =∑i,j∈Z

(−1)iRankk(Hi,j(L, k))qj

which in turn is given by the graded Euler characteristic of the Khovanov com-plex

χ(L) =∑i,j∈Z

(−1)iRankk(CHKi,jk (L))qj

which is a standard result in homological algebra. Taking away the normaliza-tion gives

χ(L) = (−1)n−qn+−2n−∑i,j∈Z

(−1)iRankk(Ci,j)qj

then from the definition of Ci,j we get

χ(L) = (−1)n−qn+−2n−∑i,j∈Z

(−1)i∑

s∈Σ(L)|i(s)=i

Rankk(Cjs)qj

where Cjs is the submodule of Cs generated by elements with quantum gradingj. Since the elements of Cjs corresponding to enhanced states are a basis for Cjsover k, we get

χ(L) = (−1)n−qn+−2n−∑i,j∈Z

(−1)i∑

s∈Σ+(L)|i(s)=i,j(s)=j

Rankk(Cjs)qj =

= (−1)n−qn+−2n−∑

s∈Σ+(L)

(−1)i(s)qj(s) = (q + q−1)JL(q)

This explicitly connects Khovanov homology to the Jones polynomial. TheJones polynomial can be defined solely in terms of the Khovanov homologygroups, so Khovanov homology is a strictly stronger oriented link invariant.

54

4.3 Khovanov homology is an oriented link invariant

This section consists of proving that Khovanov homology is invariant underthe three basic Reidemeister moves. Our main tools for doing so will be thefollowing homological algebra results.

Theorem 4.8. Suppose we have a short exact sequence of chain complexes

0→ A→ B → C → 0

Then there is a corresponding long exact sequence of homology groups

0→ H0(A)→ H0(B)→ H0(C)→ H1(A)→ . . .

This is a very standard result in homological algebra and so the proof is omitted.

Corollary 4.9. Let B be a chain complex and A ⊆ B a sub-chain complex.

1. If A is acyclic then A is quasi-isomorphic to B/A.

2. If B/A is acyclic then A is quasi-isomorphic to B.

Proof. We have a short exact sequence of chain complexes

0→ A→ B → B/A→ 0

which gives a long exact sequence of homology groups

0→ H0(A)→ H0(B)→ H0(B/A)→ H1(A)→ . . .

Suppose first that A is acyclic. Then Hi(A) = 0 for all i so the above sequencebecomes

0→ 0→ H0(B)→ H0(B/A)→ 0→ . . .

and so Hi(B) = Hi(B/A) for all i. If instead B/A is acyclic, then the abovesequence becomes

0→ H0(A)→ H0(B)→ 0→ H1(A)→ . . .

and so Hi(A) = Hi(B) for all i.

Lemma 4.10. Suppose we have a commutative diagram of chain complexes

0 → A → B → C → 0↑ f ↑ g ↑ h

0 → X → Y → Z → 0

so that f and h are quasi-isomorphisms, and the rows are exact. Then g is alsoa quasi-isomorphism.

55

Proof. We stare at the induced long exact sequences

0 → H0(A) → H0(B) → H0(C) → H1(A) . . .|| || ↑ g0 || ||0 → H0(X) → H0(Y ) → H0(Z) → H1(X) . . .

By use of the five lemma, gi must be an isomorphism for all i.

It would also be nice if we could have a recursive construction of the Khovanovcomplex, just as there is a recursive construction of the Jones polynomial. Itturns out there is.

Let X be a diagram of graded abelian groups, and suppose there are no mapsbetween groups with identical gradings. Then we may safely direct sum anyabelian groups with a particular grading, and get induced maps (via projec-tion and inclusion) between the resulting abelian groups. This produces a newdiagram F (X). We call F the flattening operator.

Proposition 4.11. Let k be a ring, L a link diagram with at least one crossingand x a crossing in L. Consider an order on the crossings of L where x is themaximal crossing. Then

[[L]] = F ([[L+x ]]

δ−→ [[L−x ]]1)Now consider an order on the crossings of L where x is the minimal crossing.Then

[[L]] = F ([[L+x ]]

δ−→ −[[L−x ]]1)where δ denotes all the induced differential maps from the smoothing change atx. We denote by −[[L−x ]]1 the chain complex [[L−x ]]1 where all differentialmaps have flipped signs.

Proof. Every state s in L is either positively or negatively smoothed at x. Asa result the corresponding k-module Cs can be found either in [[L+

x ]] or [[L−x ]].The gradings in [[L+

x ]] match with those in [[L]]. However in [[L−x ]] the quantumgrading is shifted down by 1 since the negative smoothing at x is not counted.As a result we shift the quantum grading up by 1 so that all gradings matchwith those in [[L]].

When x is a maximal crossing, the signs of the partial differentials is unaffectedupon its removal and so they match with those in [[L]].

When x is a minimal crossing, its removal affects the sign of every differentialand so we must flip all the signs in [[L]].

Proposition 4.12. Let L be a link diagram with an order on its crossings, k aring and V = k[x]/(x2). Then

[[L t U ]] = V ⊗ [[L]]

56

Proof. The unknot U will be present in every complete smoothing of LtU , andwill contribute a single factor of V . It will not be involved in any smoothingchange and so the only maps it will induce are the identity with itself.

Lemma 4.13. The following chain complex is acyclic

F (Aid−→ −A)

where −A denotes the chain complex A whose differentials have flipped signs.

Proof. The differential of the above chain complex looks like(δi+1 0id −δi

)when acting on Ai+1 ⊕Ai, where δi are the differential maps. Suppose(

δi+1 0id −δi

)(ab

)=

(δi+1aa− δib

)= 0

then we deduce a = δib and so(ab

)=

(δi 0id −δi−1

)(b0

)It follows that the complex is acyclic.

Theorem 4.14. Khovanov homology is invariant under the first Reidemeistermove.

Proof. If we order the crossings so that the kink is minimal, we have

C = [[ ]] = F ([[ ]]δ−→ −[[ ]]1)

by proposition 4.11. Moreover, δ must locally look like the multiplication mapm. Also

F ([[ ]]δ−→ −[[ ]]1) = F (V ⊗k [[ ]]

m−→ −[[ ]]1)

by proposition 4.12. There is a subchain complex

B = F ([[ ]]1 id−→ −[[ ]]1)

given by the subspace of V spanned by 1, which acts via identity under mul-tiplication. This chain complex is acyclic by lemma 4.13. As a result C isquasi-isomorphic to the chain complex C/B given by

C/B = F ([[ ]]−1) = [[ ]]−1

57

from the subspace of V spanned by x. The negative quantum shifting will beremoved in the normalization, since the kink contributes 1 to n+ irrespective ofthe orientation chosen for the corresponding link component.

Theorem 4.15. Khovanov homology is invariant under the second Reidemeis-ter move.

Proof. Let us order the crossings so that the top crossing is minimal, and thebottom crossing is second in the order. Then by smoothing the top crossing weget

C = [[ ]] = F ([[ ]]δ−→ −[[ ]]1)

where δ are the differentials induced by the smoothing change at the top. Nowsmoothing the bottom crossing in both sub-chain complexes yields

C = F

−[[ ]]1 δ2−→ [[ ]]2↑ δ1 ↑ −m

[[ ]]∆−→ −[[ ]]1

where δ1 and δ2 are induced differential maps. We know the exact form of theright and bottom arrows, and for each subchain-complex the first two crossingsin the order have determined smoothing type, so we know the signs of all theinner differentials. This chain complex becomes

C = F

−[[ ]]1 δ2−→ [[ ]]2↑ δ1 ↑ −m

[[ ]]∆−→ −V ⊗ [[ ]]1

and we consider the sub-chain complex

B = F

0 → [[ ]]2↑ ↑ −id

0 → −[[ ]]2

with the subspace of V spanned by 1 in the bottom right complex. This is thesame chain complex considered in lemma 4.13, up to sign, and so it is acyclic.Therefore C is quasi-isomorphic to the following chain complex.

C/B = F

−[[ ]]1 → 0↑ δ1 ↑

[[ ]]∆−→ −[[ ]]

58

Here ∆ becomes an isomorphism, since it has fixed image onto the subspace ofV generated by x. Again through lemma 4.13 the following sub-chain complexis acyclic.

A = F

0 → 0↑ ↑

[[ ]]∆−→ −[[ ]]

As a result, C is quasi-isomorphic to the following chain complex.

(C/B)/A = F

−[[ ]]1 → 0↑ ↑0 → 0

= −[[ ]][1]1

By carrying out this move we have decreased both n+ and n− by 1, irrespectiveof the orientation chosen for the link diagram. As a result −n− and n+ − 2n−both increase by 1, and this will cancel out with the shifts above. Keep in mindthat there is also a quasi-isomorphism

−[[ ]][1]1 i−→ [[ ]]

induced by inclusion.

Theorem 4.16. Khovanov homology is invariant under the third Reidemeistermove.

Proof. Choose an order on the crossings so that the middle crossing in both ofthe following diagrams in minimal. Then we have

[[ ]] = F ([[ ]]→ −[[ ]]1)

[[ ]] = F ([[ ]]→ −[[ ]]1)We use the same strategy as in the proof for the Jones polynomial. The sub-

chain complexes [[ ]] and [[ ]] are quasi-isomorphic by two applicationsof the previous theorem. More importantly we may write the commutativediagram

[[ ]] → −[[ ]]1↑ i1 ||

−[[ ]][1]1 → −[[ ]]1↓ i2 ||

[[ ]] → −[[ ]]1

59

where both i1 and i2 are inclusion maps induced from the previous theorem,and are thus quasi-isomorphisms. Let C denote the flattening of the middlerow. We have the commutative diagram

0 → [[ ]] → [[ ]] → −[[ ]][1]1 → 0↑ i1 ↑ ||

0 → −[[ ]][1]1 → C → −[[ ]][1]1 → 0

with exact rows and quasi-isomorphisms on the right and left sides. As a re-sult the middle map is also a quasi-isomorphism by lemma 4.10. Thus C and

[[ ]] are quasi-isomorphic. By a symmetric argument C and [[ ]] are quasi-

isomorphic. As a result [[ ]] and [[ ]] have matching homology and thecorresponding oriented links have matching n− and n+ irrespective of the ori-entation chosen. We are done.

We have proven that Khovanov homology is an oriented link invariant. Wecan now refer to the Khovanov homology of an oriened link, rather than anoriented link diagram, and all notions related to Khovanov homology can alsobe redefined for oriented links.

4.4 Properties of Khovanov homology

We follow [23] but adapt the result to our own formulation. It turns out thatKhovanov homology has a lot of nice properties akin to other homology theories.

Proposition 4.17. Let L,M be two oriented links, and k a ring. Then

Hi,j(L tM,k) = Hi,j(L, k)⊗k Hi,j(M,k)

for all i, j ∈ Z.

Proof. We note first that we have

[[L tM ]] = [[L]]⊗ [[M ]]

since the differential maps will never map between [[L]] and [[M ]], and completesmoothings of L tM correspond to complete smoothings of L and M . Thetensor above is applied along homological gradings. The quantum grading isadditive along tensors and since

n+(L tM) = n+(L) + n+(M)

and

n−(L tM) = n−(L) + n−(M)

then normalizing yields

60

CHK•,•k (L tM) = [[L tM ]][−n−(L tM)]n+(L tM)− 2n−(L tM) =

= [[L]]⊗ [[M ]][−n−(L) +−n−(M)]n+(L) + n+(M)− 2n−(L)− 2n−(M) =

[[L]][−n−(L)]n+(L)− 2n−(L) ⊗ [[M ]][−n−(M)]n+(M)− 2n−(M) =

= CHK•,•k (L)⊗ CHK•,•k (M)

Taking homology of this chain complex yields the result.

Theorem 4.18. Let k be a ring. Let L be an oriented link and let x be acrossing for some projection of L. If x has positive orientation then there is along exact sequence in Khovanov homology

. . .→ Hi,j−1(L+x , k)→ Hi,j(L, k)→ Hi−1−ω,j−2−3ω(L−x , k)→ Hi+1,j−1(L+

x , k)→ . . .

for all j ∈ Z, where some orientation of L−x is chosen and ω = n−(L−x )−n−(L).The orientation on L+

x is inherited from that of L. If x has negative orientationthen there is a long exact sequence in Khovanov homology

. . .→ Hi−c,j−1−3c(L+x )→ Hi,j(L)→ Hi,j+1(L−x )→ H1+i−c,j−1−3c(L+

x )→ . . .

for all j ∈ Z, where some orientation of L+x is chosen and c = n−(L+

x )−n−(L).The orientation on L−x is inherited from that of L.

Proof. By proposition 4.11 we have the relation

[[L]] = F ([[L+x ]]

δ−→ [[L−x ]]1)

by choosing x to be maximal in the order. This gives a short exact sequence

0→ [[L+x ]]→ [[L]]→ [[L−x ]][1]1 → 0

For the case that x has positive orientation, it is important to see that

n+(L) = n+(L+x ) + 1

n−(L) = n−(L+x )

n+(L)− n+(L−x ) = (n+(L) + n−(L))− (n+(L−x ) + n−(L−x )) + ω = 1 + ω

61

so applying the shift [−n−(L)]n+(L) − 2n−(L) to the above short exact se-quence gives

0→ CHK•,•k (L+x )1 → CHK•,•k (L)→ C → 0

where

C = CHK•,•k (L−x )[1+n−(L−x )−n−(L)]1+n+(L)−n+(L−x )+2(n−(L−x )−n−(L)) =

= CHK•,•k (L−x )[1 + ω]2 + 3ω

Taking the corresponding long exact sequence in homology gives us the firstresult.Now asume x has negative orientation. it is important to note that

n+(L) = n+(L−x )

n−(L) = n−(L−x ) + 1

n+(L)− n+(L+x ) = (n+(L) + n−(L))− (n+(L+

x ) + n−(L+x )) + c = 1 + c

so applying the shift [−n−(L)]n+(L) − 2n−(L) to the above short exact se-quence gives

0→ B → CHK•,•k (L)→ CHK•,•k (L−x )−1 → 0

where

B = CHK•,•k (L+x )[n−(L+

x )− n−(L)]n+(L)− n+(L+x ) + 2(n−(L+

x )− n−(L)) =

= CHK•,•k (L+x )[c]1 + 3c

Taking the corresponding long exact sequence in homology gives us the secondresult.

These two results are the main computational tools used to compute Khovanovhomology. A very deep fact is that integral Khovanov homology can detect theunknot. The following theorem states this in detail. Its proof may be found in[18].

62

Theorem 4.19. Let L be an oriented link with integral Khovanov homologygiven by

Hi,j(L,Z) =

Z i = 0, j = 1

Z i = 0, j = −1

0 else

Then L is equivalent to the unknot.

We finish by computing the Khovanov homology of the Hopf link. Let L be theHopf link, with the following orientation

Considering the top right crossing x, we give its negative smoothing the followingorientation with one negative crossing

This gives the following value

ω = n−(L−x )− n−(L) = 1

and the long exact sequence

. . .→ Hi,j−1( , k)→ Hi,j( , k)→ Hi−2,j−5( , k)→ . . .

Here both L−y and L+y are unknots, and we know the Khovanov homology of

the unknot. From this we deduce that the Khovanov homology of the Hopf linkis given by the table

63

j\i 0 1 26 0 0 k4 0 0 k2 k 0 00 k 0 0

64

5 The knot group

5.1 Definition and basic properties

Definition 5.1. The knot group of a knot K is the fundamental group of thecomplement of K in R3 or S3.

We must show that this is well-defined:

Theorem 5.2. If K is a knot in R3 or S3 then R3 −K and S3 −K is path-connected.

Proof. By the generalized Jordan curve theorem (Theorem 2B.1, Hatcher 2001)Hi(S

3 − K) = Z for i = 3 − 1 − 1 = 1 and 0 otherwise, so in particularH0(S3 −K) = 0. By (Remark below Proposition 2.8, Hatcher 2001) H0(S3 −K) = 0 ⊕ Z = Z, so by (Proposition 2.7, Hatcher 2001) S3 −K has exactly 1path-connected component.For R3 − K, suppose the ‖x‖ < R − 1 for every x ∈ K. This is true for Rsufficiently large. Then let A = x ∈ R3 : ‖x‖ ≥ R, so that A ⊂ R3 −K and(R3 −K)/A = (R3/A)−K = S3 −K. Let X = R3 −K. (X,A) is a good pairas defined by (Theorem 2.13, Hatcher 2001), so by the same theorem we have along exact sequence:

· · · → H1(X/A)→ H0(A)→ H0(X)→ H0(X/A)→ 0

Substituting what we know:

· · · → H1(S3 −K)→ H0(A)→ H0(R3 −K)→ H0(S3 −K)→ 0

A is non-empty and path-connected, so H0(A) = Z by (Proposition 2.7, Hatcher2001), so H0(A) = 0:

· · · → Z→ 0→ H0(R3 −K)→ 0→ 0

By exactness we must then have H0(R3 −K) = 0, whence H0(R3 −K) = Z, soR3 −K has exactly one path-connected component by the same argument, soR3 −K is path-connected.

We must then show that the two complements give the same fundamental group:

Theorem 5.3. If K ⊆ R3 is a knot then π1(R3 −K) ∼= π1(S3 −K).

Proof. Since K is compact, it must be bounded in R3, by say R. So let N :=x ∈ R3 ∪ ∞ : ‖x‖ > R ⊂ S3 −K. We have these facts:, so (R3 −K)∪N =S3 −K and (R3 −K) ∩N deformation retracts to S2. Therefore, we deduce:

• (R3 −K) ∪N = S3 −K.

• (R3−K)∩N deformation retracts to S2, so π1((R3−K)∩N) = π1(S2) = 1by (Proposition 1.14, Hatcher 2001).

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• N is contractible, so π1(N) = 1

Finally, by van Kampen’s theorem (Theorem 1.20, Hatcher 2001) we have:

π1(S3 −K) = π1(R3 −K) ∗π1((R3−K)∩N) π1(N)= π1(R3 −K) ∗1 1= π1(R3 −K)

Theorem 5.4. If K1 and K2 are equivalent knots then π1(R3−K1) = π1(R3−K2).

Proof. The fundamental group is homotopy-invariant, so a fortiori homeomorphism-invariant. Equivalent knots have homeomorphic complements by definition.

The previous two theorems show that the knot group is well-defined, i.e. thatit is a knot invariant.Homology theory immediately gives us a property about the knot group:

Theorem 5.5. The abelianization of the knot group is Z.

Proof. By (Theorem 2A.1, Hatcher 2001) the abelianization of the knot groupis H1(S3 −K) which is Z as seen in the proof of Theorem 5.2.

We will have another proof of this theorem later.

Theorem 5.6. The knot group of a knot and its mirror image, i.e. reflectionacross the plane, are isomorphic.

Proof. Reflection is a homeomorphism, and homeomorphic spaces have isomor-phic fundamental groups.

Example 5.7. We compute the knot group of the unknot. The unknot is embed-ded in R3 as N := (cos θ, sin θ, 0) | θ ∈ R, so we need to compute π1(R3−N).We use van Kampen’s theorem (Theorem 1.20, Hatcher 2001) in the followingway:

• Let Ak := (R3−N)∩

(r cos θ, r sin θ, z)∣∣∣r ≥ 0, θ ∈

[2kπ

3 , 2(k+1)π3

], z ∈ R

.

• Each Ak is homeomorphic to R × (C − 0), which deformation retractsto S1, so π1(Ak) = π1(S1) = Z, with generator being a loop around theremoved unknot.

• Let ak be a generator of π1(Ak) = Z.

• A0 ∪A1 ∪A2 = R3 −N .

• Each Ak has a neighbourhood Uk in R3 −N that deformation retracts toAk and that each Ui ∩ Uj and Ui ∩ Uj ∩ Uk are path-connected.

• So Φ : π1(A0) ∗ π1(A1) ∗ π1(A2) → π1(R3 − N) is surjective, with kernelbeing the normal subgroup generated by a0a

−11 , a1a

−12 , and a2a

−10 .

• Therefore, π1(R3 −N) = 〈a0, a1, a2 | a0 = a1 = a2〉 = 〈a |〉 = Z.

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5.2 Wirtinger presentation

This is a powerful and simple algorithm to compute a group presentation of theknot group given a diagram. This can be found in (Exercise 1.2.22, Hatcher2001).First we reformulate van Kampen’s theorem in terms of group presentation:

Theorem 5.8. Suppose:

• X is a topological space.

• Ui ⊆ X is a family of open subsets in X.

• Uii cover X.

• For each i, j, k, Ui ∩ Uj ∩ Uk is path-connected.

• Each π1(Ui) is presented as 〈aipp | fiq(aipp)q〉.

• Each π1(Ui ∩ Uj) is presented as 〈aijsp | fijq(aijss)q〉.

• Each aijs is identified as fijs1(aipp) in π1(Ui) and as fijs2(ajpp) inπ1(Uj) under the maps φij : π1(Ui ∩ Uj)→ π1(Ui) and φji respectively.

Then, π1(X) is presented as:

π1(X) = 〈aipi,p | fiq(aippi,q, fijs1(aipp)fijs2(ajpp)−1i,j,s〉

Proof. From van Kampen’s theorem (Theorem 1.20, Hatcher 2001) we have:

π1(X) = (∗iπ1(Ui))/N

where N is the normal subgroup generated by:

φij(g)φji(g)−1 | g ∈ π1(Ui ∩ Uj)i,j

Step 1

∗iπ1(Ui) = ∗i〈aipp | fiq(aipp)q〉= ∗i(〈aipp |〉/NSfiq(aipp)q)= (∗i〈aipp |〉)/NSfiq(aipp)i,q= 〈aipi,p |〉/NSfiq(aipp)i,q

where NS means “normal subgroup generated by”.Step 2We claim that N is the same as the normal subgroup N2 generated by:

φij(aijs)φji(aijs)−1i,j,s

It is obvious that each φij(aijs)φji(aijs)−1 is in N , so it suffices to show that

each φij(g)φji(g)−1 is in N2. We know that g can be written as the product of

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a finite number of aijs or a−1ijs. We induct on the number of the multiplicands.

If g = g1g2 where φij(g1)φji(g1)−1, φij(g2)φji(g2)−1 ∈ N2, then:

φij(g)φji(g)−1 = φij(g1g2)φji(g1g2)−1

= φij(g1)φij(g2)φji(g2)−1φji(g1)−1

∈ N2

where the last step is because normal subgroups is closed under conjugation.So it suffices to show g ∈ N2 when g = aijs or g = a−1

ijs. The former case istrivial. For the latter case:

φij(g)φji(g)−1 = φij(a−1ijs)φji(a

−1ijs)−1

= (φji(aijs)φij(aijs))−1

∈ N2

where the last step follows from the fact that subgroups is closed under inverse.Step 3

π1(X)= (∗iπ1(Ui))/N= (〈aipi,p |〉/NSfiq(aipp)i,q)/N2

= 〈aipi,p |〉/NS(fiq(aipp)i,q ∪ φij(aijs)φji(aijs)−1i,j,s)= 〈aipi,p |〉/NS(fiq(aipp)i,q ∪ fijs1(aipp)fijs2(ajpp)−1i,j,s)= 〈aipi,p | fiq(aipp)i,q, fijs1(aipp)fijs2(ajpp)−1i,j,s〉

We are now ready for the Wirtinger presentation.

Theorem 5.9. Let K be a knot with diagram D. Pick an orientation on K.Let the arcs in D be labelled αi. At each double-point j of D let βj1 = α±1

j1

be the arc that passes over, and let βj2 = α±1j2 and βj3 = α±1

j3 be the two arcsthat pass under, where the positions and signs are chosen to match the followingdiagram:

βj2

βj3βj1

Figure 21: Naming of arcs near a crossing.

Then, π1(R3 −K) = 〈αii | βj1βj2β−1j1 = βj3j〉.

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Proof. We first divide the diagram into regions where there are either no arcs,one non-closed arc, or two non-closed arcs crossing each other. Edges connectingtwo regions must intersect the knot at most once. For example, here is the trefoilknot divided into 7 regions (including the big region outside the red hexagon):

Figure 22: Diagram of trefoil knot divided into 7 regions.

We lay the knot flat on the xy-plane except at crossings, where the over-passingarcs are lifted up a little bit to form a circular arc, as demonstrated in thisfigure:

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Figure 23: Diagram of crossing in 3 dimensions.

We build a surface that R3 −K deformation retracts to, as follows:

• Include the xy-plane as part of the surface.

• For regions with no arcs, leave it as is.

• For regions with one arc, build a small tunnel around the arc as shown inFigure 24a.

• For regions with two arcs, build a small tunnel around each arcs as shownin Figure 24b.

• Make sure that the surfaces glue nicely between regions.

(a) Surface for one-arc regions (b) Surface for two-arc regions

Figure 24: Surfaces for different regions

Then, the complement of the knot deformation retracts to the surface we justbuilt. In fact, since the outermost region contains no arcs and extends to infinity,we can deformation retract the whole surface to remove the outermost region.This makes our surface compact. In the example, this corresponds to removingthe region 7.Let the regions be A1, · · · , An. We have neighbourhoods Ui supAi such thatUi deformation retracts to Ai and Ui ∩ Uj deformation retracts to Ai ∩ Aj forevery i, j. To ensure path-connectedness of each intersection, we can thickenthe xy-plane on our surface by connecting the highest point on each columnwith fixed x and y coordinates to the point (0, 0, 1) ∈ R3. This top-thickenedsurface deformation retracts to our original surface and on each component,so it does not change the homotopy. Then, we can extend each Ai to containenough points above the xy-plane that each Ai∩Aj ∩Ak is still path-connected.

70

The point (0, 0, 1) can also be thought the position of the eye which observesthe surface using a bird’s eye view. We will also use (0, 0, 1) as the base point.

If Ai is a region with no arcs, then the surface assigned is the xy-plane, whichis contractible, so π1(Ui) = 〈|〉, i.e. the trivial group.If Ai is a region with one arc, then the surface assigned deformation retractsto S1 embedded as a loop encircling the arc, so π1(Ui) = Z with a generatordetermined by the right hand rule (for consistency). This means that one putsout the right hand with the thumb pointing to the direction of the loop, andthen curl the remaining 4 fingers to create a thumbs up; then the direction inwhich the 4 fingers curl is the orientation of the loop encircling the arc. Wewrite π1(Ui) = 〈αj |〉, where αj is the one arc in this region, and we identify thearc with the loop encircling it.If Ai is a region with two arcs, then the surface assigned deformation retracts toa space homotopy equivalent to S1 ∨S1, where the two copies S1 are embeddedas two loops, one encircling each arc. Therefore, π1(Ui) = Z ∗ Z with the twogenerators again being determined by the right hand rule. We now name thearcs in 3-dimensions according to Figure 21, and draw the loops correspondingto the generators:

Figure 25: Loops corresponding to generators in the Wirtinger presentation.

In the figure above, the pink loop corresponds to βj1, grey to βj2, and green toβj3. As stated above, π1(Ui) is generated by βj1 and βj2. Therefore, we mustexpress βj3 in terms of βj1 and βj2.Having labelled the four regions of the xy-plane as A, B, C, and D as in thefigure above, we have:

• βj1 is passing through B and going up through C, or passing through Aand going up through D.

• βj2 is passing through C and going up through D.

• βj3 is passing through B and going up through A.

Therefore, there are two ways to pass through B and go up through D:

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1. Pass through B; use βj3 to go up through A; pass through A again; useβj1 to go up through D.

2. Pass through B; use βj1 to go up through C; pass through C again; useβj2 to go up through D.

Therefore, βj3βj1 = βj1βj2, i.e. βj3 = βj1βj2β−1j1 .

Therefore, π1(Ui) = 〈αj1, αj2, αj3 | βj1βj2β−1j1 = βj3〉.

If Ai ∩Aj does not intersect the knot, then π1(Ui ∩Uj) = 〈|〉, the trivial group.If Ai ∩ Aj intersects the knot at the arc αk, then π1(Ui ∩ Uj) = 〈αk |〉. Thisserves to identify the copies of αk in the fundamental groups of the surfacesassigned to Ai and Aj .Therefore, gathering every region, Theorem 5.8 shows us that π1(R3−K) has apresentation that has one generator for every arc in the diagram, and at everycrossing we add the relation βj1βj2β

−1j1 = βj3, which is the result desired.

Example 5.10. We compute the knot group of the trefoil knot. First we orientthe knot and label the arcs as shown:

Figure 26: Diagram of trefoil knot oriented and crossings labelled.

Then, using Theorem 5.9, we compute:

π1(R3 −K) = 〈a, b, c | aba−1 = c, bcb−1 = a, cac−1 = b〉= 〈a, b | b(aba−1)b−1 = a, (aba−1)a(aba−1)−1 = b〉= 〈a, b | baba−1b−1 = a, abab−1a−1 = b〉= 〈a, b | baba−1b−1a−1 = 1, abab−1a−1b−1 = 1〉= 〈a, b | bab = aba, aba = bab〉= 〈a, b | aba = bab〉

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We are back to Theorem 5.5, that the abelianization of the knot group is Z.

Alternative proof. Recall that from Theorem 5.9, the knot group is presentedas:

〈αii | βj1βj2β−1j1 = βj3j〉

The abelianization corresponds to interpreting the group presentation as abelian,so conjugation fixes everything, and the presentation becomes:

〈αii | βj2 = βj3j〉

This means that at every crossing, the two under-passing arcs are identified.Therefore, as we travel along the knot, we always stay on the same arc, since:

• If we pass over a crossing, then we trivially stay on the same arc.

• If we pass under a crossing, then we are still on the same arc, since thetwo under-passing arcs are identified.

Therefore, the presentation becomes 〈α |〉, i.e. Z.

Theorem 5.11. The knot group is not a complete invariant, i.e. there areinequivalent knots K1 and K2 with the same knot group.

Proof. See (3D12, Rolfsen 1976). In particular, the two knots are the grannyknot and the square knot. They are both connected sums of trefoil, but thegranny knot is formed by taking the connected sum of two identical trefoils,while the square knot is formed by taking the connected sum of a trefoil withits mirror image.The knot groups are both 〈a, b1, b2 | ab1a = b1ab1, ab2a = b2ab2〉, as it is 〈a1, b1 |a1b1a1 = b1a1b1〉 and 〈a2, b2 | a2b2a2 = b2a2b2〉 with a1 and a2 identified.

However:

Theorem 5.12. If a prime knot has the same knot group as another knot, thenthe other knot is equivalent to the prime knot or its mirror image.

Proof. (Main result in Whitten 1985).

5.3 Meridian and longitude

If K ⊂ S3 is a knot then by Theorem 1.12 there is a neighbourhood V ⊃ K thatis homeomorphic to S1×B2, whereK corresponds to S1×0. By restricting theradius in B2, one obtains a smaller neighbourhood U ⊂ V still homeomorphicto S1 ×B2. Then:

• The interiors of V and S3 − U cover S3.

• V ∩ S3 − U deformation retracts to ∂V ⊂ S3.

• V deformation retracts to K ∼= S1.

73

• S3 − U deformation retracts to S3 − V

the interiors of V and S3 − U cover S3, and the intersection V ∩ S3 − U defor-mation retracts to ∂V ⊂ S3. Therefore, by the Mayer-Vietoris sequence (P.150Hatcher 2003), we have a long exact sequence:

H3(∂V ) → H3(K)⊕H3(S3 − v) → H3(S3) →H2(∂V ) → H2(K)⊕H2(S3 − v) → H2(S3) →H1(∂V ) → H1(K)⊕H1(S3 − v) → H1(S3) →H0(∂V ) → H0(K)⊕ H0(S3 − v) → H0(S3) → 0

Substituting in what we know:

0 → 0 → Z →H2(∂V ) → 0 → 0 →H1(∂V ) → Z⊕ Z → 0 →

0 → 0 → 0 → 0

We see that H2(∂V ) = H3(S3) = Z and H1(∂V ) = H1(V ) ⊕ H1(S3 − V ) =Z⊕ Z.Therefore, we can pick m, l ∈ H1(∂V ) such that m corresponds to 0 ∈ H1(V )and ±1 ∈ H1(S3 − V ); and such that l corresponds to ±1 ∈ H1(V ) and 0 ∈H1(S3 − V ). We can also represent m and l with simple closed curves. Since∂V is homeomorphic to a torus, we can make m and l intersect at exactly 1point.

• Since m = 0 ∈ H1(V ), we see that m is null-homotopic in V .

• Since l = 0 ∈ H1(S3 − V ), we see that l is null-homotopic in S3 − V .

• Since l = ±1 ∈ H1(V ) = H1(K) = Z = 〈K〉, we see that l is homotopicto an orientation of the knot K.

It can further be proved that L(m,K) = 1 and L(l,K) = 0, where L is thelinking number (Definition 3.5).m and l are called the meridian and longitude of the knot respectively, and twocurves satisfying the properties above are unique up to isotopy (Theorem 3.1,Burde, Zieschang, and Heusener, 2013).

Example 5.13. For the unknot, such a tubular neighbourhood is the solid torus,where a meridian is shown in red below and a longitude is shown in green below.

74

Figure 27: Meridian and longitude on torus encircling the unknot.

Although the knot group is not a complete invariant (Theorem 5.11), the knotgroup together with the data of the meridian and longitude turns out to be acomplete invariant:

Theorem 5.14. If two knots K1, K2 with knot groups G1 and G2, meridians m1

and m2, longitudes l1 and l2, are such that there is an isomorphism ϕ : G1 → G2

taking [m1], [l1] ∈ G1 to [m2], [l2] ∈ G2 respectively, where [−] is the homotopyclass in the fundamental group, then K1 and K2 are equivalent.

Proof. (Theorem 3.15, Burde, Zieschang, and Heusener, 2013).

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6 Bibliography

References

[1] 1954. Edwin E. Moise. Affine structures in 3-manifolds. VIII. Invariance ofthe knot-types; local tame imbedding. Ann. of Math. https://doi.org/10.2307/1969837

[2] 1958. Yoko Hashizume. On the uniqueness of the decomposition of a link.https://projecteuclid.org/euclid.ojm/1200689438

[3] 1960. Donald E. Sanderson. Isotopy in 3-manifolds. III. Connectivity ofspaces of homeomorphisms. Proc. Amer. Math. Soc. https://doi.org/10.1090/S0002-9939-1960-0112128-9

[4] 1971. Robert D. Edwards and Robion C. Kirby. Deformations of Spaces ofImbeddings. Ann. of Math. https://doi.org/10.2307/1970753

[5] 1976. Dale Rolfsen. Knots and Links. https://bookstore.ams.org/

chel-346-h/

[6] 1976. Morris W. Hirsch. Differential Topology Springer. https://doi.org/10.1007/978-1-4684-9449-5

[7] 1985. Wilbur Whitten. Knot Complements and Groups. Topology. https://doi.org/10.1016/0040-9383(87)90019-X

[8] 1993. Maxim Kontsevich. Vassiliev knot invariants. http://pagesperso.

ihes.fr/~maxim/TEXTS/VassilievKnot.pdf

[9] 1995. Dror Bar-Natan. The Fundamental Theorem of Vassiliev Invariants.https://arxiv.org/pdf/q-alg/9702009.pdf

[10] 1996. Kunio Murasugi. Knot Theory and Its Applications. Springer. https://doi.org/10.1007/978-0-8176-4719-3

[11] 2000. Michael Sullivan. Knot factoring. https://www.tandfonline.com/doi/abs/10.1080/00029890.2000.12005198

[12] 2001. Allen Hatcher. Algebraic Topology. https://pi.math.cornell.edu/

~hatcher/AT/AT.pdf

[13] 2002. Dror Bar-Natan. On Khovanov’s categorification of the Jones poly-nomial. https://arxiv.org/abs/math/0201043

[14] 2003. John M. Lee. Introduction to Smooth Manifolds. Springer. https://doi.org/10.1007/978-0-387-21752-9

[15] 2004. Colin Adams. The Knot Book. http://math.harvard.edu/~ctm/

home/text/books/adams/knot_book/knot_book.pdf

76

https://doi.org/10.2307/1969837

https://doi.org/10.2307/1969837

https://projecteuclid.org/euclid.ojm/1200689438

https://doi.org/10.1090/S0002-9939-1960-0112128-9

https://doi.org/10.1090/S0002-9939-1960-0112128-9

https://doi.org/10.2307/1970753

https://bookstore.ams.org/chel-346-h/

https://bookstore.ams.org/chel-346-h/

https://doi.org/10.1007/978-1-4684-9449-5

https://doi.org/10.1007/978-1-4684-9449-5

https://doi.org/10.1016/0040-9383(87)90019-X

https://doi.org/10.1016/0040-9383(87)90019-X

http://pagesperso.ihes.fr/~maxim/TEXTS/VassilievKnot.pdf

http://pagesperso.ihes.fr/~maxim/TEXTS/VassilievKnot.pdf

https://arxiv.org/pdf/q-alg/9702009.pdf

https://doi.org/10.1007/978-0-8176-4719-3

https://doi.org/10.1007/978-0-8176-4719-3

https://www.tandfonline.com/doi/abs/10.1080/00029890.2000.12005198

https://www.tandfonline.com/doi/abs/10.1080/00029890.2000.12005198

https://pi.math.cornell.edu/~hatcher/AT/AT.pdf

https://pi.math.cornell.edu/~hatcher/AT/AT.pdf

https://arxiv.org/abs/math/0201043

https://doi.org/10.1007/978-0-387-21752-9

https://doi.org/10.1007/978-0-387-21752-9

http://math.harvard.edu/~ctm/home/text/books/adams/knot_book/knot_book.pdf

http://math.harvard.edu/~ctm/home/text/books/adams/knot_book/knot_book.pdf

[16] 2004. Josephine Yu. Alexander polynomial of knots. https://math.

berkeley.edu/~hutching/teach/215b-2004/yu.pdf

[17] 2005. Edward Long. Topological invariants of knots: three routesto the Alexander Polynomial. https://www.ucl.ac.uk/~ucbpeal/

alexandermac.pdf/

[18] 2010. P. B. Kronheimer and T. S. Mrowka Khovanov homol-ogy is an unknot-detector. http://abel.math.harvard.edu/~kronheim/

khovanov-unknot.pdf

[19] 2011. S. Chmutov, S. Duzhin, J. Mostovoy. Introduction to Vassiliev knotinvariants https://arxiv.org/pdf/1103.5628.pdf

[20] 2012. Rob Schneiderman. Simple Whitney towers, half-gropes and the Arfinvariant of a knot. https://arxiv.org/abs/math/0310304

[21] 2013. Gerhard Burde, Heiner Zieschang, Michael Heusener. Knots. DeGruyter. https://www.degruyter.com/view/product/179871

[22] 2013. Nancy Scherich. The Alexander Polynomial.https://nancyscherich.files.wordpress.com/2018/01/

the-alexander-polynomial.pdf

[23] 2014. Paul Turner. A hitchhiker’s guide to Khovanov homology. https:

//arxiv.org/abs/1409.6442

[24] 2015. Louis H. Kauffman. An Introduction to Khovanov Homology. http://homepages.math.uic.edu/~kauffman/IntroKhovanov.pdf

[25] 2015. Justin Roberts. Knot Knotes. http://math.ucsd.edu/~justin/

Papers/knotes.pdf

[26] 2016. Nate Bade. Lecture 18: Properties of the Alexander Polynomial.https://web.northeastern.edu/beasley/MATH7375/Lecture18.pdf

[27] 2016. Mark Lackenby. Elementary knot theory. https://arxiv.org/abs/1604.03778

[28] 2016. Zach Haney. Lecture 12: Prime factorization of knots. https://web.northeastern.edu/beasley/MATH7375/Lecture12.pdf

[29] 2019. J. C. Cha and C. Livingston. KnotInfo: Table of Knot Invariants.https://www.indiana.edu/~knotinfo/

77

https://math.berkeley.edu/~hutching/teach/215b-2004/yu.pdf

https://math.berkeley.edu/~hutching/teach/215b-2004/yu.pdf

https://www.ucl.ac.uk/~ucbpeal/alexandermac.pdf/

https://www.ucl.ac.uk/~ucbpeal/alexandermac.pdf/

http://abel.math.harvard.edu/~kronheim/khovanov-unknot.pdf

http://abel.math.harvard.edu/~kronheim/khovanov-unknot.pdf

https://arxiv.org/pdf/1103.5628.pdf

https://arxiv.org/abs/math/0310304

https://www.degruyter.com/view/product/179871

https://nancyscherich.files.wordpress.com/2018/01/the-alexander-polynomial.pdf

https://nancyscherich.files.wordpress.com/2018/01/the-alexander-polynomial.pdf

https://arxiv.org/abs/1409.6442


http://homepages.math.uic.edu/~kauffman/IntroKhovanov.pdf

http://homepages.math.uic.edu/~kauffman/IntroKhovanov.pdf

http://math.ucsd.edu/~justin/Papers/knotes.pdf

http://math.ucsd.edu/~justin/Papers/knotes.pdf

https://web.northeastern.edu/beasley/MATH7375/Lecture18.pdf





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