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 developing some basic notions about knots and links, we will step into basic knot arithmetic and operations, exploring fundamental theorems such as Rei- demeister Theorem and Schubert’s Theorem. Next, it would be a natural manner to step into the world of invariants. We will introduce various integer-valued and polynomial invariants, before wrapping up this chapter with Vassiliev’s masterpiece of family of invariants. After that, we will explore Khovanov homology, which is the categoriﬁcation of Jones’ polynomial. It is indeed a powerful invariant, as we will see it solves the unknot recognition problem. Finally, we will switch our focus to the knot group and its Wirtinger presentation to end.

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

1 Contents

1 Introduction 3 1.1 What is a knot? ...... 3 1.2 When are two knots the same? ...... 4 1.3 Making ﬁnite and discrete models ...... 5 1.4 Projections ...... 8 1.5 Links ...... 9 1.6 Diagrams ...... 11

2 Reidemeister’s Theorem and Seifert surfaces 13 2.1 Reidemeister’s Theorem ...... 13 2.2 Basic knot arithmetic ...... 14 2.3 Seifert surfaces ...... 15 2.4 Schubert’s Theorem ...... 18

3 Integer-valued and polynomial invariants 21 3.1 An overview ...... 21 3.2 3-colouring ...... 24 3.3 p-colouring ...... 30 3.4 Seifert matrix and Alexander polynomial ...... 32 3.5 Jones polynomial ...... 35 3.6 Vassiliev invariants ...... 44

4 From Jones’ polynomial to Khovanov homology 48 4.1 Alternative version of Jones polynomial ...... 48 4.2 Khovanov homology ...... 49 4.3 Khovanov homology is an oriented link invariant ...... 55 4.4 Properties of Khovanov homology ...... 60

5 The knot group 65 5.1 Deﬁnition and basic properties ...... 65 5.2 Wirtinger presentation ...... 67 5.3 Meridian and longitude ...... 73

6 Bibliography 76

2 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:

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

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

3 Since S3 := {(x, y, z, w) ∈ R4 | x2 + y2 + z2 + w2 = 1}, the unit 3-sphere in R4, is the one-point compactification of R3, hence by noting that the image of a knot is already compact , one can instead define an oriented knot to be an injective continuous map f : S1 → S3 without loss of generality, and similarly for 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 be a homeomorphism; but S1 and S3 are not homeomorphic, as one can remove two points from S3 and still get a connected space, yet the removal of any two points from S1 results in a disconnected space. Therefore, from an oriented knot S1 → S3 we can pick a point not in the image, and then use stereographic projection 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 switch the definition of an (oriented) knot freely between an injective continuous map S1 → 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 all the translations, non-zero dilations, and rotations. Deformations should be continuous, which leads to this attempted deﬁnition:

Attempted definition: Two knots f, g : S1 → R3 are equivalent if 1 3 there is a family of knots Fs : S → R indexed by s ∈ [0, 1] such 1 that F0 = f, F1 = g, and [0, 1] × S :(s, t) 7→ Fs(t) is continuous. This definition is similar to the definition of a homotopy, but here we require 1 3 each intermediate function Fs : S → R to be injective as well. However, one can 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. Then 1 f ∼ h is witnessed by Hs where Hs := F2s for s ∈ 0, 2 and Hs := G2s−1 1 for s ∈ 2 , 1 . It is continuous because F1 = G0 = g. The problem with this definition of equivalence, however, is that every knot is equivalent to the unknot under this definition. To see this, take any knot f : S1 → R3. Since S1 is compact, there is t such that the x-coordinate of f(t) is maximum. Using that point as reference, break open the knot from that point while connecting the two resulting points with an arc growing in size to the right, then shrink everything from the left to the reference point, gradually re-parametrising the function so that the limit is still injective, and then finally use a translation to turn it into the standard unknot. The problem can be fixed by considering the whole space instead of just the knot:

4 Deﬁnition 1.6. Two oriented knots f, g : S1 → R3 are equivalent if they 3 3 are ambient isotopic, i.e. there is a family of homeomorphisms Fs : R → R for s ∈ [0, 1] with F0(x) = x for all x and F1(f(t)) = g(t) for all t, such that 3 3 [0, 1] × R → R :(s, x) 7→ Fs(x) is continuous. 3 Deﬁnition 1.7. Two knots K1,K2 ⊂ R are equivalent if there is a family 3 3 of homeomorphisms Fs : R → R for s ∈ [0, 1] with F0(x) = x for all x and 3 3 F1(K1) = K2, such that [0, 1] × R → R :(s, x) 7→ Fs(x) is continuous. Remark 1.8. • These two notions of equivalence are both equivalence relations, using a similar argument.

• If two oriented knots are equivalent, then clearly their underlying knots are 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 nice enough and their underlying knots are equivalent, then the two oriented knots are either equivalent, or one is equivalent to the orientation-reversal of the other (Theorem 1.16 below).

3 3 Given a continuous family of homeomorphisms Fs : R → R where F0 is the identity, we say that F1 is isotopic to the identity. A homeomorphism R3 → 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 to 3 3 3 be orientation-preserving iff for every x ∈ R the induced map h∗ : H3(R , R − 3 3 {x}) → H3(R , R −{f(x)}) is the same map as the map induced by translation 3 hx(y) := y − x + h(x) for all x ∈ R , where H•(A, B) is the relative singular homology. To facilitate computations, one notes that for any x ∈ R3, if h is differentiable at x and the derivative at that point is non-singular, then h preserves orientation at x iff det(Dhx) > 0. This gives us a criterion that is simpler at times:

Theorem 1.9. Two oriented knots f, g : S1 → R3 are equivalent iﬀ there is an orientation-preserving homeomorphism h : R3 → R3 such that h ◦ f = g. 3 Similarly, two knots K1,K2 ⊂ R are equivalent iﬀ there is an orientation- 3 3 preserving homeomorphism h : R → R 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 discrete version of knots. Unfortunately our current definition of knots is too broad which allows for some pathological examples called wild knots, which we do not prefer to compute with.

Deﬁnition 1.10. A piecewise-linear knot is an oriented knot f : S1 → R3 1 such that there are t1, ··· , tn ∈ S with tn = t1 such that f restricted to [ti, ti+1] is linear for every i = 1 → n − 1.

5 Its underlying knot is called a polygonal knot, and is characterized by the 3 existence of p1, ··· , pn ∈ R 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 that the 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 there is an open neighbourhood N 3 p with a homeomorphism h : N → B3 taking N ∩ 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 : k~vk < 1}. Proof. 1 =⇒ 2: Suppose f is equivalent to a piecewise-linear knot p. Around each corner of p one can draw small neighbourhoods such that the intersection of the knot and each small neighbourhood is the two line segments touching the corner. Then one can use bump functions (Lemma 2.14, Lee 2003) to perturb p inside those small neighbourhoods and deform the knot into a smooth

6 embedding using a deformation that satisfies the hypothesis of the topological isotopy extension theorem (Corollary 1.4, Edwards and Kirby 1971), so this deformation extends to an ambient isotopy, showing that p is equivalent to a smooth knot, so f is also equivalent to a smooth knot. 2 =⇒ 3: Suppose f is equivalent to a smooth knot s, with F being the orientation-preserving homeomorphism. The image of s is a 1-dimensional sub- manifold of R3, and has a tubular neighbourhood by (Theorem 5.1, Chapter 4, 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} to the 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 is called a tame knot. Remark 1.14. From the fourth equivalent definition we see that an oriented knot 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 underlying knot. Then f is either equivalent to g or to the orientation-reversal (Definition 1.3) of g. Proof. Consider s : S1 → S1 such that f(t) = g(s(t)). It can be shown that s is a well-defined homeomorphism. Label the circle [0, 2π]. Lift s to a continuous function s : [0, 2π] → R. Then s is either strictly increasing or strictly decreasing. If s is strictly decreasing, then replacing g(t) with g(−t) to begin with would make s strictly increasing, so we now assume s is strictly increasing and show that 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 the intermediate knots have the same underlying knot. The deformation can be made to satisfy the hypothesis of the topological isotopy extension theorem (Corollary 1.4, Edwards and Kirby 1971), so this deformation extends to an ambient isotopy, showing that f is equivalent to g.

7 Corollary 1.16. Suppose f and g are tame knots with equivalent underlying knots. Then f is either equivalent to g or to the orientation-reversal (Definition 1.3) of g. We also need a finite and discrete version of equivalence. A slight deformation can be approximated by replacing one side of a given external triangle with the remaining two sides, as defined below: Definition 1.17. An elementary isotopy applied to a polygonal knot p with vertices p1, ··· , pn, specified by an index i and a triangle [pi, pi+1, q] which only intersects the knot at [pi, pi+1], is a polygonal knot formed by replacing the line segment [pi, pi+1] with the two line segments [pi, q] and [q, pi+1]. We let the elementary isotopies generate an equivalence between polygonal knots: Definition 1.18. Two polygonal knots p and q are said to be isotopic, also known as combinatorially equivalent, if there is a finite sequence of polygonal knots p = p0, p1, ··· , pn = q, such that each pi and pi+1 are related by an elementary isotopy or the inverse of an elementary isotopy. One wonders whether this finite and discrete version of equivalence is the same as 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, we must 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, we say that p has regular projection to E if finitely many points on E are double points, i.e. mapped by two distinct points on p, and the other points on the image are single points, i.e. mapped by only one point on p, and no vertex is projected 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 that p has regular projection to E is an open dense subset of all the 2-dimensional subspaces 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; and if E admits regular projection of p, then any subspace sufficiently close to E admits a regular projection of p. Proof. See (Proposition 1.12, Burde, Zieschang, and Heusener, 2013).

8 Remark 1.22. On a double point, we use a solid line to denote the point on top and a broken line to denote the point below, after choosing a direction of height. 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 subsection without proof. Deﬁnition 1.23. An oriented link with n components is an injective contin- `n 1 3 `n 1 3 uous map f : i=1 S → R or f : i=1 S → S . An oriented link decomposes into n oriented knots.

Deﬁnition 1.24. A link is a subset of R3 or S3 that is homeomorphic to `n 1 i=1 S . A link decomposes into n path-connected components, each of which is 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, called the 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 the number of components. Remark 1.27. The underlying link “forgets the orientation”, since each orientation- reversal has the same underlying link.

`n 1 3 Deﬁnition 1.28. Two oriented links f, g : i=1 S → R are equivalent if they are ambient isotopic.

3 Deﬁnition 1.29. Two link L1,L2 ⊂ R are equivalent if there is a family 3 3 of homeomorphisms Fs : R → R for s ∈ [0, 1] with F0(x) = x for all x and 3 3 F1(L1) = L2, such that [0, 1] × R → R :(s, x) 7→ Fs(x) is continuous.

9 Remark 1.30. • These two notions of equivalence are both equivalence relations. • If two oriented links are equivalent, then their underlying knots are also equivalent.

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

`n 1 3 Theorem 1.31. Two oriented link f, g : i=1 S → R are equivalent iff there is an orientation-preserving homeomorphism h : R3 → R3 such that h ◦ f = g. 3 Similarly, two links L1,L2 ⊂ R are equivalent iff there is an orientation- 3 3 preserving homeomorphism h : R → R such that h(L1) = L2. Definition 1.32. A piecewise-linear link is an oriented link with each component piecewise-linear. Its underlying link is called a polygonal link, and is characterized by that each component is polygonal. Definition 1.33. A link L is said to be locally flat if every component is locally flat.

`n 1 Theorem 1.34. The following are equivalent for an oriented link f : i=1 S → 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. `n 1 3 `n 1 3. f : i=1 S → R extends to an injective continuous map f : i=1 S × B2 → R3. 4. The underlying link of f is locally ﬂat.

Deﬁnition 1.35. An oriented link satisfying the above equivalent conditions is called 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 oriented link 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 underlying links. Then f is either equivalent to g or to the orientation-reversal (Definition 1.26) of g.

10 Definition 1.39. An elementary isotopy applied to a polygonal link p is an elementary isotopy applied to one of its component, except that the triangle cannot touch other components. Definition 1.40. Two polygonal links p and q are said to be isotopic, also known as combinatorially equivalent, if there is a finite sequence of polygonal links p = p0, p1, ··· , pn = q, such that each pi and pi+1 are related by an elementary 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, we say that p has regular projection to E if finitely many points on E are double points, i.e. mapped by two distinct points on p, and the other points on the image are single points, i.e. mapped by only one point on p, and no vertex is projected 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 that p has regular projection to E is an open dense subset of all the 2-dimensional subspaces 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; and if E admits regular projection of p, then any subspace sufficiently close to E admits a regular projection of p.

1.6 Diagrams Theorem 1.44. Up to planar isotopy, there are only finitely many diagrams with 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 certain number of crossings and assigned names to the knots according to the number st of crossings. For example, the knot 31 is the 1 minimal knot with 3 crossings up to mirroring, and is the trefoil knot (refer to Figure 1). In fact it is the only minimal knot with 3 crossings up to mirroring, where minimal means that any diagram of any equivalent knot has at least 3 crossings. In a diagram, at a crossing one can switch the two lines and produce a diagram of another link. For example, the diagrams of the unlink and of the Hopf link in Figure 2 are related by such a switching. Definition 1.46. A switching of a crossing in a diagram of a link is another diagram formed by replacing the solid line by a broken line and the broken line by a solid line at the crossing in the diagram of the link. Theorem 1.47. Switching all the crossings corresponds to reflecting the knot along the plane. Proof. Geometrically clear.

11 Definition 1.48. The mirror image of a knot or link is the knot or link produced 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.

12 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 diagram of the knot. Unfortunately, any knot can be represented by inﬁnitely many diﬀerent 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 links are isotopic (Deﬁnition 1.40) if and only if D1 = M1M2...MnD2, where Mi is either 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 of generality, 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 the new vertex to be q. Without loss of generality, we can assume all projections involved 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 zero or one Ω1 move. Note the interior of this triangle would overlap various branches in D2. To simplify the case, in the diagram, we can triangulate [pi,1, pi,2, q] into elementary pieces of four different types. (The explicit cases are left to the reader as an exercise.) After drafting them out, we can observe that each case is in fact one of Ω2,Ω3 and planar isotopy. Hence, we showed that each elementary isotopy

13 can be decomposed into a ﬁnite sequence of planar isotopy and Reidemeister moves.

Remark 2.2. We note that Ω1, Ω2, Ω3 is not the full list of Reidemeister moves in the most formal sense. Each of them has their own variations, where the number of variations depends on if we count orientations or not. However, the proof above applies to all variations, because these extra cases can be discarded by 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 any small band in R3 which meets the knots only at its ends. “Connect” the two knots by cutting the two ends of K1,K2, and joining the other two sides of the band. 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 that cuts the knot at exactly two distinct points. In this case, we say S factors the knot. Remark 2.7. Note that the connected sum has to be defined under oriented knots as if we do not give an orientation, as it is not guaranteed that the operation is well-defined. Take the two pretzel knots P (3, 5, 7) and P (3, 5, 9) as an example, the four possibilities fall into four different isotopy classes. However,

14 we can omit the orientation if we define the connected sum under the set of invertible knots, knots with both orientations equivalent to each other. This is the reason why Figure 5 has no orientation shown, since all knots with crossing number ≤ 7 are invertible. Wikipedia has the number of invertible knots up to crossing number 16. As an additional sidenote, we can also now define the disconnected sum of two knots as the disjoint union of them. We notate this by K1 t K2. We will also construct some more notions related to divisibility of links. Definition 2.8. Given knots L, K. We say L divides K iff there exists a knot M such that K = L#M. Definition 2.9. For a non-trivial knot K, we call it prime if there exists no non-trivial L and M such that K = L#M; otherwise we call K a composite knot. 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, where U 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 we find 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 we established 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 it provides an alternative way of dealing with knot theory in a more topological manner. Theorem 2.10. For every oriented knot (or link) K, there exists an orientable and 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 at every crossing jump to the other arc, still walking along the orientation. 4. Repeat this process for other starting points, and obtain a disjoint union of circles, called the Seifert circles. 5. We note that the above process can be seen as taking the natural smooth at every crossing of the knot. 6. For concentric circles, raise the inner circles upward.

15 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 corresponding discs with a half-twisted band, so that the boundary of the band gives the original crossing.

9. The resulting surface is the surface required.

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

Part 3: Showing that the resulting surface is orientable For 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, colour the top of the disc with red and the bottom of the disc with blue (pretend that the disc is two atoms thick). In the second case, colour the top of the disc with blue and the bottom with red. Note that travelling along the band ﬂips the colouring 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 right of the boundary; for concentric circles, this property is preserved. Therefore, the colouring is consistent as we travel along the bands from discs to discs, which shows that the surface is orientable, as it has two distinct sides.

Part 4: Showing that the boundary of the resulting surface is the original knot At each crossing, we carefully used a half-twisted band to connect the two circles so that the boundary of the band gives the original crossing. Filling in the circles then 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 orientable two-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, hence there 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 link diagram. Deﬁnition 2.14. The genus g(K) of a knot K is the minimal genus across all the possible Seifert surfaces for K.

16 Fact. A knot has genus 0 iﬀ it is the unknot. Assume K is the unknot. Then K bounds a disc, which obviously has genus 0. For the other direction, suppose a knot 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 polygonal disc 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 of genus (Theorem 2.17 below). To any Seifert surface we can assign a unique knot simply by considering the boundary. However, given a knot K, there may exist more than one Seifert surface associated with the knot. In fact, given a knot K, and one of its Seifert surface S which is genus g, we can undergo topological surgery to obtain a new Seifert surface S0 with genus g + 1. This is the reason why in particular we have to set genus to be minimal in the deﬁnition above.

Genus turns out to be a integer-valued knot invariant, meaning if K and L are two equivalent knots, we have g(K) = g(L). Theorem 2.15. Let K, L be two equivalent knots. Then, the genus of K is equal to that of L. Proof. It suffices to prove g(K) ≥ g(L). Let S be a Seifert surface for K such that g(S) = g(K). Let Fs be the ambient isotopy bringing K to L. Now we have F1(S) is a Seifert surface for L. S = F0(S) is homeomorphic to F1(S), so they 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 t K2. If H and S meet transversely, then there exists an isotopy of H that the intersection of it and S is a simple arc. Proof. Note H ∩ S is a set consisting a simple arc and possibly finitely many disjoint circles. Take any circle σ. σ must bound a disc on H ∩ S, since if this is not the case, we can construct a new Seifert surface with one less genus. If σ 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 attach two discs along their boundaries to the cuts. This surface also contradicts minimality of genus. Finally, an isotopy of H is given by shrinking H inside interior 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, consider their Seifert surfaces with minimal genera. Assuming without loss of generality that 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 connected sum. 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 surface H of K1#K2 of minimal genus. By Remark 2.6, we have a sphere S that factors K1#K2, i.e. splitting K1#K2 into K1 and K2. We now consider the intersection of H and S, where we assume transversality. Now H ∩ S consists of an arc that connects the two points of factoring and a set of pairwise disjoint circles (note it is possible that this set is empty). However, all the disjoint circles can 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. An inverse of knot K (name it by L) would satisfy g(K) + g(L) = g(U) = 0, which is a contradiction since genus is non-negative. Corollary 2.19. Let K be a non-trivial knot, L be a knot. Then K#L is non-trivial.

Proof. We note g(−) is a non-negative function. Suppose K#L is the trivial knot. 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, as well 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 factorization of K. Proof. If K is prime then it is obvious. So we assume K is composite, by its deﬁnition there are non-trivial knots L, M such that K = L#M. We can again check if L, M are prime knots and repeat the process above. Note this procedure cannot repeat more than g(K) times, since the only knot with genus 0 is the trivial knot. Hence, we will always get a ﬁnite sequence of knots with connected sum K.

To prove uniqueness, We would need some additional notions and lemmas. Note the following deﬁnition is very uninsightful, i.e. it has no use other than proving this theorem, but nevertheless we put it here. Deﬁnition 2.21. Let K be a knot. Let S be a set of pairwise disjoint spheres {S1,S2, ..., Sr}, such that:

18 • 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 remainder of this section we will abuse notation of S both referring to the finite amount of balls and the (infinite amount of) points that made up the spheres for the sake of simplicity. Lemma 2.22. Let S be a dss for K. Then the factoring of K consists of |S| + 1 prime 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 must exist since all spheres are disjoint. Now we define a new knot K˜ by the union of γ and K complement Bl, where γ is the arc on Sl joining the two points of 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 induction hypothesis) 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 disjoint 0 to all elements in S, and notate its associated balls T1,T2. Deﬁne S := (S \ 0 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 is equivalent to S0.

Proof. We notate the two (associated) interior and exterior balls of St as Bt,Ct respectively, for all t from 1 to r. Suppose j 6= k. Without loss of generality Bk ⊂ T1, and we can split into two cases, i.e. whether there exists l such that Bk ⊂ Bl ⊂ T1 or not. If the answer is positive, then balls Bk,Ck represent the knot Kk and “K \ Kk” w.r.t. both systems. Otherwise, Bk,Ck represent the 0 same 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 S0 share the same representation of factor knots. Remark 2.25. We should be aware that in any non-empty dss S, we have at least 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 of S2. This is because for this to not happen, all balls of S2 must be contained in A, i.e. A contains some prime knots which is not contained in S2, arriving to a 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 is equivalent to S2.

19 Proof. We will do induction on |S1 ∩ S2|. Base case: Induction on |S1|+|S2|. (Note: Base cases of this induction is trivial to check, we would not state it here.) Assume true for |S1| + |S2| < n. Note any sphere in S1 has no intersection with S2. Hence by remark 2.25, we have some sphere S1p maximal within S2q. Deﬁne S3 := (S2 \ S2q) ∪ S1p. Hence S2 is equivalent to S3 by construction and Lemma 2.24. Then, it suﬃces to prove 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 is non-empty, hence by Remark 2.25, we have a innermost ball Q with respect to both S1 and S2. We name its respective sphere R. Note we can assume R and S2 has non-empty intersection, or else we can proceed with a manner similar to the base step, using remark 2.25. Hence there exists an “innermost” curve µ of R ∩ S2. This induces a disk D with µ as its boundary, and its interior having empty intersection with S2. Considering the definition of a dss, we have µ ⊂ F for some F ∈ S2. Hence D ⊂ E, where E is the ball with respect to F . Note E represents a prime knot, hence if we slice F into two balls by D, only one ball would contain the prime knot (the other will represent the unknot). We let V be that ball, W be the respective sphere. Now we “ -shrink” W to W 0 such that W 0 still contains the prime knot, but no longer has µ as boundary. We assume 0 0 W is transverse to all other spheres in S2. Now we define S4 := (S2 \ F ) ∪ W . 0 By Lemma 2.24, S4 is equivalent to S2. Since W ∩ R does not contain µ, but F ∩ 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 of Theorem 2.20 and an easy corollary of Theorem 2.26. Note a similar theorem holds for links without reservation. The proof is much longer, and it is due to [Hashizume 1958].

20 3 Integer-valued and polynomial invariants

3.1 An overview One of the major issues mathematicians are trying to solve in knot theory is the recognition problem, i.e. to determine whether two given knots are related by a finite sequence of elementary isotopies. Reidemeister Theorem in the previous chapter did give us a insightful pathway. In this chapter, we will introduce another 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 function F 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 F from 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 and an invariant F , if we can show that F (K1) 6= F (K2), then we can conclude that K1 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 we will cover some of those in the following subsections. We would ﬁrst introduce three elementary link invariants and some properties about them.

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

Deﬁnition 3.5. Let D be a diagram of an oriented link N. We give each crossing 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 numbers up, then dividing the sum by 2. Corollary 3.12 below explains why it is indeed an integer-valued invariant.

21 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 Reidemeister moves. First, a crossing resolvable by an Ω1 move is a self-intersection. If we assume the two strings in an Ω2 move comes from different components of the link 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 number as well, as the type of the central crossing is unchanged, and the types of the other two crossings would not change either, since Ω3 simply involves shifting the 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 with all orientations reversed. Then L(M) = L(P ) = −L(N). Similar results still hold for n-component links with n > 2. Definition 3.8. The unknotting number u(K) of a knot K is defined as the minimum natural number n such that there is a diagram D of K and a collection of n crossings of D which, when switched, transforms D into a diagram of the unknot. Note we can expand this definition to the notion of unlinking number, which describes 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 minimum number of crossings across all diagrams of K. Fact. Note two knots need not be equivalent if some of their invariants are equal. The map that sends all knots to zero, with unknot and trefoil, serves as an 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 a finite amount of crossings.

Proof. Take any knot K and some diagram of knot K, D. Draw a line parallel to y-axis and tangent to the knot such that all points on the knot has x-coordinate larger than that of any point on the line except points of tangency. Now choose a 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. We construct a new knot by t 7→ (x(t), y(t), t), and then connect the two endpoints by drawing a line parallel to z-axis instead. This is obviously the unknot, as it is contained in the xz-plane after some stretching and pulling. We end the proof by noting the new and old knots are related by switching all crossings with its two corresponding points ((x(u), y(u), z(u)), (x(v), y(v), z(v))) having the following relationship: (u − v)(z(u) − z(v)) < 0.

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

c(K) Proof. From the proof above, u(K) ≤ c(K). If we switch more than 2 crossings to obtain the unknot by the mechanism above, we use the alternative map below instead: t 7→ (x(t), y(t), 1 − t), then connecting the two endpoints by drawing a line parallel to z-axis. By similar steps to the proof above, we know this alternative map is also the unknot, and number of switchings required is c(K) less than c(K) − 2 . 1 Corollary 3.12. Despite the factor of 2 , the linking number is always an integer. Proof. We would assume Theorem 3.10 is true for links. (The philosophy of the proof is the same.) Let K be a link. Note that for every change of crossing, the linking 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 unknotting number 0. Its linking number is 0.

2. The left trefoil. It has crossing number 3: since 31 is a diagram of the trefoil, the crossing number is bounded above by 3; all diagrams with crossing number less than 3 must be diagrams of the unknot. This can be verified 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 unknot by Theorem 3.23 below, so it is non-zero. To see it is actually 1, switch any crossing and label the nearer crossing (in terms of an anti-clockwise path) C1. Then reflect the upper part along a tangent at C1, and twist accordingly to get an unknot. Its linking number is 0 since all crossings are 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 you just choose one the resultant is the Hopf link.) Its linking number is 0 regardless the orientation of the two components. 4. Hopf link. Its crossing number is 2 and its unlinking number is 1. Its linking number is 1 or −1 depending on the orientation.

23 (a) Unknot (b) Trefoil

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 connection with linear algebra enables us to distinguish links with a high number of crossings. To learn this material, [25] was used but ultimately we presented our own formulation 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 each arc 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 coloured differently. 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 exists a non-trivial assignment of colours A ∈ T (D), i.e. an assignment that uses at least 2 colours. Example 3.16. Example of a possible 3-coloured trefoil:

24 Figure 8: 3-coloured trefoil

Theorem 3.17. Tricolourability and the number of 3-colourings τ(D) is a link invariant. Proof. By Theorem 3.3, it suﬃces to show that the number of 3-colourings is preserved under each Reidemeister move. To do this, we establish a bijection between the 3-colourings of a diagram before applying Reidemeister move and the 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 the crossing are the same colour a. So for this crossing to satisfy the tricolourability property, 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 that the 3 distinct arcs on the right must be the same colour (i.e case 1 below). If a 6= b, then the third arc on the right hand side must be colour c distinct from

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

(a) Case 1

(b) Case 2

Figure 10: Invariance under R2

For R3, we deﬁne a colouring on the left hand side and look for a corresponding bijection 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 colourings constant, 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 the same colour. Then x can take the values 0,1,2,3. However, suppose x = 2 then the third crossing has at least two incidents arcs with the same colour and so x must be 3 to satisfy the tricolourability property. Therefore x can only take the values 0 (case a), 1 (case b,c,d : depending on which crossing x is applied to) and 3 (case e). In all 5 cases we have a well-deﬁned bijection satisfying the 3-colouring property.

26 (a) Case a

(b) Case b

(d) Case d

(e) Case e

Figure 11: Invariance under R3

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

27 Indeed, if we define our 3 colours to be 0, 1 and 2 in the field F3 then we can observe that a crossing satisfying the 3-colouring property is equivalent to xi + xj + xk = 0 where xi, xj, xk ∈ {0, 1, 2} is defined as the assigned colour to the incident arc Ai,Aj,Ak at the crossing Ca. m Theorem 3.18. T (D) = {(x1, x2, ..., xk) ∈ F3 : xi + xj + xk = 0 at each crossing involving arcs Ai,Aj,Ak} Theorem 3.19. Let D be the diagram of a link. Then, T (D) is a vector space and τ(D) = 3dim T (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 have m arcs and n crossings. Then define A to be an m × n matrix encoding the m m linear equations of T (D) such that T (D) = {x ∈ F3 : Ax = 0}. Then τ(D) = 3dim T (D) = 3dim(Ker(A)). Example 3.20. The first example we shall study is the trefoil where we have labelled 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 1 1 1 1 1 1 1 1 1 1 1 1 1 Noting that Ax = 0 ⇐⇒ 1 1 1 x = 0 ⇐⇒ 0 0 0 x = 0, we see 1 1 1 0 0 0 dim Ker(A) that dim Ker(A) = 2 and τ(31) = 3 = 9.

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

Figure 13: 51 knot with labelled arcs and crossings

This gives rise to the following system of equations:

Ax = 0 0 1 1 0 1 0 1 0 1 1   ⇐⇒ 1 1 0 1 0 x = 0   1 0 1 1 0 1 0 1 0 1 1 1 0 1 0  0 −1 1 0 0    ⇐⇒ 0 0 0 −1 1  x = 0   0 0 0 −1 1  0 0 1 0 −1 1 1 0 1 0 0 −1 1 0 0   ⇐⇒ 0 0 −1 0 1 x = 0   0 0 0 −1 1 0 0 0 0 0

So dim Ker(A) = 1 and τ(51) = 3. Example 3.22. The ﬁnal example is the ﬁgure 8 knot labelled as follows :

29 Figure 14: 41 knot with labelled arcs and crossings

0 1 1 1 1 1 0 1  1 1 0 1 1 1 0 1 0 1 1 1  0 1 1 1   ∼   ∼   1 1 1 0 0 −1 1 0  0 0 −1 1 1 0 1 1 0 0 1 −1 0 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, we are unable to draw any conclusions concerning the others.

Theorem 3.23. The trefoil 31 is distinct from the unknot 01, the ﬁgure-8 knot 41, and the cinquefoil knot 51.

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

Deﬁnition 3.24. Let D be a diagram of a link L, then Tp(D) is the set of colourings 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 have 2xi − xj − xk ≡ 0 (mod p), where each xi ∈ Fp is the colour assigned to Ai.

30 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 knots 01, 41 and 51 which we failed to do with 3-colouring. We can already assume that τ(01) = 5 as the unknot only has the 5 trivial colourings.

Figure 15: 51 knot with labelled arcs and crossings

This gives rise to the matrix A:

 0 −1 −1 0 2  1 0 0 0 −1  0 2 0 −1 −1 0 1 0 0 −1     −1 −1 0 2 0  ∼ 0 0 1 0 −1      2 0 −1 −1 0  0 0 0 1 −1 −1 0 2 0 −1 0 0 0 0 0 so dim ker(A) = 2 and τ5(51) = 25

31 Example 3.27.

Figure 16: 51 knot with labelled arcs and crossings

This gives rise to the matrix :

 0 −1 −1 2  −1 2 −1 0  −1 2 −1 0   2 −1 0 −1  0 −2 3 −1  0 −2 3 −1   ∼   ∼   −1 2 −1 0   0 −1 −1 2   0 −1 −1 2  −1 0 2 −1 0 3 −2 −1 0 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 distinguish the 51 and the 41 knot as τ5(51) 6= τ5(41). However, we were still unable to distinguish the 41 knot and the unknot and shall try to explore this in the next invariants.

3.4 Seifert matrix and Alexander polynomial The Alexander polynomial is the first polynomial invariant, and it is the only polynomial invariant since its discovery for over sixty years. Not only does it provide much more information as an invariant, but the fact that a polynomial corresponds to an infinite sequence of integers also allows for easier recognition. There are multiple ways to derive the Alexander polynomial, and we would choose 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 will only state important facts without proving them. Also note from this point onwards 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)−1 loops 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 identify the inner horizontal equator and any vertical loop around it (i.e. meridian and longitude). For each additional component, we identify the boundary of those components as a loop (note knots are simple closed polygonal curves). (Equivalently, we choose any basis of the first homology group of T .) We name these curves as li, i = 1, 2, ..., 2g(L) + µ(L) − 1(= c).

32 Next, thicken the Seifert surface by S × [0, 1], where by convention we denote S 0 to be S × {0}. Project li onto S × {1}, and call these loops li. We can finally define the Seifert matrix: Definition 3.28. The Seifert matrix M of a Seifert surface (for a link) is 0 the c × c matrix with entries defined by Mij := L(li ∪ lj). The Seifert matrix of the trivial knot is defined as the empty matrix by convention.

Definition 3.29. The Alexander polynomial ∆A(L) of link L is a polynomial T in indeterminate t, defined by ∆A(L)(t) := det(M − tM ). Note it is only well- defined as a link invariant up to normalization by multiplying some Laurent monomial, such that the constant term is positive. Proving the Alexander polynomial is actually a link invariant is out of the scope of this document, hence we omit it as well. [22] pp.4-5 gives a clear construction of this proof.

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

The Alexander polynomial is very powerful (at its time), but it is difficult to compute without other techniques. It is also difficult to actually locate the loops mentioned above when the knots become more complicated. This changed when John Conway discovered that a variation of the Alexander polynomial (now called 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 diagrams except for one crossing, with diﬀerences explained by the notation.

Remark 3.30. The property linking the classical Alexander polynomial and 1 − 1 Alexander-Conway polynomial is ∆A(L)(t) = ∆C (L)(t 2 − t 2 ). (Proof of this relation can be found in Kunio, pp.110-112) This relation is well-deﬁned since Alexander 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 of interesting properties. We will prove some of them below.

Theorem 3.31. ∆A(L) satisﬁes:

−1 1. ∆A(L)(t) = ∆A(L)(t ), 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.

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

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

2. Note by the remark above, we have ∆A(L)(1) = ∆C (L)(0). Remains to prove 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 unknot, and the property assures us that the Alexander polynomial when t = 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 diagrams 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 symmetric Laurent polynomial p with p(1) = 1, there exists a knot with Alexander polynomial p. Equivalently, for all polynomials q in t2 with constant term 1, there exists a knot with Alexander-Conway polynomial q. The proof requires too much machinery to be put here, one can refer to [Rolfsen pp.171] for a proof. Basic knowledge 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 do not intend to introduce the notion of equivalence of Seifert matrices. Interested readers should go through Murasugi’s Knot Theory and its Applications Ch.5-6 for a better understanding. 1. This is proved by the claim that the Seifert matrix of K is transpose to that of K. To see why the claim is true, take regular diagrams of K, K respectively. Name them D, D respectively. D and D are related by all orientations reversed as well. Consider the c loops we have constructed at the beginning of this section. Since all orientations are reversed, we have 0 0 L(li ∪ lj) = L(lj ∪ li).

34 2. Similarly, this is proved by the claim that the Seifert matrix of K˜ is the negative 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 proving it 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 focus our study of Seifert matrices and Alexander polynomials onto prime knots only.

Theorem 3.34. Given two oriented links K1,K2. Suppose they have Seifert matrices 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. Call them F1,F2. We now construct a Seifert surface of K1#K2 by adding a small band. (Refer to Definition 2.4) The first part is now trivial by considering that any loop in F1 and F2 are disjoint if we assume F1 and F2 are disjoint. The second part is a corollary of the first part, which is trivial by considering the definition 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 and link diagrams are unoriented unless stated otherwise. Let L be an link diagram. + We denote by Lx the smoothing 7→ about some crossing x in L. This smoothing is done locally, producing a new − link diagram. Denote by Lx similarly the smoothing 7→ The well-deﬁnedness of these smoothing processes is clear. Let h(L) denote the number of crossings in L. Since smoothing removes a crossing from the link diagram, it follows that

+ h(Lx ) = h(L) − 1 − h(Lx ) = h(L) − 1 for any crossing x in L. Therefore applying the smoothing process successively must terminate at some link L0 satisfying h(L0) = 0. From previous results this means that L0 is a disjoint union of unknots.

35 The Jones polynomial is constructed recursively, keeping track of these smoothings 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 Kauffman bracket is defined as a function

h·i : LinkDiagrams → Z[A] satisfying

1. hUi = −(A2 + A−2)

+ −1 − 2. hLi = AhLx i + A hLx i 3. hL t Ui = hUi · hLi for any link diagram L and any crossing x in L. To show that this definition is complete and well-defined, we will prove that it is equivalent to a closed form of the Kauffman bracket.

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

Theorem 3.35. For any link diagram L we have the explicit formula X hLi = Aw(s)(−A2 − A−2)|s| s∈Σ(L) and this is an equivalent deﬁnition for the Kauﬀman bracket.

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

hLi = hUi|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 the explicit formula holds for all link diagrams with crossing numbers less that h(L).

36 hLi + −1 − = AhLx i + A hLx i X X = A Aw(s)(−A2 − A−2)|s| + A−1 Aw(s)(−A2 − A−2)|s| + − s∈Σ(Lx ) s∈Σ(Lx ) X X = Aw(s)+1(−A2 − A−2)|s| + Aw(s)−1(−A2 − A−2)|s| + − s∈Σ(Lx ) s∈Σ(Lx )

for some arbitrary crossing x in L. Let s be a state in L. Then s is either positive 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 sign at x in w(s). As a result

X hLi = Aw(s)(−A2 − A−2)|s| s∈Σ(L)

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

hUi = (−A2 − A−2)

Let x be a crossing in L. Then any state s that is positive at x restricts ∗ + ∗ to a state s on Lx with w(s ) = w(s) − 1. Likewise, if s were negative at x, we would have w(s∗) = w(s) + 1. It follows that

X X hLi = Aw(s)(−A2 − A−2)|s| = A Aw(s)(−A2 − A−2)|s|+ + s∈Σ(L) s∈Σ(Lx )

−1 X w(s) 2 −2 |s| + −1 − +A A (−A − A ) = AhLx i + A hLx i − s∈Σ(Lx )

To prove the third step, Let L = L∗ t U. Then states on L are in one to one correspondence with states on L∗, since U has no crossings and does not interact with L∗. However, every complete smoothing of L will have one more unknot than the corresponding smoothing on L∗. It follows that

hLi = (−A2 − A−2)hL∗i

This completes the equivalence.

37 Deﬁnition 3.36. Let L be an oriented link diagram. A crossing will henceforth be 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+ minus the number of negative crossings n−. It is easy to see that the orientation chosen does not matter if L were a knot. Hence the writhe of an (unoriented) knot is well defined. Every oriented link (diagram) has an underlying unoriented link (diagram). We define the Kauffman bracket hLi of an oriented link diagram L to be the Kauffman bracket of the underlying unoriented link diagram. Definition 3.37. Let L be an oriented link. Let L∗ be an oriented link diagram for L. Define

∗ ∗ hL i f (A) = (−A3)−wr(L ) L hUi The Jones polynomial of L is given by

−1/4 JL(t) = fL(t ) The Jones polynomial is in fact a Laurent series in the variable t1/4. This is because fL(A) is a Laurent series in A by theorem 3.35, which demonstrates that hL∗i 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. h i = Ah i + A−1h i =

= A(−A2 − A−2)h i + A−1h i =

= −A3h i

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

38 2. h i = Ah i + A−1h i =

= A2h i + h i + h i + A−2h i =

h i + (A2 + A−2)h i + (−A2 − A−2)h i =

= h i

So the Kauﬀman bracket itself is invariant under this move. It follows that the Jones polynomial is invariant under this move, since the writhe is unchanged by the second Reidemeister move for any chosen orientation. 3. h i = Ah i + A−1h i

However

h i = h i = h i

by two applications of the second Reidemeister move. Finally

h i = Ah i + A−1h i

so the required equality follows, since the writhe is unchanged by the third Reidemeister 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 diagram below. The first step is to compute the Kauffman Bracket. We choose to first compute the smoothing of the top crossing.

39 + − (b) Vertical Smoothing Lx (a) Horizontal Smoothing Lx

+ −1 − hLi = AhLx i + A hLx i

= A + A−1 + −1 Now by smoothing out the bottom crossing of both Lx and Lx we get 4 new knots :

40 + + (a) Vertical Smoothing Lx (1) (b) Horizontal Smoothing Lx (2)

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

Applying recursively the Kauﬀman Bracket property, we have:

+ + −1 + hLx i = AhLx (1)i + A hLx (2)i

− − −1 − hLx i = AhLx (1)i + A hLx (2)i

= 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:

+ −1 − hLi = AhLx i + A hLx i

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 crossings have a negative orientation we have : wr(L) = −2

A6 + A2 + A−2 + A−6 f (A) = (−A3)2 = A6(−A4 − A−4) = −A10 − A2 L −A2 − A−2

−1 So by taking A = t 4 , the Jones polynomial of our oriented Hopf Link is

−5/2 −1/2 JL(t) = −t − t

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

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

where JL+ (t),JL− (t),JL0 (t) are the Jones polynomial of the link L which only diﬀer locally at a crossing according to the following 3 diagrams:

(a) L+ (b) L−

42 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 polynomial of the 41 knot. We consider the central crossing highlighted below :

)

(a) L+

(b) L−

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 in Example 3.38 with the reverse orientation. By a similar argument to Example 3.38 we ﬁnd that the Hopf Link with the orientation of L− has Jones polynomial −5/2 1/2 JL− (t) = −t − t . Applying the skein relation:

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

−1 1/2 −1/2 −5/2 1/2 t (t) − tJL− (t) = (t − t )(−t − t ) 3 2 −1 −tJL− (t) = −t + t − t + 1 − t 2 1 −1 −2 JL− (t) = t − t + 1 − t + t 2 −1 −2 So J41 (t) = t − t + 1 − t + t .

43 To conclude, we could for instance use the Jones polynomial to distinguish the unknot 01 and the 51 knot which was impossible with the previous invariants we had and in particular the p-colouring invariant. 51 has Jones polynomial 2 4 5 6 7 J51 (t) = t + t − t + t − t whereas J51 (t) = 1 so we can indeed distinguish these two knots. The Jones polynomial nevertheless isn’t ﬂawless as 51 and 10132 mirror both 2 4 5 6 7 have the Jones polynomial J01 (t) = t +t −t +t −t so we cannot distinguish them with this invariant.

3.6 Vassiliev invariants Vassiliev invariants are one of many major advancements in low-dimensional topology, as well as knot theory. Rather than one single invariant, it represents a whole class of invariants. In this final subsection, we would first define what is a Vassiliev invariant, and our final aim is to investigate into Fundamental Theorem 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 we mention 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, and the 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 geometric intuition we can say two singular knots are equivalent if one can continuously deform into another without affecting any new self-intersections and the transversality of the self-intersections. It remains to formalize this concept. We would need the definition of a flat disk. Definition 3.45. Given an α-point A of a singular knot K. Let B be a small closed neighbourhood around A. Let γ ⊂ B be a curve such that its interior is “flat” (i.e. is subset of some plane) and contains B ∩ K. Then we say the interior 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 flat disks as F (K). Definition 3.46. Given two singular knots K, L. They are said to be equivalent if 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 singular diagrams D, E. K and L are equivalent iff D = M1M2...MnE, where Mi is either planar isotopy or Reidemeister moves not involving α-points, or Ω3 with an α-point as central crossing. We notate this special case Ω. Proof. Omitted. Previous workings have shown that skein relations are crucial to calculation of invariants. For singular knots, we introduce the Vassiliev skein relation as follows: 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 knots with exactly n α-points by applying the Vassiliev skein relation n times. Note we can also view the skein relation as resolving an α-point to a β-point. Such generalization is independent of the order of α-points chosen, since one can express all such generalizations into summation of the same 2n terms, and addition is commutative. Note we can denote the generalization of the knot invariant to singular 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 if V (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 the coefficients of Alexander-Conway polynomial and that of Jones polynomial, are Vassiliev invariants. We would include the former as an example here. Theorem 3.50. The nth coefficient of the Alexander-Conway polynomial is a Vassiliev 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 its generalizations as well) Suppose we have a singular knot Q with k α-points. Let k R be the complete smoothing of Q. Then ∆C (Q) = t ∆C (R). So if k ≥ (n+1), n+1 n ∆C (Q) would be multiple of t , hence coefficient of t becomes 0.

Deﬁnition 3.51. Note singular knots are functions from S1 to R3, so for each singular knot there exists a parametrizing circle. A chord diagram of an singular knot K, σ(K), is deﬁned as marking n pairs of points on the parametrizing circle, then drawing n chords to connect those points two by two such that the endpoints 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 singular 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 deduce that there is a one-to-one correspondence of α-points of K1 and K2. Now we 3 place K1 and K2 in R such that their α-points coincide and a small neighbourhood of each α-point coincide. Now we can construct a deformation of K1 to K2 preserving the position of all α-points and their respective neighbourhoods. If there are (a finite number of) α-points produced in the process of deformation, we can assume generic conditions on them, i.e. they occur at different 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 is unchanged.

We now denote Vn as the R-module of all Vassiliev invariants of order ≤ n, and An 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 n α-points K. By deﬁnition, we have Ker δn = Vn−1. Hence δn descends to an 0 injective linear transformation δn : 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 τ be a chord in the diagram that does not intersect with any other chord. We call τ as an isolated chord.

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

Definition 3.56. A weight system of order n is a function f ∈ RAn that satisfies (4T). An unframed weight system of order n is a weight system of order n that satisfies (1T). Remark 3.57. From Definition 3.56, we denote the subspace of weight systems f of order n Wn , and the subspace of unframed weight systems of order n Wn. f Immediately we have Wn ⊆ Wn ⊆ 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 the four components, then note that under the alternating sum, these components cancel each other out.

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

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

Part 1: 1T Let K be a singular knot where its chord diagram σ(K) has an isolated chord τ. Let the α-point corresponding to τ be . We can hence divide the knot into two parts such that they do not have α-points in common, since τ is isolated. By switching crossings, we obtain an alternative form of K, K˜ such that the two halves lie on two sides of some plane respectively. Performing Vassiliev skein relation on , we have V (K) = V (K˜ ) = V (K˜2) − V (K˜1) = 0, where K˜1 and K˜2 represent 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 of this proof is to allocate a singular knot Ki with partial knot diagrams Mi (as in Lemma 3.58), satisfying σ(Ki) = Di. This is a trivial step since Mi are designed to suit this purpose. We ﬁnish the proof by directly applying the lemma and noting Theorem 3.52. Note for a total proof of the image, we have to prove the converse statement as well, but we will skip this since this is out of scope of this document. Interested readers 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 the two 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 derivative of f to be identically zero. Note by Theorem 3.10, K can be transformed to U by a finite number of switching crossings.

Since 0 = V ( ) = V ( ) − V ( ) (ﬁrst equality is valid since f has ﬁrst 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 unoriented unless stated otherwise. We continue to follow [24, p. 4-5]. We give an alternate definition of Jones’ polynomial. Define the modified Kauffman bracket as a function

h·i2 : LinkDiagrams → Z[q] satisfying the following

−1 1. hUi2 = q + q

+ − 2. hLi2 = hLx i2 − qhLx i2

3. hL t Ui2 = hUi2 · hLi2 Carry out the substitution A2 = −q−1 in the original Kauﬀman bracket. Then it is clear that (1) and (3) match with those in the original Kauﬀman bracket. However

+ − + −2 − −1 + −1 − hLx i2 − qhLx i2 = hLx i2 + A hLx i2 = A (AhLx i2 + A hLx i2) so (2) differs by a factor of A−1. By induction the two versions of the Kauffman bracket differ by A−1 raised to the power of the crossing number of the link. As a result

−h(·) h·i2 = A h·i

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

hLi hLi2 J (q) = (−A3)−wr(L) = (−A3)−n++n− · An++n− = L hUi q + q−1

hLi2 hLi2 = (−1)n−−n+ A4n−−2n+ = (−1)n−−n+ (−q)n+−2n− = q + q−1 q + q−1

hLi2 = (−1)n− qn+−2n− q + q−1 We will leave the Jones polynomial in this form. There is a corresponding explicit 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 together with a label ±1 for every unknot in the complete smoothing of L induced by the 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 and let 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 underlying state. Theorem 4.1. Let L be a link diagram. Then

X i(s) j(s) hLi2 = (−1) q s∈Σ+(L) 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

−h(L) X w(s) 2 −2 |s| X −2i(s) −1 |s| hLi2 = A A (−A − A ) = A (q + q ) = s∈Σ(L) s∈Σ(L) X = qi(s)(−1)i(s)(q + q−1)|s| s∈Σ(L) Now in the above expression (q+q−1)|s| is a generating function for all enhanced states with underlying state s, where the coeﬃcient to qt is the number of such enhanced states where λ = t. All such states will have the same value of i(s) and so it follows combinatorially that

X i(s) j(s) hLi2 = (−1) q s∈Σ+(L)

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

4.2 Khovanov homology The Jones polynomial misses some information about the enhanced states of a link diagram. Khovanov developed a homology theory inspired by the Jones polynomial which captures more data about these enhanced states whilst remaining an oriented link invariant. In fact the coeﬃcients of the Jones polynomial 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 corresponding chain complex of graded k-modules called the Khovanov complex. Fix an ordering for the crossings in L. Lets say there are n crossings. A state can now be represented by a positively ordered sequence (x1 . . . xn) of elements ±1 that represent 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. We will have a map δst from Cs into Ct for two states t, s if t is given by switching a positive smoothing in s to a negative one at some crossing x in L. As a result i(t) = i(s) + 1. We refer to i(·) as the homological grading.

⊗|s| 2 The k-modules are Cs = V where V = k[x]/(x ). This gives a factor of V for every unknot in the complete smoothing of L induced by s. The elements 1, x will 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 enhanced states with underlying state s, and these elements will span Cs.

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

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

sx In which case δst is given by the local map (−1) m 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

sx In which case δst is given by the local map (−1) ∆ 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 in the smoothing change and is given by identity. It is clear that δst is a k-linear morphism in both cases.

We will now construct the Khovanov complex. We deﬁne

i M C = Cs s∈Σ(L)|i(s)=i as the k-module with homological grading i. We have the maps

i i+1 δi : C → C given by X δi = δst i(s)=i,i(t)=i+1 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 and λ 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 check that δst does so for every pair of states s, t. For this it suﬃces to check that m and ∆ 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 0 belongs 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 quantum grading j (as well as the element 0). The maps δi descend down to maps

i,j i+1,j δi,j : C → C by the previous proposition. This will give a graded chain complex of L denoted by [[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]]{j} denote the shifting of the quantum grading of [[L]] by +j.

We now turn to oriented links. The chain complex [[·]] for an oriented link diagram is defined for the underlying unoriented link diagram. Definition 4.3. Let L be an oriented link diagram together with an order on the crossings of L. The Khovanov complex of L with coefficient ring k is defined as

•,• CHKk (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 the negative smoothing of s at xi, t the negative smoothing of s at xj and u the negative 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 , and

so it satisﬁes the associative identity. Moreover, txi = sxi and 1+sxj = rxj and so

r s r +s (−1) xj m ◦ ((−1) xi m ⊗ id) = (−1) xj xi m ◦ (m ⊗ id) = s +t t s = −(−1) xj xi m ◦ (id ⊗ m) = −(−1) xi m ◦ ((−1) xj id ⊗ m)

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

r s r +s ((−1) xj ∆ ⊗ id) ◦ (−1) xi ∆ = (−1) xj xi (∆ ⊗ id) ◦ ∆ = s +t t s = −(−1) xj xi (id ⊗ ∆) ◦ ∆ = −((−1) xi id ⊗ ∆) ◦ (−1) xj ∆

3. We consider the maps

∆ ◦ m :1 ⊗ 1 7→ 1 ⊗ x + x ⊗ 1 1 ⊗ x 7→ x ⊗ x (1) x ⊗ 1 7→ x ⊗ x x ⊗ x 7→ 0

52 and (id ⊗ m) ◦ (∆ ⊗ id) :1 ⊗ 1 7→ 1 ⊗ x + x ⊗ 1 1 ⊗ x 7→ x ⊗ x (2) x ⊗ 1 7→ x ⊗ x x ⊗ x 7→ 0

These are the same maps. Now adding the signs yields

r s t s (−1) xj ∆ ◦ (−1) xi m = −((−1) xi id ⊗ m) ◦ ((−1) xj ∆ ⊗ id)

and

t s r s (−1) xi ∆ ◦ (−1) xj m = −((−1) xj id ⊗ m) ◦ ((−1) xi ∆ ⊗ id)

It can be shown completely symmetrically that

r s t s (−1) xj ∆ ◦ (−1) xi m = −((−1) xi m ⊗ id) ◦ ((−1) xj id ⊗ ∆)

and

t s r s (−1) xi ∆ ◦ (−1) xj m = −((−1) xj m ⊗ id) ◦ ((−1) xi 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 L with coefficient ring k are defined as the homology groups of the chain complex •,• CHKk (L) for any choice of order on the crossings of L. That is, ker(δ ) Hi,j(L, k) = i+n−,j−n++2n− im(δi+n−−1,j−n++2n− ) One can check that the ordering chosen for the crossings in the link diagram do not affect the Khovanov homology groups. This is because −1 will always be a unit in k, and choosing a different ordering will simply alter the partial differential maps δst by ±1. This has no effect on the homology of the chain complex. Definition 4.6. Let M be a module over an integral domain k. We define the rank of M over k as

Rankk(M) = dimfrac(k)(frac(k) ⊗ M)

53 Theorem 4.7. We have the following relation between the Jones polynomial and the graded Euler characteristic of Khovanov homology with coeﬃcients in an integral domain.

−1 X i i,j j (q + q )JL(q) = χ(L) = (−1) Rankk(H (L, k))q i,j∈Z Proof. The graded Euler characteristic of Khovanov homology is given by the polynomial

X i i,j j χ(L) = (−1) Rankk(H (L, k))q i,j∈Z which in turn is given by the graded Euler characteristic of the Khovanov complex

X i i,j j χ(L) = (−1) Rankk(CHKk (L))q i,j∈Z which is a standard result in homological algebra. Taking away the normalization gives

n− n+−2n− X i i,j j χ(L) = (−1) q (−1) Rankk(C )q i,j∈Z then from the deﬁnition of Ci,j we get

n− n+−2n− X i X j j χ(L) = (−1) q (−1) Rankk(Cs )q i,j∈Z s∈Σ(L)|i(s)=i j where Cs is the submodule of Cs generated by elements with quantum grading j j j. Since the elements of Cs corresponding to enhanced states are a basis for Cs over k, we get

n− n+−2n− X i X j j χ(L) = (−1) q (−1) Rankk(Cs )q = i,j∈Z s∈Σ+(L)|i(s)=i,j(s)=j

n− n+−2n− X i(s) j(s) −1 = (−1) q (−1) q = (q + q )JL(q) s∈Σ+(L)

This explicitly connects Khovanov homology to the Jones polynomial. The Jones polynomial can be deﬁned solely in terms of the Khovanov homology groups, 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 under the three basic Reidemeister moves. Our main tools for doing so will be the following 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 sequence becomes

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 above sequence 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 also a 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 Khovanov complex, just as there is a recursive construction of the Jones polynomial. It turns out there is.

Let X be a diagram of graded abelian groups, and suppose there are no maps between groups with identical gradings. Then we may safely direct sum any abelian groups with a particular grading, and get induced maps (via projection and inclusion) between the resulting abelian groups. This produces a new diagram F (X). We call F the ﬂattening operator. Proposition 4.11. Let k be a ring, L a link diagram with at least one crossing and x a crossing in L. Consider an order on the crossings of L where x is the maximal crossing. Then

+ δ − [[L]] = F ([[Lx ]] −→ [[Lx ]]{1}) Now consider an order on the crossings of L where x is the minimal crossing. Then

+ δ − [[L]] = F ([[Lx ]] −→−[[Lx ]]{1}) where δ denotes all the induced differential maps from the smoothing change at − − x. We denote by −[[Lx ]]{1} the chain complex [[Lx ]]{1} where all differential maps have flipped signs. Proof. Every state s in L is either positively or negatively smoothed at x. As + − a result the corresponding k-module Cs can be found either in [[Lx ]] or [[Lx ]]. + − The gradings in [[Lx ]] match with those in [[L]]. However in [[Lx ]] the quantum grading 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 match with those in [[L]].

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

When x is a minimal crossing, its removal affects the sign of every differential and 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 a ring 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, and will contribute a single factor of V . It will not be involved in any smoothing change and so the only maps it will induce are the identity with itself. Lemma 4.13. The following chain complex is acyclic

F (A −→−id 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 0 id −δi i+1 i when acting on A ⊕ A , where δi are the diﬀerential maps. Suppose

δ 0 a δ a i+1 = i+1 = 0 id −δi b a − δib then we deduce a = δib and so

a δ 0 b = i b id −δi−1 0 It follows that the complex is acyclic. Theorem 4.14. Khovanov homology is invariant under the ﬁrst Reidemeister move. 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 map m. Also

δ m F ([[ ]] −→−[[ ]]{1}) = F (V ⊗k [[ ]] −→−[[ ]]{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 multiplication. This chain complex is acyclic by lemma 4.13. As a result C is quasi-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 be removed in the normalization, since the kink contributes 1 to n+ irrespective of the 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 the bottom crossing is second in the order. Then by smoothing the top crossing we get

C = [[ ]] = F ([[ ]] −→−δ [[ ]]{1}) where δ are the differentials induced by the smoothing change at the top. Now smoothing the bottom crossing in both sub-chain complexes yields   δ2 −[[ ]]{1} −→ [[ ]]{2}    C = F  ↑ δ1 ↑ −m    [[ ]] −→∆ −[[ ]]{1} where δ1 and δ2 are induced differential maps. We know the exact form of the right and bottom arrows, and for each subchain-complex the first two crossings in the order have determined smoothing type, so we know the signs of all the inner differentials. This chain complex becomes   δ2 −[[ ]]{1} −→ [[ ]]{2}    C = F  ↑ δ1 ↑ −m    [[ ]] −→∆ −V ⊗ [[ ]]{1} and we consider the sub-chain complex   0 → [[ ]]{2}    B = F ↑ ↑ −id    0 → −[[ ]]{2} with the subspace of V spanned by 1 in the bottom right complex. This is the same 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.   −[[ ]]{1} → 0    C/B = F  ↑ δ1 ↑    [[ ]] −→∆ −[[ ]]

58 Here ∆ becomes an isomorphism, since it has ﬁxed image onto the subspace of V generated by x. Again through lemma 4.13 the following sub-chain complex is acyclic.

 0 → 0   ↑ ↑  A = F   [[ ]] −→∆ −[[ ]] As a result, C is quasi-isomorphic to the following chain complex.   −[[ ]]{1} → 0   (C/B)/A = F  ↑ ↑ = −[[ ]][1]{1} 0 → 0

By carrying out this move we have decreased both n+ and n− by 1, irrespective of 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 mind that there is also a quasi-isomorphism

−[[ ]][1]{1} −→i [[ ]] induced by inclusion. Theorem 4.16. Khovanov homology is invariant under the third Reidemeister move. Proof. Choose an order on the crossings so that the middle crossing in both of the 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 applications of the previous theorem. More importantly we may write the commutative diagram

[[ ]] → −[[ ]]{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 middle row. 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 result 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 the corresponding oriented links have matching n− and n+ irrespective of the orientation chosen. We are done. We have proven that Khovanov homology is an oriented link invariant. We can now refer to the Khovanov homology of an oriened link, rather than an oriented link diagram, and all notions related to Khovanov homology can also be redeﬁned for oriented links.

4.4 Properties of Khovanov homology We follow [23] but adapt the result to our own formulation. It turns out that Khovanov 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

i,j i,j i,j H (L t M, k) = H (L, k) ⊗k H (M, k) for all i, j ∈ Z. Proof. We note ﬁrst that we have

[[L t M]] = [[L]] ⊗ [[M]] since the diﬀerential maps will never map between [[L]] and [[M]], and complete smoothings of L t M correspond to complete smoothings of L and M. The tensor above is applied along homological gradings. The quantum grading is additive along tensors and since

n+(L t M) = n+(L) + n+(M) and

n−(L t M) = n−(L) + n−(M) then normalizing yields

60 •,• CHKk (L t M) = [[L t M]][−n−(L t M)]{n+(L t M) − 2n−(L t M)} =

= [[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)} =

•,• •,• = CHKk (L) ⊗ CHKk (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 a crossing for some projection of L. If x has positive orientation then there is a long exact sequence in Khovanov homology

i,j−1 + i,j i−1−ω,j−2−3ω − i+1,j−1 + ... → H (Lx , k) → H (L, k) → H (Lx , k) → H (Lx , k) → ...

− − for all j ∈ Z, where some orientation of Lx is chosen and ω = n−(Lx )−n−(L). + The orientation on Lx is inherited from that of L. If x has negative orientation then there is a long exact sequence in Khovanov homology

i−c,j−1−3c + i,j i,j+1 − 1+i−c,j−1−3c + ... → H (Lx ) → H (L) → H (Lx ) → H (Lx ) → ...

+ + for all j ∈ Z, where some orientation of Lx is chosen and c = n−(Lx )−n−(L). − The orientation on Lx is inherited from that of L. Proof. By proposition 4.11 we have the relation

+ δ − [[L]] = F ([[Lx ]] −→ [[Lx ]]{1}) by choosing x to be maximal in the order. This gives a short exact sequence

+ − 0 → [[Lx ]] → [[L]] → [[Lx ]][1]{1} → 0 For the case that x has positive orientation, it is important to see that

+ n+(L) = n+(Lx ) + 1

+ n−(L) = n−(Lx )

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

61 so applying the shift [−n−(L)]{n+(L) − 2n−(L)} to the above short exact sequence gives

•,• + •,• 0 → CHKk (Lx ){1} → CHKk (L) → C → 0 where

•,• − − − − C = CHKk (Lx )[1+n−(Lx )−n−(L)]{1+n+(L)−n+(Lx )+2(n−(Lx )−n−(L))} =

•,• − = CHKk (Lx )[1 + ω]{2 + 3ω} Taking the corresponding long exact sequence in homology gives us the ﬁrst result. Now asume x has negative orientation. it is important to note that

− n+(L) = n+(Lx )

− n−(L) = n−(Lx ) + 1

+ + + n+(L) − n+(Lx ) = (n+(L) + n−(L)) − (n+(Lx ) + n−(Lx )) + c = 1 + c so applying the shift [−n−(L)]{n+(L) − 2n−(L)} to the above short exact sequence gives

•,• •,• − 0 → B → CHKk (L) → CHKk (Lx ){−1} → 0 where

•,• + + + + B = CHKk (Lx )[n−(Lx ) − n−(L)]{n+(L) − n+(Lx ) + 2(n−(Lx ) − n−(L))} =

•,• + = CHKk (Lx )[c]{1 + 3c} Taking the corresponding long exact sequence in homology gives us the second result. These two results are the main computational tools used to compute Khovanov homology. A very deep fact is that integral Khovanov homology can detect the unknot. 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 homology given by  Z i = 0, j = 1 i,j  H (L, Z) = 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 the Hopf link, with the following orientation

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

This gives the following value

− ω = n−(Lx ) − n−(L) = 1 and the long exact sequence

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

− + Here both Ly and Ly are unknots, and we know the Khovanov homology of the unknot. From this we deduce that the Khovanov homology of the Hopf link is given by the table

63 j\i 0 1 2 6 0 0 k 4 0 0 k 2 k 0 0 0 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 the complement 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) ˜ 3 Hi(S − K) = Z for i = 3 − 1 − 1 = 1 and 0 otherwise, so in particular 3 3 H˜0(S − K) = 0. By (Remark below Proposition 2.8, Hatcher 2001) H0(S − K) = 0 ⊕ Z = Z, so by (Proposition 2.7, Hatcher 2001) S3 − K has exactly 1 path-connected component. For R3 − K, suppose the kxk < R − 1 for every x ∈ K. This is true for R suﬃciently large. Then let A = {x ∈ R3 : kxk ≥ R}, so that A ⊂ R3 − K and (R3 − K)/A = (R3/A) − K = S3 − K. Let X = R3 − K.(X,A) is a good pair as deﬁned by (Theorem 2.13, Hatcher 2001), so by the same theorem we have a long exact sequence:

· · · → H˜1(X/A) → H˜0(A) → H˜0(X) → H˜0(X/A) → 0

Substituting what we know:

˜ 3 ˜ ˜ 3 ˜ 3 · · · → H1(S − K) → H0(A) → H0(R − K) → H0(S − K) → 0

A is non-empty and path-connected, so H0(A) = Z by (Proposition 2.7, Hatcher 2001), so H˜0(A) = 0: ˜ 3 · · · → Z → 0 → H0(R − K) → 0 → 0 ˜ 3 3 By exactness we must then have H0(R − K) = 0, whence H0(R − K) = Z, so R3 − K has exactly one path-connected component by the same argument, so R3 − K is path-connected. We must then show that the two complements give the same fundamental group: 3 3 ∼ 3 Theorem 5.3. If K ⊆ R is a knot then π1(R − K) = π1(S − K). Proof. Since K is compact, it must be bounded in R3, by say R. So let N := {x ∈ R3 ∪ {∞} : kxk > 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. 3 2 3 2 • (R −K)∩N deformation retracts to S , so π1((R −K)∩N) = π1(S ) = 1 by (Proposition 1.14, Hatcher 2001).

65 • N is contractible, so π1(N) = 1 Finally, by van Kampen’s theorem (Theorem 1.20, Hatcher 2001) we have:

3 3 π (S − K) = π ( − K) ∗ 3 π (N) 1 1 R π1((R −K)∩N) 1 3 = π1(R − K) ∗1 1 3 = π1(R − K)

3 3 Theorem 5.4. If K1 and K2 are equivalent knots then π1(R − K1) = π1(R − 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. that it 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 group 3 is H1(S − 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. reflection across the plane, are isomorphic. Proof. Reflection is a homeomorphism, and homeomorphic spaces have isomorphic fundamental groups. Example 5.7. We compute the knot group of the unknot. The unknot is embed- 3 3 ded in R as N := {(cos θ, sin θ, 0) | θ ∈ R}, so we need to compute π1(R − N). We use van Kampen’s theorem (Theorem 1.20, Hatcher 2001) in the following way: n h i o • Let A := ( 3−N)∩ (r cos θ, r sin θ, z) r ≥ 0, θ ∈ 2kπ , 2(k+1)π , z ∈ . k R 3 3 R

• Each Ak is homeomorphic to R × (C − {0}), which deformation retracts 1 1 to S , so π1(Ak) = π1(S ) = Z, with generator being a loop around the removed unknot.

• Let ak be a generator of π1(Ak) = Z. 3 • A0 ∪ A1 ∪ A2 = R − N. 3 • Each Ak has a neighbourhood Uk in R − N that deformation retracts to Ak and that each Ui ∩ Uj and Ui ∩ Uj ∩ Uk are path-connected. 3 • So Φ: π1(A0) ∗ π1(A1) ∗ π1(A2) → π1(R − N) is surjective, with kernel −1 −1 −1 being the normal subgroup generated by a0a1 , a1a2 , and a2a0 . 3 • Therefore, π1(R − N) = ha0, a1, a2 | a0 = a1 = a2i = ha |i = Z.

66 5.2 Wirtinger presentation This is a powerful and simple algorithm to compute a group presentation of the knot group given a diagram. This can be found in (Exercise 1.2.22, Hatcher 2001). 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.

•{ Ui}i cover X.

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

• Each π1(Ui) is presented as h{aip}p | {fiq({aip}p)}qi.

• Each π1(Ui ∩ Uj) is presented as h{aijs}p | {fijq({aijs}s)}qi.

• Each aijs is identiﬁed as fijs1({aip}p) in π1(Ui) and as fijs2({ajp}p) in π1(Uj) under the maps φij : π1(Ui ∩ Uj) → π1(Ui) and φji respectively.

Then, π1(X) is presented as:

−1 π1(X) = h{aip}i,p | {fiq({aip}p}i,q, {fijs1({aip}p)fijs2({ajp}p) }i,j,si

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:

−1 {φij(g)φji(g) | g ∈ π1(Ui ∩ Uj)}i,j

Step 1

∗iπ1(Ui) = ∗ih{aip}p | {fiq({aip}p)}qi = ∗i(h{aip}p |i/NS{fiq({aip}p)}q) = (∗ih{aip}p |i)/NS{fiq({aip}p)}i,q = h{aip}i,p |i/NS{fiq({aip}p)}i,q where NS means “normal subgroup generated by”. Step 2 We claim that N is the same as the normal subgroup N2 generated by:

−1 {φij(aijs)φji(aijs) }i,j,s

−1 It is obvious that each φij(aijs)φji(aijs) is in N, so it suﬃces to show that −1 each φij(g)φji(g) is in N2. We know that g can be written as the product of

67 −1 a ﬁnite number of aijs or aijs. We induct on the number of the multiplicands. −1 −1 If g = g1g2 where φij(g1)φji(g1) , φij(g2)φji(g2) ∈ N2, then:

−1 −1 φij(g)φji(g) = φij(g1g2)φji(g1g2) −1 −1 = φij(g1)φij(g2)φji(g2) φji(g1) ∈ N2 where the last step is because normal subgroups is closed under conjugation. −1 So it suﬃces to show g ∈ N2 when g = aijs or g = aijs. The former case is trivial. For the latter case:

−1 −1 −1 −1 φij(g)φji(g) = φij(aijs)φji(aijs) −1 = (φji(aijs)φij(aijs)) ∈ N2 where the last step follows from the fact that subgroups is closed under inverse. Step 3

π1(X) = (∗iπ1(Ui))/N = (h{aip}i,p |i/NS{fiq({aip}p)}i,q)/N2 −1 = h{aip}i,p |i/NS({fiq({aip}p)}i,q ∪ {φij(aijs)φji(aijs) }i,j,s) −1 = h{aip}i,p |i/NS({fiq({aip}p)}i,q ∪ {fijs1({aip}p)fijs2({ajp}p) }i,j,s) −1 = h{aip}i,p | {fiq({aip}p)}i,q, {fijs1({aip}p)fijs2({ajp}p) }i,j,si

We are now ready for the Wirtinger presentation. Theorem 5.9. Let K be a knot with diagram D. Pick an orientation on K. ±1 Let the arcs in D be labelled αi. At each double-point j of D let βj1 = αj1 ±1 ±1 be the arc that passes over, and let βj2 = αj2 and βj3 = αj3 be the two arcs that pass under, where the positions and signs are chosen to match the following diagram:

βj3 βj1

βj2

Figure 21: Naming of arcs near a crossing.

3 −1 Then, π1(R − K) = h{αi}i | {βj1βj2βj1 = βj3}ji.

68 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 connecting two regions must intersect the knot at most once. For example, here is the trefoil knot 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-passing arcs are lifted up a little bit to form a circular arc, as demonstrated in this ﬁgure:

69 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 in Figure 24a. • For regions with two arcs, build a small tunnel around each arcs as shown in 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 just built. 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 removing the region7. Let the regions be A1, ··· ,An. We have neighbourhoods Ui sup Ai such that Ui deformation retracts to Ai and Ui ∩ Uj deformation retracts to Ai ∩ Aj for every i, j. To ensure path-connectedness of each intersection, we can thicken the xy-plane on our surface by connecting the highest point on each column with ﬁxed x and y coordinates to the point (0, 0, 1) ∈ R3. This top-thickened surface deformation retracts to our original surface and on each component, so it does not change the homotopy. Then, we can extend each Ai to contain enough 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 observes the 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, which is contractible, so π1(Ui) = h|i, i.e. the trivial group. If Ai is a region with one arc, then the surface assigned deformation retracts 1 to S embedded as a loop encircling the arc, so π1(Ui) = Z with a generator determined by the right hand rule (for consistency). This means that one puts out the right hand with the thumb pointing to the direction of the loop, and then curl the remaining 4 ﬁngers to create a thumbs up; then the direction in which the 4 ﬁngers curl is the orientation of the loop encircling the arc. We write π1(Ui) = hαj |i, where αj is the one arc in this region, and we identify the arc with the loop encircling it. If Ai is a region with two arcs, then the surface assigned deformation retracts to a space homotopy equivalent to S1 ∨ S1, where the two copies S1 are embedded as two loops, one encircling each arc. Therefore, π1(Ui) = Z ∗ Z with the two generators again being determined by the right hand rule. We now name the arcs in 3-dimensions according to Figure 21, and draw the loops corresponding to 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 to βj3. As stated above, π1(Ui) is generated by βj1 and βj2. Therefore, we must express βj3 in terms of βj1 and βj2. Having labelled the four regions of the xy-plane as A, B, C, and D as in the ﬁgure above, we have:

• βj1 is passing through B and going up through C, or passing through A and 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:

71 1. Pass through B; use βj3 to go up through A; pass through A again; use βj1 to go up through D.

2. Pass through B; use βj1 to go up through C; pass through C again; use βj2 to go up through D. −1 Therefore, βj3βj1 = βj1βj2, i.e. βj3 = βj1βj2βj1 . −1 Therefore, π1(Ui) = hαj1, αj2, αj3 | βj1βj2βj1 = βj3i.

If Ai ∩ Aj does not intersect the knot, then π1(Ui ∩ Uj) = h|i, the trivial group. If Ai ∩ Aj intersects the knot at the arc αk, then π1(Ui ∩ Uj) = hαk |i. This serves to identify the copies of αk in the fundamental groups of the surfaces assigned to Ai and Aj. 3 Therefore, gathering every region, Theorem 5.8 shows us that π1(R − K) has a presentation that has one generator for every arc in the diagram, and at every −1 crossing we add the relation βj1βj2βj1 = βj3, which is the result desired. Example 5.10. We compute the knot group of the trefoil knot. First we orient the knot and label the arcs as shown:

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

Then, using Theorem 5.9, we compute:

72 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 presented as: −1 h{αi}i | {βj1βj2βj1 = βj3}ji The abelianization corresponds to interpreting the group presentation as abelian, so conjugation ﬁxes everything, and the presentation becomes:

h{αi}i | {βj2 = βj3}ji

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 the two under-passing arcs are identiﬁed.

Therefore, the presentation becomes hα |i, i.e. Z. Theorem 5.11. The knot group is not a complete invariant, i.e. there are inequivalent knots K1 and K2 with the same knot group. Proof. See (3D12, Rolfsen 1976). In particular, the two knots are the granny knot and the square knot. They are both connected sums of trefoil, but the granny 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 with its mirror image. The knot groups are both ha, b1, b2 | ab1a = b1ab1, ab2a = b2ab2i, as it is ha1, b1 | a1b1a1 = b1a1b1i and ha2, b2 | a2b2a2 = b2a2b2i with a1 and a2 identiﬁed. However: Theorem 5.12. If a prime knot has the same knot group as another knot, then the 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 that is homeomorphic to S1×B2, where K corresponds to S1×{0}. By restricting the radius in B2, one obtains a smaller neighbourhood U ⊂ V still homeomorphic to 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 deformation retracts to ∂V ⊂ S3. Therefore, by the Mayer-Vietoris sequence (P.150 Hatcher 2003), we have a long exact sequence:

3 3 H3(∂V ) → H3(K) ⊕ H3(S − v) → H3(S ) → 3 3 H2(∂V ) → H2(K) ⊕ H2(S − v) → H2(S ) → 3 3 H1(∂V ) → H1(K) ⊕ H1(S − v) → H1(S ) → 3 3 H˜0(∂V ) → H˜0(K) ⊕ H˜0(S − v) → H˜0(S ) → 0

Substituting in what we know:

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

3 3 We see that H2(∂V ) = H3(S ) = Z and H1(∂V ) = H1(V ) ⊕ H1(S − V ) = Z ⊕ Z. Therefore, we can pick m, l ∈ H1(∂V ) such that m corresponds to 0 ∈ H1(V ) 3 and ±1 ∈ H1(S − V ); and such that l corresponds to ±1 ∈ H1(V ) and 0 ∈ 3 H1(S − 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 1 point.

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

3 3 • Since l = 0 ∈ H1(S − V ), we see that l is null-homotopic in S − V .

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

It can further be proved that L(m, K) = 1 and L(l, K) = 0, where L is the linking number (Deﬁnition 3.5). m and l are called the meridian and longitude of the knot respectively, and two curves 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 knot group together with the data of the meridian and longitude turns out to be a complete 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 homotopy class in the fundamental group, then K1 and K2 are equivalent. Proof. (Theorem 3.15, Burde, Zieschang, and Heusener, 2013).

75 6 Bibliography References

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