MMarkov’s Theorem
Matteo BaruccoNirvana CoppolaNovember 12, 2019 a r X i v : . [ m a t h . G T ] N ov ontents hapter 1 Introduction
This survey consists of a detailed proof of Markov’s Theorem based on [1] and Carlo Petronio’sclasses. It was part of an exam project in A.Y. 2016/2017 for the course Knot Theory.Let us first introduce some basic definitions and known properties about braids.
Definition 1. A n -braid or braid over n strings is the image of a function: α : n a j =1 [0 , → R × [0 , α j is the restriction of α on the j -th copy of [0 , τ ∈ S n with the following properties: α ( j ) (0) = (0 , , j ); α ( j ) (1) = (0 , , τ ( j )); α ( j )3 is strictly increasing for every j ; α is a proper embedding.Figure 1.1: A 4-braid β Definition 2.
Two n -braids α and α are equivalent if the maps defining them are isotopic througha family of maps with the same properties. Note that two equivalent braids have the same associatedpermutation. 5 IntroductionDefinition 3.
The group B n of n -braids is the set of the equivalence classes of n -braids as definedabove, with the operation of juxtaposition. Given two n -braids α and β , their juxtaposition is α ∗ β ( j ) i ( t ) = ( α ( j ) i (2 t ) if 0 ≤ t ≤ / β ( τ ( j )) i (2 t −
1) if 1 / ≤ t ≤ i = 1 , α ∗ β ( j )3 ( t ) = ( / α ( j )3 (2 t ) if 0 ≤ t ≤ / / / β ( τ ( j ))3 (2 t −
1) if 1 / ≤ t ≤ . Definition 4.
The closure of a braid is the link obtained by connecting the corresponding endsthrough concentric arcs (Figure 1.2).Figure 1.2: The closure ˆ β of the braid β of Figure 1.1 Fact 5.
The group of n -braids has the following presentation: B n = h σ , . . . σ n − | σ i σ j = σ j σ i if | i − j | ≥ , σ i σ i +1 σ i = σ i +1 σ i σ i +1 i , where the generator σ i is shown in Figure 1.3 and its inverse is in Figure 1.4. Remark 6.
The first kind of relation in the presentation of the n -braids group is easy to understand,and we can notice that the second kind of relation translates into a Reidemeister move R III , asshown in figure 1.5. ......
Figure 1.3: The generator σ i ... ... Figure 1.4: The inverse σ − i of σ i Figure 1.5: The relation σ i σ i +1 σ i = σ i +1 σ i σ i +1 Introduction
We now "algebraically" define two deformation moves that are crucial for Markov’s Theorem.
Definition 7.
Conjugation move (Figure 1.6): B m c w − cw ∈ B m , where w ∈ B m . Figure 1.6: Conjugation by σ i is trivial on the closure Definition 8.
Stabilization move: B m c cσ m ∈ B m +1 where we can see c in B m +1 adding a string as in Figure 1.7.Figure 1.7: Stabilization move is trivial on the closureWe can now state the main theorem: Theorem 9 (Markov) . Two braids whose closures are equivalent as links are related to each otherby a sequence of conjugation and stabilization moves. hapter 2
Tools
In order to prove Markov’s Theorem, we need to introduce some other deformation moves. Let L be a PL link, with an edge [ a, c ] (where in general [ a , . . . , a n ] is the convex hull of the points a , . . . , a n ). We will write ab to indicate the edge [ a, b ] oriented from a to b .In this chapter, we suppose that L is a link, r is an axis such that the projection P of L on theplane r ⊥ is a link diagram and L inherits the orientation from the axis r , so all of its edges has awell defined sign. Note that L is in braid position if and only if all of its edges are positive. Definition 10.
The move that generates PL link isotopy, given by L L \ [ a, c ] ∪ [ a, b ] ∪ [ b, c ] , is called E ba,c . This move is licit if L ∩ [ a, b, c ] = [ a, c ].Figure 2.1: The move E ba,c Definition 11.
Given a negative edge e = p p m of L , m + 1 basepoints p , p , . . . , p m ∈ e , suchthat p i , p i +1 is negative, and m vertices q , . . . , q m / ∈ e , the sawtooh move S q ,...,q m p ,...,p m = E q m p m − ,p m ◦ · · · ◦ E q p ,p is the composition of type- E moves, each of which is licit when we operate it, and such that all[ p i , q i +1 ] and [ q i +1 , p i +1 ] are positive (Figure 2.2).9 Figure 2.2: The sawtooth move S q ,...,q p ,...,p Lemma 12 (Existence of a Sawtooth) . Given a negative edge e of L , there exists a sawtooth S on e . Moreover:1. if Ω is a convex set such that Ω ∩ e is either empty or a vertex of e , then S can be chosen suchas to avoid Ω; formally, if p , . . . , p m are the basepoints and q , . . . , q m are the vertices of S ,then for every i Ω ∩ [ p i , p i +1 , q i +1 ] ⊆ Ω ∩ e ;2. if e is another edge of L and e ∩ e is one of the vertices of e , than S can be chosen such that(using the notation above) [ p , . . . , p m , q , . . . , q m ] ∩ e is exactly that vertex. Proof.
We first consider the case where the link L has a crossing point over the edge e . Let c ∈ e and d on another edge f be the two points that project to the same point on r ⊥ . We will assumethat c ∈ r ⊥ , for simplicity reasons. Let π be the plane containing c and r and let r be the parallelto r containing c (Figure 2.3). Figure 2.3: Near a crossing pointThe link L intersects π in a finite number of points, because it is compact and in general positionwith respect to r ⊥ . Hence r ∩ L only contains c and d , while f ∩ π = { d } , otherwhise the rotation axis r would not induce any sign on it. The plane r ⊥ divides the space into two semi-spaces (positiveand negative), according to the orientation on r . So, in one of the two subspaces, we can drawa half-line lying on the plane π , starting from c , that does not intersect L in any other point: ifsuffices that the angle it forms with r is small enough (and different from 0). Then the half-lineis almost parallel to the line r . Let us take the points ˜ q ad q on this half-line, such that ˜ q is theintersection between it and r , while q does not belong to the line segment c ˜ q (Figure 2.4).Figure 2.4: Construction of ˜ q and q Let p and p be two points on e such that pp is negative and c is internal to it. We can choose p and p close enough to c , and q close enough to ˜ q , such that [ p, q, p ] ∩ L = [ p, p ]. So the move E qp,p is licit and the edges pq and qp are positive; it will be a tooth in our sawtooth (Figure 2.5).Figure 2.5: Construction of a toothWe now consider the case where e = [ p , p ] has no crossing points. Let Ξ be the strip definedby the parallels to r that intersect e . Then Ξ ∩ L only contains e , hence we can rotate the striparound e by a sufficiently small angle φ , so that the image Ξ of the strip Ξ intersects r in a point˜ q , and still only intersects L in e . Let us consider the triangle whose basis is e and whose vertex isa point q close to ˜ q , belonging to the same half-strip and such that p q and q p are positive linesegments. So the move E q p ,p is licit and is a sawtooth. (Figure 2.6). Figure 2.6: No crossing pointsLast, we adapt the proof in order to show 1 . and 2 . :1. If Ω is as stated, e has a crossing point c and r is as above, then Ω can intersect only one ofthe two half-lines that c divides r into: if it did not, being a convex set it would intersectalso e in its internal point c , a contradiction. Then, if Ω does not intersect the positive (resp.negative) half-line, we can choose c ˜ q to be in the positive (resp. negative) half-space and theproof above continues to work. Also, if there are not crossing points, it suffices to take oneof the half-strips of Ξ that does not intersect Ω (it is possible thanks to the convexity of Ω)and rotate it as above in order to find an intersection point ˜ q with r . If the rotation angle issmall enough, the half-strip still does not intersect Ω.2. Let { a } = e ∩ e . We can still consider the strip Ξ over the edge e defined as above. Since thelink is in general position with respect to r ⊥ , Ξ only intersects e in a . Now we can execute allthe steps above taking all the vertices of the sawtooth in the same half-space (no matter which)and far from r ⊥ and e , so that near e the convex hull of the sawtooth is indistinguishablefrom its projection on the strip Ξ. Then it only intersects e in a , too.This concludes the proof.Some other useful deformation moves are the following. Definition 13.
The move R ba,c is given by E ba,c with the edges ac , ab and bc all positive. Definition 14.
The move W b,ca,d is defined when ab , bc , cd , ad , ca and db are all positive and isgiven by the composition: W b,ca,d = E cb,d ◦ E ba,d where the two type- E moves are licit. Remark 15.
We can obtain the same link using another definition of W b,ca,d , namely: E ba,c ◦ E ca,d Figure 2.7: Crossing type- W moveFigure 2.8: Non-crossing type- W move Remark 16.
There are two possible type- W moves (Figures 2.7 and 2.8). We can see this usingthe diagram obtained by projecting the link and all the involved line segments on r ⊥ . Let p be theintersection between r and r ⊥ . The point b belongs to one of the two half-planes delimited by theline containing a and d ; if this is the half-plane containing p , then we get the crossing type- W move,otherwise we get the non-crossing type- W move. Proposition 17.
Two braids whose closures are connected by a type- R move are conjugate to eachother. Proof.
Let the R move be some R ba,c .First, we can notice that, up to subdivision, we can consider take the triangle [ a, b, c ] to bearbitrarily small. Indeed, if x ∈ [ a, c ], y ∈ [ a, b ], z ∈ [ b, c ] the following equality holds (Figure 2.9): R ba,c = ( R zb,c ) − ◦ ( R ya,b ) − ◦ R by,z ◦ ( R xy,z ) − ◦ R zx,c ◦ R ya,x ◦ R xa,c . Figure 2.9: Factorization of a type- R move So, after a finite number of subdivisions, each obtained by some type- R moves, we can separatelywork on smaller triangles.From this moment on, we will only consider braid position links. Let us fix an angle θ andconsequently a half-line θ = θ , which is the starting point of every braid position diagram.Since the triangle [ a, b, c ] involved in the type- R move is small, we may assume that only one ofthe following happens, explained in Figure 2.10:• the internal part of [ a, b, c ] only contains one vertex of L ;• the internal part of [ a, b, c ] only contains one crossing of L ;• only an edge of L passes through the internal part of [ a, b, c ], while the half-line θ = θ doesnot;• an edge of L and the half-line θ = θ pass through the internal part of [ a, b, c ].Figure 2.10: The 4 cases for the type- R In the first case, we simply have an equivalence of braids; to see that, it suffices to read thebraids corresponding to L and to R ( L ) (that is the link L after the move R ), starting from θ = θ :the only difference between the two of them may be near the edge [ a, c ] in L and the edges [ a, b ] and[ b, c ] in R ( L ). In the diagram, the link intersects the boundary of the triangle [ a, b, c ] twice, and italways passes over the triangle or under it. Reading the braid as an element of B n , we find exactlythe same generators, as in Figure 2.11.Figure 2.11: The equivalence of L and R ( L )The second and third cases are similar: we have another equivalence of braids. Let us focuson the fourth case. We may assume that the half-line θ = θ also crosses L in the internal part of [ a, b, c ]; otherwise, up to subdivision, we can get the third case. We can notice that both the edge of L and the half-line θ = θ intersect twice the boundary of [ a, b, c ], obtaining four different points(otherwise the edge of L would have no sign) as showed in Figure 2.12.Figure 2.12: Type- R move, fourth caseSo, the braids we get before and after doing the move R can be (in any order): wσ i and σ i w or wσ − i and σ − i w (Figure 2.13).Figure 2.13: Correspondent braid diagramIn each case, they are conjugate in B n . Definition 18.
The move W b,ca,d is said to be empty if [ a, b, c, d ] ∩ L = [ a, d ]. Proposition 19.
A type- W move factors through two type- R moves and an empty type- W move. Proof.
Let us write W b,ca,d = E cd,b ◦ E ba,d , with ad , bc , cd , ab and db all positive. We could also write W b,ca,d = E bc,a ◦ E ca,d , but the proof would be the same, role-reversing a and d , and b and c . We willonly consider the first case. By hypothesis: L ∩ [ a, d, b ] = [ a, d ] L ∩ [ b, c, d ] = { b } . There exists a neighbourhood U of d in [ a, d ] such that for every x ∈ U :• [ x, d, b, c ] ∩ L = [ x, d ], since L is compact;• xb has the same sign as db . Figure 2.14: Factorization of W b,ca,d Let us take x in this neighbourhood. So the move factors as follows (Figure 2.14): W b,ca,d = ( E xa,b ) − ◦ W b,cx,d ◦ E xa,d . All the involved edges are positive:• ad , ab , bc and cd by hypothesis;• ax and xd have the same sign as ad , since x ∈ [ a, d ];• xb has the same sign as db by the choice of x .So the two type- E moves are actually of type- R and the type- W move is well defined; moreoverit is empty, since [ x, d, b, c ] ∩ L = [ x, d ].Recall that P is the projection on r ⊥ . Definition 20.
The move W b,ca,d is said to be really empty if P ([ a, b, c, d ]) ∩ P ( L ) = P ([ a, d ]). Proposition 21.
An empty type- W move can be decomposed into some type- R moves and onereally empty crossing type- W move. Proof.
Let us suppose that W = W b,ca,d is non-crossing. We will show that it can be decomposedinto two type- R moves and an empty crossing type- W move (Figure 2.15). Let us consider thehalf-line originated from p that contains d , then let us rotate it by a small negative angle (withrespect to the orientation given by r ) on r ⊥ . Let x be a point on it, in the half-plane delimited by[ a, b ] not containing d , close to [ a, b ], so that the line segments ax , xb , cx and xd are all positive.Hence the move W b,ca,x , operated after R xa,d is licit, non-crossing and: W b,ca,d = ( R xc,d ) − ◦ W b,ca,x ◦ R xa,d . Moreover, the move W b,ca,x is still empty, since the convex hull [ a, b, c, x ] is almost contained in[ a, b, c, d ].So we may focus on the case W is crossing. We can apply a homothety of ratio t >
1, withrespect to a center q ∈ r ∩ int[ a, b, c, d ], to M = L \ [ a, d ]. If we choose t large enough, then theimage M t of M will be disjoint from [ a, b, c, d ] and its projection will be disjoint from P ([ a, b, c, d ]).Let ˜ a and ˜ d be a perturbation of the images of a and d respectively, such that ˜ aa and d ˜ d are bothpositive.Then: L t = [˜ a, a ] ∪ [ a, d ] ∪ [ d, ˜ d ] ∪ M t and L t = [˜ a, a ] ∪ [ a, b ] ∪ [ b, c ] ∪ [ c, d ] ∪ [ d, ˜ d ] ∪ M t are joined by the really empty crossing move W b,ca,d Figure 2.16). Finally, L → L t and L t → L aresmall perturbations of a homothety, hence factor through a sequence of type- R moves.By Proposition 17, type- R moves are conjugation moves, and by Proposition 19 and Proposition21 a type- W move can be decomposed into other conjugation moves and a really empty type- W move, that is a stabilization. In order to prove Markov’s Theorem it is enough to show the following: Theorem 22.
Two braids whose closures are equivalent as links, are related to each other by asequence of type- R and W moves.Another useful deformation move is the following: Definition 23.
Given a , b , c , d such that ab , ac , bd , cd are all positive, while ad is negative, we set(Figure 2.17) T a,c,da,b,d = E ca,d ◦ ( E ba,d ) − . Proposition 24.
A type- T move can be decomposed into type- R and W moves. Proof.
Let us take a point a ∈ [ a, b, d ] close to a and very close to [ a, b ], and d ∈ [ a, c, d ] close to d and very close to [ c, d ]. So the line segments aa and a b are positive, since ab is, and cd , d d arealso positive, since cd is. Consequently, the moves E a a,b and E d c,d are type- R moves.Now, the move W c,d a,a is well defined, because aa is positive as noticed above, ac and cd arepositive by hypothesis, d a has the same sign as da , so it is positive too, and finally both theauxiliary edges a c and d a are positive since ac and da are. Similarly, W a ,bd ,d is well defined.So the following holds (Figure 2.18) T a,c,da,b,d = ( R d c,d ) − ◦ ( W a ,bd,d ) − ◦ W c,d a,a ◦ R a a,b which concludes the proof. Proposition 25.
Two sawteeth S and S erected on the same negative edge e are connected by achain of type- R and W moves. Proof.
We first claim that given a point d ∈ e and a sawtooth S = S b ,...,b m a ,...,a m on e , we can constructa finer sawtooth with d among its basepoints. If d is already among the basepoints of S , there isnothing to show. Otherwise, let us suppose that d belongs to the internal part of [ a i , a i +1 ], where a i and a i +1 are two basepoints of S . We can assume, up to moving the vertex b i +1 using a licit type- T move, that the edge [ b i +1 , d ] is positive or negative. Let us consider the first case (the other issimilar). Let c be a point close to the triangle [ a i , a i +1 , b i +1 ] and such that dc and ca i +1 are positive.Hence the move W d,cb i +1 ,a i +1 is licit and produces a finer sawtooth with d among its basepoints, asshown in Figure 2.19.Getting to the statement, by the claim we may assume that S and S have the same basepoints.We operate a type- T move to each of the type- E moves that define S , thus obtaining the sawtooth S . This is permitted, since after every operation of type T , we get another intermediate sawtooth.Finally, thanks to Proposition 24, the type- T moves can be decomposed into type- R and W moves. Figure 2.15: The decomposition of a non-crossing type- W moveFigure 2.16: The decomposition in a really empty type- W move and R moves Figure 2.17: A type- T moveFigure 2.18: Decomposition of a type- T moveFigure 2.19: Including d in a sawtooth hapter 3 Markov’s Theorem: the proof
Given a
P L link L in general position with respect to the rotation axis r , so that every edge hasa sign, we can define the height h ( L ) of L as the number of negative edges of L . The link is inbraid position if and only if its height is 0, if and only if it is the closure of a braid. In general,two isotopic P L links are joined by a chain of type- E , R , S , W or T moves, so each operation maychange the height. Remark 26.
If two
P L links L and L are the closure of two braids and are joined by a chain ofdeformation moves of the above mentioned types, and if the height of each link of the chain is 0,then the deformation moves which occur in the chain can only be of type- R and W , therefore inorder to prove Markov’s Theorem, we only need to show that we can find a chain of deformationmoves from L to L that preserve the height.This will easily follow from the next two lemmas, to the proof of which this chapter is devoted. Lemma 27.
If two links L and L are joined by a single deformation move of type- E , R , S , W or T and h ( L ) = h ( L ) >
0, then there is a chain of links L = L → L → · · · → L n = L each obtained from the previous one using a deformation move of the above mentioned type, suchthat h ( L i ) < h ( L ) for every i = 1 , . . . , n − Proof.
The move joining L and L can only be of type R , W or E sending a negative edge to apositive and a negative one.Since h ( L ) >
0, then L has a negative edge, namely xy . If the move is of type R or W , then itdoes not apply to xy , because it only involves positive edges. In each case we can erect a sawtoothon xy that avoids the convex hull of the points involved in the move; formally, if the sawtooth is S q ,...,q n p ,...,p n :• if R = R ca,b , then for each i : [ a, b, c ] ∩ [ p i , p i +1 , q i +1 ] = [ p i , p i +1 ];• if W = W b,ca,d , then for each i : [ a, b, c, d ] ∩ [ p i , p i +1 , q i +1 ] = [ p i , p i +1 ]: this is possible becauseby the previous results we can assume that the move is of empty type.21 If h ( L ) >
1, and the move is of type E , we can erect a sawtooth on an edge which is not involvedin the type- E move and, as above:• if E = E ca,b , then for each i : [ a, b, c ] ∩ [ p i , p i +1 , q i +1 ] = [ p i , p i +1 ].Now let us suppose that h ( L ) = 1 and fix E = E ba,c . Without loss of generality, ac and bc arenegative, while ab is positive. First, we suppose that a and b are close enough so that it is possibleto erect a sawtooth on both ac and bc with the same vertices (and the same number of basepoints),all the involved edges have a sign and all the involved pyramids intersect the link only in the edgecontained in [ a, c ] and do not intersect with each other (Figure 3.1).Figure 3.1: Factorization of E ba,c Let S da = S d ,...,d m a ,...,a m and similarly S db the sawteeth on ac and bc respectively. We will show that E ba,c can be decomposed as: E ba,c = ( S db ) − ◦ F ◦ F ◦ · · · ◦ F m − ◦ S da (3.1)where the F i s are defined as follows:• If a m − b m − is positive, then the moves R b m − a m − ,d m and ( R a m − d m − ,b m − ) − are licit on the edge a m − b m − after applying S da and we define F m − = ( R a m − d m − ,b m − ) − ◦ R b m − a m − ,d m ;• If a m − b m − is negative, then the moves R b m − d m − ,a m − and ( R a m − b m − ,d m ) − are licit and we define F m − = ( R a m − b m − ,d m ) − ◦ R b m − d m − ,a m − . The case where a m − b m − is positive is shown in Figure 3.2.In the same way we define the operations F m − , . . . , F . Finally we need to define F . Since ab is positive, we simply take F = R ba,d . Figure 3.2: The move F m − , case a m − b m − positive Let us study the chain (3.1). The first move is a sawtooth, and decreases height by 1. Thetype- F moves are compositions of type- R moves, and preserve height. The final S − operationincreases the height by 1 again.In general, the existence of a sawtooth on a negative edge xy given a set of vertices { q , . . . , q m } isguaranteed if there exist some points p = x , p , . . . , p m = y on xy such that p i q i +1 and q i +1 p i +1 areall positive and [ p i , p i +1 , q i +1 ] ∩ L = [ p i , p i +1 ] for every i . This is an open condition on y . Moreover,if xy is another negative edge, and y is close to y , then we can erect a sawtooth on xy with the samevertices { q , . . . , q m } and basepoints { p , . . . , p m } such that [ p i , p i , q i +1 , p i +1 , p i +1 ] ∩ L = [ p i , p i +1 ];this is an open condition on y .Now, given z ∈ [ a, b ] and fixed a random sawtooth S on zc , let U z be an open connectedneighbourhood containing only points y ∈ [ a, b ] such that on yc can be erected a sawtooth with thesame vertices as S and all the pyramids avoid the link in the same sense as above. Hence S z ∈ [ a,b ] U z =[ a, b ] and since [ a, b ] is compact, there exists a finite subcover, given by: z = a, z , . . . , z n = b . Wecan assume that the points z i are ordered on the edge ab . Since each U z i is an open line segment,then U z i ∩ U z i +1 is an open line segment too; let x i ∈ U z i ∩ U z i +1 for each i (Figure 3.3).Figure 3.3: The finite open cover U z , . . . , U z n Now we can erect a sawtooth with common vertices on z i c and x i c and also on x i c and z i +1 c for every i , and write the move E ba,c as a composition of moves of type (3.1). Finally we changeevery S ◦ S move that appears using Proposition 25, thus obtaining an initial move of type S − followed by the composition of type- R and W moves and finally another sawtooth. That meansthat with the first operation we decrease height, then we keep it constant until the last move, whichincreases it again. Lemma 28.
If two links L and L are joined by a chain of deformation moves L → L → L such that h ( L ) > h ( L ) and h ( L ) > h ( L ), then there is a chain of links L = L → L → · · · → L n = L such that h ( L i ) < h ( L ) for each i . The moves which increase height are the following:• type- S − moves, that create a negative edge;• some of the type- E moves (Figure 3.4), precisely, if E = E ba,c :1. the move sends the positive edge ac to a positive and a negative one (without loss ofgenerality, we assume bc is the negative edge);2. the move sends the positive edge ac to two negative edges;3. the move sends the negative edge ac to two negative edges.Figure 3.4: The three possible type- E moves.There is also another type- E move that increases height, namely ( E ba,c ) − , where ac is negativewhile ab and bc are both positive. Anyway this is a particular case of sawtooth, so we do not needto separately consider it.The possible chains are the following:(a) first we apply a type- S − move, and then a S move;(b) first we apply a type- E move, and then a S move (or viceversa first we apply a type- S − move,and then a E − move);(c) first we apply a type- E move and then a E − move.It is easy to see that the proof of the lemma in case (c) follows by case (b). Indeed if the firstmove is E ba,c , where bc is negative, and ( E yx,z ) − is the second one, then thanks to case 2 of Lemma12 we can erect a sawtooth S on bc which avoids ac , and so E ba,c ◦ ( E yx,z ) − = (cid:0) E ba,c ◦ S (cid:1) ◦ (cid:0) S − ◦ ( E yx,z ) − (cid:1) . Using case (b) separately on E ba,c ◦ S and S − ◦ ( E yx,z ) − we can prove case (c).In case (a), let S − = ( S q ,...,q m p ,...,p m ) − : L → L be the first sawtooth and S = S t ,...,t k r ,...,r k : L → L be the second one. If they are erected on the same edge, then by Lemma 25 we can factor S ◦ S − using type- R and W moves, thus obtaining a chain of links with the same height. Now let us suppose Figure 3.5: Two sawteeth of different edgesthat the two sawteeth are constructed on two different edges, namely e and e . Then the chain is asin Figure 3.5.Now we will decompose this chain without reaching the level h ( L ) of height, as shown in Figure3.6. Thanks to Lemma 12, there exists a sawtooth on e that avoids the first tooth of S , namely[ p , p , q ]. By applying it, we decrease height by 1 and now we can apply the move ( E q p ,p ) − . Thereexists another sawtooth on e which avoids the second tooth of S . We can change the sawtooth withtype- R and W moves, by Lemma 25, preserving height. Now applying Lemma 12 again we erect asmall sawtooth on [ p , p ] in order to decrease height again, and apply a S − move destroying thewhole sawtooth on [ p , p ], thus obtaining a single negative edge. This move is licit, since the smallsawtooth on [ p , p ] has been erected after the one erected on e , so this latter avoids every tooth ofthe small sawtooth and the triangle [ p , p , q ] by construction. Iterating this process, we obtain thenegative edge e and a sawtooth on e . Changing the sawtooth again with type- R and W moves, weget the link L . As we have already noticed, the height is lower than h ( L ) during the whole process.For case (b), it suffices to consider the chain L E −→ L
00 S −→ L the other being identical. Fix E = E ba,c and S = S q ,...,q m p ,...,p m . If the sawtooth is erected on an edge of L that also belongs to L (that means not ab nor bc ), there exists another sawtooth S on the sameedge, avoiding the triangle [ a, b, c ]. So we can apply the following chain of deformation moves to L :( S ◦ S ) ◦ E ◦ S obtaining L . Since S ◦ S can be decomposed into type- R and W moves, the height is lower than h ( L ) during the whole process.If the sawtooth S is erected on ab or bc (we can assume that it is on bc ), we proceed as follows.First let us suppose that ac is positive. Let d be a point on [ a, c ], close enough to c so that dp i is apositive line segment for every i . First we notice that R da,c is licit. Then we erect the sawtooth on[ b, c ] tooth by tooth using type- W moves as follows (Figure 3.7).• Since d is close to c , the triangles [ c, d, p m − ] and [ c, p m − , q m ] avoid the link; so we can applythe move W p m − ,q m d,c ;• similarly, for every i > W p i − ,q i d,p i .We now operate the following deformation chain: W b,q d,p ◦ · · · ◦ W p m − ,q m − d,p m − ◦ W p m − ,q m d,c ◦ R da,c Finally, we apply ( E da,b ) − ; since ad and db are positive, this moves increases height by 1 at most.The result is L . Again, the height is lower than h ( L ) during the whole process.If the edge ac is negative, then necessarily also ab and bc are negative. Without loss of generality,the sawtooth S = S q ,...,q m p ,...,p m is erected on bc . Let us proceed as follows (Figure 3.8).First, we construct a sawtooth S t ,...,t k r ,...,r k on the edge ac , which avoids the convex hull of S q ,...,q m p ,...,p m (this is possible thanks to Lemma 12, case 1 . ). Let d be a point on [ t k , c ], close to c , so that dp i ispositive for every i . We apply a sequence of type- W moves: W b,q d,p ◦ · · · ◦ W p m − ,q m − d,p m − ◦ W p m − ,q m d,c . Now, let x be a point on ac close to c ; let us apply the move T t k ,x,bt k ,d,b , which can be decomposedinto type- R and W moves, so again the height is preserved. Last we can operate ( S t ,...,t k r ,...,r k − ,x ) − (which increases height by 1) and finally the move ( E xa,b ) − (which preserves height), thus obtainingthe link L . Also in this case, the height is lower than h ( L ) during the whole process. Proof of Markov’s Theorem.
Let L = ˆ β and L = ˆ β . Since L and L are in braid position, theirheight is 0. Let L = L → L → · · · → L n = L be a chain of deformation moves from L to L . We will prove the theorem by induction on themaximum height appearing in the chain.As showed in Remark 26, if height is constantly 0 along the whole chain, then the deformationmoves which occur in the chain can only be of type R and W , that means the braids β and β arejoined by a chain of conjugation and stabilization moves.Let us suppose that height is not constantly 0, and let h MAX > i such that h ( L i ) = h MAX , then h ( L i − ) < h MAX and h ( L i +1 ) < h MAX . Indeed, if h ( L i ) = h ( L i +1 ) then we may join L i and L i +1 by a chain ofdeformation moves L i = M → M → · · · → M k = L i +1 where h ( M j ) < h ( L i ) for every j = 1 , . . . k − L i − → L i → L i +1 such that h ( L i ) = h MAX , thusobtaining subchains L i − = N → N → · · · → N l = L i +1 where h ( N j ) < h ( L i ) for every j . In this way we have produced a new chain of deformationmoves joining L and L , whose maximum height is lower than h MAX . By induction hypothesis weconclude.
Figure 3.6: Factorization of the chain with two sawteeth on different edges Figure 3.7: Decomposition of the chain when ac is positiveigure 3.8: Decomposition of the chain when ac is negative ibliographyibliography