aa r X i v : . [ m a t h . G T ] J u l MINIMAL GENERATING SETS OF REIDEMEISTERMOVES
MICHAEL POLYAK
Abstract.
It is well known that any two diagrams representingthe same oriented link are related by a finite sequence of Reidemeis-ter moves Ω1, Ω2 and Ω3. Depending on orientations of fragmentsinvolved in the moves, one may distinguish 4 different versions ofeach of the Ω1 and Ω2 moves, and 8 versions of the Ω3 move.We introduce a minimal generating set of 4 oriented Reidemeistermoves, which includes two Ω1 moves, one Ω2 move, and one Ω3move. We then study which other sets of up to 5 oriented movesgenerate all moves, and show that only few of them do. Somecommonly considered sets are shown not to be generating. Anunexpected non-equivalence of different Ω3 moves is discussed. Introduction
A standard way to describe a knot or a link in R is via its diagram ,i.e. a generic plane projection of the link such that the only singularitiesare transversal double points, endowed with the over/undercrossinginformation at each double point. Two diagrams are equivalent if thereis an orientation-preserving diffeomorphism of the plane that carriesone diagram to the other diagram. A classical result of Reidemeister[6] states that any two diagrams of isotopic links are related by a finitesequence of simple moves Ω1, Ω2, and Ω3, shown in Figure 1.
Ω1 Ω1 Ω2 Ω3 Ω Figure 1.
Reidemeister movesHere we assume that two diagrams D , D ′ related by a move coincideoutside an oriented embedded disk C ⊂ R (with an orientation of C induced by the standard orientation of R ), called the changing disk , Mathematics Subject Classification.
Key words and phrases.
Reidemeister moves, knot and link diagrams. Our notation makes no distinction between a move and the inverse move. and look as a corresponding pair R , R ′ of arc diagrams in Figure 1inside C . In other words, there are two orientation-preserving diffeo-morphisms f, f ′ : C → B of C to the standard oriented 2-disk B ,such that f ∂C = f ′ ∂C and f ( C ∩ D ) = R , f ′ ( C ∩ D ′ ) = R ′ .To deal with oriented links we consider oriented diagrams. Depend-ing on orientations of fragments involved in the moves, one may distin-guish four different versions of each of the Ω1 and Ω2 moves, and eightversions of the Ω3 move, see Figures 2, 3, and 4 respectively. Ω1 b Ω1 b Ω1 c Ω1 a Ω1 d Figure 2.
Oriented Reidemeister moves of type 1 Ω2 a Ω2 c Ω2 b Ω2 d Figure 3.
Oriented Reidemeister moves of type 2When one checks that a certain function of knot or link diagramsdefines a link invariant, it is important to minimize the number ofmoves. We will call a collection S of oriented Reidemeister moves a generating set , if any oriented Reidemeister move Ω may be obtainedby a finite sequence of isotopies and moves from the set S inside thechanging disk of Ω.While some dependencies between oriented Reidemeister moves arewell-known, the standard generating sets of moves usually include sixdifferent Ω3 moves, see e.g. Kauffman [3]. For sets with a smallernumber of Ω3 moves there seems to be a number of different, oftencontradictory, results. In particular, Turaev [7, proof of Theorem 5.4]introduces a set of five oriented Reidemeister moves with only one Ω3move. There is no proof (and in fact we will see in Section 3 thatthis particular set is not generating), with the only comment being areference to a figure where, unfortunately, a move Ω2 which does notbelong to the set is used. Wu [9] uses the same set of moves citing[7], but additionally incorrectly puts the total number of oriented Ω3moves at 12 (instead of 8). Kaufmann [3, page 90] includes as anexercise a set of all Ω1 and Ω2 moves together with two Ω3 moves. INIMAL GENERATING SETS OF REIDEMEISTER MOVES 3 Ω3 a Ω3 c Ω3 e Ω3 g Ω3 d Ω3 f Ω3 h Ω3 b Figure 4.
Oriented Reidemeister moves of type 3Meyer [4] uses a set with four Ω1, two Ω2, and two Ω3 moves andstates (again without a proof) that the minimal number of needed Ω3moves is two. The number of Ω3 moves used by ¨Ostlund [5] is alsotwo, but his classification of Ω3 moves works only for knots and is non-local (depending on the cyclic order of the fragments along the knot).Series of exercises in Chmutov et al. [1] (unfortunately without proofs)suggest that only one Ω3 suffices, but this involves all Ω2 moves. Thesediscrepancies are most probably caused by the fact that while manypeople needed some statement of this kind, it was only an auxiliarytechnical statement, a proof of which would be too long and wouldtake the reader away from the main subject, so only a brief commentwas usually made. We believe that it is time for a careful treatment.In this note we introduce a simple generating set of four Reidemeistermoves, which includes two Ω1 moves, one Ω2 move and one Ω3 move: Ω1 a Ω1 b Ω2 a Ω3 a Figure 5.
A generating set of Reidemeister moves
MICHAEL POLYAK
Theorem 1.1.
Let D and D ′ be two diagrams in R , representing thesame oriented link. Then one may pass from D to D ′ by isotopy anda finite sequence of four oriented Reidemeister moves Ω1 a , Ω1 b , Ω2 a ,and Ω3 a , shown in Figure 5. This generating set of Reidemeister moves has the minimal numberof generators. Indeed, it is easy to show that any generating set shouldcontain at least one move of each of the types two and three; Lemma2.2 in Section 3 implies that there should be at least two moves of typeone. Thus any generating set of Reidemeister moves should contain atleast four moves.Our choice of the move Ω3 a as a generator may look unusual, sincethis move (called a cyclic Ω3 move, see e.g. [3]) is rarely included inthe list of generators, contrary to a more common move Ω3 b , whichis the standard choice motivated by the braid theory . The reason isthat, unexpectedly, these moves have different properties, as we discussin detail in Section 3. Indeed, Theorem 1.2 below implies that anygenerating set of Reidemeister moves which includes Ω3 b has at leastfive moves. If we consider sets of five Reidemeister moves which containΩ3 b , then it turns out that out of all combinations of Ω1 and Ω2 moves,only 4 sets generate all Reidemeister moves. The only freedom is inthe choice of Ω1 moves, while Ω2 moves are uniquely determined: Theorem 1.2.
Let S be a set of at most five Reidemeister moves whichcontains only one move, Ω3 b , of type three. The set S generates allReidemeister moves if and only if S contains Ω2 c and Ω2 d and containsone of the pairs ( Ω1 a , Ω1 b ), ( Ω1 a , Ω1 c ), ( Ω1 b , Ω1 d ), or ( Ω1 c , Ω1 d ). Ω1 a Ω1 c Ω3 b Ω2 c Ω2 d Figure 6.
A generating set of Reidemeister moves con-taining Ω3 b One of these generating sets is shown in Figure 6. It is interestingto note that while (by Markov theorem) the set Ω1 a , Ω1 c , Ω2 a , Ω2 b This is the only Ω3 move with all three positive crossings.
INIMAL GENERATING SETS OF REIDEMEISTER MOVES 5 and Ω3 b shown in Figure 7 allows one to pass between any two braidswith isotopic closures, this set is not sufficient to connect any pair ofgeneral diagrams representing the same link. This means that someextra moves should appear in the process of transforming a generallink diagram into a closed braid. And indeed, in all known algorithmsof such a transformation the additions moves occur. For example, inVogels algorithm [8] the moves Ω2 c and Ω2 d are the main steps of thealgorithm. Ω1 a Ω1 c Ω3 b Ω2 a Ω2 b Figure 7.
This is not a generating setEven more unexpected is the fact that all type one moves togetherwith Ω2 a , Ω2 c (or Ω2 d ) and Ω3 b are also insufficient, see Figure 8 . Ω1 a Ω1 b Ω1 b Ω1 c Ω1 d Ω3 b Ω2 c Ω2 a Figure 8.
Another set which is not generating
Remark 1.3.
For non-oriented links a sequence of Reidemeister movescan be arranged in such a form that first a number of Ω1 moves areperformed, then Ω2 moves are performed, after this Ω3 moves are per-formed, and finally Ω2 moves have to be performed again, see [2]. Itwould be interesting to find such a theorem for oriented case.All our considerations are local, and no global realization restrictionsare involved. Therefore all our results hold also for virtual links.
MICHAEL POLYAK
Section 2 is dedicated to the proof of Theorem 1.1. In Section 3 wediscuss various generating sets which contain Ω3 b and prove Theorem1.2We are grateful to O. Viro for posing the problem and to S. Chmutovfor valuable discussions. The author was supported by an ISF grant1261/05 and by the Joseph Steiner family foundation.2. A minimal set of oriented Reidemeister moves
In this section we prove Theorem 1.1 in several easy steps. The firststep is to obtain Ω2 c , Ω2 d : Lemma 2.1.
The move Ω2 c may be realized by a sequence of Ω1 a , Ω2 a and Ω3 a moves. The move Ω2 d may be realized by a sequence of Ω1 b , Ω2 a and Ω3 a moves.Proof. Ω1 a Ω2 a Ω3 a Ω1 b Ω2 a Ω3 a Ω1 a Ω1 b (cid:3) Now the remaining moves of type one may be obtained as in [5]:
Lemma 2.2 ([5]) . The move Ω1 c may be realized by a sequence of Ω1 b and Ω2 d moves. The move Ω1 d may be realized by a sequence of Ω1 a and Ω2 c moves.Proof. Ω1 b Ω2 d Ω1 a Ω2 c (cid:3) This concludes the treatment of all Ω1 and Ω2 moves, except forΩ2 b ; we will take care of it later. Having in mind Section 3, where wewill deal with Ω3 b instead of Ω3 a , we will first consider Ω3 b : INIMAL GENERATING SETS OF REIDEMEISTER MOVES 7
Lemma 2.3.
The move Ω3 b may be realized by a sequence of Ω2 c , Ω2 d , and Ω3 a moves.Proof. Ω2 c Ω3 a Ω2 d (cid:3) To deal with Ω2 b we will need another move of type three: Lemma 2.4.
The move Ω3 c may be realized by a sequence of Ω2 c , Ω2 d , and Ω3 a moves.Proof. Ω2 c Ω3 a Ω2 d (cid:3) At this stage we can obtain the remaining move Ω2 b of type two: Lemma 2.5.
The move Ω2 b may be realized by a sequence of Ω1 d , Ω2 c and Ω3 c moves.Proof. Ω1 d Ω3 c Ω1 d Ω2 c (cid:3) To conclude the proof of Theorem 1.1, it remains to obtain Ω3 d –Ω3 h . Since by now we have in our disposal all moves of type two, thisbecomes an easy exercise: Lemma 2.6.
The moves Ω3 d – Ω3 h of type three may be realized bya sequence of type two moves, Ω3 a , and Ω3 b . MICHAEL POLYAK
Proof.
We realize moves Ω3 d -Ω3 h as shown in rows 1-5 of the figurebelow, using Ω3 f to get Ω3 g , and Ω3 g to get Ω3 h : Ω2 a Ω3 b Ω2 b Ω2 a Ω3 b Ω2 b Ω2 d Ω3 a Ω2 c Ω2 a Ω3 f Ω2 b Ω2 c Ω3 g Ω2 d (cid:3) Remark 2.7.
There are other generating sets which include Ω3 a .In particular, Ω1 a , Ω1 b , Ω2 b and Ω3 a also give a generating set. Toadapt the proof of Theorem 1.1 to this case, one needs only a slightmodification of Lemma 2.1. All other lemmas do not change. INIMAL GENERATING SETS OF REIDEMEISTER MOVES 9 Other sets of Reidemeister moves
In this section we discuss other generating sets and prove Theorem1.2. Unexpectedly, different Ω3 moves have different properties as faras generating sets of Reidemeister moves are concerned. Let us studythe case of Ω3 b in more details, due to its importance for braid theory.In a striking contrast to Theorem 1.1 which involves Ω3 a , Theorem1.2 implies that there does not exist a generating set of four moveswhich includes Ω3 b . It is natural to ask where does the proof in Section2 breaks down, if we attempt to replace Ω3 a with Ω3 b .The only difference between Ω3 a and Ω3 b may be pinpointed toLemma 2.1: it does not have an analogue with Ω3 b replacing Ω3 a , aswe will see in the proof of Lemma 3.8 below.An analogue of Lemma 2.3 is readily shown to exist. Indeed, Ω3 a may be realized by a sequence of Ω2 c , Ω2 d and Ω3 b moves, as illustratedbelow: Ω2 c Ω3 b Ω2 d Using this fact instead of Lemma 2.3, together with the rest of Lemmas2.2-2.6, implies that Ω1 a and Ω1 b , taken together with Ω2 c , Ω2 d , andΩ3 b , indeed provide a generating set. Moreover, a slight modification ofLemma 2.2 shows that any of the other three pairs of Ω1 moves in thestatement of Theorem 1.2 may be used instead of Ω1 a and Ω1 b . Thuswe see that all sets described in Theorem 1.2 are indeed generating andobtain the “if” part of the theorem. It remains to prove the “only if”part of Theorem 1.2, i.e., to show that other combinations of four Ω1and Ω2 moves, taken together with Ω3 b , do not result in generatingsets. We will proceed in three steps: Step 1.
Prove that any such generating set should contain at leasttwo Ω1 moves and to eliminate two remaining pairs (Ω1 a ,Ω1 d ) and (Ω1 b , Ω1 c ) of Ω1 moves. Step 2.
Prove that any such generating set should contain at leasttwo Ω2 moves and to eliminate pairs (Ω2 a ,Ω2 c ), (Ω2 a ,Ω2 d ),(Ω2 b ,Ω2 c ), and (Ω2 b ,Ω2 d ). Step 3.
Eliminate the remaining pair (Ω2 a ,Ω2 b ).The remainder of this section is dedicated to these three steps. Step1 is the simplest and is given by Lemma 3.1 below. Step 2 is the most complicated; it is given by Corollaries 3.5 and 3.6. Step 3 is relativelysimple and is given by Lemma 3.8.To show that a certain set of Reidemeister moves is not generating,we will construct an invariant of these moves which, however, is notpreserved under the set of all Reidemeister moves. The simplest clas-sical invariants of this type are the writhe w and the winding number rot of the diagram. The winding number of the diagram grows (respec-tively drops) by one under Ω1 b and Ω1 d (respectively Ω1 a and Ω1 c ).The writhe of the diagram grows (respectively drops) by one under Ω1 a and Ω1 b (respectively Ω1 c and Ω1 d ). Moves Ω2 and Ω3 do not change w and rot . These simple invariants suffice to deal with moves of typeone (see e.g. [5]): Lemma 3.1 ([5]) . Any generating set of Reidemeister moves containsat least two Ω1 moves. None of the two pairs ( Ω1 a , Ω1 d ) or ( Ω1 b , Ω1 c ), taken together with all Ω2 and Ω3 moves, gives a generating set.Proof. Indeed, both Ω1 a and Ω1 d preserve w + rot , so this pair (or anyof them separately) together with Ω2 and Ω3 moves cannot generateall Reidemeister moves. The case of Ω1 b and Ω1 c is obtained by thereversal of an orientation (of all components) of the link. (cid:3) This concludes Step 1 of the proof. Let us proceed with Step 2. Herethe situation is quite delicate, since the standard algebraic/topologicalinvariants, reasonably well behaved under compositions, can not beapplied. The reason can be explained on a simple example: supposethat we want to show that Ω2 d cannot be obtained by a sequence ofReidemeister moves which includes Ω2 c . Then our invariant should bepreserved under Ω2 c and distinguish two tangles shown in Figure 9a.However, if we compose them with a crossing, as shown in Figure 9b,we may pass from one to another by Ω2 c . Thus the invariant shouldnot survive composition of tangles. Ω2 c Ω2 c ba Figure 9.
Composition destroys inequivalenceInstead, we will use a certain notion of positivity, which is indeeddestroyed by such compositions. It is defined as follows. Let D be a INIMAL GENERATING SETS OF REIDEMEISTER MOVES 11 (2 , D , D .Decorate all arcs of both components of D with an integer weight by thefollowing rule. Start walking on D along the orientation. Assign zeroto the initial arc. Each time when we pass an overcrossing (we don’tcount undercrossings) with D , we add a sign (the local writhe) of thisovercrossing to the weight of the previous arc. Now, start walking on D along the orientation. Again, assign zero to the initial arc. Each timewhen we pass an undercrossing (now we don’t count overcrossings) with D , we add a sign of this undercrossing to the weight of the previousarc. See Figure 10a. Two simple examples are shown in Figure 10b,c. x yy+1 x+1 x yy x x yxy y xy−1x−1 00
00 0−10 −1 a b c Figure 10.
Weights of diagramsWe call a component positively weighted , if weights of all its arcs arenon-negative. E.g., both components of the (trivial) tangle in Figure10b are positively weighted. None of the components of a diagram inFigure 10c are positively weighted (since the weights of the middle arcson both components are − S b the set which consists of all Ω1 moves and Ω3 b . Lemma 3.2.
Let D be a (2 , -tangle diagram with both positivelyweighted components. Then both components of a diagram obtainedfrom it by a sequence of moves which belong to S b ∪ Ω2 a are also posi-tively weighted.Proof. Indeed, an application of a first Reidemeister move does notchange this property since we count only intersections of two differentcomponents. An application of Ω2 a adds (or removes) two crossings oneach component in such a way, that walking along a component we firstmeet a positive crossing and then the negative one, so the weights ofthe middle arcs are either the same or larger than on the surroundingarcs, see Figure 11a. An application of Ω3 b preserves the weights sinceΩ3 b involves only positive crossings. (cid:3) x yx y y xy x x+1 x yy+1 x yyx xyy xx−1 x yy+1 x yx y xyy xx+1 x yy−1 a cb Figure 11.
Weights and Reidemeister moves of type two
Lemma 3.3.
Let D be a (2 , -tangle diagram with a positively weightedsecond component. Then any diagram obtained from it by Ω2 c also hasa positively weighted second component.Proof. An application of Ω2 c may add (or remove) two undercrossingson D , but in such a way that we first meet a positive undercrossingand then the negative one, so the weight of a middle arc is larger thanon the surrounding arcs, see Figure 11b. (cid:3) Lemma 3.4.
Let D be a (2 , -tangle diagram with a positively weightedfirst component. Then any diagram obtained from it by Ω2 d also has apositively weighted first component.Proof. An application of Ω2 d may add (or remove) two overcrossingson D , but in such a way that we first meet a positive overcrossing andthen the negative one, so the weight of a middle arc is larger than onthe surrounding arcs, see Figure 11c. (cid:3) Comparing Figures 10b and 10c we conclude
Corollary 3.5.
None of the two sets S b ∪ Ω2 a ∪ Ω2 c and S b ∪ Ω2 a ∪ Ω2 d generates Ω2 b . The reversal of orientations (of both components) of the tangle inthe above construction gives
Corollary 3.6.
None of the two sets S b ∪ Ω2 b ∪ Ω2 c and S b ∪ Ω2 b ∪ Ω2 d generates Ω2 a . Remark 3.7.
In [7, Theorem 5.4] (and later [9]) the set S b ∪ Ω2 a ∪ Ω2 c is considered as a generating set. Fortunately (V. Turaev, personalcommunication), an addition of Ω2 d does not change the proof of theinvariance in [7, Theorem 5.4].Note that the above corollaries imply that any generating set S whichcontains only one move, Ω3 b , of type three, should contain at least twoΩ2 moves. This concludes Step 2 of the proof. INIMAL GENERATING SETS OF REIDEMEISTER MOVES 13
Since at the same time such a set S should contain at least two Ω1moves by Lemma 3.1, we conclude that if S consists of five moves,there should be exactly two Ω2 moves and two Ω1 moves. This simpleobservation allows us to eliminate the last remaining case: Lemma 3.8.
Let S be a set which consists of two Ω1 moves, Ω2 a , Ω2 b , and Ω3 b . Then S is not generating.Proof. Given a link diagram, smooth all double points of the diagramrespecting the orientation, as illustrated in Figure 12. smooth smooth smooth
Figure 12.
Smoothing the diagram respecting the orientationCount the numbers C − and C + of clockwise and counter-clockwiseoriented circles of the smoothed diagram, respectively. Note that Ω2 a ,Ω2 b , and Ω3 b preserve an isotopy class of the smoothed diagram, thuspreserve both C + and C − . On the other hand, Ω1 b and Ω1 d add one to C + , and Ω1 a , Ω1 c add one to C − . Thus if S contains Ω1 a and Ω1 c , allmoves of S preserve C + . The case of Ω1 b and Ω1 d is obtained by thereversal of an orientation (of all components) of the link. If S containsΩ1 a and Ω1 b , all moves of S preserve C + + C − − w . Similarly, if S contains Ω1 c and Ω1 d , all moves of S preserve C + + C − + w . In all theabove cases, moves from S can not generate Ω2 c , Ω2 d , since each ofΩ2 c and Ω2 d may change C + as well as C + + C − ± w (while preserving w and C + − C − = rot ). (cid:3) This concludes the proof of Theorem 1.2.
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