aa r X i v : . [ m a t h . G T ] J u l A Note on Topologically-Trivial Braids
Orlin Stoytchev ∗ Abstract
We give a simple characterization of braids that can be unplaited keeping separately theirupper ends and their lower ends tied togetherConsider Artin’s [1] braid group B n and its subgroup of pure braids P n ⊂ B n . Each representativeof P n can be regarded as a geometric object embedded in R in the following way: two pieces ofhorizontal planes, one being a vertical translate of the other, and n strands each connecting a pointon the lower plane with the translate of that point on the upper plane. The strands of course do notintersect each other and in addition the vertical coordinate of each strand is a monotonic functionof the parameter of the strand. Now consider those elements of P n that are topologically trivial asembedded in R , namely there are isotopies bringing them to the trivial braid. These are not justthe usual isotopies for Artin braids where the two parallel planes are fixed in R . We allow moveslike rotations of the planes, flipping a strand above (below) and around one of the planes, passingthe plane through some part of the braid. The elements of P n satisfying this property obviouslyform a subgroup R n ⊂ P n . This subgroup was studied by Shepperd [4] using algebraic methods.By bringing each pure braid to a certain canonical form, the author provides a (rather complicated)algorithm of deciding whether or not this braid can be unplaited keeping its ends tied together,i.e., whether or not it is topologically trivial in the above sense.In the present note we exhibit a simple and intuitive solution of that same problem. It can bedescribed in words as follows: Pick one of the strands, e.g., the n -th strand, and pass the wholebraid, starting with the upper common end, through itself along the n -th strand, then straighten itback up again. You get a (pure) braid in which the n -th strand runs straight, without crossing anyof the other strands. Then the original braid is topologically trivial if and only if by removing the n -th strand we get an n − P n − or, equivalently, represents somepower of the full twist (see below) of n − k strands and the remaining n − k strands and pull the common end back up again. Depending on the direction of rotation of thecommon end, the two groups of strands will be twisted each around itself in opposite directions.Let us denote by b k the element of P n obtained in this way from the trivial braid. In terms of theArtin’s generators s i we have: b k = ( s k − · · · s s ) − k ( s n − · · · s k + s k + ) n − k . (1) ∗ American University in Bulgaria, 2700 Blagoevgrad, Bulgaria;Institute for Nuclear Research and Nuclear Energy, 1784 Sofia , Bulgaria;[email protected] he element b ≡ d is a full (clockwise) twist of the whole braid, the element b is a braid inwhich the first strand is straight, while the remaining n − b n ≡ d − ≡ b − is a full twist in the opposite (counterclockwise) direction. One mayconsider a seemingly more general move in which one picks arbitrarily k strands from the braid,pulls them to the left, then passes the common end between them and the remaining n − k strands.However this more general move is equivalent to first applying a suitable element of B n whichrearranges the strands, then splitting the strands into the first k and the remaining n − k , thenpassing the common end between these groups and finally applying the inverse element of B n . Aswe shall see, the subgroup generated by { b k } is normal in B n , so it is enough to apply the simplermoves, producing the elements b k . For the same reason it is enough to apply these operationsonly to the upper end of the braid — in other words you do not get anything more by passing thecommon end through a nontrivial braid somewhere in the middle. Thus the question of findingthe topologically trivial braids reduces to the study of the subgroup R n , generated by the elements { b k } . Note.
The statement that there are no more general moves than the ones described above, allowingone to unplait a braid while keeping its ends tied together, should be taken as a (very plausible)conjecture. It is such also in the work of Shepperd [4].For our purposes it is more convenient to use a different set of elements, generating the samesubgroup R n , namely d : = ( s n − · · · s s ) n , r : = s s · · · s n − s n − s n − · · · s , r : = s s · · · s n − s n − s n − · · · s , r i : = s i − · · · s s s · · · s n − s n − s n − · · · s i , i = , , . . . n − , r n : = s n − s n − · · · s s s · · · s n − s n − . (2)We call d a full twist (Fig.1) and we call r i flips (Fig.2). It is a simple matter to see that the Figure 1: The full twist d in the case n = following identity relates the generators b k to the generators r i and d : b k = d ( r k r k − · · · r ) − = ( r k r k − · · · r ) − d . (3)Notice that the move described by r k r k − · · · r is similar to the one corresponding to r but per-formed to the whole bunch of the first k strands. Using algebra (Artin’s braid relations) one seesthat a flip of all n strands gives the same effect as two full twists: ( r n r n − · · · r r ) = d = ( r n − r n − · · · r r n ) = · · · = ( r r n · · · r r ) . (4)The last equation is a mathematical expression of the so-called “belt trick” demonstrating thewell-known fact in topology that a complete rotation by 720 ◦ is homotopic to the identity. (Anexplanation of the simple connection between the fundamental group of SO ( ) and braids can be Figure 2: The flip r i found in [5].) Equation 4 shows that the subgroup R n ⊂ P n is actually generated by { r i } i = ,... n − and d . We have the following Lemma 1
The subgroup R n is normal in B n . Proof:
Since d is central in B n it suffices to exhibit explicit formulas for the conjugates of all flips r i . The following identities can be checked directly: s j r i s − j = s − j r i s j = r i , i − j > j − i > , s i − r i s − i − = r i r i − r − i , s − i − r i s i − = r i − , s i r i s − i = r i + , i ≤ n − s − i r i s i = r − i r i + r i , i ≤ n − (cid:3) We will denote by R ′ n the subgroup generated by r i , i = , . . . , n . (This is obviously normal in B n and contains d but not d .) Let us observe that the n elements r i , i = , . . . n satisfy the cyclicityrelations contained in Equation 4, but if we limit ourselves to the (first) n − r i , they areprobably free. The next proposition formalizes the process, described in the second paragraph —passing a braid through itself along the n -th strand: Proposition 1
Each class in the factorgroup P n / R ′ n contains a representative in which the n-th strand runsstraight, without being entangled with the others. This representative is unique up to a full twistof the first n − strands. Proof:
To prove the existence we describe algebraically a process (see examples below) throughwhich from each braid b ∈ P n we get a new braid s ( b ) with the prescribed property by multiplying b with elements from R ′ n . Let b be written as some word in the Artin generators s i , i = , . . . , n − b = s ± i s ± i · · · s ± i k , i j ∈ { , , . . . n − } . (6)Now we track the original n -th strand, starting from the right-hand end of the word, and markthose letters, which correspond to this strand passing in front of another strand. There are twopossibilities — the original n -th strand is at i -th position and we come across a letter s i − (thecurrent i -th strand passes in front of the ( i − ) -st) or we come across a letter s − i (the current i -thstrand passes in front of the ( i + ) -st). In the first case we insert on the left of the marked letteran element r − i , in the second case we insert on the left of the marked letter an element r i . We callthe new braid obtained after all insertions s ( b ) . In this braid the original n -th strand passes always ehind the other strands, so the word can be simplified to represent a pure braid in which the n -thstrand runs straight, or, equivalently, the corresponding word does not contain s n − .In order to prove uniqueness and for the further development we need the concept of a spher-ical braid — several distinct points on a sphere and the same number of points, in the samepositions, on a smaller sphere, connected by strands in such a way that the radial coordinate ofeach strand is monotonic in the parameter of that strand. When one considers the obvious mul-tiplication for spherical braids and isotopy classes one obtains the so called braid group of thesphere [3], which algebraically is B n / { r i } i = ,... n . This is known as the mapping-class group ofthe sphere (with 0 punctures and 0 boundaries). Let’s denote this by S n . Using a stereographicprojection one can map an Artin braid to a spherical braid and vice-versa (choosing an axis for theprojection not intersecting any strand). In the forward direction (from Artin braids to sphericalones) this induces a homomorphism between the braid groups: p : B n → S n and we have precisely Ker p = R ′ n . The moves described by the generators r i (see Eq. 2) for the corresponding spherical braid repre-sent flipping the i -th strand around the inner sphere (which is an isotopy for the spherical braid)while d is a rotation of the inner sphere by 360 ◦ around some axis, e.g., the one we used to de-fine the stereographic projection). Once we have obtained a representative in which the n -th strandruns straight (as in the existence part of the proof) we see that the only moves of the correspondingspherical braid, leaving this strand straight, are rotations by 360 ◦ around this strand and isotopiesin the usual (Artin) sense of the n − (cid:3) Applying topological arguments we see that the following is true:
Proposition 2
A braid b ∈ P n is topologically trivial (i.e., can be unplaited keeping separately its lower andupper ends tied together) if and only if the corresponding topologically equivalent braid s ( b ) constitutes a full twist of the first n − strands. The topologically nonequivalent braids are in 1-1correspondence with the elements of P n − / { d } , i.e., the elements of the pure braid group of n − strands modulo its center. Proof:
We consider our braid as a spherical braid. It is topologically trivial if and only if it canbe brought to the trivial spherical braid by isotopies of the strands and possibly full rotationsaround arbitrary axes. In fact, taking into account the topology of SO ( ) and its connectionswith spherical braids (see [5]) it is enough to take a single rotation by 360 ◦ around some fixedaxis (or no rotation at all). Next we observe that if there is an isotopy bringing a spherical braidto the trivial one then there is another isotopy which first straightens the n -th strand and then,keeping it straight brings the remaining n − s : [ , ] × [ , ] → SO ( ) satisfying certain boundary conditions. To avoidtoo many notations we prefer to skip the details. Note that even though there may be an isotopybringing directly the initial braid to the trivial one, by first straightening the n -th strand we mayend up with a braid for which the remaining n − n -th strand. (cid:3) Example 1 – the Standard three-strand braid:
The standard three-strand braid as well as any three-strand braid can be unplaited with its endstied together, as is well-known. This is a simple corollary of Proposition 2 since P — the purebraid group of 2 strands — is the cyclic group generated by the corresponding full twist d . Example 2 – the English sennit:
Consider the braid with 5 strands given by the following word (Fig.3(a)): b = ( s s s − s − ) . (7)We expand the power and put asterisks above those letters, corresponding to the fifth strand pass-ing in front of another strand: b = s s s − s − s s ∗ s − ∗ s − s s s − s − s s s − s − ∗ s ∗ s s − s − . (8)The transformed braid s ( b ) is obtained by inserting appropriate elements r ± i to the left of themarked letters: s ( b ) = s s s − s − s s r s − r s − s s s − s − s s s − s − r − s r − s s − s − = s s s − s s s s s s s s s s s s s s s s − s − s s s − s − ×× s − s − s − s − s − s − s − s − s − s − s − s − s − . In principle one can simplify directly the above word and show that it represents the trivial braid.It is somewhat easier to remove the fifth strand and work with the remaining braid with 4 strands.Indeed, tracking the fifth strand in s ( b ) we see that it crosses all other strands always behind, sothe braid s ( b ) is equivalent to the braid obtained by removing the fifth strand and then adding itrunning straight at fifth position. The braid obtained by removing the fifth strand from s ( b ) can bedescribed as follows: Suppose that the fifth strand is currently at j -th position and you encountera letter s ± i with i = j , j + j -th strand).Define a new letter m j ( s ± i ) = ( s ± i if i < j s ± i − if i > j + s ( b ) cross out those letters thatcorrespond to a move of the original fifth strand. The rest of the word splits into pieces for whichthat strand does not move. Apply the transformation m j with the appropriate j to each piece. Inour example we obtain the following braid: s ′ ( b ) = s × s s − × s m ( s s ) × s m ( s s s ) × s m ( s s s s s s s s − ) × s − m ( s s ) ×× × s − m ( s − s − ) × s − s − s − s − × s − s − s − s − s − s − s − s − = s s − s s s s s s s s s s s s s − s s s − s − s − s − s − ×× s − s − s − s − s − s − s − . ow, using Artin’s relations after somewhat lengthy calculations we get s ′ ( b ) = Id , which means that b ∈ R or in other words the English sennit is topologically trivial. Example 3 – the braided theta:
Consider the braid with 6 strands given by the following word (Fig. 3(b)): b = ( s − s − s − s − s s s s ) (9)This is like the normal three-strand braid except that each strand in the latter is replaced by twostrands running together. Therefore we can think of each such pair as a ribbon. Thus we havesomething like the classical three-strand braid but plaited out of ribbons, which run flat withoutbeing twisted. You can take a strip of paper, cut two longitudinal slits to form a “theta” and thentry to plait it to obtain the “braided theta” (see Bar-Natan’s gallery of knotted objects [2] fromwhich the example was borrowed). We have b = s − s − s − s − s s s s s − s − ∗ s − ∗ s − s s s s s − s − s − s − ∗ s s ∗ s s , s ( b ) = s − s − s − s − s s s s s − s − r s − r s − s s s s s − s − s − s − r − s s r − s s = s − s − s − s − s s s s s − s − s s s s s s s s s s s s s s ×× s s s s s − s − s − s − s − s − s − s − s − s − s − s − s − s − s − s − s − . Removing the sixth strand produces the following: s ′ ( b ) = s − s − s − s − s s × s × s m ( s − s − s s s ) × s m ( s s s s ) × s m ( s s s s s s ×× s s s ) × s − m ( s − ) × s − m ( s − s − ) × s − s − s − s − s − × s − s − s − s − s − s − s − = s − s − s − s − s s s − s − s s s s s s s s s s s s s s s s ×× s − s − s − s − s − s − s − s − s − s − s − s − s − = Id (10)This shows that what we have called “braided theta” is topologically trivial as a braid of 6 strandswith the ends tied together. In order to show that the “braided theta” is trivial when considered asplaited out of three ribbons we need more — we need to check that all transformations involvedmove the strands in pairs. This is indeed the case. Alternatively we can work directly with theribbons (pairs of strands) and introduce ribbon flips R , R , R and their inverses, which are similarto the ones in Figure 2 but performed on the 3 ribbons. By definition we have R i : = r i r i − andthe effect of a flip R i is similar to that of the usual flip r i except that it twists the i -th ribbon by720 ◦ (counterclockwise). It is easier to find experimentally, rather than doing the algebra, that the“braided theta” in Figure 3 (b) is the product R R − R − R . References [1] E. Artin,
Theory of Braids , Math. Ann. ∼ drorbn/Gallery/KnottedObjects/BraidedThetas/[3] E. Fadell and J. Van Buskirk, The Braid Groups of E and S , Duke Math. J. (1962),243–257.[4] J. A. H. Shepperd, Braids Which Can be Plaited with Their Threads Tied Together at EachEnd , Proc. Royal Soc. of London. Ser. A, , No. 1321 (1962), 229-244.[5] V. Stojanoska and O. Stoytchev,
Touching the Z in Three-Dimensional Rotations , Mathemat-ics Magazine, , No. 5 (2008), 345–357., No. 5 (2008), 345–357.