Planar pure braids on six strands
PPLANAR PURE BRAIDS ON SIX STRANDS
JACOB MOSTOVOY AND CHRISTOPHER ROQUE-M ´ARQUEZ
Abstract.
The group of planar (or flat) pure braids on n strands, also known as the pure twin group, is thefundamental group of the configuration space F n, ( R ) of n labelled points in R no three of which coincide.The planar pure braid groups on 3, 4 and 5 strands are free. In this note we describe the planar pure braidgroup on 6 strands: it is a free product of the free group on 71 generators and 20 copies of the free abeliangroup of rank two. Introduction
The group of planar braids on n strands B n is the Coxeter group with the generators σ , . . . , σ n − andthe relations σ i = 1 for all 1 ≤ i < n ; σ i σ j = σ j σ i for all 1 ≤ i, j < n with | i − j | > . Elements of B n can be drawn as planar braids on n strands; these are collections of n smooth descendingarcs in R which join n distinct points on some horizontal line with n points directly below them on anotherhorizontal line; two arcs are allowed to intersect, although intersections of three or more arcs at the same pointare forbidden. The arcs are considered up to a smooth isotopy of the plane which keeps the arcs descendingand preserves the endpoints of the arcs, and modulo a local move similar to the second Reidemeister moveon knots.In particular, the generator σ i is represented by the following “elementary” braid:and the relations have the following form:One obtains the symmetric group S n from B n by adding the relation( σ i σ i +1 ) = 1 for all 1 ≤ i < n − . The quotient homomorphism B n → S n assigns to a planar braid the permutation obtained by following thestrands of a braid; this is entirely similar to the case of the usual braids. The kernel of this homomorphismis the planar pure braid group P n ; it consists of the planar braids each of whose strands ends directly belowits initial point.The planar pure braid group is the fundamental group of the complement in R n to the union of all thesubspaces x i = x j = x k , i < j < k. This space, which we denote here by F n, ( R ), may be thought of as the configuration space of n orderedparticles in R no three of which are allowed to coincide. A planar pure braid is simply a graph of a closedcurve in this space. Khovanov showed in [8] that F n, ( R ) is a classifying space for P n : its homotopy groupsare trivial in dimensions greater than one. a r X i v : . [ m a t h . G R ] J un he planar braid group appears under the names cartographical group in [14, 13] and twin group in [8, 9].In [11] planar braids are called flat braids (this term was later used in [7] to denote an entirely differentobject). The space F n, ( R ), also called the no-3-equal manifold of R , was studied in [2, 3, 12] and variousother papers.The planar pure braid groups on 3, 4 or 5 strands are free. In this note we explicitly identify the planarpure braid group on six strands.A few words about the conventions. We assume that for braids a and b , the product ab means “ a on topof b ”. We shall speak about the “number of a strand” for not necessarily pure braids: the strands will becounted by their upper endpoints from left to right.2. The structure of P , P and P Proposition 1.
The groups of pure planar braids on less than six strands are free: P and P are trivial, P = Z , P = F and P = F .Proof. It can be easily seen that F , ( R ) and F , ( R ) are contractible. The space F , ( R ) is a complement toa line in R and, hence is, homotopy equivalent to a circle. A direct calculation shows that P is generatedby the following braid:The space F , ( R ) has a free action of R by translations so it retracts to its intersection with the hyperplane x + x + x + x = 0 which is a complement in R to 4 lines passing through the origin; in turn, this spaceis homotopy equivalent to a one-point union of 7 circles. The generators of P are as follows:Finally, F , ( R ), considered up to translations and positive rescalings is the complement in S to aconfiguration of curves. This configuration is symmetric with respect to the centre of S ; its intersectionwith one hemisphere can be thought of as the union of all the straight lines in a three-dimensional ballpassing through certain five points:This can be seen to be the complement to the one-point union of 31 unlinked and unknotted circles. See [5]for a proof along different lines. (cid:3) Pure planar braids on 6 strands
The structure of P . It is known that for n ≥ P n is never free since the homology of F n, ( R ) in this case does not vanish in higher dimensions [3]. heorem 2. The group P is isomorphic to the free product 71 copies of the infinite cyclic group and 20copies of the free abelian group F ab2 of rank two. Let us give an explicit description of the free factor ( F ab2 ) ∗ .There is an “obvious” pair of commuting elements in P , namely, the two braids g and h as in Figure 1.For each integer triple 1 ≤ i < i < i ≤ q i i i ∈ B as the planar braid whose strands are segments Figure 1. of straight lines and whose underlying permutation is the (3 , k to i k for k = 1 , ,
3. Theelements q i i i ∈ B are precisely those which can be drawn as braids whose first three strands, as well asthe last three strands, do not intersect each other. For instance, here is the braid q :There are exactly 20 triples 1 ≤ i < i < i ≤
6. Under the isomorphism P = F ∗ ( F ab2 ) ∗ that weexhibit in the proof of Theorem 2, the 20 pairs of braids ( q − i i i · g · q i i i , q − i i i · h · q i i i ) generate the 20copies of F ab2 .Theorem 2 was initially established with the help of a computer implementation of the Reidemeister-Schreier method (as applied in [6] to the case of the usual pure braids), followed by some ad hoc fiddling.The proof presented here also follows the Reidemeister-Schreier type argument. We consider the sequenceof subgroups P ⊂ H ⊂ B where H consists of the planar braids whose underlying permutation sends theset { , , } to itself (and, therefore, maps the set { , , } to itself as well). The group H can be thoughtof as the group of bicoloured braids, say, with the first three strands red and the last three black . It turnsout that H is a free product of two copies of the same group which we describe explicitly; this, in turn,produces a description of P . We should mention that the planar pure braids have been studied using theReidemeister-Schreier method before; see [1].3.2. The structure of H . Consider the group A generated by a , . . . , a subject to two kinds of relations: • a i = 1 for all i ; • a i a j = a j a i whenever the vertices i and j on the graph in Figure 3 are connected by an edge. Remark 3.
A partially commutative group defined in this fashion is called a graph product ; in this case, of8 groups of order two, see [4].Let A (cid:48) and A (cid:48)(cid:48) be two copies of A with the generators a (cid:48) , . . . a (cid:48) of A (cid:48) and a (cid:48)(cid:48) , . . . a (cid:48)(cid:48) of A (cid:48)(cid:48) correspondingto the generators a , . . . , a of A . Proposition 4.
The group of bicoloured braids H is isomorphic to the free product A (cid:48) ∗ A (cid:48)(cid:48) . The bicolouredbraids which correspond to the generators a (cid:48) , . . . a (cid:48) and a (cid:48)(cid:48) , . . . , a (cid:48)(cid:48) are shown in Figure 2. In particular, conjugating A (cid:48) by q inside H we obtain A (cid:48)(cid:48) . in a nod to [10]. igure 2. The generators of H . Proof.
Assigning to each a i a planar braid as in Figure 2 we get a homomorphism of the group described inthe statement of the Proposition to H . Let us construct its inverse.First, given b = σ j . . . σ j m ∈ H we show how to write b as the product of the a (cid:48) i and a (cid:48)(cid:48) j . Write s x for thepermutation defined by x ∈ B and set q k = q i i i where { i , i , i } = { s σ j ...σ jk (1) , s σ j ...σ jk (2) , s σ j ...σ jk (3) } . Then, b = ( q σ j q − )( q σ j q − ) . . . ( q m − σ j m q − m ) , since q = q m = 1.For each of the factors q k − σ j k q − k there are two possibilities: σ j k may either involve two strands ofdifferent colours or two strands of the same colour. If the colours of the strands are different, we have q k − (cid:54) = q k and q k − σ j k q − k = 1. Indeed, in this case, if q k − can be represented by a braid whose first three,as well as the last three, strands are disjoint, the same is true for q k − σ j k ; this implies q k − σ j k = q k .On the other hand, when the factor σ j k involves a crossing of two strands of the same colour, we have q k − = q k . Aside from this crossing, the first three strands, as well the last three strands of the braid q k σ j k q − k do not cross each other. Figure 2 lists all such braids; these are the generators a (cid:48) i and a (cid:48)(cid:48) j . Indeed,Table 1 shows all the pairs ( q i i i , σ m ) such that the factor σ m in the braid q i i i σ m q − i i i has both strandsof the same colour.Now, assume that b (cid:48) differs from b by applying one of the relations in B . If b (cid:48) is obtained from b byexchanging σ j k with σ j k +1 with | j k +1 − j k | > b = . . . ( q k − σ j k q − k )( q k σ j k +1 q − k +1 ) . . . Figure 3.
The graph of relations in A . able 1. Generators and relations for H . The intersection of the row i i i with thecolumn σ k shows the non-trivial values of q i i i · σ k · q − i i i . σ σ σ σ σ commutation relations123 a (cid:48) a (cid:48) a (cid:48) a (cid:48) [ a (cid:48) , a (cid:48) ] , [ a (cid:48) , a (cid:48) ] , [ a (cid:48) , a (cid:48) ] , [ a (cid:48) , a (cid:48) ]124 a (cid:48) a (cid:48) [ a (cid:48) , a (cid:48) ]125 a (cid:48) a (cid:48) [ a (cid:48) , a (cid:48) ]126 a (cid:48) a (cid:48) a (cid:48) [ a (cid:48) , a (cid:48) ] , [ a (cid:48) , a (cid:48) ]134 a (cid:48) a (cid:48) [ a (cid:48) , a (cid:48) ]135136 a (cid:48) a (cid:48) a (cid:48) [ a (cid:48) , a (cid:48) ]235 a (cid:48) a (cid:48) a (cid:48) a (cid:48) [ a (cid:48) , a (cid:48) ] , [ a (cid:48) , a (cid:48) ]456 a (cid:48)(cid:48) a (cid:48)(cid:48) a (cid:48)(cid:48) a (cid:48)(cid:48) [ a (cid:48)(cid:48) , a (cid:48)(cid:48) ] , [ a (cid:48)(cid:48) , a (cid:48)(cid:48) ] , [ a (cid:48)(cid:48) , a (cid:48)(cid:48) ] , [ a (cid:48)(cid:48) , a (cid:48)(cid:48) ]356 a (cid:48)(cid:48) a (cid:48)(cid:48) [ a (cid:48)(cid:48) , a (cid:48)(cid:48) ]346 a (cid:48)(cid:48) a (cid:48)(cid:48) [ a (cid:48)(cid:48) , a (cid:48)(cid:48) ]345 a (cid:48)(cid:48) a (cid:48)(cid:48) a (cid:48)(cid:48) [ a (cid:48)(cid:48) , a (cid:48)(cid:48) ] , [ a (cid:48)(cid:48) , a (cid:48)(cid:48) ]256 a (cid:48)(cid:48) a (cid:48)(cid:48) [ a (cid:48)(cid:48) , a (cid:48)(cid:48) ]246245 a (cid:48)(cid:48) a (cid:48)(cid:48) a (cid:48)(cid:48) [ a (cid:48)(cid:48) , a (cid:48)(cid:48) ]146 a (cid:48)(cid:48) a (cid:48)(cid:48) a (cid:48)(cid:48) a (cid:48)(cid:48) [ a (cid:48)(cid:48) , a (cid:48)(cid:48) ] , [ a (cid:48)(cid:48) , a (cid:48)(cid:48) ]and b (cid:48) = . . . ( q k − σ j k +1 q (cid:48) k − )( q (cid:48) k σ j k q − k +1 ) . . . . Then, b and b (cid:48) , as words in the a (cid:48) i and a (cid:48)(cid:48) j , differ by a non-trivial relation only if q k − = q k = q (cid:48) k = q k +1 .Such a relation states that two of the generators a (cid:48) i or a (cid:48)(cid:48) j commute: inspection shows that these relationsare the same for a (cid:48) i and for a (cid:48)(cid:48) j (never involving generators of both kinds) and coincide with the commutationrelations in A ; see Table 1. In the same fashion, the relations σ j = 1 produce the relations a (cid:48) i = a (cid:48)(cid:48) j = 1. (cid:3) The kernel of a free product of morphisms.Lemma 5.
Let f (cid:48) : G (cid:48) → S and f (cid:48)(cid:48) : G (cid:48)(cid:48) → S be surjective homomorphisms of groups with kernels K (cid:48) and K (cid:48)(cid:48) respectively. Then, the kernel of the induced homomorphism f (cid:48) ∗ f (cid:48)(cid:48) : G (cid:48) ∗ G (cid:48)(cid:48) → S is isomorphic to K (cid:48) ∗ K (cid:48)(cid:48) ∗ F | S |− .Proof. Choose representatives z (cid:48) α ∈ G (cid:48) and z (cid:48)(cid:48) α ∈ G (cid:48)(cid:48) for each element α ∈ S assuming that e ∈ S isrepresented by the neutral elements in G (cid:48) and G (cid:48)(cid:48) .Consider the free group F | S |− whose generators x α are indexed by α ∈ S −{ e } . We have a homomorphism F | S |− → G (cid:48) ∗ G (cid:48)(cid:48) which sends x α to z (cid:48) α z (cid:48)(cid:48)− α . Together with the natural inclusions K (cid:48) (cid:44) → G (cid:48) and K (cid:48)(cid:48) (cid:44) → G (cid:48)(cid:48) it gives a homomorphism φ : K (cid:48) ∗ K (cid:48)(cid:48) ∗ F | S |− → ker ( f (cid:48) ∗ f (cid:48)(cid:48) ) ⊂ G (cid:48) ∗ G (cid:48)(cid:48) . Any y ∈ ker ( f ∗ f ) can be written as y = r s . . . r m s m with r i ∈ G (cid:48) and s j ∈ G (cid:48)(cid:48) . Write α k = ( f (cid:48) ∗ f (cid:48)(cid:48) )( r s . . . r k )and β k = ( f (cid:48) ∗ f (cid:48)(cid:48) )( r s . . . r k s k ) . Then we have y = (cid:0) r z (cid:48)− α (cid:1) · (cid:0) z (cid:48) α z (cid:48)(cid:48)− α (cid:1) · (cid:0) z (cid:48)(cid:48) α s z (cid:48)(cid:48)− β (cid:1) · (cid:0) z (cid:48)(cid:48)− β z (cid:48) β (cid:1) · . . . · (cid:0) z (cid:48) β m − r m z (cid:48)− α m (cid:1) · (cid:0) z (cid:48) α m z (cid:48)(cid:48)− α m (cid:1) · (cid:0) z (cid:48)(cid:48) α m s m (cid:1) n this expression, ( z (cid:48) β k − r k z (cid:48)− α k ) ∈ K (cid:48) while ( z (cid:48)(cid:48) α k s k z (cid:48)(cid:48)− β k ) ∈ K (cid:48)(cid:48) . Moreover, ( z (cid:48) α k z (cid:48)(cid:48)− α k ), if non-trivial, is theimage of the generator x α k ∈ F | S |− and, therefore, replacing ( z (cid:48) α k z (cid:48)(cid:48)− α k ) with x α k (and ( z (cid:48)(cid:48)− β k z (cid:48) β k ) with x − β k ),we obtain an element ˜ y ∈ K (cid:48) ∗ K (cid:48)(cid:48) ∗ F | S |− such that φ (˜ y ) = y . Call this element the rewriting of the product r s . . . r m s m .We claim that the rewriting ˜ y depends only on y . Indeed, replace s k in the product r s . . . r m s m by s k es k where s k s k = s k and e is considered as an element of G (cid:48) . In the rewriting, this will result inreplacing ( z (cid:48)(cid:48) α k s k z (cid:48)(cid:48)− β k ) with (cid:0) z (cid:48)(cid:48) α k s k z (cid:48)(cid:48)− γ (cid:1) · (cid:0) z (cid:48)(cid:48) γ s k z (cid:48)(cid:48)− β k (cid:1) where γ = ( f (cid:48) ∗ f (cid:48)(cid:48) )( r s . . . r k s k );this does not alter the rewriting as an element of K (cid:48) ∗ K (cid:48)(cid:48) ∗ F | S |− . The same is true if r k is replaced by r k er k with r k r k = r k .It remains to observe that for w ∈ K (cid:48) ∗ K (cid:48)(cid:48) ∗ F | S |− , the rewriting of φ ( w ) coincides with w . Thiscan easily be established by induction on the length of w considered as a word in t (cid:48) ∈ K (cid:48) , t (cid:48)(cid:48) ∈ K (cid:48)(cid:48) andthe x α . Therefore, the rewriting gives a two-sided inverse to φ and the kernel of f (cid:48) ∗ f (cid:48)(cid:48) is isomorphic to K (cid:48) ∗ K (cid:48)(cid:48) ∗ F | S |− . (cid:3) Lemma 5 also has an elegant topological proof, pointed out to the authors by Omar Antol´ın Camarena;we sketch it here very briefly.Replace the groups in question by their classifying spaces. Consider the pushout square ∗ → BG (cid:48)(cid:48) ↓ ↓ BG (cid:48) → BG (cid:48) ∨ BG (cid:48)(cid:48) Each of its vertices maps to BS by means of the trivial map, Bf (cid:48) , Bf (cid:48)(cid:48) and B ( f (cid:48) ∗ f (cid:48)(cid:48) ), respectively. Thehomotopy fibres of these maps also form a pushout squareΩ BS = S → BK (cid:48)(cid:48) ↓ ↓ BK (cid:48) → B (ker f (cid:48) ∗ f (cid:48)(cid:48) )Therefore, B (ker f (cid:48) ∗ f (cid:48)(cid:48) ) is homotopy equivalent to the BK (cid:48) connected to BK (cid:48)(cid:48) by | S | intervals; that is, to BK (cid:48) ∨ BK (cid:48)(cid:48) ∨ ( S ) | S |− .3.4. Proof of Theorem 2.
Consider the homomorphism H → S × S which sends a braid into its under-lying permutation; its kernel is P . Since H = A ∗ A and S × S has 36 elements, Theorem 2 will be provedas soon as we establish the following: Proposition 6.
The kernel Q of the map A → S × S which sends the generator a i into the permutationof the braid a (cid:48) i is isomorphic to the free product of F and 10 copies of F ab2 . The rest of this section is dedicated to the proof of this statement.The cosets of Q in A are pairs of permutations on 3 letters. The set Z × Z (cid:48) where Z := { , a , a , a a , a a , a a a } and Z (cid:48) := { , a , a , a a , a a , a a a } is a system of representatives for these cosets.Consider b = a j . . . a j m ∈ Q and let µ k ∈ Z × Z (cid:48) ⊂ A be the representative of the coset of a j . . . a j k ∈ A ,with µ = 1. Then, we have b = ( µ a j µ − ) . . . ( µ m − a j m µ − m ) . Let
X ⊂ Q be the subset consisting of the non-trivial elements of the form µa j ν − with µ, ν ∈ Z × Z (cid:48) and0 < j ≤ Lemma 7. If µa j ν − = µ (cid:48) a k ν (cid:48)− ∈ X , then j = k . As a corollary, X = (cid:116) X j where X j consists of the elements of the form µa j ν − . roof. First, note that X = X = ∅ . Indeed, for each µ ∈ Z × Z (cid:48) we have that µa ∈ Z × Z (cid:48) and thus µa ν − ∈ Q with ν ∈ Z × Z (cid:48) implies µa ν − = 1. The same is true for a instead of a .In order to see that the sets X j are disjoint we observe that the index j can be recovered from theintersection properties of the braids representing the elements µa j ν − ∈ X . Indeed, µa j ν − ∈ X can bedrawn as a braid whose black strands do not intersect each other if and only if j ∈ { , , } .When this happens, we can distinguish X as consisting of the elements which can be drawn so that theblack strands do not intersect the red strands and X consists of the elements which can only be drawn insuch a way that each red strand intersects some black strand.Similarly, when j ∈ { , , } , all the red strands are disjoint from each other and the same argumentapplies. (cid:3) The group Q is generated inside A by the elements of X subject to two kinds of relations. The relationsof the first kind come from the relations a j = 1 and are of the form( µa j ν − )( νa j µ − ) = 1 , where µ and ν satisfy µa j ν − ∈ P . In particular, the set X j contains, together with each element, itsinverse.The relations of the second kind come from the relations a j a k = a k a j shown in the graph on Figure 3;these are of the form ( λa j µ − )( µa k ν − ) = ( λa k ˜ µ − )(˜ µa j ν − ) . Recall that the set X is trivial; the relations that involve µa ν − are all of the form µa i ν − = ( µσ ) a i ( νσ ) − , where i ∈ { , , } . Recall that we also have µa i ν − = ( νa i µ − ) − . When i = 7, there are no otherrelations involving the elements of X and the four elements µ, µσ , ν and νσ are all different; therefore, | Z × Z (cid:48) | / X contributes | Z × Z (cid:48) | / X which consists of braids that are mirror reflections of those in X .Similarly, one sees that X and X produce 9 generators each. For each of the generators coming from X , there is one generator coming from X with which it commutes; there are no other relations involvingthem. Therefore, we see that X and X contribute a free factor isomorphic to ( F ab2 ) ∗ . Computation showsthat the commuting pairs ( µa ν − , µa ν − ) are all of the form ( q · g · q − , q · h · q − ) where q = q i i i and( i , i , i ) varies over the set of triples(1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (1 , , , (2 , ,
4) (2 , , , (2 , , , and g and h are shown on Figure 1.When i = 3, the set X consists of two mutually inverse elements, namely, the braid h and its inverse.Similarly, the set X consists of the braid g and its inverse. The only relation that these braids satisfy statesthat they commute; this gives another free factor of the form F ab2 . This exhausts the list of generators andProposition 6 is proved.Note that the commuting pairs of generators of P that lie in A (cid:48)(cid:48) ⊂ H are the conjugations by q of thecommuting pairs in A (cid:48) ⊂ H . This shows that the commuting pairs of generators in P can be chosen to beof the form ( q · g · q − , q · h · q − ) where q = q i i i for all possible 1 ≤ i < i < i ≤ Acknowledgments.
We would like to thank Omar Antol´ın Camarena, Fred Cohen and Jes´us Gonz´alez foruseful conversations. The authors would like to thank for hospitality the Higher School of Economics (J.M.)and the Samuel Gitler Collaboration Center (C.R.-M.).
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