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Asian Journal of Algebra

Year: 2008 | Volume: 1 | Issue: 1 | Page No.: 1-9
DOI: 10.3923/aja.2008.1.9
Retraction of Braid and Braid Group
M. El-Ghoul and M.M. Al-Shamiri

Abstract: In this study we introduce the retraction and conditional retraction of braids and braid groups, we show the retraction of braid group is not necessary a braid group also a retraction of a singular braid is not necessary a singular braid. We prove that a retraction of a braid is a braid and every retraction of a braid group is a monoid also we prove that a retraction is a braid invariant. The limit of all types of retraction is described.

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How to cite this article
M. El-Ghoul and M.M. Al-Shamiri, 2008. Retraction of Braid and Braid Group. Asian Journal of Algebra, 1: 1-9.

Keywords: Braid, braid group and retraction

INTRODUCTION

A braid theory introduced by E. Artin at 1925 and the concept of braid theory was found to have applications in other fields after the 1950s and this gave fresh impetus to the study of braids. More studies on braid theory are studied by many researches by this braid theory has gradually been prospected, refined and polished. In mathematics it is, now, recognized from one of the basic theories and is of benefit in such branches as topology and algebraic geometry. Also, it is of profound use in other areas of the sciences physics, statistical mechanics, chemistry and biology. The braid group was took an important role in this field. The iridescent hue of this concepts flowering into full bloom and activity occurred in 1984, when V. Jones put into action with inordinate success the original aim of Artin. i.e. the application of braids to knot theory. In (El-Ghoul et al., 2006, 2007) we introduced a new direction on knot theory called folding and retraction of knot. More studies of retraction in El-Ghoul (1985, 1995, 1998, 2002). In this study our intention introduce the concepts of retraction on the braid theory and braid group, continuations to the two articles above, we study the effect of retraction and conditional retraction on braid and braid group and singular braid.

DEFINITIONS

Here we will show some definitions and basic concepts which we will use it in the main results.

Definition 1

Let D be a unit cube, so D = {(x, y, z): 0 ≤ x, y, z ≤ 1} on the top face of cube place n points, a1, a2, ..., an and similarly, place n points on the bottom face b1, b2, ..., bn Now,join the points a1, a2, ..., an with b1, b2, ..., bn by means of n arcs d1, d2, ..., dn (as smooth curves), this arcs are mutually disjoint and each d1 connects some aj to bk (J ≠K or J = K) not connect aj to aK or bj to bK. Each plane Es, such that z = s, 0 ≤ s ≤ 1 (parallel to xy-plane intersections each arc di at one and only one point.

A configuration of n arcs d1, d2, ..., dn with end points a1, a2, ..., an and b1, b2, ..., bn is called n-braid or a braid with n strings Fig. 1 (Murasugi, 1996).

Fig. 1: Representation of braid in unit cube

Fig. 2: Product of β1 and β2

Definition 2

Let Bn be a set of all n-braids and β1, β2 ε Bn, we may create a third n-braid from them which we shall call their product and denoted by β1β2 as follows:

Glue the bottom arcs of β1 with the top arcs of β2 (Murasugi and Kurpita, 1999) Fig. 2.

Remark 1

For β1 ε Bn there is β1-1 such that β1β1-1 = e ε Bn Fig. 3.
β1β2 ≠β2β1
The product of braids is associative, i.e., (β1β2312β2)

Theorem 1

The set of all n-braids, Bn forms a group. This group is usually called the n-braid group or Artin's n-braid group (Murasugi and Kurpita, 1999; Gemein, 2001).

Theorem 2

For any n ≤ 1 the n-braid group Bn has the following presentation:

Where, σi denotes the standard generator of the braid group (Kauffman, 1991; Murasugi and Kurpita, 1999; Gemein, 2001) Fig. 4.

Fig. 3: Product of β1 and β1-1 and β1 ° β2 ≠β2 ° β1

Fig. 4: Standard generator σi

Definition 3

An n-dimensional manifold is a Housdorff topological space M, such that every point of M has a neighborhood homeomorphic to open set U ⊂ Rn (Munkers, 1975).

Definition 4

Let A be a subset of a topological space X. A continuous map r: X →A is said to be retraction if r(a) = a for all a ε A (Massay, 1967; Munkers, 1975).

Definition 5

A subset A ⊂ X is deformations retract of X if there is a retraction r: X →A such that i ° r homotopic to the identity map. That is, there exists a continuous function f: X x [0, 1] →X such that for x ε X, f (x, 0) = x and f (x, 1) = r (x) and for all a ε A and all t ε [0, 1], f (a, t) = a (Massay, 1967).

Definition 6

Let β1, β2 be two n-braids in a cube D, we say β1 ambient isotopic to β2, denoted by β1 ≈ β2, if there exists a homeomorphism H: Dx[0,1]→Dx [0,1], such that H(x,t) = ht((x),t), tε[0,1], ht: D→D and h01) = β1, ht2) = β2 (Murasugi and Kurpita, 1999).

Definition 7

Let β be a n-braid and suppose the ith string di of β joins ai to bj(i) for i = 1, 2, 3, ….n. Define g:

Bn →Sn (the set af all permutations of the set {1, 2, 3, 4, ….., n} as, the permutation for a given braid is called a braid permutation and denoted by π(β), if

then β is called a pure n-braid (Murasugi and Kurpita, 1999).

Remark 2

If β1 ≈ β2 ⇒ π(β1) = π(β2).

Theorem 3
Let Bn be a n-braid group and Sn the symmetry group of n elements, then there exist a natural surjective homomorphism f from Bn onto Sn, takes any braid β to the permutation determined by β, f (β) = π (β).

The kernel of f is a pure n-braid group, denoted by Pn and Bn/Pn is isomorphic to Sn, the index of Pn in Bn is a finite, [Bn:Pn] = n! (Murasugi and Kurpita, 1999).

THE MAIN RESULTS

Let β = σi be a n-braid and r a retraction from β-{a} onto β', where a is a point on one arc of β, then we have two cases:

β' is a trivial n-1-braid, if a ε di or a ε di+1.
β' = σi as a n-1-braid if a ε dj, j ≠i, i + 1 Fig. 5.

If the retraction r from β-{a1, a2, ..., an} onto β' by remove all points from top level or bottom level or together, then we have two cases:

β' is a trivial n-braid.
β' = σi as a n-braid Fig. 6.

Theorem 4
A retract of any braid by remove a point or (points) from any arc or (arcs) is a braid.

Proof
Let β be a n-braid, β = {d1, d2, ..., dn} where di is a string and r be a continuous map from β-{a} onto β' where a is a point on di, i = 1, 2, ..., n defined by , then r(di) = di ∀di ∈ β' hence r is a retraction and β' is a n-1-braid.

Theorem 5
Let β = σ1σ2.....σk be a n-braid and r: (β-a) →β' be a retraction then β' takes three cases:

Trivial n-1-braid, if a ε d1.
β' = σ1σ2....σk-1 as a n-1-braid, if a εdi, 1 < i ≤ k + 1.
β' = σ1σ2....σk as a n-1-braid, if a ε di, i > k + 1.

Fig. 5: Two cases of retraction of σi by remove one point

Fig. 6: Two cases of retraction of σi by remove more then one point

Proof

Since d' 1 of σ1 glue with d1'' of σ2, also d1'' glue with d1''' of σ3…., d1k-1 of σk-1 glue with d1k of σk and since d1 = d1'd1''d1'''....d1k-1d1k hence if we remove d1, then all σ1, σ2, ..., σk are finished.
Since di, 1 < i ≤ k + 1 is not glue with any arc of any σ, then the only σ which finished is σi-1 and σi takes its place and represents σi in the retract.
Since all di, i > k + 1 are straight strings then remove it not finish any σ, hence the retract β' = σ1σ2...σk.

Corollary 1
The limit of retractions of i) in Theorem 5 is a 1-braid group, also the limit of the retraction of ii) is a 1-braid, but the limit of the retractions of iii) is σ1σ2....σk.

Lemma 1
Let β, β1, β2 be a three elements of a n-braid group Bn and r be a retraction from Bn-{a} onto , then not necessary

r(β1 ° β2) = r(β1) ° r(β2),
r(β-1) = (r(β))-1, β-1 be the inverse element of β.

We can show the lemma by the following examples:

Let β1 = σ3σ1 and β2 = σ1σ2σ3σ1 be a two elements of 4-braid group B4, r be a retraction from B4-{a} onto a ∈ d1, then β1 ° β2 = σ3σ12σ2σ3σ1, r(β1) = σ2 is a 3-braidas, r(β2) = σ1 is a 3-braid, r(β1 ° β2) = σ2σ1σ2 is a 3-braid and r(β1) ° r(β2) = σ2σ1 as a 3-braid, hence r(β1 ° β2) ≠r(β1) ° r(β2) Fig. 7.
Let β = σ2σ1σ3 be an element of 4-braid group B4 and β-1 = σ3-1σ1-1σ2-1 be its inverse, r be a retraction from B4-{a} onto a ∈ d1, then r(β) = σ2σ3 as a 3-braid, hence (r(β))-1 = σ3-1σ2-1 and r(β-1) = σ1-1 as a 3-braid, hence r(β-1) ≠(r(β))-1 Fig. 8.

Fig. 7: r(β1 ° β2) ≠r(β1) ° r(β2)

Theorem 6
The retraction of a n-braid group Bn is not necessary a braid group.

Proof
From the lemma above we can say easy there is not exist for all element in r(Bn-{a}) an inverse element.

Theorem 7
Let Bn be a n-braid group and M = {σ1, σ2, ..., σn-1} be a set of its generators, r be a retraction from Bn-{a} onto , then i) if a ε d1 or a ε dn then r(M) is a set of the generators of Bn-1, ii) if a ε di, 1 < i < n, then r(M) = {σ1, σ2, ..., σi-1, σi+1, σn-1}

Proof

If we remove an arc d1 from all the generators, then σ1 was vanished, also if we remove dn, then σn-1 was vanished and σ2 becomes σ1, etc.
Proof ii) came directory from the proof i).

Theorem 8
Let β = σ2σ4σ6...σ2k-2 be a n-braid, then β' = σ1σ3...σ2k-1 as a n-1-braid is a retract of β.

Proof
Let β = σ2σ4σ6...σ2k-2 be a n-braid, {d1, d2, ..., dn} be an arcs of β and r be a continuous map from β-{a} onto β' where a is a point on di.i =1, 2, ..., n defined, by , then r(di) = di ∀ di ε β' hence r is a retraction and β' = σ1σ3...σ2k-1 is a n-1-braid.

Theorem 9
Let β = σ1σ3...σ2k-1 be a n-braid, then β = σ2σ4σ6...σ2k-2 as a n-1-braid is a retract of β.

Fig. 8: r(β-1) ≠(r(β))-1

Proof
The proof is similar to the proof of Theorem 8.

Theorem 10
Let Bn be a pure n-braid group generated by a1, a2, ..., an-1, where, ai = (σn-1σn-2, ..., σi+1) and r:(ai-a) →bi be a retraction then

bi is a trivial n-1-braid if a ε di or a ε dn.

if a ε dj, i < j < n.
bi = ai-1 in n-1-braid, if a ε dj, j < i

Proof

The proof is clear.

Theorem 11
Let Bn be a n-braid group and r be a retraction from Bn-{a} onto , then is a monoid.

Proof
Let Bn be a n-braid group and r be a retraction from Bn-{a} onto , then the all elements of are n-1-braids, hence by definition the product of any two elements is also an element in , also the associative law is holds and we can see easy the identity element is exists, hence is a monoid.

Theorem 12
A retract of a singular braid not necessary a singular braid.

Proof

We can show that by the following Example:

Example 1

Let β = σ1σ3σ2 be a singular 4-braid and r: (β-{a}) → be a retraction and {a} be a singular point then r(β-{a}) = σ1 which is not singular braid Fig. 9.

Fig. 9: Singular braid not necessary a singular braid

Fig. 10: r(β1) = r(β2) but β1 β2/td>

Fig. 11: β1, β2, β3 are equivalents and has the same r(βi-{a}), i = 1, 2, 3

Theorem 13
A retraction of any isotopic braids is an invariant.

Proof
Let β1, β2 be a two n-braids such that β1 ≈ β2 and a Γ be a set of all n-braids and r be a retraction from Γ-{a} onto , we want to prove r(β1) = r(β2), since β1 ≈ β2 then π(β1) = π(β2), this means the ends points of the arcs d1, d2, ..., dn of β1 and β2 have the same location, by elementary move Ω or Ω-1 on the arcs the ends points can not be changed, but the σ`s which components of β1 and β2 are exchanged the places, hence by remove any arc from β1 and β2 gives the same result which means r(β1) = r(β2).

Remark 1

If r(β1) = r(β2), then not necessary β1 ≈ β2.

By the following example we show that:

Example 2

Let , β2 = σ1σ2σ3 be a two 4-braids, then r(β1) = r(β2) but β1 β2 Fig.10.

Example 3

Let β1, β2, β3 be a three 4-braids, easy we can get one from the another by some elementary moves, hence this three 4-braids are equivalents, if we apply the retraction on the three 4-braids, r:(βi-{a}) → , a ε d1 we get = (σ2σ1)3, (Fig. 11).

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