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✓ 7 results · all verified · 5 also independently AI-judged
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Geometric Braids and Artin Generators

1 · Prerequisites

2 · Summary

This page sets up geometric braids on n strands as the level-preserving object: with h=14(n+1)∈(0,12) and the base configuration qj=((2j−n−1)h,0) on the horizontal diameter of the open unit disc D∘, a braid based at Q=(q1,…,qn) is an n-tuple of continuous motions zj ⁣:I→D∘ whose values are pairwise distinct at every height, with bottom values zj(0)=qj and top endpoint set {z1(1),…,zn(1)}={q1,…,qn}. Labels follow the bottom points, so the top matching is a permutation π(β), the endpoint permutation, and the setwise condition is exactly what makes that permutation well defined; a braid is pure when the permutation is the identity. A braid isotopy is a jointly continuous family of braids with the bottom points and the top endpoint set held fixed throughout the deformation; the endpoint permutation is constant along such isotopies, and no deformation moving a bottom point, freeing a strand from being a graph over the height, or abandoning collision-freeness is permitted.

The page then develops the algebra of geometric braids from the pictures. Stacking first-under-second is a braid when its two factors are, rescaling the two height intervals and joining the strand of label π(β)(j), and it is well defined and associative on isotopy classes; with the stationary braid as identity and the time-reversal of a braid, relabelled by its own endpoint permutation, as inverse, the isotopy classes Gn form a group with [γ⋆β]=[γ][β]. The elementary half twist σi rotates the two adjacent base points qi,qi+1 by a half turn inside the support disc Ui containing no other base point, the label i passing below the midpoint in the anticlockwise sense; this fixes the sign convention for signed crossings once and for all, σi− is its opposite, and [σi−]=[σi]−1. Two geometric relations are proved as actual braid isotopies: half twists with disjoint supports commute, and σi⋆σi+1⋆σi∼σi+1⋆σi⋆σi+1, the latter by an explicit interpolation of the three-point configuration through a rigid rotation of the triple by the angle π.

The second half of the page reduces an arbitrary braid to that algebra. Every braid admits a generic polygonal representative: finitely many straight segments, only simple transverse crossings at pairwise distinct interior heights and never at a breakpoint, and always a uniform positive clearance from the collisions and from the boundary of the disc. Ordering the crossings by height and straightening the braid between consecutive crossings inside the convex chambers on which the left-to-right order of the strands is constant shows that every braid is braid-isotopic to a stacking of signed half twists, with the explicit crossing reading (lowest crossing rightmost, sign read from which strand passes below). Consequently the classes [σ1],…,[σn−1] generate Gn. The final proposition sends the abstract Artin generator xi of ⟨x1,…,xn−1∣braid and far-commutation relations⟩ to the geometric class [σi]: the relators are exactly the two geometric relations just proved, so von Dyck's theorem extends the assignment to a unique homomorphism Bn→Gn, and the crossing decomposition makes it surjective, including the degenerate cases n≤1 where both groups are trivial.

Surjectivity is the only claim about the Artin map made here. Injectivity — presentation completeness — is deliberately not proved on this page and is not assumed anywhere: the geometric group is defined from isotopies of motions, the abstract group from a presentation, and the page establishes only that the former is generated by the images of the latter's generators. Nothing on the page uses a choice principle: the base configuration, the half twist, every isotopy and every polygonal approximation are given by explicit formulas, and all finite choices involved are made from explicitly displayed data. The companion examples page carries out the two-strand winding computation, the three-strand relation in explicit coordinates, and the two counterexamples showing why the endpoint condition is setwise and why the monotonicity clause in the definition of braid isotopy is not redundant.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Geometric braids in the disc with setwise endpoints

Definition

Throughout this page n∈N is a natural number (The natural numbers N (von Neumann)) with the labels 1,…,n, and

I:=[0,1]⊆R

is the unit interval (Intervals of R: the nine order-convex forms, nondegeneracy, and length). Points of R2 are written as vectors and are added and scaled coordinatewise. The closed unit disc is

D:=B‾2(0,1)={w∈R2:∥w∥2≤1}

and its interior is D∘:={w∈R2:∥w∥2<1} (Euclidean spheres and closed balls as subspaces of Rn), with the subspace topology inherited from R2 (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace); the cylinder D∘×I carries the product topology (The product set ∏i∈IXi of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space) and its subspace topology. Continuous means continuous with respect to these topologies (Continuity of a map of topological spaces at a point and globally).

The base configuration. Put

h:=14(n+1),qj:=((2j−n−1)h, 0)∈R2(1≤j≤n),

and let Q:=(q1,…,qn). The points q1,…,qn lie in D∘, listed strictly left to right, and are equally spaced:

∥qj∥2=(n+1−2j)h for j≤n+12,∥qj∥2=(2j−n−1)h for j≥n+12,

so that ∥qj∥2≤(n−1)h<14 and qj+1−qj=(2h,0) for every j. In particular qi≠qj for i≠j, and Q is an ordered tuple of pairwise distinct points of D∘. The configuration Q is fixed once and for all on this page and is not part of the data of a braid.

The label set. The labels 1,…,n are part of the data. Throughout this page and its companion they are identified with the set n={0,1,…,n−1} of predecessors of n (The natural numbers N (von Neumann)) by the bijection κ(i):=i−1, and it is through κ that the symmetric group Sn=Sym⁡(n) of The finite symmetric group Sn, one-line notation, and cycle notation acts on the labels, with composition read with the right-hand factor first. Thus a symbol such as (i i+1) denotes the transposition exchanging the labels i and i+1 for 1≤i≤n−1, the symbol id⁡ denotes the identity permutation of the labels, and a bijection of {1,…,n} is regarded as an element of Sn through κ.

Geometric braid. A geometric braid on n strands based at Q, or simply a braid, is an n-tuple

β=(z1,…,zn)

of continuous maps zj ⁣:I→D∘ (Continuity of a map of topological spaces at a point and globally) such that

  1. zi(t)≠zj(t) whenever i≠j and t∈I;
  2. zj(0)=qj for every j;
  3. {z1(1),…,zn(1)}={q1,…,qn}.

The j-th strand of β is the graph {(zj(t),t):t∈I}⊆D∘×I, and the second coordinate t is its height. The defining conditions say that each strand meets every horizontal slice R2×{t} in exactly one point, that no two strands meet, that the bottom endpoints are the labelled base points q1,…,qn, and that the top endpoints form the base configuration setwise. A braid is called pure when in addition zj(1)=qj for every j.

This parametrised, level-preserving presentation is the object used in this page: the zj are the point motions, and the strands are recovered as their graphs. It is not an unqualified tame link in the cylinder. The moving points stay in the interior D∘ of the disc, and the base configuration is chosen in D∘: this interior convention gives every motion a positive distance from the boundary circle ∂D=S1, which the later polygonal approximation uses.

Endpoint permutation. Let β=(z1,…,zn) be a braid. For each j the setwise condition (3) produces at least one index π(j) with zj(1)=qπ(j), and the pairwise distinctness of q1,…,qn makes it unique; moreover j↦π(j) is injective, because qπ(j)=zj(1) shows that distinct j give distinct points qπ(j). Hence j↦π(j) is a bijection of {1,…,n}, that is, a permutation (The finite symmetric group Sn, one-line notation, and cycle notation). It is called the endpoint permutation of β and is written π(β)∈Sn:

zj(1)=qπ(β)(j)(1≤j≤n).

Labels are transported from the bottom: the j-th strand is the one that starts at qj, and π(β)(j) records where it ends. Thus β is pure exactly when π(β)=id⁡, while the setwise condition (3) alone allows π(β)≠id⁡. The adjective setwise in the title refers to condition (3); it is not a purity assumption.

Elementary cases and the trivial braid. For n=0 the tuple is empty, the conditions are vacuous, and there is exactly one braid, the empty tuple; its endpoint permutation is the unique element of S0. For n=1 we have q1=(0,0), and a braid is exactly a continuous path z1 ⁣:I→D∘ with z1(0)=z1(1)=(0,0), the endpoint permutation being the identity of S1. For every n the trivial braid e is the tuple of constant motions zj(t):=qj; it is pure, since ej(1)=qj=qid⁡(j) for every j.

Slicing is continuous by construction. Because each zj is a genuine function of the height with zj(0)=qj, the bottom endpoints are fixed pointwise, and the top condition is imposed only on the set of top endpoints. This is the distinction used by Braid isotopy relative to the top and bottom endpoints: an isotopy of braids must keep each bottom point fixed and the top configuration setwise equal to Q, but it need not return each strand to its own starting point.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Braid isotopy relative to the top and bottom endpoints

Definition

Let n∈N and let Q=(q1,…,qn) be the base configuration of Geometric braids in the disc with setwise endpoints, so that braids based at Q are tuples of continuous maps zj ⁣:I→D∘ satisfying conditions (1)--(3) of that definition. Write I=[0,1] (Intervals of R: the nine order-convex forms, nondegeneracy, and length) and give I×I the product topology (The product set ∏i∈IXi of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space).

Let β=(z1,…,zn) and β′=(z1′,…,zn′) be braids based at Q. A braid isotopy from β to β′ relative to the top and bottom endpoints, or simply a braid isotopy, is an n-tuple

Z=(Z1,…,Zn),Zj ⁣:I×I⟶D∘,

of continuous maps (Continuity of a map of topological spaces at a point and globally) such that

  1. for every s∈I, the tuple Z(s,⋅):=(Z1(s,⋅),…,Zn(s,⋅)) is a braid based at Q, i.e. t↦Zj(s,t) is continuous into D∘, the n values Zj(s,t) are pairwise distinct for every t, the bottom condition Zj(s,0)=qj holds for every s and j, and {Z1(s,1),…,Zn(s,1)}={q1,…,qn} for every s;
  2. Zj(0,t)=zj(t) and Zj(1,t)=zj′(t) for all j and t.

The first variable s is the isotopy parameter and the second variable t is the height; a family of motions depending on s is a braid isotopy exactly when the map (s,t)↦Zj(s,t) is jointly continuous, not merely continuous in s for each fixed t. When such a Z exists we say that β and β′ are braid-isotopic and write β∼β′.

Relation to homotopy. A braid is a continuous map I→(D∘)n with additional properties, and a braid isotopy is a homotopy of such maps (Homotopies of continuous maps, homotopies relative to a subspace, and path homotopies relative to the endpoints) that is relative to the bottom subset {0}⊆I, in the following sense: the homotopy condition Zj(s,0)=zj(0)=zj′(0) for all s says that the whole bottom point is fixed throughout the deformation. At the top, the condition is imposed setwise: {Z1(s,1),…,Zn(s,1)}={q1,…,qn}. Joint continuity then forces each individual top endpoint to remain constant as s varies, since a continuous map from an interval into this finite discrete set is constant.

Endpoint permutations of the slices. Let Z be a braid isotopy from β to β′. Each slice Z(s,⋅) is a braid and hence has an endpoint permutation π(Z(s,⋅))∈Sn (Geometric braids in the disc with setwise endpoints, The finite symmetric group Sn, one-line notation, and cycle notation). This definition does not separately impose that these slice permutations agree: it requires each slice to be a braid and the family to be jointly continuous. The slice permutations do agree, and therefore the endpoint permutation is an invariant of braid isotopy, with π(β)=π(β′); this is proved on this page, in the stacking proposition, by a connectedness argument for the height interval I.

Isotopy of the ambient cylinder is not enough. The definition constrains the deformation to the product D∘×I through height-preserving motions; it is neither an isotopy of an arbitrary embedded link nor a free homotopy of the tuple of paths. Deformations that make a strand meet a horizontal plane more than once, or that move bottom points, are not braid isotopies; a counterexample is recorded on the companion examples page.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Stacking of geometric braids is a well-defined associative operation on isotopy classes

Statement

Let n∈N and let Q=(q1,…,qn) be the base configuration of Geometric braids in the disc with setwise endpoints, with h:=14(n+1) and qj=((2j−n−1)h,0). Let β=(z1,…,zn) and γ=(w1,…,wn) be braids based at Q with endpoint permutations π(β) and π(γ) (Geometric braids in the disc with setwise endpoints), and let ∼ denote braid isotopy relative to the top and bottom (Braid isotopy relative to the top and bottom endpoints).

(a) The stacked tuple is a braid. Define γ⋆β by

(γ⋆β)j(t):={zj(2t),0≤t≤12,wπ(β)(j)(2t−1),12≤t≤1.

Then γ⋆β is a braid based at Q. It is the stacking of β below γ: the motion β runs during the first half of the height interval and the motion γ during the second half, after the strand of label π(β)(j) has been joined.

(b) The endpoint permutation multiplies. π(γ⋆β)=π(γ)∘π(β), with (π∘σ)(j)=π(σ(j)).

(c) The operation descends to isotopy classes. If β∼β′ and γ∼γ′ then γ⋆β∼γ′⋆β′.

(d) The operation is associative. If δ is a further braid based at Q, then (δ⋆γ)⋆β∼δ⋆(γ⋆β).

(e) The endpoint permutation is constant along isotopies. If β∼β′ then π(β)=π(β′); equivalently π is constant on each braid isotopy class, so that π is a function of the class [β] alone.

Consequently the stacking of isotopy classes, [γ] [β]:=[γ⋆β], is a well-defined associative binary operation on the set of braid isotopy classes based at Q, with π(δ⋆γ⋆β)=π(δ)π(γ)π(β) for any three braids.

Facts & Assumptions

Given: A natural number n, the base configuration Q=(q1,…,qn) with qj=((2j−n−1)h,0) and h=14(n+1), and braids β=(zj), β′, γ=(wj), γ′, δ based at Q.

[F1]

A braid based at Q is a tuple (u1,…,un) of continuous maps uj ⁣:I→D∘ with ui(t)≠uj(t) for i≠j, uj(0)=qj, and {u1(1),…,un(1)}={q1,…,qn}; its endpoint permutation π(u)∈Sn is the unique permutation with uj(1)=qπ(u)(j) for all j; the points q1,…,qn are pairwise distinct, lie in the interior of the closed unit disc, and are listed with strictly increasing first coordinates (Geometric braids in the disc with setwise endpoints).

[F2]

A braid isotopy from β to β′ is a tuple Z=(Z1,…,Zn) of jointly continuous maps Zj ⁣:I×I→D∘ such that each slice Z(s,⋅) is a braid based at Q and Zj(0,t)=zj(t), Zj(1,t)=zj′(t) for all j,t (Braid isotopy relative to the top and bottom endpoints).

[L3]

Composites of continuous maps are continuous, and a function whose domain is covered by finitely many closed sets on each of which it is continuous is continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous).

[L5]

The sets [0,12] and [12,1] are closed subsets of I carrying the subspace topology, their union is I, and likewise the two closed halves {t≤12} and {t≥12} of the square I×I are closed and cover it (Intervals of R: the nine order-convex forms, nondegeneracy, and length, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

Proof

technique · direct
1.1

(a) With (γ⋆β)j given by the two-branch formula of the statement, the two branches agree at t=12, because the first gives zj(1)=qπ(β)(j) and the second gives wπ(β)(j)(0)=qπ(β)(j) by [F1]; both branches are composites of continuous maps with the rescalings t↦2t and t↦2t−1, hence continuous, and the two closed halves of I cover I, so t↦(γ⋆β)j(t) is continuous by [L3] and [L5] and takes values in D∘; it is collision-free because on each half the tuple is the collision-free tuple of time-slices of a braid reparametrised by an injective continuous map and the halves meet only at t=12; the bottom values are (γ⋆β)j(0)=zj(0)=qj, and the top values are (γ⋆β)j(1)=wπ(β)(j)(1)=qπ(γ)(π(β)(j)), which run through the set {q1,…,qn}; hence γ⋆β is a braid based at Q.

F1L3L5
1.2

Braid isotopy is transitive: given an isotopy Z from β to β′ and an isotopy W from β′ to β′′, put Uj(s,t):=Zj(2s,t) for s≤12 and Uj(s,t):=Wj(2s−1,t) for s≥12; the branches agree at s=12 because both equal zj′(t), each is jointly continuous, and the two closed halves of the square cover it, so U is jointly continuous by [L3]; every slice of U is a slice of Z or of W, hence a braid based at Q, and the boundary slices are β and β′′, so U is an isotopy from β to β′′.

F2L3L5
1.3

(e) Let Z be an isotopy from β to β′ and fix j; if n=0 there is no endpoint label and the unique endpoint permutation is fixed; otherwise, for each k put Ak:={s∈I:Zj(s,1)=qk}. The sets Ak are pairwise disjoint and cover I, since every Zj(s,1) is one of the pairwise distinct points qk of [F1]; each Ak is closed in I, being the preimage under the continuous map s↦Zj(s,1) of the closed set {qk} by [L4]; each Ak is also open in I, because for s0∈Ak for n=1 one already has A1=I; for n≥2 the distance δ:=min⁡{∣ql−qk∣:l≠k} is positive by [F1] and continuity at s0 yields a neighbourhood V of s0 in I with ∣Zj(s,1)−qk∣<δ/2 for s∈V, which forces Zj(s,1)=qk and so s∈Ak. Hence the nonempty Ak are pairwise disjoint nonempty clopen subsets of the connected space I by [L4], and if some Ak were nonempty and proper, its open complement ⋃l≠kAl would separate I from it; so exactly one Ak is all of I. Thus Zj(s,1), and with it the endpoint permutation of the braid Z(s,⋅), is independent of s by [F2], and in particular π(β)=π(β′).

F1F2L4
2.1

(b) The top values computed in step 1.1 satisfy (γ⋆β)j(1)=q(π(γ)∘π(β))(j) for every j, so the bijection j↦(π(γ)∘π(β))(j) has the defining property of the endpoint permutation of γ⋆β in [F1]; that permutation is unique, whence π(γ⋆β)=π(γ)∘π(β).

F1step 1.1
2.2

Triple concatenations. Let σ:=π(β) and τ:=π(γ), and for 0<a<b<1 let Θa,b be the tuple that equals zj(t/a) for 0≤t≤a, equals wσ(j)(t−ab−a) for a≤t≤b, and equals vτσ(j)(t−b1−b) for b≤t≤1, where vj are the motions of δ; the three branches agree at the break points by [F1], each is continuous, and the three closed pieces of I cover I, so each Θja,b is continuous by [L3] and [L5]; collision-freeness and the endpoint conditions are checked exactly as in step 1.1, so Θa,b is a braid based at Q, and reading off the two bracketings gives (δ⋆γ)⋆β=Θ1/2, 3/4 and δ⋆(γ⋆β)=Θ1/4, 1/2, since in both the three motions occur in the order β,γ,δ with couplings σ then τ and only the two height breaks differ.

F1step 1.1L3L5
2.3

(c) Let Z be an isotopy from β to β′ and W an isotopy from γ to γ′, and put Uj(s,t):=Zj(s,2t) for t≤12 and Uj(s,t):=Wπ(β)(j)(s,2t−1) for t≥12; the two branches agree at t=12 because Zj(s,1)=qπ(β)(j)=Wπ(β)(j)(s,0) for every s by [F1] and [F2], and step 1.3 shows that the coupling π(β) is the endpoint permutation of every braid Z(s,⋅), so no s-dependent relabelling is needed; joint continuity follows from [L3] and [L5], each slice U(s,⋅) is the stacking of the braids Z(s,⋅) and W(s,⋅), hence a braid by step 1.1, and the boundary slices are γ⋆β and γ′⋆β′; hence γ⋆β∼γ′⋆β′.

F1F2step 1.1step 1.3L3L5
3.1

Moving the breaks is an isotopy. Let s↦(as,bs) be continuous with 0<as<bs<1, a0=12, b0=34, a1=14, b1=12, for instance the linear interpolation of the two pairs; then (s,t)↦Θjas,bs(t) is jointly continuous, because the three regions {t≤as}, {as≤t≤bs}, {t≥bs} are closed and cover the square and on each the formula is a composite of continuous maps with the positive denominators as, bs−as, 1−bs bounded below on the compact parameter interval; each slice is a braid by step 2.2 and the boundary slices are Θ1/2, 3/4 and Θ1/4, 1/2, so those two braids are braid-isotopic.

step 2.2F2L3L5
4.1

(d) Steps 2.2 and 3.1 exhibit the two bracketings (δ⋆γ)⋆β and δ⋆(γ⋆β) of the triple as braid-isotopic representatives, and step 2.3 shows that the isotopy class of a stacking depends only on the isotopy classes of its two factors; hence (δ⋆γ)⋆β∼δ⋆(γ⋆β) for the given braids, and the induced operation on classes is associative.

F2step 2.2step 2.3step 3.1
5.1

Assertions (a), (b), (c), (d) and (e) are steps 1.1, 2.1, 2.3, 4.1 and 1.3, so the stacking of classes [γ][β]=[γ⋆β] is a well-defined associative binary operation on the set of braid isotopy classes based at Q, and π(δ⋆γ⋆β)=π(δ)π(γ)π(β) follows by applying step 2.1 twice. ∎

step 1.1step 1.3step 2.1step 2.3step 4.1

Remarks

  • The reparametrisation used in step 3.1 is the only place where associativity needs work: the two bracketings of a threefold stack differ by how the height interval is cut, and a continuous family of cuts is an isotopy because the underlying sequence of motions, and all label couplings, are unchanged.
  • Clause (e) is what makes the endpoint permutation a function of the isotopy class rather than of the representative; it is the connectedness of the height interval, through [L4], that rules out a strand ending at a different base point at the two ends of an isotopy.
  • Nothing in the proposition uses a choice principle: the base configuration, and every reparametrisation, is given by an explicit formula, and the two-element cover of I in step 1.1 is finite.
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

The isotopy classes of geometric braids based at Q form a group, and the endpoint permutation is a homomorphism

Statement

Let n∈N and let Q=(q1,…,qn) be the base configuration of Geometric braids in the disc with setwise endpoints. Write

Gn:={[β]:β a braid based at Q}

for the set of braid isotopy classes relative to the top and bottom (Braid isotopy relative to the top and bottom endpoints), and let [γ][β]:=[γ⋆β] be the stacking of Stacking of geometric braids is a well-defined associative operation on isotopy classes. Then:

(a) Gn is a group (Group and abelian group) with this operation. Its identity is the class [e] of the trivial braid ej(t):=qj, and the inverse of [β], for β=(z1,…,zn) with endpoint permutation π(β), is the class of the reversed braid

β‾j(t):=zπ(β)−1(j)(1−t)(t∈I, 1≤j≤n).

(b) The endpoint permutation map

π ⁣:Gn⟶Sn,[β]⟼π(β),

is a well-defined group homomorphism (The finite symmetric group Sn, one-line notation, and cycle notation).

The construction is choice-free: all motions and reparametrisations used are given by explicit formulas.

Facts & Assumptions

Given: A natural number n, the base configuration Q=(q1,…,qn), braids β=(zj), γ, δ based at Q, the trivial braid e, and isotopy classes as above.

[F1]

A braid based at Q is a tuple (u1,…,un) of continuous maps uj ⁣:I→D∘ with ui(t)≠uj(t) for i≠j, uj(0)=qj, and {u1(1),…,un(1)}={q1,…,qn}; the endpoint permutation π(u)∈Sn is the unique permutation with uj(1)=qπ(u)(j); a braid is pure exactly when π(u)=id⁡, and the trivial braid ej(t)=qj is pure (Geometric braids in the disc with setwise endpoints, The finite symmetric group Sn, one-line notation, and cycle notation).

[F2]

Stacking [γ][β]=[γ⋆β] of Stacking of geometric braids is a well-defined associative operation on isotopy classes is well defined on isotopy classes and associative, π(γ⋆β)=π(γ)∘π(β), and the endpoint permutation is constant along braid isotopies: if β∼β′ then π(β)=π(β′).

[F3]

A braid isotopy from β to β′ is a tuple Z=(Z1,…,Zn) of jointly continuous maps Zj ⁣:I×I→D∘ such that every slice Z(s,⋅) is a braid based at Q and Zj(0,t)=zj(t), Zj(1,t)=zj′(t) for all j,t; isotopy implies homotopy of the strands relative to the endpoints of the motions, in the sense of Homotopies of continuous maps, homotopies relative to a subspace, and path homotopies relative to the endpoints (Braid isotopy relative to the top and bottom endpoints).

[L4]

A group is a set with an associative binary operation, a two-sided identity and two-sided inverses (Group and abelian group).

[L5]

Composites of continuous maps are continuous, continuity on a finite closed cover pastes, and the interval I=[0,1] carries the subspace topology in which [0,12] and [12,1] are closed and cover I (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, Intervals of R: the nine order-convex forms, nondegeneracy, and length).

Proof

technique · direct
1.1

(a), the identity. Write μ(t):=max⁡(0,2t−1) and ν(t):=min⁡(2t,1); since the right factor of a stacking runs during the first half of the height interval, (e⋆β)j(t)=zj(2t)=zj(ν(t)) for t≤12 and (e⋆β)j(t)=eπ(β)(j)(2t−1)=qπ(β)(j)=zj(1)=zj(ν(t)) for t≥12, while (β⋆e)j(t)=ej(2t)=qj=zj(0)=zj(μ(t)) for t≤12 and (β⋆e)j(t)=zπ(e)(j)(2t−1)=zj(2t−1)=zj(μ(t)) for t≥12; so e⋆β and β⋆e are the reparametrisations zj∘ν and zj∘μ of the tuple β. Both μ and ν are continuous nondecreasing maps of I onto I fixing 0 and 1 by [L5], so for s∈I the maps μs(t):=(1−s)t+sμ(t) and νs(t):=(1−s)t+sν(t) are again of that kind, and Zj(s,t):=zj(μs(t)), Wj(s,t):=zj(νs(t)) are jointly continuous by [L5]; each slice (zj(μs(t)))j is a braid based at Q, because μs(0)=0, μs(1)=1 and the collision-freeness and continuity conditions of [F1] are inherited from β, and likewise for νs; the boundary slices are β, e⋆β and β⋆e by μ0=ν0=id⁡, μ1=μ, ν1=ν. Hence e⋆β∼β and β⋆e∼β, so [e] is a two-sided identity for the operation of [F2].

F1F2F3L5
1.2

(a), the inverse is a braid. For β‾j(t):=zπ(β)−1(j)(1−t) each β‾j is a composite of continuous maps with values in D∘, and β‾i(t)≠β‾j(t) for i≠j because π(β)−1 is injective and the zk are collision-free by [F1]; its bottom values are β‾j(0)=zπ(β)−1(j)(1)=qπ(β)(π(β)−1(j))=qj, using the defining property of π(β) in [F1], and its top values are β‾j(1)=zπ(β)−1(j)(0)=qπ(β)−1(j), which run through the set {q1,…,qn}; hence β‾ is a braid based at Q with π(β‾)=π(β)−1.

F1
1.3

(b). The map π is well defined on classes [β] by the constancy of the endpoint permutation along isotopies in [F2]; it satisfies π(γ⋆β)=π(γ)∘π(β) by [F2], and π(e)=id⁡ because e is pure by [F1]; a map of groups that preserves the operation and the identity is a group homomorphism into the symmetric group Sn of The finite symmetric group Sn, one-line notation, and cycle notation, so the formula π ⁣:Gn→Sn, [β]↦π(β), defines a group homomorphism once Gn is known to be a group.

F1F2
2.1

(a), β‾⋆β∼e. By the stacking formula of [F2] and step 1.2, (β‾⋆β)j(t)=zj(2t) for t≤12 and (β‾⋆β)j(t)=β‾π(β)(j)(2t−1)=zπ(β)−1(π(β)(j))(2−2t)=zj(2−2t) for t≥12; that is, β‾⋆β is the out-and-back reparametrisation zj∘λ of β with λ(t):=2t for t≤12 and λ(t):=2−2t for t≥12. For s∈I put λs(t):=(1−s)λ(t); then λs is continuous with λs(0)=λs(1)=0, so Zj(s,t):=zj(λs(t)) is jointly continuous by [L5], each slice (zj(λs(t)))j is a braid based at Q because it is a reparametrisation of the collision-free tuple β with all bottom and top values equal to qj, and the boundary slices are β‾⋆β at s=0 and e at s=1; hence β‾⋆β∼e.

F1F2F3step 1.2L5
3.1

(a), β⋆β‾∼e. By the same computation with the roles of the two factors exchanged, (β⋆β‾)j(t)=zπ(β)−1(j)(1−2t) for t≤12 and (β⋆β‾)j(t)=zπ(β)−1(j)(2t−1) for t≥12, which is again an out-and-back parametrisation of β with the labels relabelled by π(β)−1; putting κ(t):=1−2t for t≤12 and κ(t):=2t−1 for t≥12, and then κs(t):=(1−s)κ(t)+s, gives a braid isotopy: κs(0)=κs(1)=1, so every slice begins and ends at zπ(β)−1(j)(1)=qj and is collision-free; at s=1 the slice is the constant braid. Thus this is a braid isotopy from β⋆β‾ to e.

F1F2step 1.2step 2.1
4.1

(a), conclusion. By steps 1.1, 2.1 and 3.1 the operation of [F2] on the isotopy classes based at Q is associative, has the two-sided identity class [e], and gives [β‾][β]=[e]=[β][β‾] for every [β]; by [L4] the set Gn of isotopy classes is therefore a group with identity [e] and [β]−1=[β‾].

L4F2step 1.1step 2.1step 3.1
5.1

Assertions (a) and (b) are steps 4.1 and 1.3, the latter now applicable because step 4.1 makes Gn a group; the group structure uses only the explicit stacking, reversal and reparametrisation formulas displayed above. ∎

step 1.3step 4.1

Remarks

  • The inverse is built from time reversal together with the relabelling π(β)−1 at the top; the relabelling is necessary because braid isotopy fixes the bottom points but only the top set, so a naive time reversal of the tuple would not return the bottom labels.
  • For n=0 the set G0 has exactly one element and S0 is trivial, so both assertions are immediate; for n=1 the group G1 consists of the isotopy classes of loops in the disc D∘ based at q1=(0,0). This page does not determine G1, and no claim about it is used later.
  • No choice principle is used: the identity isotopies are the explicit reparametrisations μs,νs,λs, and the inverse is the explicit formula β‾j(t)=zπ(β)−1(j)(1−t).
DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-sol)audited 2026-09-27Open item page →

The elementary geometric half twist, its support disc, and its opposite

Definition

Let n∈N and let Q=(q1,…,qn) be the base configuration of Geometric braids in the disc with setwise endpoints, with h=14(n+1) and qj=((2j−n−1)h,0). Fix an index i with 1≤i≤n−1. The two adjacent base points are

qi=((2i−n−1)h,0),qi+1=((2i+1−n)h,0),

their midpoint is mi:=(qi+qi+1)/2=((2i−n)h,0)=qi+(h,0), and their support disc is the open disc

Ui:={w∈R2:∥w−mi∥2<32h}.

The support disc contains the two base points qi,qi+1, whose distance from mi is h, and it contains no other base point: for k<i one has ∥qk−mi∥2=(2(i−k)+1)h≥3h and for k>i+1 one has ∥qk−mi∥2=(2(k−i)−1)h≥3h, both exceeding 32h. Moreover Ui⊆D∘, because every point of Ui has norm at most ∥mi∥2+32h≤(n−2)h+32h<(n+1)h<1. Finally Ui∩Uj=∅ whenever ∣i−j∣>1, since then the midpoints are at distance 2∣i−j∣h≥4h>3h, twice the radius.

The positive half twist. Define ρ ⁣:I→R2, the anticlockwise diamond path, by

ρ(t):={(2th−h, −2th),0≤t≤12,(2th−h, 2th−2h),12≤t≤1.

Thus ρ(0)=(−h,0), ρ(12)=(0,−h), ρ(1)=(h,0), and ρ is continuous, piecewise linear, and satisfies ∥ρ(t)∥2≤h for all t with ρ(t)≠0 throughout. The elementary half twist at i, written σi, is the tuple of motions

(σi)i(t):=mi+ρ(t),(σi)i+1(t):=mi−ρ(t),(σi)k(t):=qk  (k∉{i,i+1}).

It is a braid based at Q (Geometric braids in the disc with setwise endpoints, Continuity of a map of topological spaces at a point and globally): the two moving points stay in Ui and are antipodal about mi, so they are distinct; every other point is fixed and lies outside Ui; and the endpoint permutation is the transposition of i and i+1, since (σi)i(1)=mi+(h,0)=qi+1 and (σi)i+1(1)=mi−(h,0)=qi.

The words positive and anticlockwise refer to the fixed orientation of the plane of the disc and to the fixed planar projection whose horizontal axis contains Q: the moving pair turns through the half turn anticlockwise, the label i passing below its midpoint and the label i+1 above. Together with the first-under-second stacking convention of Stacking of geometric braids is a well-defined associative operation on isotopy classes this fixes the sign convention for every signed crossing on this page and its companion.

The opposite half twist. The opposite (clockwise) half twist at i is defined by the reflected path

ρ−(t):=(ρ1(t),−ρ2(t)),

through the motions (σi−)i(t):=mi+ρ−(t), (σi−)i+1(t):=mi−ρ−(t) and (σi−)k(t):=qk for k∉{i,i+1}. It is again a braid supported in Ui with endpoint permutation the transposition of i and i+1; its pair turns through the same half turn clockwise, the label i passing above the midpoint. The two motions are not equal, and they are not related by a reparametrisation of the height: they are the two signed crossings of the pair.

The opposite motion is the group inverse. For a braid β the reversed braid β‾j(t)=zπ(β)−1(j)(1−t) represents [β]−1 in the group Gn (The isotopy classes of geometric braids based at Q form a group, and the endpoint permutation is a homomorphism). For β=σi one computes σi‾=σi−: reversing the height and relabelling by the transposition of i and i+1 sends ρ(t)=mi+(ρ1,ρ2) to the motion with relative path −ρ(1−t), which is the reflected path ρ− traversed from t=0 to t=1. Hence

[σi−]=[σi]−1in Gn.

Elementary cases. For n≤1 there is no index i with 1≤i≤n−1, and there are no elementary half twists; the assertions above are vacuous. For n≥2 and 1≤i≤n−1 both [σi] and [σi−] are nonidentity elements of Gn, since their endpoint permutations are transpositions, which are not the identity permutation.

Two letters do not commute in general. For adjacent indices the supports Ui and Ui+1 overlap and no commutation is asserted; Far commutativity of elementary geometric half twists proves commutativity only for disjoint supports.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Far commutativity of elementary geometric half twists

Statement

Let n∈N and let Q=(q1,…,qn) be the base configuration of Geometric braids in the disc with setwise endpoints, with h=14(n+1). Let i,j be indices with 1≤i,j≤n−1 and ∣i−j∣>1; then the two pairs {i,i+1} and {j,j+1} are disjoint, and such indices exist only for n≥4, so for n≤3 the assertions below are vacuous. Write mi=qi+(h,0) and let ρ be the diamond path, so that the half twists σi, σj and their supports Ui, Uj are as in The elementary geometric half twist, its support disc, and its opposite, with Ui∩Uj=∅.

(a) The simultaneous braid. Define Σij, the simultaneous execution of the two half twists, by

(Σij)i:=mi+ρ,(Σij)i+1:=mi−ρ,(Σij)j:=mj+ρ,(Σij)j+1:=mj−ρ,

and (Σij)k(t):=qk for the remaining labels k. Then Σij is a braid based at Q, and both stackings of the two half twists are braid-isotopic to it. Here ∼ denotes braid isotopy (Braid isotopy relative to the top and bottom endpoints) and ⋆ the stacking of Stacking of geometric braids is a well-defined associative operation on isotopy classes:

σi⋆σj ∼ Σij ∼ σj⋆σi.

(b) Far commutativity. Consequently [σi][σj]=[σj][σi] in the group Gn of The isotopy classes of geometric braids based at Q form a group, and the endpoint permutation is a homomorphism: far-away half twists commute.

The isotopy is explicit and no choice principle is used.

Facts & Assumptions

Given: A natural number n, the base configuration Q=(q1,…,qn), indices i,j with 1≤i,j≤n−1 and ∣i−j∣>1, and the half twists σi,σj based at Q.

[F1]

A braid based at Q is a tuple (u1,…,un) of continuous maps uj ⁣:I→D∘ with ui(t)≠uj(t) for i≠j, uj(0)=qj, and {u1(1),…,un(1)}={q1,…,qn}; the endpoint permutation π(u) is the unique permutation with uj(1)=qπ(u)(j) (Geometric braids in the disc with setwise endpoints).

[F2]

The half twist at i is (σi)i=mi+ρ, (σi)i+1=mi−ρ and (σi)k=qk for k∉{i,i+1}, where mi=qi+(h,0) and ρ(0)=(−h,0), ρ(12)=(0,−h), ρ(1)=(h,0) with ∥ρ(t)∥2≤h and ρ(t)≠0 for every t; σi is a braid based at Q with endpoint permutation the transposition of i and i+1; its support disc Ui contains qi,qi+1 and no other base point, satisfies Ui⊆D∘, and Ui∩Uj=∅ whenever ∣i−j∣>1 (The elementary geometric half twist, its support disc, and its opposite).

[F3]

Stacking is (γ⋆β)j(t)=zj(2t) for t≤12 and (γ⋆β)j(t)=wπ(β)(j)(2t−1) for t≥12, with π(γ⋆β)=π(γ)∘π(β); it descends to isotopy classes and is associative, and braid isotopy is the equivalence relation generated by the jointly continuous families of Braid isotopy relative to the top and bottom endpoints (Stacking of geometric braids is a well-defined associative operation on isotopy classes).

[F4]

A braid isotopy from β to β′ is a tuple Z=(Z1,…,Zn) of jointly continuous maps Zj ⁣:I×I→D∘ such that every slice Z(s,⋅) is a braid based at Q and Zj(0,t)=zj(t), Zj(1,t)=zj′(t) for all j,t (Braid isotopy relative to the top and bottom endpoints).

[F5]

Gn is a group whose operation is induced by stacking, so [γ][β]=[γ⋆β], and [β]=[β′] whenever β∼β′ (The isotopy classes of geometric braids based at Q form a group, and the endpoint permutation is a homomorphism, Stacking of geometric braids is a well-defined associative operation on isotopy classes).

[L6]

Composites of continuous maps are continuous and continuity on the two closed halves of a square pastes, the interval I=[0,1] and its two closed halves carrying the subspace topology (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, Continuity of a map of topological spaces at a point and globally, Intervals of R: the nine order-convex forms, nondegeneracy, and length).

Proof

technique · direct
1.1

(a), Σij is a braid. Each motion of Σij is either the constant qk or one of mi±ρ, mj±ρ, hence continuous with values in Ui respectively Uj, and Ui∪Uj⊆D∘ by [F2]; the only nonconstant pairs are {i,i+1}, which are antipodal about mi and therefore distinct because ρ(t)≠0 for every t by [F2], and {j,j+1}, likewise antipodal about mj; the two supports are disjoint and the remaining motions are the constant qk with qk∉Ui∪Uj, so all n motions are pairwise distinct at every height; the bottom values are mi+ρ(0)=mi+(−h,0)=qi, mi−ρ(0)=qi+1 and likewise for j, the remaining values being the base points themselves; the top values are mi+ρ(1)=qi+1, mi−ρ(1)=qi and likewise for j, so the top values run through {q1,…,qn} and the endpoint permutation of Σij is the product of the transpositions of i,i+1 and of j,j+1.

F1F2
1.2

The interpolation of the two time windows. For s∈I put αs(t):=0 for t≤1−s2 and αs(t):=t−(1−s)/2(1+s)/2 for t≥1−s2, and βs(t):=t(1+s)/2 for t≤1+s2, βs(t):=1 for t≥1+s2; the two branches of each definition agree at the switch point, so αs,βs are continuous, nondecreasing and map I onto I with αs(0)=βs(0)=0 and αs(1)=βs(1)=1, the map (s,t)↦(αs(t),βs(t)) is jointly continuous because the switch points depend continuously on s and the two branches agree there, and α0(t)=max⁡(0,2t−1), β0(t)=min⁡(2t,1), α1=β1=id⁡I.

F2L6
1.3

Pasting two isotopies. If Z is a braid isotopy from β to β′ and W one from β′ to β′′, then Uj(s,t):=Zj(2s,t) for s≤12 and Uj(s,t):=Wj(2s−1,t) for s≥12 is jointly continuous by [L6] because the branches agree at s=12 and the two closed halves of the square cover it, every slice of U is a slice of Z or of W and hence a braid based at Q by [F4], and its boundary slices are β and β′′; so braid isotopy is transitive.

F4L6
2.1

The isotopy. Define Zk(s,t):=mi+ρ(αs(t)) for k=i, Zk(s,t):=mi−ρ(αs(t)) for k=i+1, Zk(s,t):=mj+ρ(βs(t)) for k=j, Zk(s,t):=mj−ρ(βs(t)) for k=j+1, and Zk(s,t):=qk otherwise. Every Zk is a composite of jointly continuous maps by step 1.2 and [F2], hence jointly continuous; for fixed s the tuple Z(s,⋅) is collision-free and takes the base values at t=0 and the setwise base values at t=1 by the same computations as in step 1.1, with αs,βs in place of the identity: at t=0 one has αs(0)=βs(0)=0 so the four moving labels sit at qi,qi+1,qj,qj+1, while at t=1 one has αs(1)=βs(1)=1 so they sit at the same four points with the two neighbouring labels interchanged.

F1F2step 1.1step 1.2
3.1

The two ends of the isotopy. At s=0 the formulas of step 1.2 give Zk(0,t)=mi+ρ(max⁡(0,2t−1)) for k=i and Zk(0,t)=mi−ρ(max⁡(0,2t−1)) for k=i+1, which is the motion of σi run during the second half of the height interval and held at qi respectively qi+1 during the first half, while Zk(0,t)=mj±ρ(min⁡(2t,1)) for the labels k=j,j+1 runs σj during the first half and holds it at the swapped base points during the second half, and all other labels are constant; comparing with the stacking formula (γ⋆β)k(t)=zk(2t) for t≤12, =wπ(β)(k)(2t−1) for t≥12 of [F3], with β=σj, γ=σi and π(σj) the transposition of j,j+1, shows Z(0,⋅)=σi⋆σj. At s=1 one has α1=β1=id⁡I by step 1.2 and step 2.1, so Z(1,⋅)=Σij by the definitions of step 1.1.

F2F3step 1.1step 1.2step 2.1
4.1

(a), first stacking. Step 2.1 exhibits Z as a tuple of jointly continuous maps whose every slice is a braid based at Q, and step 3.1 identifies its boundary slices as σi⋆σj and Σij; hence Z is a braid isotopy from σi⋆σj to Σij in the sense of [F4], that is σi⋆σj∼Σij.

F4step 2.1step 3.1
4.2

(a), second stacking. Interchanging the roles of the two pairs, that is replacing (αs,i,Ui) by (βs,j,Uj) and conversely throughout steps 1.2, 2.1 and 3.1, yields in the same way a braid isotopy whose first boundary slice is σj⋆σi, the pair {i,i+1} now executing its half twist during the second half of the height interval and the pair {j,j+1} during the first, and whose second boundary slice is again Σij; hence σj⋆σi∼Σij.

F1F2F3F4step 1.1step 1.2step 2.1step 3.1
5.1

Steps 4.1 and 4.2 give σi⋆σj∼Σij∼σj⋆σi, so transitivity of braid isotopy in step 1.3 yields σi⋆σj∼σj⋆σi; passing to isotopy classes with [F5] gives [σi][σj]=[σi⋆σj]=[σj⋆σi]=[σj][σi], which is (b), while (a) is steps 1.1, 4.1 and 4.2. ∎

F5step 4.1step 4.2step 1.3step 1.1

Remarks

  • The only geometric input is that the two supports are disjoint: the pairs of moving labels are distinct, so the two half turns never see each other, and the two time windows can be slid past one another.
  • The isotopy of step 2.1 is not a reparametrisation of the height in the sense of items 1 to 4: the two pairs are reparametrised by different functions αs and βs, and this is legitimate because each pair is unaffected by the other.
  • Assertion (b) is a statement in the group Gn of isotopy classes, not an equality of the braids σi⋆σj and σj⋆σi themselves; the two stackings are distinct parametrised tuples whenever n≥4, since their height windows differ.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

The geometric three strand braid relation

Statement

Let n≥3 and let i be an index with 1≤i≤n−2. Let σi and σi+1 be the elementary half twists of The elementary geometric half twist, its support disc, and its opposite based at Q, and let ⋆ be the stacking of Stacking of geometric braids is a well-defined associative operation on isotopy classes. Then

σi⋆σi+1⋆σi ∼ σi+1⋆σi⋆σi+1,

and consequently, in the group Gn of The isotopy classes of geometric braids based at Q form a group, and the endpoint permutation is a homomorphism,

[σi] [σi+1] [σi]=[σi+1] [σi] [σi+1].

The proof exhibits an explicit intermediate braid: writing κ for the point reflection w↦2qi+1−w and rot for the rotation of the three points qi,qi+1,qi+2 about qi+1 by the angle πu at height u, with the remaining strands fixed, the bracketing σi⋆(σi+1⋆σi) is braid-isotopic to rot, which is a braid based at Q with endpoint permutation the transposition of i and i+2, and the reflection κ, followed by the relabelling of i and i+2, turns that bracketing into σi+1⋆(σi⋆σi+1), so that bracketing is braid-isotopic to rot as well; the two bracketings of each word are themselves braid-isotopic by the associativity of stacking. All constructions are explicit and no choice principle is used.

Facts & Assumptions

Given: A natural number n≥3, an index i with 1≤i≤n−2, the base configuration Q=(q1,…,qn) with qj+1−qj=(2h,0) and h=14(n+1), and the half twists σi,σi+1 based at Q.

[F1]

A braid based at Q is a tuple (u1,…,un) of continuous maps uj ⁣:I→D∘ with uk(t)≠ul(t) for k≠l, uk(0)=qk, and {u1(1),…,un(1)}={q1,…,qn}; its endpoint permutation π(u) is the unique permutation with uk(1)=qπ(u)(k); the points qi,qi+1,qi+2 are collinear and equally spaced, so in the coordinates centred at qi+1 they are (−2h,0),(0,0),(2h,0), and ∣qj−qi+1∣≥4h for every j∉{i,i+1,i+2} (Geometric braids in the disc with setwise endpoints).

[F2]

The half twist at k is (σk)k=mk+ρ, (σk)k+1=mk−ρ and (σk)j=qj otherwise, where mk=qk+(h,0); the diamond path ρ satisfies ρ(0)=(−h,0), ρ(12)=(0,−h), ρ(1)=(h,0), ∥ρ(v)∥2≤h, and ρ2(v)≤0 with ρ2(v)=0 only for v∈{0,1} and ρ(v)≠0 for every v∈I; and π(σk) is the transposition of k and k+1, while the support disc contains qk,qk+1 and no other base point (The elementary geometric half twist, its support disc, and its opposite).

[F3]

A braid isotopy is a tuple of jointly continuous maps Zj ⁣:I×I→D∘ whose every slice is a braid based at Q and whose boundary slices are the two given braids (Braid isotopy relative to the top and bottom endpoints).

[F4]

Stacking places its right factor in the lower half of the height interval and its left factor in the upper half: writing zj for the strands of the right factor β and wj for those of the left factor γ, one has (γ⋆β)j(t)=zj(2t) for t≤12 and (γ⋆β)j(t)=wπ(β)(j)(2t−1) for t≥12; stacking of braids is a braid, it descends to isotopy classes, π(γ⋆β)=π(γ)∘π(β), and the two bracketings of a threefold stacking are braid-isotopic, (δ⋆γ)⋆β∼δ⋆(γ⋆β) (Stacking of geometric braids is a well-defined associative operation on isotopy classes).

[F5]

Gn is a group with operation [γ][β]=[γ⋆β], so equal isotopy classes have equal products (The isotopy classes of geometric braids based at Q form a group, and the endpoint permutation is a homomorphism).

[L6]

Composites of continuous maps are continuous and continuity pastes over the two closed halves of a square; the interval I=[0,1] carries the subspace topology in which the points 0,12,1 cut it into closed pieces (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, Continuity of a map of topological spaces at a point and globally, Intervals of R: the nine order-convex forms, nondegeneracy, and length).

Proof

technique · direct
1.1

(The two bracketings, in local coordinates.) Put c:=qi+1, p1:=qi, p2:=qi+1, p3:=qi+2, m:=mi=qi+(h,0) and m′:=mi+1=qi+1+(h,0), so that the centred vectors p1−c=(−2h,0), p2−c=(0,0), p3−c=(2h,0), while m−c=(−h,0) and m′−c=(h,0); since π(σi) is the transposition of i,i+1 and π(σi+1) that of i+1,i+2 by [F2] and π multiplies by [F4], the composite π(σi⋆σi+1)=π(σi)∘π(σi+1) is the 3-cycle i↦i+1↦i+2↦i and π(σi+1⋆σi) is its inverse i↦i+2↦i+1↦i, so applying the stacking formula of [F4] twice shows that the strands i,i+1,i+2 of W0:=σi⋆(σi+1⋆σi) are (m+ρ(4u), m−ρ(4u), p3) for 0≤u≤14, (m′+ρ(4u−1), p1, m′−ρ(4u−1)) for 14≤u≤12 and (p3, m+ρ(2u−1), m−ρ(2u−1)) for 12≤u≤1, while the same computation with the roles of i and i+1 interchanged shows that the strands i,i+1,i+2 of W1:=σi+1⋆(σi⋆σi+1) are (p1, m′+ρ(4u), m′−ρ(4u)) for 0≤u≤14, (m+ρ(4u−1), p3, m−ρ(4u−1)) for 14≤u≤12 and (m′+ρ(2u−1), m′−ρ(2u−1), p1) for 12≤u≤1, every remaining strand of either tuple being constantly at its base point; hence W0 and W1 are braids based at Q by [F4], with π(W0)=(i i+1)∘(i+1 i+2)∘(i i+1) and π(W1)=(i+1 i+2)∘(i i+1)∘(i+1 i+2), both the transposition of i and i+2, and by the associativity clause of [F4] W0 is braid-isotopic to σi⋆σi+1⋆σi and W1 to σi+1⋆σi⋆σi+1.

F1F2F4L6
1.2

(The rotation braid.) Let Rθ be the linear rotation of R2 about the origin through angle θ, and define vk:=qk−c and rotk(u):=c+Rπuvk for k∈{i,i+1,i+2}; put rotk(u):=qk for all other labels. The motions are continuous, the two outer centred vectors are antipodal because vi+2=−vi, and vi+1=0, so the middle strand stays at c and all three remain pairwise distinct. Their distance from the origin is at most ∥c∥2+2h≤(n+1)h<1; every other base point is at distance at least 4h from c by [F1], so no moving strand meets a constant one. At u=0 the triple has values c+vk=qk, and at u=1 it has values c−vk=q2i+2−k; thus rot is based at Q and has endpoint permutation (i i+2), shared by W0 and W1.

F1F2L6
2.1

(The interpolation family.) For s∈I define Zk(s,u):=(1−s) (W0)k(u)+s rotk(u) for every label k, where (W0)k and rotk are the motions of step 1.1 and step 1.2; each Zk is jointly continuous, being a sum of products of continuous functions, and satisfies ∥Zk(s,u)∥2≤(1−s)∥(W0)k(u)∥2+s∥rotk(u)∥2<1, so it maps I×I into D∘; at u=0 one has Zk(s,0)=qk for every s because (W0)k(0)=rotk(0)=qk, and at u=1 one has Zk(s,1)=qπ(k) for every s because (W0)k(1)=rotk(1)=qπ(k) for the transposition π of i and i+2 shared by both braids; so each slice Z(s,⋅) satisfies the endpoint conditions of [F1].

F1F3step 1.1step 1.2L6
2.2

(The collision criterion.) Fix s∈(0,1) and u∈I and let k<l be two labels of the triple {i,i+1,i+2}; since rotk(u)−rotl(u)=Rπu(qk−ql) by step 1.2 and qk−ql=−2hλkl(1,0) with λkl:=l−k>0 by [F1], the equality Zk(s,u)=Zl(s,u) is equivalent to (1−s) ((W0)k(u)−(W0)l(u))=2hs λkl (cos⁡πu,sin⁡πu), hence to the statement that (W0)k(u)−(W0)l(u) is a positive multiple of (cos⁡πu,sin⁡πu); so a slice Z(s,⋅) with s∈(0,1) has no collision between two labels of the triple as soon as no difference (W0)k−(W0)l with k<l in the triple is a positive multiple of (cos⁡πu,sin⁡πu) at any height u.

F1F2step 1.1step 1.2
2.3

(Phase one: 0≤u≤14, v=4u.) In this and the next two phase calculations, a subscript a∈{1,2,3} on (W0)a denotes the local position in the active triple, namely the global strand (W0)i+a−1; all other strands remain fixed. Here (W0)1−(W0)2=2ρ(v) has second coordinate ≤0, while (W0)1−(W0)3=(−3h,0)+ρ(v) and (W0)2−(W0)3=(−3h,0)−ρ(v) have first coordinate at most −2h<0; since cos⁡(πu)≥cos⁡(π4)>0 and sin⁡(πu)≥0 on this phase, a positive multiple of (cos⁡πu,sin⁡πu) has positive first coordinate, which excludes the pairs (1,3) and (2,3), while for the pair (1,2) the second coordinates would have to agree, forcing 2ρ2(v)=λsin⁡(πu)≥0 and hence ρ2(v)=0, that is v∈{0,1}; at v=0 one has (W0)1−(W0)2=2ρ(0)=(−2h,0), a negative multiple of (cos⁡0,sin⁡0)=(1,0), and at v=1 one has (W0)1−(W0)2=2ρ(1)=(2h,0), which is not a multiple of (cos⁡π4,sin⁡π4) at all.

F2step 1.1
2.4

(Phase two: 14≤u≤12, v=4u−1.) Here (W0)1−(W0)2=(2h+2hv,−2hv) respectively (2h+2hv,2hv−2h) according to whether v≤12 or v≥12, (W0)1−(W0)3=2ρ(v) and (W0)2−(W0)3=(2hv−4h,−2hv) respectively (2hv−4h,2hv−2h); all three have second coordinate at most 0, while sin⁡(πu)≥sin⁡(π4)>0 on this phase, so no one of them is a positive multiple of (cos⁡πu,sin⁡πu), whose second coordinate is positive.

F2step 1.1
2.5

(Phase three: 12≤u≤1, v=2u−1.) Here (W0)1−(W0)2=(3h,0)−ρ(v) and (W0)1−(W0)3=(3h,0)+ρ(v) have first coordinate at least 2h>0, while cos⁡(πu)≤0 on this phase, so neither is a positive multiple of (cos⁡πu,sin⁡πu); the remaining difference (W0)2−(W0)3=2ρ(v) has second coordinate at most 0, and a positive multiple of (cos⁡πu,sin⁡πu) has second coordinate ≥0 because sin⁡(πu)≥0 on this phase, so a coincidence would force the common second coordinate to vanish, that is ρ2(v)=0 and sin⁡(πu)=0, which gives v∈{0,1} and u∈{12,1}; with v=2u−1 this leaves the two candidates (u,v)=(12,0), where (W0)2−(W0)3=(−2h,0) is not a multiple of (cos⁡π2,sin⁡π2)=(0,1), and (u,v)=(1,1), where (W0)2−(W0)3=(2h,0) is a negative multiple of (cos⁡π,sin⁡π)=(−1,0) rather than a positive one.

F2step 1.1
3.1

(The isotopy from W0 to rot.) By steps 2.2, 2.3, 2.4 and 2.5 no two labels of the triple can meet in any slice Z(s,⋅) with s∈(0,1) and any height u, and at s=0 the slice is W0 and at s=1 it is rot, which are braids with pairwise distinct strands by steps 1.1 and 1.2; the triple values (W0)k(u) and rotk(u) all lie within distance 2h of c by steps 1.1 and 1.2 and [F2], so ∥Zk(s,u)−c∥2≤(1−s)∥(W0)k(u)−c∥2+s∥rotk(u)−c∥2≤2h puts every triple strand of every slice in the ball of radius 2h about c, while the remaining strands are constantly at base points of distance at least 4h from c by [F1], so no triple strand ever meets one of them; with the endpoint computations of step 2.1 this makes Z a braid isotopy from W0 to rot in the sense of [F3].

F1F2F3step 1.1step 1.2step 2.1step 2.2step 2.3step 2.4step 2.5
4.1

(The reflection.) Let κ(w):=2c−w be the point reflection in c and let τ be the transposition of i and i+2; for k∈{i,i+1,i+2} put Zk′(s,u):=κ(Zτ(k)(s,u)) and Zk′(s,u):=qk for the remaining labels; each Zk′ is jointly continuous and, by step 3.1 and ∥κ(w)−c∥2=∥w−c∥2, every value lies within distance 2h of c and hence in D∘, because ∥c∥2+2h≤(n+1)h<1 as in step 1.2; within the triple κ is injective, so distinct reflected strands stay distinct, and they stay within distance 2h of c and hence at distance at least 2h from the constant strands, which are at distance at least 4h from c by [F1]; so every slice Z′(s,⋅) is a braid based at Q; at s=1 one has Z′=rot, because κ(rotτ(k)(u))=c−Rπuvτ(k)=c+Rπuvk=rotk(u) for the three points p1,p2,p3, whose labelling by τ only exchanges the two outer centred vectors and satisfies vτ(k)=−vk; at s=0 the reflected slice is W1, because the reflection identities κ(m+ρ(v))=m′−ρ(v), κ(m−ρ(v))=m′+ρ(v), κ(p1)=p3 and κ(p3)=p1 turn the window values of W0 of step 1.1 into those of W1: on [0,14] the reflected values are (κ(p3),κ(m−ρ(4u)),κ(m+ρ(4u)))=(p1,m′+ρ(4u),m′−ρ(4u)), on [14,12] they are (κ(m′−ρ(4u−1)),κ(p1),κ(m′+ρ(4u−1)))=(m+ρ(4u−1),p3,m−ρ(4u−1)) and on [12,1] they are (κ(m−ρ(2u−1)),κ(m+ρ(2u−1)),κ(p3))=(m′+ρ(2u−1),m′−ρ(2u−1),p1), which are exactly the window values of W1; hence Z′ is a braid isotopy from W1 to rot.

F1F2F3step 1.1step 1.2step 3.1L6
5.1

(Conclusion.) Steps 3.1 and 4.1 show that W0 and W1 are both braid-isotopic to rot, hence braid-isotopic to each other; step 1.1 identifies W0 as the bracketing σi⋆(σi+1⋆σi) and W1 as the bracketing σi+1⋆(σi⋆σi+1), and the associativity clause of [F4] shows that each is braid-isotopic to the corresponding word with the other bracketing, so that σi⋆σi+1⋆σi∼σi+1⋆σi⋆σi+1; passing to isotopy classes with [F5] gives [σi][σi+1][σi]=[σi+1][σi][σi+1] in Gn. ∎

F4F5step 1.1step 3.1step 4.1

Remarks

  • The proof is local: only the three strands i,i+1,i+2 move, and all formulas are those of the three-strand picture with base points (−2h,0),(0,0),(2h,0) spaced 2h apart, which is why the lemma holds for every n≥3 and every i with 1≤i≤n−2.
  • The moving strands stay within distance 2h of qi+1 throughout the interpolation and the reflection, while every other base point is at distance at least 4h from qi+1; this clearance is what makes the local computation an isotopy of n-strand braids.
  • The braid relation is the geometric statement that three consecutive half twists of a triple can be deformed into the same three half twists performed by rotating the triple rigidly by π about its middle point; the point reflection in that middle point exchanges the two outer strands and turns one word into the other.
  • The bracketing enters the formulas but not the conclusion: the proof computes the bracketings σi⋆(σi+1⋆σi) and σi+1⋆(σi⋆σi+1), and the associativity clause of Stacking of geometric braids is a well-defined associative operation on isotopy classes supplies the isotopy to the bracketings (σi⋆σi+1)⋆σi and (σi+1⋆σi)⋆σi+1.
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Geometric braids admit generic polygonal representatives

Statement

Let n∈N and let β=(z1,…,zn) be a braid based at Q (Geometric braids in the disc with setwise endpoints). Then there is a braid β′=(β1′,…,βn′) based at Q with β′∼β (Braid isotopy relative to the top and bottom endpoints) such that:

  1. Polygonal. There are m≥1 and 0=t0<t1<⋯<tm=1 such that every βj′ is affine on each of the closed intervals [tk−1,tk]; for n=0 the tuple is empty and m:=1 is understood.
  2. General position. Writing ξj:=π1∘βj′ for the first coordinate of the j-th strand, no two strands meet in the projection at a breakpoint, ξi(tk)≠ξj(tk) for all i≠j and 1≤k≤m−1, and every interior coincidence of first coordinates is a simple coincidence of exactly one pair: for all i≠j and t∈(0,1) with ξi(t)=ξj(t), the height t lies in the interior of one of the affine pieces, the difference ξi−ξj changes sign at t, and ξl(t)≠ξi(t) for every l∉{i,j}.
  3. Consequences. The set C:={t∈(0,1):ξi(t)=ξj(t) for some i≠j} is finite, say C={c1<⋯<cm′} with m′≥0 (the empty list for m′=0), and at each cr exactly one pair of strands has equal first coordinates, that pair exchanging its two positions across cr.
  4. Boundary clearance. There is a real b′>0 with ∥βj′(t)∥2≤1−b′ for every j and every t∈I; that is, the representative stays at a uniform positive distance b′ from the boundary circle ∂D. For n=0 the condition is vacuous.

Thus a braid can be replaced by a polygonal one whose projected crossings are transversal, occur two at a time, and occur at pairwise distinct interior heights, and which keeps a uniform positive distance from the boundary circle. The construction is explicit and only finitely many choices are made; no choice principle is used.

Facts & Assumptions

Given: A natural number n and a braid β=(z1,…,zn) based at Q, with base configuration Q=(q1,…,qn), h=1/(4(n+1)) and qj=((2j−n−1)h,0).

[F1]

A braid based at Q is a tuple of continuous maps zj ⁣:I→D∘ with zi(t)≠zj(t) for i≠j, zj(0)=qj and {z1(1),…,zn(1)}={q1,…,qn}; the base points are pairwise distinct, lie in D∘ with ∥qj∥2≤(n−1)h<1, satisfy qj+1−qj=(2h,0) and have pairwise distinct first coordinates (Geometric braids in the disc with setwise endpoints, Intervals of R: the nine order-convex forms, nondegeneracy, and length, Continuity of a map of topological spaces at a point and globally).

[F2]

A braid isotopy from α to γ is a tuple of jointly continuous maps Zj ⁣:I×I→D∘ whose every slice Z(s,⋅) is a braid based at Q and whose slices at s=0,1 are α,γ (Braid isotopy relative to the top and bottom endpoints).

[F3]

I=[0,1] is a nonempty compact metric space, so every continuous real-valued function on it is bounded and attains a least value, every continuous map from I to a metric space is uniformly continuous, and every closed bounded subset of R is compact; a finite union of closed bounded pieces of R is closed and bounded (Heine-Borel in Rn: with the Euclidean metric a subset of Rn is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous, Intervals of R: the nine order-convex forms, nondegeneracy, and length).

[F5]

For a nonzero polynomial f over the integral domain R of degree d the set of its real roots has at most d elements, evaluation of a formal polynomial at a real point is a ring homomorphism and a polynomial taking a nonzero value is not the zero polynomial, while products of nonzero polynomials over R are nonzero; consequently, if a polynomial in N variables does not vanish at every point of a nonempty open box then, viewed as a polynomial in the last variable over the polynomial ring in the remaining variables, at least one of its coefficient polynomials does not vanish at every point of the projection box, since a point of the box at which every coefficient polynomial took the value 0 would make the evaluation of the polynomial zero (A nonzero polynomial of degree n over an integral domain has at most n distinct roots, Evaluation and roots of a polynomial in a commutative target ring, The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution, Over an integral domain, degrees add under multiplication of nonzero polynomials).

[L7]

Finitely many nonvacuous choices may be made: if S1,…,SN are nonempty sets then there is a function picking an element of each Si, and a finite nonempty set of positive reals has a positive least element (Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

Proof

technique · direct
1.1

The uniform margins of β. For n=0 the empty braid itself satisfies the statement with m=1, so assume n≥1. For each label j the continuous function t↦1−∥zj(t)∥2 attains a positive minimum on I by [F1], [F3] and [F4]; let b>0 be the least of these finitely many values. If n≥2, each continuous separation function t↦∥zi(t)−zj(t)∥2 for i<j likewise attains a positive minimum; let M>0 be the least of these finitely many pair minima.

F1F3F4L7
1.2

The polygonal approximation. Since each zj is continuous on the compact metric space I, it is uniformly continuous by [F3], so by [L7] we may choose, for the finitely many labels j, a real δj>0 with ∥zj(t)−zj(s)∥2<ε whenever ∣t−s∣<δj, where ε:=min⁡(M/4,b/2) for n≥2 and ε:=b/2 for n=1; fix an integer m≥2 with 1/m<min⁡jδj, possible because every sufficiently large integer satisfies the bound, and put tk:=k/m, and define pj ⁣:I→R2 to be affine on each [tk−1,tk] with pj(tk)=zj(tk). Then each pj is continuous by [F4], and for t∈[tk−1,tk] the point pj(t)=(1−λ)zj(tk−1)+λzj(tk) with λ=m(t−tk−1)∈I satisfies ∥pj(t)−zj(t)∥2≤(1−λ)∥zj(tk−1)−zj(t)∥2+λ∥zj(tk)−zj(t)∥2<ε because ∣t−tk−1∣,∣tk−t∣≤1/m<δj.

F1F3F4L7
2.1

First conclusion: the interpolation is a braid isotopy to the polygonal braid. The tuple p=(p1,…,pn) of step 1.2 has the same bottom and top values as β. Define Hj(s,t):=(1−s)zj(t)+s pj(t); each map is jointly continuous by [F4]. If n≥2, then for every i≠j and (s,t) the estimate from steps 1.1 and 1.2 gives ∥Hi(s,t)−Hj(s,t)∥2≥M−2ε>0; for n=1 the collision condition is vacuous. In either case, ∥Hj(s,t)∥2≤1−b+ε<1, so every slice is a braid based at Q and H is a braid isotopy from β to the polygonal braid p. For n=1 there are no projected crossings or pair conditions, so p already satisfies the full statement, with clearance 1−∥p1(t)∥2≥b−ε=b/2>0. Henceforth n≥2; then ∥pi(t)−pj(t)∥2≥M−2ε=:M1>0 and 1−∥pj(t)∥2≥b−ε=:b1>0.

F1F2F4step 1.1step 1.2
3.1

The perturbation box and its uniformity. Let V be the set of interior vertices (j,k) with 1≤j≤n, 1≤k≤m−1, and choose η∗>0 with M1−4η∗>0 and b1−2η∗>0, for instance η∗:=min⁡(M1/8,b1/4); for each interior vertex let ηj,k∈(−η∗,η∗) be a real parameter and let p(η) be the tuple obtained from p by moving the vertex pj(tk) to pj(tk)+(ηj,k,0), all other data unchanged. At every height t∈[tk−1,tk] the point pj(η)(t) is the convex combination, with weight λ=m(t−tk−1), of the possibly moved vertices at tk−1 and tk, so ∥pj(η)(t)−pj(t)∥2≤η∗; hence ∥pi(η)(t)−pj(η)(t)∥2≥M1−2η∗>0 and ∥pj(η)(t)∥2≤1−b1+η∗<1 for all i≠j and t, so every choice of parameters gives a braid p(η) based at Q whose bottom points are the base points and whose top points are those of p; moreover for two parameter values η,η′ the interpolation p(sη+(1−s)η′) is a braid isotopy by the same estimates, so all these braids are braid-isotopic to p and hence to β.

F1F2F4step 2.1L7
4.1

Bad configurations are polynomial conditions. For the parameter-dependent braid p(η) of step 3.1, put Xj,k:=π1(pj(η)(tk)); thus Xj,k=π1(pj(tk))+ηj,k for 1≤k≤m−1, while Xj,0 and Xj,m are the fixed endpoint coordinates. For a pair i<j and a piece [tk−1,tk] put A:=Xi,k−1−Xj,k−1 and B:=(Xi,k−Xi,k−1)−(Xj,k−Xj,k−1), so that the first-coordinate difference of the pair at height t=tk−1+λ/m equals A+λB and, if A≠0, B≠0 and A+B≠0, the pair has equal first coordinates at a unique height inside the piece, at which the difference changes sign, if and only if A(A+B)<0, the height being λ=−A/B. Consequently (a) a pair has equal first coordinates at a breakpoint tk, 1≤k≤m−1, exactly when Xi,k−Xj,k=0, with the parameter-dependent coordinates just defined, and (b) if two distinct pairs of strands have equal first coordinates at the same height t∗∈(0,1) that is not one of the breakpoints t1,…,tm−1, then, the interiors of distinct pieces being disjoint, both coincidences lie in the interior of one and the same piece [tk−1,tk], and for the two pairs on that common piece, with the common local parameter λ:=mt∗−(k−1)∈(0,1), the two coincidences give Ar+λBr=0 for r=1,2 and hence A1B2=A2B1.

F1F4step 3.1
5.1

Each bad condition is avoided on a box. All coordinates Xi,k are affine functions of the parameters ηl,κ by step 3.1, so each equation Xi,k−Xj,k=0 of step 4.1(a) is the zero set of a polynomial in the parameters that is nonzero, since it restricts to the nonzero affine function ηj,k↦(constant)−ηj,k when only ηj,k varies (here Xi,k does not involve ηj,k because i≠j); likewise, for a piece [tk−1,tk] and two distinct pairs of strands with the common piece of step 4.1(b), the equation A1B2−A2B1=0 is the zero set of the parameter polynomial Φ:=A1B2−A2B1, and Φ is not the zero polynomial, so the bad configurations of step 4.1(b) are confined to a proper algebraic condition: take a label l of the first pair that is not a label of the second (it exists because the pairs are distinct); if k≤m−1, then the coordinate Xl,k occurs in B1 only, with coefficient ±1, so the formal partial derivative ∂Φ/∂Xl,k, equivalently the derivative with respect to the parameter ηl,k, equals ∓A2, and this is a nonzero polynomial because the other pair's coefficient A2 is the difference of the first coordinates of the vertices (i2,k−1) and (j2,k−1) of that pair: either k−1=0 and it is the nonzero constant given by the distinct first coordinates of qi2,qj2 of [F1], or k−1≥1 and it involves the two independent parameters ηi2,k−1 and ηj2,k−1 with coefficients +1 and −1; if instead k=m, so that both pairs cross in the piece [tm−1,tm], then Br=Cr−Ar with Cr:=Ar+Br the top first-coordinate difference of the pair, a number independent of the parameters, so that Φ=A1C2−A2C1 with C1≠0 and C2≠0 because the top configuration has pairwise distinct first coordinates by [F1], and the partial derivative ∂Φ/∂Xl,m−1=±C2 is nonzero, the vertex (l,m−1) being a parameter because m≥2; hence in every case the required polynomial is not the zero polynomial, and the finite family of these conditions, over all pairs of strands, all breakpoints and all pieces, is avoided below with [F5] and [F6].

F1F4F5F6step 4.1
6.1

Avoiding finitely many proper algebraic conditions. Let J:=∏l,κ(−η∗,η∗) be the box of parameters and let Φ1,…,Φs be the finitely many parameter polynomials of step 5.1, each of which is not the zero polynomial; then J contains a point at which all Φ1,…,Φs are nonzero, by induction on the number N of parameters (for N=0, the box has one empty parameter tuple and each nonzero polynomial is a nonzero constant, so the claim holds): for N=1 each Φi is a nonzero polynomial in one variable, so its root set has at most deg⁡Φi elements by [F5], the bad set is a union of finitely many finite sets inside the nonempty interval J, and an interval is not a subset of a finite set by [F6]; for N>1 write each Φi as a polynomial in the last parameter with coefficient polynomials in the remaining parameters, for each Φi retain one coefficient polynomial that is formally nonzero, and apply the induction hypothesis to these finitely many nonzero coefficient polynomials, and the box J′ to choose the first N−1 parameters so that none of them vanishes at that point, and then avoid, in the last coordinate interval, the finitely many roots of the resulting nonzero polynomials in the last parameter, again by [F5] and [F6].

F5F6step 5.1
7.1

Conclusion. Choose the parameters by step 6.1. Then no pair of strands has equal first coordinates at a breakpoint tk with 1≤k≤m−1 by step 4.1(a), and no two pairs of strands have equal first coordinates at the same interior height by step 4.1(b); since the first-coordinate difference of a pair restricted to one affine piece is affine and is not identically zero on that piece (it is nonzero at each end of the piece: at the base and top heights because the first coordinates of distinct strands are then distinct by [F1], and at an interior breakpoint because the chosen parameters avoid step 4.1(a)), each pair realises at most one interior coincidence in each piece, and such a coincidence lies in the interior of the piece, changes the sign of the difference, and involves no third strand (a third strand with the same first coordinate at that height would be a second pair meeting at the same height); hence the set C of interior coincidences is finite, and each of its elements is a crossing of exactly one pair which exchanges the two positions of that pair across the crossing height. The resulting braid p(η) is polygonal with breakpoints t0<⋯<tm by step 3.1, satisfies the general-position clauses by the above, keeps the uniform boundary clearance ∥pj(η)(t)∥2≤1−b1+η∗ with b1−η∗>0 by step 3.1, and is braid-isotopic to β by steps 2.1 and 3.1, which is the assertion. ∎

F1F2F4step 2.1step 3.1step 4.1step 6.1

Remarks

  • The hypothesis that the strands move in the interior D∘ of the disc is what produces the uniform boundary margin b>0 of step 1.1; a strand touching the boundary would make the approximation ∥pj∥2<1 fail, and an additional inward push would be required.
  • Only the first coordinate is used in the general-position clauses: a crossing in this lemma means a coincidence of first coordinates of two strands, not a collision of points. Collisions are excluded throughout by the uniform distance bound M1−2η∗>0, which is positive by construction and is the reason the perturbation keeps every slice a braid.
  • The perturbation moves one coordinate per vertex; the verification that each excluded condition is a nonzero polynomial in the parameters is where step 5.1 uses that the two pairs of strands are distinct, so that some label occurs in only one of them, and that the moved vertices carry independent parameters, whose coefficients witness the nonvanishing of the relevant partial derivative; and the induction of step 6.1 is the only place where the root bound for polynomials enters.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

Every geometric braid is isotopic to a stacking of signed elementary half twists

Statement

Let n∈N and let β be a braid based at Q (Geometric braids in the disc with setwise endpoints), with Q=(q1,…,qn), h=14(n+1) and qj=((2j−n−1)h,0). Write σi+1:=σi and σi−1:=σi− for the two elementary half twists at i (The elementary geometric half twist, its support disc, and its opposite), and write e for the trivial braid ej(t):=qj. Then there are an integer M≥0, indices i1,…,iM∈{1,…,n−1} and signs ε1,…,εM∈{+1,−1} such that

β ∼ σi1ε1⋆σi2ε2⋆⋯⋆σiMεM,

where ⋆ is the stacking of Stacking of geometric braids is a well-defined associative operation on isotopy classes and ∼ is braid isotopy relative to the top and bottom (Braid isotopy relative to the top and bottom endpoints); for M=0 the word displayed above is the empty word and its value is e. Consequently, in the group Gn of The isotopy classes of geometric braids based at Q form a group, and the endpoint permutation is a homomorphism,

[β]=[σi1]ε1[σi2]ε2⋯[σiM]εM,

so the classes [σ1],…,[σn−1] generate Gn; for n≤1 there is no index i, the family of generators is empty, and Gn is the trivial group generated by the empty family.

The word read off from the crossings. The proof has the following more precise content, which is the form used in the rest of this page and its companion. Let β′ be a generic polygonal representative of β (Geometric braids admit generic polygonal representatives) with crossing heights c1<⋯<cm′, and for each k let pk be one more than the number of strands whose first coordinate at height ck is strictly smaller than the common first coordinate of the two crossing strands at ck. Then pk∈{1,…,n−1} and β is braid-isotopic to the word σpm′εm′⋆⋯⋆σp2ε2⋆σp1ε1 in which εk:=+1 when the strand that occupies position pk just below the crossing ck has smaller second coordinate than its partner at height ck, and εk:=−1 otherwise. So the crossings of a generic polygonal representative, read from the lowest height to the highest, give the factors of the word read from right to left, the lowest crossing contributing the rightmost factor. All constructions are explicit, only finitely many choices are made, and no choice principle is used.

Facts & Assumptions

Given: A natural number n, the base configuration Q=(q1,…,qn) with qj=((2j−n−1)h,0) and h=14(n+1), a braid β=(z1,…,zn) based at Q, and, when n≥2, the half twists σ1,…,σn−1 based at Q.

[F1]

A braid based at Q is a tuple (u1,…,un) of continuous maps uj ⁣:I→D∘=∥⋅∥2-open unit disc, with ui(t)≠uj(t) for i≠j, uj(0)=qj, and {u1(1),…,un(1)}={q1,…,qn}; its endpoint permutation π(u) is the unique permutation with uj(1)=qπ(u)(j); the base points are pairwise distinct with ℜq1<⋯<ℜqn, qj+1−qj=(2h,0), and ∥qj∥2≤(n−1)h<1; π is multiplicative, π(γ⋆β)=π(γ)∘π(β) (Geometric braids in the disc with setwise endpoints, Intervals of R: the nine order-convex forms, nondegeneracy, and length, Continuity of a map of topological spaces at a point and globally, The product set ∏i∈IXi of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, The finite symmetric group Sn, one-line notation, and cycle notation).

[F2]

A braid isotopy from β to β′ is a tuple Z=(Z1,…,Zn) of jointly continuous maps Zj ⁣:I×I→D∘ such that every slice Z(s,⋅) is a braid based at Q, with Zj(0,t)=zj(t) and Zj(1,t)=zj′(t); we then write β∼β′ (Braid isotopy relative to the top and bottom endpoints).

[F3]

Stacking first-under-second is (γ⋆β)j(t)=zj(2t) for t≤12 and =wπ(β)(j)(2t−1) for t≥12, where zj are the strands of β and wj those of γ; γ⋆β is a braid based at Q; π(γ⋆β)=π(γ)∘π(β); the operation descends to isotopy classes, so β∼β′ and γ∼γ′ give γ⋆β∼γ′⋆β′; it is associative up to braid isotopy; and π is constant on braid isotopy classes (Stacking of geometric braids is a well-defined associative operation on isotopy classes).

[F4]

The elementary half twist at i is (σi)i=mi+ρ, (σi)i+1=mi−ρ and (σi)j=qj for j∉{i,i+1}, where mi=qi+(h,0) is the midpoint of qi,qi+1 and the diamond path ρ satisfies ρ(0)=(−h,0), ρ(12)=(0,−h), ρ(1)=(h,0), is affine on each of [0,12] and [12,1], and has ρ(t)≠0; the opposite half twist uses ρ−(t):=(ρ1(t),−ρ2(t)); both σi and σi− are braids based at Q with endpoint permutation the transposition of i and i+1, and [σi−]=[σi]−1 in Gn (The elementary geometric half twist, its support disc, and its opposite).

[F5]

Every braid β based at Q admits a braid-isotopic representative β′ that is polygonal with breakpoints 0=t0<t1<⋯<tm=1, has no two strands meeting in the projection at a breakpoint, and has a finite set C={c1<⋯<cm′} of interior crossing heights such that at each cr exactly one pair of strands has equal first coordinates, that pair lying in the interior of one affine piece with the difference of first coordinates changing sign there, and no third strand has that first coordinate (Geometric braids admit generic polygonal representatives).

[F6]

Gn is a group with operation [γ][β]=[γ⋆β], identity [e], and inverses [β]−1=[β‾]; the endpoint permutation is a homomorphism π ⁣:Gn→Sn (The isotopy classes of geometric braids based at Q form a group, and the endpoint permutation is a homomorphism).

[F7]

D∘ is an open ball of R2. For x,y∈D∘ and 0≤s≤1, the triangle inequality gives ∥(1−s)x+sy∥2≤(1−s)∥x∥2+s∥y∥2<1; hence D∘ and its finite Cartesian powers are convex. An order chamber in (D∘)n is obtained by imposing strict linear inequalities on first coordinates, which every segment between two of its points retains. Such a segment stays collision-free (A convex subset of Rm contains every line segment between two of its points, Open ball, closed ball and sphere in a metric space).

[L9]

Induction on the natural numbers: if a statement holds for 0 and holds for M whenever it holds for every m<M, then it holds for every M∈N (The principle of mathematical induction, The natural numbers N (von Neumann)).

Proof

technique · direct
1.1

The cases n≤1. If n≤1 there is no index i with 1≤i≤n−1, so the family of generators is empty and its empty word has value e; moreover every strand satisfies zj(1)=qj, because for n=1 the top set {z1(1)}={q1} forces z1(1)=q1 and for n=0 there is no strand; then Hj(s,t):=(1−s)zj(t)+sqj is a braid isotopy from β to e, since it is jointly continuous, every value is a convex combination of two points of the convex set D∘, each slice has bottom qj and top set {qj}, and a slice has at most one strand, so no two strands of a slice can meet. Hence β∼e, the conclusion holds with M=0, and for the rest of the proof n≥2.

F1F2F4F7F8
1.2

Reduction to generic polygonal representatives, and the structure of their crossings. For M≥0, let A(M) assert that every generic polygonal braid based at Q with exactly M projected crossing heights is isotopic to the signed half-twist word read from those crossings as in the statement, with chronological first crossing on the right in stacking order. By [F5] there is a generic polygonal braid β′ based at Q with β′∼β, and it is enough to prove the assertion for β′, because ∼ is transitive (two braid isotopies that meet end to end paste to a jointly continuous family by [F8]) and because the class in Gn is unchanged; for such a β′ the differences ℜzi−ℜzj of first coordinates are continuous functions of the height, so on each of the intervals [0,c1),(c1,c2),…,(cm′,1] the left-to-right order of the labels is constant; on [0,c1) that order is 1,2,…,n, because ℜβ1′(0)<⋯<ℜβn′(0) by [F1]; at t=c1 the two strands with equal first coordinate are therefore adjacent in that order, and because no third strand has that first coordinate at c1 the two of them are precisely the strands labelled p and p+1, where p is one more than the number of strands whose first coordinate at c1 is strictly smaller than the common first coordinate χ:=ℜβp′(c1) of the pair.

F1F5F8
1.3

Reparametrisation of heights is a braid isotopy. Let φ ⁣:I→I be continuous and nondecreasing with φ(0)=0 and φ(1)=1, and let δ be a braid based at Q; then δjφ(t):=δj(φ(t)) is a braid based at Q, and δφ∼δ: indeed φs(t):=(1−s)t+sφ(t) is again continuous and nondecreasing with φs(0)=0, φs(1)=1, and Hj(s,t):=δj(φs(t)) is jointly continuous with every slice a braid, because its bottom values are δj(0)=qj, its top values are δj(1), and the values δj(φs(t)) are pairwise distinct for each s,t as they are values of δ at the single height φs(t).

F1F2F8
1.4

One-crossing braids: the hypotheses and their endpoint configuration. Let P abbreviate the following hypothesis on a braid α based at Q: there are p∈{1,…,n−1} and c∈(0,1) such that (i) the horizontal order of labels for t<c is 1,…,n; (ii) for t>c it is the sequence 1,…,p−1,p+1,p,p+2,…,n, omitting the left or right block if p=1 or p=n−1; and (iii) at c only the pair p,p+1 has equal first coordinates, its points are distinct, and its common first coordinate lies strictly between those of the neighboring labels whenever those neighbors exist. Under P the top configuration is forced: at height 1 the left-to-right order of labels is the sequence in (ii), while that of q1,…,qn is 1,…,n by [F1]. Thus αj(1)=qj for j∉{p,p+1}, αp+1(1)=qp and αp(1)=qp+1; writing τ:=(p p+1), we have π(α)=τ and α(1)=Qτ, where (Qτ)j:=qτ(j).

F1F2F4
2.1

One-crossing braids: straightening the motion on each side of the crossing. Assume P of step 1.4, and define μj(t):=(1−t/c)qj+(t/c)αj(c) for t≤c and νj(t):=1−t1−cαj(c)+t−c1−cαj(1) for t≥c; for s∈I let αj(s)(t):=(1−s)αj(t)+sμj(t) for t≤c and αj(s)(t):=(1−s)αj(t)+sνj(t) for t≥c. Each α(s) is a braid based at Q: its values lie in the convex set (D∘)n, it is continuous on each of the two closed pieces by [F8] and hence continuous, its bottom values are αj(s)(0)=qj=μj(0) and νj(1)=αj(1)=qτ(j), so its top set is {q1,…,qn}; and collisions are impossible, since for t<c the differences of first coordinates are ℜαj(s)(t)−ℜαi(s)(t)=(1−s)(ℜαj(t)−ℜαi(t))+s(ℜμj(t)−ℜμi(t)) with the first summand positive for i<j by (i) and the second equal to (1−t/c)2h(j−i)+(t/c)(ℜαj(c)−ℜαi(c))≥0 by [F1] and (iii), while for t>c and i preceding j in the order of (ii) the corresponding expression with ν has second summand t−c1−c(ℜqτ(j)−ℜqτ(i))>0 by [F1] and (ii), and at t=c both formulas give the collision-free configuration α(c). Hence s↦α(s) is a braid isotopy from α to the two-piece affine braid α(1) with αj(1)(t)=μj(t) for t≤c and αj(1)(t)=νj(t) for t≥c, so α∼α(1).

F1F2F4F7F8step 1.4
2.2

The induction base: braids with no crossings. Let δ be a generic polygonal braid based at Q with m′=0, so that no two strands ever have equal first coordinates; then by [F1] and step 1.2 the left-to-right order of the labels is the constant order 1,…,n, so δj(1)=qj for every j and Hj(s,t):=(1−s)δj(t)+sqj is a braid isotopy from δ to e: it is jointly continuous, every value lies in the convex set D∘, each slice has bottom qj and top qj, and ℜHj(s,t)−ℜHi(s,t)=(1−s)(ℜδj(t)−ℜδi(t))+s 2h(j−i)>0 for i<j and every s, so no slice has a collision. Hence δ∼e, the empty word, and assertion A(0) of step 1.2 holds.

F1F2F3F7F8step 1.2
3.1

One-crossing braids: the wall of configurations with the pair vertically aligned. Assume P of step 1.4, write αp(c)=(χ,y1) and αp+1(c)=(χ,y2) with y1≠y2 and χ=ℜαp(c), put mp:=qp+(h,0)=((2p−n)h,0), and choose the sign ϵ:=+1 if y1<y2 and ϵ:=−1 if y1>y2; let Tϵ be the configuration with Tjϵ:=qj for j∉{p,p+1}, Tpϵ:=mp+(0,−ϵh) and Tp+1ϵ:=mp+(0,ϵh), and put Ps:=(1−s)α(c)+sTϵ for s∈I. Each Ps is a collision-free configuration whose first coordinates are increasing with the single tie ℜPs[p]=ℜPs[p+1]: the values lie in (D∘)n by [F7]; for j<p one has ℜPs[j]=(1−s)ℜαj(c)+sℜqj<(1−s)χ+sℜmp=ℜPs[p], and symmetrically ℜPs[j]>ℜPs[p+1] for j>p+1, because both endpoint inequalities are strict, so the pair strands never meet the others; the other strands keep their strict relative order for the same reason; and Ps[p]−Ps[p+1]=i((1−s)(y1−y2)−2ϵhs) is purely imaginary and nonzero for every s, because for ϵ=+1 both summands are ≤0 and vanish simultaneously only if s=1 and y1=y2 or s=0 and y1=y2, and symmetrically for ϵ=−1; consequently the family βs that equals the affine path from Q to Ps on [0,c] and the affine path from Ps to Qτ on [c,1] consists of braids depending jointly continuously on (s,t): for t<c the difference of the j-th and i-th first coordinates is (1−t/c)2h(j−i)+(t/c)(ℜPs[j]−ℜPs[i])>0 for i<j, for t>c it is 1−t1−c(ℜPs[j]−ℜPs[i])+t−c1−c(ℜqτ(j)−ℜqτ(i))>0 for i preceding j in the order of (ii), and at t=c the configuration is Ps, collision-free by the above; since β0=α(1) and β1 is the affine two-piece path through Tϵ, this gives α(1)∼β1.

F1F2F4F7F8step 1.4step 2.1
4.1

One-crossing braids: the end of the family is a signed half twist. In the notation of step 3.1, let φ ⁣:I→I be the nondecreasing piecewise affine map with φ(t)=t/(2c) for t≤c and φ(t)=12+t−c2(1−c) for t≥c; comparing the formulas of [F4] with the two-piece affine motion of step 3.1, whose pair moves affinely from (qp,qp+1) to (mp+(0,−ϵh),mp+(0,ϵh)) and then affinely to (qp+1,qp) while every other strand stays at its base point, gives β1,j(t)=σp,jϵ(φ(t)) for every j and t: for ϵ=+1 the pair motions of β1 are those of mp±ρ on the two halves reparametrised by φ, and for ϵ=−1 those of mp±ρ−. Hence β1=σpϵ∘φ in the sense of step 1.3 and therefore β1∼σpϵ.

F4step 1.3step 3.1
5.1

The one-crossing claim. Under hypothesis P of step 1.4 the braid α is braid-isotopic to σpϵ, where ϵ=+1 when the strand labelled p has the smaller second coordinate at the crossing height c and ϵ=−1 otherwise: this is α∼α(1) from step 2.1, α(1)∼β1 from step 3.1 and β1∼σpϵ from step 4.1, composed with the transitivity of braid isotopy.

F2step 2.1step 3.1step 4.1
6.1

The induction step: cutting at a height between the two lowest crossings. Let β′ be generic polygonal with m′≥1 crossing heights c1<⋯<cm′, let p1 be as in step 1.2 for the crossing c1, and let ϵ1=+1 when the strand labelled p1 has the smaller second coordinate at height c1 and ϵ1=−1 otherwise; if m′=1 then β′ satisfies hypothesis P of step 1.4 with p:=p1 and c:=c1, so step 5.1 gives β′∼σp1ϵ1 and A(1) holds. If m′≥2, choose heights θ1<θ<θ2 with c1<θ1<θ<θ2<c2, let l1,…,ln be the labels in left-to-right order at height θ (constant on (c1,c2) and hence on [θ1,θ2]), and let Qσ be the configuration with (Qσ)lk:=qk for every k, so that Qσ has the same left-to-right label order as β′(θ) and both lie in the convex set C:={u:ℜul1<⋯<ℜuln}; replacing β′ on [θ1,θ2] by the two-piece affine path through Qσ, by the same interpolation as in steps 2.1 and 3.1, gives a braid isotopy from β′ to a generic polygonal braid β∗ with β∗(θ)=Qσ, because the interpolation of two points of the convex set C∩(D∘)n stays in it, so no slice acquires a collision and the values stay in the disc; then αj(u):=βj∗(θu) and γk(v):=βlk∗(θ+v(1−θ)) are braids based at Q, because αj(0)=qj and γk(0)=qk, while their top configurations are permutations of Q; the braid α satisfies hypothesis P with the crossing c1/θ and the same index p1, since below that crossing the order is the base order and above it the order is the transposed order of (ii); and β∗=γ⋆α up to the monotone reparametrisation φ(t)=t/(2θ) for t≤θ, φ(t)=12+t−θ2(1−θ) for t≥θ of step 1.3, so β′∼γ⋆α; finally γ is generic polygonal with the m′−1 crossing heights (cr−θ)/(1−θ) for r≥2. Relabelling the upper braid by its current horizontal rank does not alter the rank pair or the vertical sign at any upper crossing, so its crossing word is exactly the suffix of the crossing word of β∗.

F1F2F3F4F5F7F8step 1.2step 1.3step 1.4step 5.1
7.1

The induction step concluded. Assume m′≥2 and the notation of step 6.1; by the induction hypothesis A(m) for every m<m′, applied to the generic polygonal braid γ with its m′−1<m′ crossings, the braid γ is isotopic to its exact signed crossing word W, and by step 5.1 the braid α satisfies α∼σp1ϵ1; hence, using that stacking respects braid isotopy in each factor, β′∼γ⋆α∼W⋆σp1ϵ1, which is again a stacking of signed elementary half twists. Together with the case m′=1 of step 6.1 and the base case A(0) of step 2.2, the induction of [L9] gives A(m′) for every m′.

F3F5L9step 5.1step 2.2step 6.1
8.1

Conclusion, and generation of Gn. For the original braid β let β′ be the generic polygonal representative of [F5]; it has finitely many crossing heights, say m′, and β∼β′; by A(m′) of step 7.1 there are M≥0, indices ik and signs εk with β′∼σi1ε1⋆⋯⋆σiMεM and hence β∼σi1ε1⋆⋯⋆σiMεM; moreover the more precise reading of the word, with pk and εk as in step 6.1 for each crossing, is exactly the one recorded in the statement. Passing to classes in Gn with [F3] and [F6] gives [β]=[σi1]ε1⋯[σiM]εM, using [σi+1]=[σi] and [σi−1]=[σi]−1 of [F4] and the empty word for M=0; since every element of Gn is the class of a braid, the classes [σ1],…,[σn−1] generate Gn, and for n≤1 step 1.1 gives the trivial group on the empty family. ∎

F3F4F5F6F8step 1.1step 7.1

Remarks

  • The argument is a crossing-by-crossing decomposition. Nothing is reproved about the classification of braids: the two ingredients are the convexity of the order chambers of the configuration space, which straightens every crossing-free stretch, and the two-dimensional fact that a pair of points whose first coordinates change sign exactly once at a crossing carries exactly one half turn of relative motion, which is computed in step 3.1 by the explicit wall of configurations in which the pair is vertically aligned.
  • The sign convention is the one fixed by The elementary geometric half twist, its support disc, and its opposite together with first-under-second stacking: the rightmost factor of the word sits in the lowest part of the cylinder, and σp+1 is the half twist in which the two strands pass with the strand labelled p below, which is also the anticlockwise half turn of the pair about its midpoint in the fixed projection.
  • Only generation is proved here: the word produced by the crossings maps onto the braid. The converse statement, that the Artin relations are a complete set of relations among the half twists, is a different theorem and is not used on this page; the presentation is only shown to surject onto the geometric braid group.
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

The Artin presentation surjects onto the geometric braid group

Statement

Let n∈N. Write Gn for the geometric braid group of The isotopy classes of geometric braids based at Q form a group, and the endpoint permutation is a homomorphism, let σ1,…,σn−1 be the geometric half twists of The elementary geometric half twist, its support disc, and its opposite, with classes [σi]∈Gn, and let Bn=⟨x1,…,xn−1∣R⟩ be the Artin braid group of The braid group by Artin presentation, whose generator written there as σi is here written xi to keep it distinct from the geometric half twist. Then the assignment

φ(xi):=[σi](1≤i≤n−1)

extends to a homomorphism φ ⁣:Bn→Gn, it does so uniquely, and φ is surjective. Consequently every element of Gn is a finite product of the classes [σi]±1 of the half twists, and the composition of φ with the endpoint permutation homomorphism π ⁣:Gn→Sn is the permutation map xi↦(i i+1) of the presented group.

Only surjectivity is asserted. Nothing here shows that φ is injective, that is, that the Artin relations are a complete set of relations for the geometric braid group; the presentation is shown to surject onto Gn only. For n≤1 the presentation has no generator and Bn and Gn are both trivial, so the assertions are vacuous.

Facts & Assumptions

Given: A natural number n, the geometric braid group Gn of The isotopy classes of geometric braids based at Q form a group, and the endpoint permutation is a homomorphism, the Artin presentation Bn=⟨X∣R⟩ of The braid group by Artin presentation with X={σ1,…,σn−1} for n≥2 and X=∅ for n≤1, and the elementary half twists of The elementary geometric half twist, its support disc, and its opposite.

[F1]

For n≥2 the group Bn is the quotient of the free group on X={σ1,…,σn−1} by the normal closure of the relations σiσi+1σi=σi+1σiσi+1 (1≤i≤n−2) and σiσj=σjσi (∣i−j∣>1), interpreted in the sense of Group presentation by generators and relations and Relators and relations; finitely generated, finitely related, and finite presentations; for n=0 and n=1 there are no generators and Bn is the trivial group of the empty presentation (The braid group by Artin presentation, Group presentation by generators and relations, Free group on a set of generators, Group and abelian group).

[F2]

In the presentation ⟨X∣R⟩ of [F1] an equation u=v is recorded by the relator u−1v in the sense of the free group on X (Relators and relations; finitely generated, finitely related, and finite presentations, Free group on a set of generators, Group presentation by generators and relations).

[F3]

Let ⟨X∣R⟩ be a presentation, H a group, and u ⁣:X→H a function. If the evaluation of every r∈R under u is eH, then there is a unique homomorphism u‾ ⁣:⟨X∣R⟩→H with u‾([x])=u(x) for every x∈X; moreover u‾ is surjective if and only if u(X) generates H (Von Dyck's theorem: maps of generators that satisfy the relators extend uniquely from a presented group).

[F4]

The half twist σi and its opposite σi− are braids based at Q with classes [σi],[σi]−1∈Gn, and their endpoint permutations are the transposition of i and i+1; the endpoint permutation is a homomorphism π ⁣:Gn→Sn (The elementary geometric half twist, its support disc, and its opposite, The isotopy classes of geometric braids based at Q form a group, and the endpoint permutation is a homomorphism).

[F5]

For ∣i−j∣>1 the half twists satisfy [σi][σj]=[σj][σi] in Gn, and for n≤3 there are no such pairs of indices, so the assertion is vacuous (Far commutativity of elementary geometric half twists).

[F6]

For 1≤i≤n−2 the half twists satisfy [σi][σi+1][σi]=[σi+1][σi][σi+1] in Gn, and for n≤2 there is no such index, so the assertion is vacuous (The geometric three strand braid relation).

[F7]

Every braid class in Gn is a finite product of the classes [σ1],…,[σn−1] and their inverses; equivalently, the set {[σ1],…,[σn−1]} generates Gn in the sense of The subgroup ⟨S⟩ generated by a subset, the cyclic subgroup ⟨g⟩, and cyclic groups, and for n≤1 the empty family generates the trivial subgroup {e} (Every geometric braid is isotopic to a stacking of signed elementary half twists, The subgroup ⟨S⟩ generated by a subset, the cyclic subgroup ⟨g⟩, and cyclic groups).

Proof

technique · direct
1.1

The relators of the Artin presentation evaluate to the identity under the half-twist assignment. Assume n≥2, let X={σ1,…,σn−1} and let u ⁣:X→Gn be the assignment u(σi):=[σi]; the relators of [F1] are, by [F2], the words (σiσi+1σi)−1(σi+1σiσi+1) for 1≤i≤n−2 and (σiσj)−1(σjσi) for ∣i−j∣>1. Their evaluations are ([σi][σi+1][σi])−1[σi+1][σi][σi+1]=e by [F6] and ([σi][σj])−1[σj][σi]=e by [F5].

F1F2F5F6
1.2

The cases n≤1. For n≤1 the presentation has no generators and defines the trivial group Bn by [F1], while ⟨∅⟩={e} is the trivial subgroup of Gn and [F7] says that this empty family generates Gn, so Gn is trivial as well; the unique map Bn→Gn is therefore a group homomorphism, it is the only homomorphism between these groups, and it is surjective because its codomain is trivial.

F1F3F7
2.1

Von Dyck extends the assignment to a homomorphism. By step 1.1 the hypothesis of [F3] is satisfied, so there is a unique homomorphism φ ⁣:Bn→Gn with φ(xi)=[σi] for every generator, and φ is surjective if and only if the set of these images generates Gn.

F1F3step 1.1
3.1

Surjectivity. The images φ({σ1,…,σn−1})={[σ1],…,[σn−1]} generate Gn by [F7], so the surjectivity criterion of [F3] applies to the homomorphism of step 2.1 and φ is surjective; consequently every element of Gn is a finite product of the classes [σi]±1, and composing the unique homomorphism with the endpoint permutation homomorphism of [F4] gives the permutation map xi↦(i i+1), because π([σi]) is that transposition.

F3F4F7step 2.1
4.1

Conclusion. Steps 2.1 and 3.1 give, for n≥2, a unique homomorphism φ ⁣:Bn→Gn with φ(xi)=[σi], and step 1.2 gives the same for n≤1; in both cases φ is surjective, and no injectivity is claimed. ∎

F1F3step 2.1step 3.1step 1.2

Remarks

5 · Examples, counterexamples and false statements

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Sources