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Oriented Links, Braid Closures, and Markov Equivalence

1 · Prerequisites

2 · Summary

This page builds the bridge between links in the three-sphere and braids. It fixes the smooth category for oriented links and their ambient isotopies, then develops regular projections, planar isotopy of diagrams and the oriented Reidemeister moves, and proves Reidemeister's theorem in the oriented category: two regular diagrams represent equivalent oriented links exactly when finitely many planar isotopies and oriented R1, R2, R3 moves connect them. The general-position perturbation of a link isotopy is the mechanism that produces the finite sequence of moves, and each local move has a ball-supported ambient realization between lifts agreeing outside the move ball. Arbitrary lifts of one diagram are compared by interpolation of their height coordinates.

The second half turns a braid into a link. The closure of a geometric braid is constructed in the standard solid torus about the braid axis, its components are the cycles of the endpoint permutation, and the closure is shown to depend only on the braid isotopy class; a free-homotopy argument identifies braid isotopies of closed braids with conjugacy in the braid group. The Yamada-Vogel algorithm then computes the height of a diagram through the coherence of its Seifert circles: a positive-height diagram has a defect region, a reducing move lowers the height by one, and a height-zero diagram can be put in closed-braid form on the sphere, which yields Alexander's theorem. Markov's theorem is the corresponding statement for closures: two closure links are equivalent exactly when the braids are related by conjugations and stabilizations, and the page develops the Traczyk factorization of an arbitrary Reidemeister sequence through braid isotopies and the four-band exchange computation that makes the peak induction terminate. The axiom of choice is carried where the annulus lemma, the transversality arguments and the ambient isotopy extension theorem require it, and the raw closure quotient, fixed framing and definition of the Markov moves are choice-free. Passing from a general continuous braid to its smooth closure class uses countable choice through smoothing and ambient isotopy extension.

3 · Logical flowchart

4 · Definitions, theorems and proofs

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

A smooth isotopy of a compact manifold extends to an ambient isotopy

Statement

Assume ACω. Let M be a compact smooth manifold without boundary, let N be a smooth manifold, and let F ⁣:M×I→N be a smooth isotopy such that Ft:=F(⋅,t) is an embedding for every t∈I. Suppose F is constant near the ends of I: there is ε∈(0,12) with F(x,t)=F(x,0) for t≤ε and F(x,t)=F(x,1) for t≥1−ε and all x∈M. Then for every open neighbourhood W of the image F(M×I) there is an ambient isotopy H ⁣:N×I→N such that

Ht∘F0=Ftfor every t∈I,

H0=idN, each Ht is a diffeomorphism of N, and Ht is the identity outside W for every t. Moreover Ht=idN for t∈[0,ε2] and Ht=H1 for t∈[1−ε2,1].

Facts & Assumptions

Given: ACω, a compact boundaryless smooth manifold M, a smooth manifold N, a smooth isotopy F ⁣:M×I→N with every Ft an embedding and F constant on M×[0,ε]∪M×[1−ε,1] for some ε∈(0,12), and an open neighbourhood W of F(M×I) in N.

[F1]

ACω is the countable axiom of choice (The Axiom of Countable Choice (ACω)).

[F2]

Assume ACω: every open cover of a smooth manifold with boundary admits a smooth partition of unity subordinate to it, and a smooth partition of unity subordinate to a cover (Uj)j∈J is a family (ϕj)j∈J of smooth functions M→[0,1] with locally finite supports, supp⁡ϕj⊆Uj and ∑jϕj=1 (Smooth partitions of unity exist on manifolds with boundary, Smooth partition of unity on a manifold with boundary).

[F4]

Assume ACω. For a smooth embedded submanifold S↪N and a smooth vector field Y along S: if S is closed in N, then there is a global smooth vector field Y^ on N with Y^∣S=Y (A vector field along an embedded submanifold extends to a neighbourhood and globally when the submanifold is closed).

[F5]

Let J⊆R be a compact interval and let Xt be a smooth time-dependent vector field on N whose union of supports over t∈J is contained in a compact subset K⊆N. Then there is a global evolution operator Ψt,s ⁣:N→N for all s,t∈J (Compactly supported time-dependent vector fields have global evolution on a compact time interval): Ψs,s=idN, the cocycle law Ψu,t∘Ψt,s=Ψu,s holds, each Ψt,s is smooth, and for fixed s and p the curve t↦Ψt,s(p) solves γ˙(t)=Xt(γ(t)).

[F6]

A smooth map F ⁣:M→N is a smooth embedding when it is injective, an immersion, and a homeomorphism onto its image with the subspace topology (Smooth embeddings).

Proof

technique · direct
1.1F1givenconstruct

Compactness of the image and a relatively compact neighbourhood. If M=∅, take Ht=idN for every t; all extension and support assertions are then immediate. Assume M≠∅. The set K:=F(M×I)⊆N is compact, being the continuous image of the compact space M×I, and it is closed in N because smooth manifolds are Hausdorff. Each point of K has a coordinate ball whose closure is compact and contained in W; finitely many of these balls cover K, and their union V is an open neighbourhood of K with compact closure V‾⊆W. Fix such a V.

1.2F6givenalgebra

The velocity field along the slices, as a field along a graph. Extend F to R×M by Ft=F0 for t<0 and Ft=F1 for t>1. This extension is smooth because the given F is constant on the full endpoint collars of width ε, and every extended slice remains an embedding. Put J:=R and consider the map Θ ⁣:J×M→J×N, Θ(t,x):=(t,Ft(x)), with image S:=Θ(J×M). Θ is injective because its first coordinate is t; its derivative at (t,x) equals (dt,∂tF(x,t) dt+d(Ft)x), which is injective because d(Ft)x is injective by [F6]; and Θ is proper: for a compact subset L⊆J×N, its time projection is compact, and Θ−1(L) is closed in the compact product of that projection with M. The inverse on the image is continuous locally by the embedding property of its slices, or globally by this properness. Its injective derivative therefore makes S a closed embedded submanifold without boundary of J×N, of dimension 1+dim⁡M. The constant time extension avoids applying a boundaryless extension theorem to a graph with boundary. The assignment Y(t,Ft(x)):=(0,∂tF(x,t))∈T(t,Ft(x))(J×N) is well defined because each Ft is injective, and it is a smooth vector field along S: near a point of S the inverse y↦x of Ft is smooth by [F6], so Y is the composite of smooth maps; along S it is everywhere tangent to the splitting of T(J×N) into the J-direction and TN.

2.1F2F4step 1.2construct

Extension and truncation. By [F4] applied to the closed embedded submanifold S of the smooth manifold J×N, the field Y extends to a global smooth vector field Y~ on J×N with Y~∣S=Y. Write Y~=(a,X) in the splitting T(J×N)≅R⊕TN, so that X is a smooth family Xt:=X(t,⋅) of vector fields on N with Xt(Ft(x))=∂tF(x,t) for all x∈M and t∈J. Choose a smooth function ψ ⁣:N→[0,1] with ψ=1 on a neighbourhood of K and supp⁡ψ⊆V: the open sets V and N∖K cover N because K is closed, so [F2] applied to this two-element cover produces ψ as the member subordinate to V, whose support lies in V. The other member has closed support contained in N∖K; its support complement is an open neighbourhood of K on which that other member vanishes and hence ψ=1. Choose a smooth β ⁣:R→[0,1] with β=1 on [ε,1−ε] and β=0 on (−∞,ε2]∪[1−ε2,∞), and put Xt′:=β(t) ψ Xt for t∈J. Every Xt′ is a smooth vector field on N with supp⁡Xt′⊆V‾, so ⋃t∈Isupp⁡Xt′⊆V‾ is compact.

3.1F5step 2.1

The ambient isotopy. Apply [F5] on the compact interval I to the family Xt′ of step 2.1 and let Ψt,s be the resulting global evolution operator. Put Ht:=Ψt,0 for t∈I. Then H0=idN, the map H ⁣:N×I→N is smooth, and each Ht is a diffeomorphism with inverse Ht−1=Ψ0,t, by the cocycle law in [F5]. Since supp⁡Xt′⊆V for every t, every trajectory of X′ starting outside V is constant, so Ht=idN on N∖V for every t; a fortiori Ht is the identity outside W, because V⊆W. Finally Xt′=0 whenever t≤ε2 or t≥1−ε2, because β vanishes there. Uniqueness of the evolution with zero velocity gives Ψt,s=idN when s,t belong to either one of those intervals. Starting at time zero therefore gives Ht=idN on the initial interval. On the terminal interval the cocycle law gives H1=Ψ1,t∘Ht=Ht, so the ambient isotopy is stationary at its final map there.

4.1F5step 2.1step 3.1algebra

The isotopy identity. Fix x∈M and consider γ(t):=Ft(x). Then γ(0)=F0(x), and for every t∈I the derivative is γ′(t)=∂tF(x,t). Since γ(t)∈K for all t, we have ψ(γ(t))=1, and therefore Xt′(γ(t))=β(t) ∂tF(x,t). If t∈[ε,1−ε] then β(t)=1 and this equals γ′(t); if t∉[ε,1−ε] then either β(t)=0 or, by the standing hypothesis that F is constant near the ends, ∂tF(x,t)=0; in both cases Xt′(γ(t))=0=γ′(t). Hence γ solves the initial-value problem γ˙=Xt′(γ), γ(0)=F0(x), and by the defining property of the evolution operator in [F5] the unique solution is t↦Ψt,0(F0(x))=Ht(F0(x)).

5.1step 3.1step 4.1∎

Conclusion. Step 4.1 gives Ht∘F0=Ft for every t∈I; step 3.1 gives that H0=idN, that every Ht is a diffeomorphism, that Ht=idN outside W, and that Ht=idN for t∈[0,ε2] and Ht=H1 for t∈[1−ε2,1]. Thus H is an ambient isotopy of N extending the isotopy F of F0(M) and supported in the prescribed neighbourhood W.

Remarks

  • The argument is the standard proof of the isotopy extension theorem: the velocity of the isotopy is read as a vector field along the image of the trajectory (t,x)↦(t,Ft(x)), extended to the ambient manifold, truncated to a prescribed neighbourhood, and integrated. The neighbourhood-retraction corollary and the Euclidean tubular-neighbourhood theorem recorded as dependencies of this item are the standard alternative suppliers of the same extension step; the proof above quotes the vector-field extension lemma directly.
  • Compactness of M is used twice: to make F(M×I) compact, so that W may be taken with compact closure and the field truncated, and to make Θ proper, so that S is closed in J×N and the extension lemma [F4] applies in its global form.
LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Every geometric braid is braid-isotopic to a smooth braid

Statement

Assume ACω. Let β=(z1,…,zn) be a geometric braid on n strands based at Q. Then for every ρ>0 there is a braid β′=(z1′,…,zn′) all of whose strand maps zj′ ⁣:I→D∘ are smooth, together with a braid isotopy Z from β to β′ such that

∣Zj(s,t)−zj(t)∣<ρfor all j, s, t.

Moreover the isotopy is relative to the endpoints in the precise sense of Braid isotopy relative to the top and bottom endpoints: every slice Z(s,⋅) is a braid based at Q, so every bottom endpoint is the fixed point Zj(s,0)=qj, and the top endpoints may be taken individually fixed, Zj(s,1)=zj(1), so that the endpoint permutation is preserved. If the given braid is already smooth on a neighbourhood of t=0 and t=1, the construction below may be centred there and the isotopy may be taken trivial on a neighbourhood of the endpoints; for an arbitrary continuous braid the endpoint values remain fixed as above, and neighbourhood agreement cannot be required unless the original strands are smooth on such a neighbourhood.

Facts & Assumptions

Given: ACω, a geometric braid β=(z1,…,zn) based at Q (Geometric braids in the disc with setwise endpoints), and a real number ρ>0.

[F1]

ACω is the countable axiom of choice (The Axiom of Countable Choice (ACω)).

[F2]

Assume ACω. Let F ⁣:M→Rk be continuous on a smooth manifold M and let ε ⁣:M→(0,∞) be a positive continuous error function. Then there exists a smooth map F~ ⁣:M→Rk with ∥F~(p)−F(p)∥<ε(p) for all p∈M (Whitney approximation for Euclidean-valued maps).

[F3]

A geometric braid on n strands based at Q is an n-tuple of continuous maps zj ⁣:I→D∘ with zi(t)≠zj(t) for i≠j, zj(0)=qj for every j, and {z1(1),…,zn(1)}={q1,…,qn} (Geometric braids in the disc with setwise endpoints).

[F4]

A braid isotopy Z=(Z1,…,Zn) from β to β′ is an n-tuple 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) (Braid isotopy relative to the top and bottom endpoints).

Proof

technique · direct
1.1F3givenalgebra

Empty, singleton and uniform margins. For n=0 choose the empty smooth braid and constant empty isotopy; every assertion is vacuous. Henceforth n≥1. The finite disk-boundary function t↦min⁡j(1−∣zj(t)∣) is positive and continuous on I, so has positive minimum. If n≥2, the finite collision function t↦min⁡i<j∣zi(t)−zj(t)∣ also has positive minimum. Choose m>0 at most both minima when n≥2, and at most the boundary minimum alone when n=1. Put η=min⁡{ρ/4,m/8}, so 3η<ρ and 3η<m/2. No minimum of an empty pair list is used.

2.1F1F2step 1.1construct

Approximate on a boundaryless domain. Extend the continuous tuple z:I→R2n to R by the constant tuple z(0) for t<0 and z(1) for t>1. This extension is continuous because its values agree at 0,1. Apply [F2] on the boundaryless smooth manifold R, with constant error η, and restrict the resulting smooth map to I. It gives g=(g1,…,gn) with ∣gj(t)−zj(t)∣<η for every j,t. Thus no boundaryless Whitney assertion is applied directly to I.

3.1F3step 2.1construct

Exact endpoint collars, including preserved smooth germs. Choose δ∈(0,1/2) so ∣zj(t)−qj∣<η on [0,δ] and ∣zj(t)−zj(1)∣<η on [1−δ,1] for all j. Take smooth cutoffs χ0,χ1 supported in these disjoint collars and equal to one on the half-sized endpoint collars. Ordinarily use the constant collar data aj0(t)=qj, aj1(t)=zj(1). If the original strands are smooth in an endpoint neighbourhood, shrink the corresponding collar into that neighbourhood and instead use aj0(t)=zj(t) or aj1(t)=zj(t) there. The products with their cutoffs extend smoothly by zero off the collars. Define hj=(1−χ0−χ1)gj+χ0aj0+χ1aj1. This is smooth with hj(0)=qj, hj(1)=zj(1). In the already-smooth case it agrees with zj throughout the smaller original endpoint neighbourhood, rather than replacing that neighbourhood by a constant.

4.1step 1.1step 2.1step 3.1algebra

All collar choices satisfy the same error bound. On either collar, the replacement error ∣ajℓ−zj∣ is less than η for constant data and zero for preserved original data. Since the cutoffs have disjoint supports and weights sum to one, ∣hj−zj∣≤(1−χ0−χ1)∣gj−zj∣+χ0∣aj0−zj∣+χ1∣aj1−zj∣<3η≤3m/8<m/2. This covers both arbitrary continuous endpoints and already-smooth endpoint neighbourhoods.

5.1F3step 1.1step 3.1step 4.1algebra

The repaired motion is a braid, with the same endpoints. For all i≠j and all t, ∣hi(t)−hj(t)∣≥∣zi(t)−zj(t)∣−∣hi(t)−zi(t)∣−∣hj(t)−zj(t)∣>m−3m8−3m8=m4>0, so the values h1(t),…,hn(t) are pairwise distinct; and ∣hj(t)∣≤∣zj(t)∣+3m8<1, so they lie in D∘. Together with hj(0)=qj and {hj(1)}={zj(1)}={qj} from [F3] and step 3.1, this shows that h is a braid based at Q, smooth by step 3.1, whose endpoint permutation equals that of β because hj(1)=zj(1) for every j.

6.1F4step 3.1step 5.1algebra

The straight-line isotopy. Define Zj(s,t):=(1−s)zj(t)+s hj(t) for (s,t)∈I×I. It is jointly continuous, Z(0,⋅)=β and Z(1,⋅)=h, and every slice is a braid based at Q: for all s,t,i≠j, ∣Zj(s,t)−zj(t)∣=s∣hj(t)−zj(t)∣<3m8,∣Zj(s,t)∣≤∣zj(t)∣+3m8<1, so the n values are pairwise distinct (their pairwise distances are at least m−3m8−3m8=m4>0) and lie in D∘; moreover Zj(s,0)=(1−s)qj+sqj=qj and Zj(s,1)=(1−s)zj(1)+s zj(1)=zj(1) for every s. Hence Z is a braid isotopy from β to the smooth braid h.

7.1F4step 1.1step 3.1step 6.1algebra∎

Conclusion. By step 6.1 the smooth braid h is braid-isotopic to β through the isotopy Z, whose deviation from β satisfies ∣Zj(s,t)−zj(t)∣=s∣hj(t)−zj(t)∣<3η<ρfor all j,s,t, by the choice of η in step 1.1. The isotopy keeps every bottom endpoint fixed pointwise and every individual top endpoint fixed, so the endpoint permutation is preserved by step 5.1. Where a smooth original collar was retained in step 3.1, h=z there, so the entire straight-line isotopy is also fixed there. Taking β′:=h proves the statement for the prescribed ρ, and ρ>0 was arbitrary.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

The closure of a geometric braid

Definition

Let n≥0 and let β=(z1,…,zn) be a geometric braid with the fixed real basepoints qj=((2j−n−1)/(4(n+1)),0) and endpoint permutation π, as in Geometric braids in the disc with setwise endpoints. Identify the disk with {x∈C:∣x∣<1} and put V=(D∘×[0,1])/((x,1)∼(x,0)). The quotient and its topology use 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 the circle R/Z of The circle as S1=R/Z with basepoint [0]. The interval images sj={[(zj(t),t)]:0≤t≤1},cl⁡(β)=⋃j=1nsj are the closure in the open solid torus. In general an individual sj is an interval image, not a closed circle: its last point is the first point of sπ(j).

Components and orientation. For a cycle (j,π(j),…,πk−1(j)) concatenate these k strand maps in that order, with parameter in [0,k]. Their endpoints agree in V, so the concatenation factors through R/kZ. It is injective on this circle: points at distinct nonintegral heights can agree only when their parameters have the same fractional part, and the distinct strands at that height have distinct disk coordinates; at integral heights only the specified adjacent endpoints agree. Thus each cycle gives an embedded circle. Different cycles give disjoint circles, and every strand belongs to exactly one cycle. These circles are precisely the connected components of the closure. The orientation is increasing concatenation parameter. In particular the number of components equals the number of cycles of π.

The standard axis and fixed framing. In S3={(z,w)∈C2:∣z∣2+∣w∣2=1}, as in Euclidean spheres and closed balls as subspaces of Rn, set A={(0,w):∣w∣=1},Pθ={(z,w)∈S3:z≠0, arg⁡z=θ}. Fix the following particular diffeomorphism, including its disk framing: φ([(x,t)])=(1−∣x∣2 e2πit,x):V⟶S3∖A. Its inverse has disk coordinate x=w and circle coordinate [t]=[arg⁡z/(2π)]. Every Pθ is consequently an open disk. The raw topological oriented closure is β^top=φ(cl⁡(β)). This explicit φ is part of the construction; the definition does not allow an arbitrary page-preserving change of framing.

Page intersections and smooth representatives. The whole closure meets every page in exactly n points. A component belonging to a π-cycle of length k meets each page in exactly k points: at any fractional height it uses exactly those k distinct strands. The page coordinate of its concatenation is [t], for t∈R/kZ, so its oriented degree is +k. If all disk-coordinate strand maps are smooth and have all positive-order derivatives zero at their endpoints, their cycle concatenations have matching jets at every seam. The page coordinate has nonzero derivative, so the result is a smooth embedding of disjoint oriented circles, in the sense of Smooth embeddings. For an endpoint-flat smooth braid, its literal φ-image is therefore a smooth oriented link and is denoted β^.

Closure in the smooth oriented-link category. The raw image need not be smooth, even if individual strands are smooth but their endpoint jets fail to match at the cycle seams. For an arbitrary continuous braid, assume ACω (The Axiom of Countable Choice (ACω)) and choose a smooth representative βs by Every geometric braid is braid-isotopic to a smooth braid. Make it endpoint-flat by composing all disk-coordinate strands with a fixed smooth nondecreasing ψ:I→I, equal to 0 near 0 and 1 near 1. Interpolation of height maps (1−u)t+uψ(t) is an endpoint-fixed braid isotopy, because every strand is evaluated at the same height. The smooth-category closure of β is the oriented-link isotopy class of this chosen model's literal φ-image; when a subset is needed, use that chosen closed-braid representative. Independence of the representative is proved by The closure depends only on the braid isotopy class ↗. Page and cycle properties refer to the chosen constructed closed braid, with the same number of strands and endpoint permutation as β.

If β already has smooth strands with matching cycle-seam jets, in particular if it is endpoint-flat, its literal raw image is a smooth link and is retained as β^. The raw quotient, fixed φ, finite cycles, page counts and trivial closures remain choice-free. Finite elementary words have explicit smooth endpoint-flat models, so forming their literal closures also requires no choice assumption; ACω is used for the general continuous-representative convention and its independence theorem.

The trivial closure. For the constant braid its circles are {(1−qj2e2πit,qj):[t]∈R/Z}. They are latitudes of the sphere S={(z,w)∈S3:Im⁡w=0}, with Re⁡w=qj. Each northern cap Cj={p∈S:Re⁡w(p)≥qj} is a smooth disk: stereographic coordinates on S give radius (1−qj)/(1+qj) for this cap. To make their spanning disks disjoint, let e be the unit vector in the Im⁡w direction, set ε=π/16 and fj(p)=ε(Re⁡w(p)−qj) on Cj, and use Fj(p)=cos⁡(fj(p))p+sin⁡(fj(p))e. The boundary is fixed. The tubular parametrization (p,s)↦cos⁡(s)p+sin⁡(s)e is injective for ∣s∣<π/2, and 0≤fj<π/8. On overlapping caps, if qi<qj then fi−fj=ε(qj−qi)>0, so their graphs are disjoint. Thus these are pairwise disjoint smooth spanning disks. The trivial n-braid closes to the oriented n-component unlink; for n=1 this is the unknot.

At n=0 all strand and cycle unions are empty, every page has zero intersections, and the closure is the empty link. The basepoint list and disk constructions above then have no entries.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

The closure depends only on the braid isotopy class

Statement

Assume ACω. If β,β′ are braid-isotopic geometric n-braids based at Q, then their closures are equivalent oriented links. More precisely, the braid isotopy induces an isotopy of the closures through closed n-braids about the standard axis A, hence an ambient isotopy of S3 carrying β^ to β′^. Consequently the closure construction is well defined on braid isotopy classes.

Facts & Assumptions

Given: ACω, two braids β,β′ based at Q (Geometric braids in the disc with setwise endpoints), a braid isotopy Z from β to β′ (Braid isotopy relative to the top and bottom endpoints), and the closure construction of The closure of a geometric braid with its diffeomorphism φ ⁣:V→S3∖A and axis A.

[F1]

Every geometric braid is braid-isotopic to a braid whose strand maps are smooth, by an isotopy arbitrarily close to it that keeps each bottom endpoint qj and each individual top endpoint fixed (Every geometric braid is braid-isotopic to a smooth braid, The Axiom of Countable Choice (ACω)). AC_omega is assumed there and is inherited here.

[F2]

Under ACω, a smooth isotopy F ⁣:M×I→N of a compact boundaryless manifold through embeddings, constant near the ends, extends to an ambient isotopy H of N with Ht∘F0=Ft, supported in any prescribed neighbourhood of the image (A smooth isotopy of a compact manifold extends to an ambient isotopy).

[F3]

Raw topological closure uses the fixed φ([(x,t)])=(1−∣x∣2e2πit,x) and one component per permutation cycle. Under ACω, the smooth closure of a general continuous braid is formed from a chosen smooth endpoint-flat representative; its selected closed-braid model has exactly n points in every page. A literal smooth matching-jet raw closure is retained (The closure of a geometric braid).

[F4]

A braid isotopy from β to β′ is an n-tuple Z=(Z1,…,Zn) of jointly continuous maps on I×I such that every slice Z(u,⋅) is a braid based at Q, with Z(0,⋅)=β and Z(1,⋅)=β′; the top endpoints Zj(u,1) are independent of u (Braid isotopy relative to the top and bottom endpoints).

Proof

technique · direct
1.1F2F3F4construct

Smooth braid isotopies give isotopies of closures. Assume the family is smooth and its disk-coordinate strands are constant on collars of height 0,1. The endpoint permutation π is independent of the family parameter by [F4]. For each π-cycle of length k, concatenate its k strands on R/kZ, exactly as in [F3]. Their fixed endpoint collars make all jets agree at the seams. Thus the source is the compact one-dimensional manifold M=⨆cycles of πR/kZ, and the concatenations followed by the fixed φ give a smooth family Fu:M→S3∖A. Each Fu is an embedding by the distinct-points condition and the component argument of [F3], with its orientation inherited from the increasing cycle parameter. Reparametrize u to make the family constant near 0,1. By [F2] this isotopy extends to an ambient isotopy of S3, supported away from A in a neighbourhood of its compact image. Each image still has exactly n intersections with every page. If a literal smooth closure has matching cycle-seam jets but is not constant in endpoint collars, interpolate its common height map from the identity to the fixed flat map of [F3]. The matching jets remain matching in every smooth parameter slice, so this gives another compact cycle embedding family and [F2] identifies that literal closure with its constant-collar model.

1.2F1F3F4givenconstruct

Smoothing a braid isotopy with all seams fixed. First choose smooth representatives at the two ends by [F1] and give them endpoint collars by the fixed height reparametrization of [F3], and concatenate their approximation homotopies with the given family. The empty family needs no approximation. For n≥1 the disk-boundary distance has positive uniform minimum on the compact parameter square; for n≥2 include the finitely many pairwise strand distances as well. Use the boundary minimum alone at n=1. Reparametrize height and family parameters to make the family constant in height collars and equal to the two smooth end braids in family-parameter collars. Approximate its finitely many real coordinate functions by tensor-product Bernstein polynomials on the square. For a continuous scalar function f on [0,1] and N≥2, with out-of-range binomial coefficients taken as zero, the binomial weights sum to one, have mean t and variance t(1−t)/N≤1/(4N), by the identities k(Nk)=N(N−1k−1) and k(k−1)(Nk)=N(N−1)(N−2k−2). Uniform continuity gives error at most ϵ on ∣k/N−t∣<δ. The remaining weight is at most 1/(4Nδ2), since its squared deviation is at least δ2 per unit weight. Thus the total error is at most ϵ+2∥f∥∞/(4Nδ2), uniformly in t, and tends to zero. Apply this to each of the finitely many real coordinates, successively in the two variables; convex averaging is a contraction for the uniform norm, so the two errors add. This proves the required uniform square approximation. Choose error smaller than one tenth of that separation. Repair each family-parameter edge by adding a smooth collar cutoff times the difference between the prescribed smooth edge and the approximant's restriction to that edge; the two family collars are disjoint, and these corrections have norm at most the approximation error. Then repair the two height edges to their fixed points by the analogous disjoint height cutoffs. On the family edges these latter corrections vanish because the prescribed end braids already have the correct height endpoints. The resulting map is smooth on the square, fixes all four edges, and differs from the collared continuous family by less than five times the chosen error, so remains in the disk with all strands distinct. Finally compose its height variable with a smooth map constant near 0,1 and equal to the identity outside the original fixed height collars, and its family variable with one constant near 0,1; this makes all height jets agree with the fixed endpoints and all family end collars constant, without changing the braids at those ends. The same uniform margin permits straight interpolation to the collared family. Thus this is a smooth braid isotopy between the chosen smooth representatives, with the one fixed endpoint permutation throughout.

2.1F1F2F3step 1.1step 1.2construct

Arbitrary chosen models and independence. Choose any smooth endpoint-flat models βs,βs′ of the given braids. Their endpoint-fixed approximation isotopies, the given Z, and the inverse approximation isotopy form a continuous braid family between them. Step 1.2 makes this into a smooth family of collared models, and step 1.1 gives ambient equivalence of their selected literal smooth closures. If either chosen model is smooth with matching jets but lacks constant collars, the explicit height interpolation of the Definition [F3] preserves all matching cycle-seam jets, and the compact-cycle argument of step 1.1 supplies the same equivalence. Thus the result holds for every permitted choice of model. Applying the same argument with β′=β proves independence of the smooth-category closure class in [F3]; it is established here, rather than assumed from the Definition. No ambient smooth isotopy of a nonsmooth raw image is used. All constructed model images remain closed n-braids about A.

3.1F1F2step 2.1∎

Conclusion. Every braid isotopy from β to β′ therefore yields an ambient isotopy of S3 carrying β^ to β′^, so the closure construction factors through the braid isotopy class; the two uses of ACω are exactly the smoothing of [F1] and the ambient isotopy extension of [F2].

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Free homotopy classes of loops are conjugacy classes

Statement

Let X be a path-connected topological space with basepoint x0, and let α,β ⁣:I→X be loops at x0. Write I=[0,1]. Two loops α,β at x0 are freely homotopic when there is a continuous map H ⁣:I×I→X with

H(s,0)=α(s),H(s,1)=β(s),H(0,t)=H(1,t)(s,t∈I),

so that the two boundary paths t↦H(0,t)=H(1,t) coincide but need not be constant. Free homotopy is an equivalence relation on the loops at x0; its classes are the free homotopy classes. Then α and β are freely homotopic if and only if their classes in π1(X,x0) are conjugate:

β is freely homotopic to α⟺[β]=[γ]−1[α][γ]  for some loop γ at x0.

Consequently the free homotopy classes of loops in X correspond bijectively to the conjugacy classes of π1(X,x0) (The conjugacy class Cl⁡G(x) and centralizer CG(x) of an element).

Facts & Assumptions

Given: A path-connected topological space X with basepoint x0, two loops α,β ⁣:I→X at x0, and a free homotopy H from α to β with boundary loop γ(t):=H(0,t)=H(1,t).

[F1]

Two based loops at x0 are equivalent when they are path-homotopic relative to the endpoints, the multiplication on loop classes is [α][β]:=[α∗β], the constant loop is cx0, and the reversed loop is αˉ(s)=α(1−s) (Based loops and the fundamental group).

[F2]

For every pointed space the product [α][β]=[α∗β] is well defined and makes π1(X,x0) a group; its identity is the class of the constant loop cx0, and [α]−1=[αˉ] (Loop classes form the group π1(X,x0) under concatenation).

[F3]

The conjugacy class of an element x of a group is Cl⁡G(x)={gxg−1:g∈G} (The conjugacy class Cl⁡G(x) and centralizer CG(x) of an element).

[F4]

A path homotopy from α to β relative to the endpoints is a homotopy H:I×I→X rel {0,1}, that is, with H(0,t)=α(0)=β(0) and H(1,t)=α(1)=β(1) for all t (Homotopies of continuous maps, homotopies relative to a subspace, and path homotopies relative to the endpoints).

[F5]

If f:X→Y and g:Y→Z are continuous then g∘f is continuous; and if F1,…,Fn are closed subsets of a space X with F1∪⋯∪Fn=X and f∣Fk is continuous for every k, then f is continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous).

Proof

technique · direct
1.1F5F6F7givenconstruct

The square homotopy that realizes the conjugation. Assume a free homotopy H from α to β is given and let γ(t):=H(0,t)=H(1,t); then γ is a loop at x0, because γ(0)=H(0,0)=α(0)=x0 and γ(1)=H(0,1)=β(0)=x0. Define the piecewise-affine path in the square B ⁣:I→I×I by B(s):=(0,1−4s) for s≤14, B(s):=(4s−1,0) for 14≤s≤12, and B(s):=(1,2s−1) for s≥12, and define φ(s,t):=(1−t)(s,1)+t B(s)∈I×I and J:=H∘φ ⁣:I×I→X. For joint continuity on a rectangle, the product topology has neighborhoods ∣s−s0∣<δ, ∣t−t0∣<δ. At a point (a0,b0), ∣(a+b)−(a0+b0)∣≤∣a−a0∣+∣b−b0∣ and ∣ab−a0b0∣≤∣a∣ ∣b−b0∣+∣b0∣ ∣a−a0∣, with ∣a∣≤∣a0∣+1 when ∣a−a0∣<1. These estimates prove joint continuity of addition and multiplication; composing them with continuous coordinates gives continuity of each polynomial expression used below. Every piece of B is affine, so B is continuous by [F5] and [F6]; on each of the three closed pieces every component of φ is a sum of products of affine coordinate functions, hence is continuous by these estimates; the pieces agree on their seams, so φ is continuous by [F5], [F6] and [F7]; hence J is continuous by [F5].

2.1F5F6F7step 1.1givenconstruct

The moving-basepoint homotopy. Assume now that γ is an arbitrary loop at x0 and define, for (s,u)∈I×I, Ru(s):=γ(u(1−4s)) for s≤14, Ru(s):=α(4s−1) for 14≤s≤12, and Ru(s):=γ(u(2s−1)) for s≥12. Each Ru is a loop at γ(u), because the three pieces join continuously at s=14 and s=12, they take the values γ(u),x0,x0,γ(u) at s=0,14,12,1, and each argument supplied to γ or α is a polynomial in (s,u) on its closed rectangle. These arguments are continuous by the estimates in step 1.1, and composing with the continuous loops is permitted by [F5]; consequently the map R ⁣:I×I→X, R(s,u):=Ru(s), is continuous by [F5], [F6] and [F7].

2.2F4step 1.1givenalgebra

The square homotopy is a path homotopy from β to (γˉ∗α)∗γ. For J of step 1.1: J(s,0)=H(φ(s,0))=H(s,1)=β(s); J(s,1)=H(B(s)) and the three pieces of B give H(B(s))=γ(1−4s)=γˉ(4s) for s≤14, H(B(s))=α(4s−1) for 14≤s≤12 and H(B(s))=γ(2s−1) for s≥12, which is exactly the loop (γˉ∗α)∗γ of [F1]; finally φ(0,t)=(1−t)(0,1)+t(0,1)=(0,1) and φ(1,t)=(1−t)(1,1)+t(1,1)=(1,1), so J(0,t)=H(0,1)=β(0)=x0 and J(1,t)=H(1,1)=β(1)=x0 for every t. Hence J is a path homotopy relative to the endpoints from β to (γˉ∗α)∗γ in the sense of [F4].

3.1F1step 2.1given

The moving-basepoint family is a free homotopy. For the family R of step 2.1 one has R(s,0)=cx0-insertions: R0(s)=x0 for s≤14 and s≥12, while R0(s)=α(4s−1) in between, so R0=cˉx0∗α∗cx0 is the loop obtained from α by adjoining constant loops; and R1=γˉ∗α∗γ. Since R(0,u)=γ(u)=R(1,u) for every u, the family R is a free homotopy from R0 to R1=γˉ∗α∗γ in the sense of the Statement.

3.2F1F2step 2.2algebra

Free homotopy implies conjugacy. By step 2.2 and [F1] the classes satisfy [β]=[(γˉ∗α)∗γ]=[γˉ][α][γ]=[γ]−1[α][γ], using associativity of the group product and [γˉ]=[γ]−1 from [F2].

4.1F1F2F5F6step 3.1givenalgebra

Conjugacy implies free homotopy. Conversely, let γ be a loop at x0 with [β]=[γ]−1[α][γ]. By [F2] the class of R0=cˉx0∗α∗cx0 is [cx0]−1[α][cx0]=[α], so by [F1] there is a path homotopy relative to the endpoints from α to R0; by step 3.1 the family R is a free homotopy from R0 to γˉ∗α∗γ; and [γˉ∗α∗γ]=[γ]−1[α][γ]=[β], so again by [F1] there is a path homotopy relative to the endpoints from γˉ∗α∗γ to β. Reparametrising the homotopy parameter by the three-part affine map t↦3t, t↦3t−1, t↦3t−2 on [0,13],[13,23],[23,1] and pasting the three homotopies, which agree on the seams, gives one continuous K ⁣:I×I→X with K(s,0)=α(s), K(s,1)=β(s) and K(0,t)=K(1,t) for every t: the pasting is licensed by [F5] and the affine reparametrisation by [F6]. So α and β are freely homotopic.

5.1F3F5step 3.2step 4.1∎

Conclusion. Step 3.2 shows that a free homotopy from α to β produces a conjugating loop γ with [β]=[γ]−1[α][γ], and step 4.1 shows conversely that each conjugating relation produces a free homotopy. Hence α and β are freely homotopic if and only if their classes are conjugate. Moreover free homotopy is an equivalence relation: it is reflexive via the constant homotopy H(s,t):=α(s), symmetric by reversing the deformation parameter, and transitive by the pasting argument of step 4.1. Therefore the free homotopy classes of loops at x0 are in bijection with the conjugacy classes of π1(X,x0), the map being induced by α↦[α].

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Two disjoint circles in the two-sphere cobound an annulus

Statement

Assume the Axiom of Choice. Let C,C′ be two disjoint smoothly embedded circles in S2 (Smooth embeddings). Then S2∖(C∪C′) has exactly three components: two of them are open disks and one of them is an open annulus whose boundary is C∪C′. Consequently C and C′ cobound a unique annulus A, and each of the two complementary disks is bounded by one of the circles.

Facts & Assumptions

Given: AC, two disjoint smoothly embedded circles C,C′ in S2, and the standard model S2=R2∪{∞} as the one-point compactification of the plane (The one-point (Alexandroff) compactification X∗=X∪{∞}, whose open sets are the open sets of X together with the complements in X∗ of the closed compact subsets of X).

[F1]

Assume AC. The image of a topological embedding S1↪S2 has exactly two complementary path components, and it is their common boundary; for an embedding into R2 there are exactly two complementary components, one bounded and one unbounded, with the curve as common boundary (Jordan–Brouwer separation). AC is inherited only from Alexander duality.

[F2]

Assume AC. Every homeomorphism between Jordan curves in R2 extends to a homeomorphism of R2 mapping the bounded complementary component of the first onto the bounded complementary component of the second; in particular each closed bounded Jordan region is a closed 2-disk (Jordan–Schönflies extension for plane curves).

[F3]

A smooth embedding of S1 is injective, an immersion, and a homeomorphism onto its image, hence a Jordan curve (Smooth embeddings).

[F4]

Assume AC. A smoothly embedded compact submanifold S⊆Rm has a tubular neighbourhood: there is a positive smooth function δ on S such that the normal addition map on {(p,v)∈N⊥S:∣v∣<δ(p)} is a diffeomorphism onto an open neighbourhood of S (The Euclidean tubular neighbourhood theorem).

[F6]

Every open connected subset of Rn is polygonally connected, hence path connected (Every connected component of an open subset of Rn is open and polygonally connected).

[F8]

Assume AC. Let a finite 2-connected graph be drawn in the plane by simple arcs meeting only at common endpoints, let one cycle C be drawn as a Jordan curve, and let every other edge be a simple polygonal arc whose relative interior lies in the bounded component of R2∖C. Then every component of the complement of the drawing has a graph cycle as its boundary, and V−E+F=2 where F is the number of complement components (Finite plane graph ear and face facts).

Proof

technique · direct
1.1F1F2F3given

Reduction to a planar picture and Jordan–Brouwer. By Jordan separation, C has two complementary components and the connected circle C′ lies in one. Choose o in the other component; in the model S2=R2∪{∞} of the Given we may suppose o=∞, since otherwise we replace the model by its image under a smooth rotation of S2 carrying o to ∞; the number of complementary components and their homeomorphism types are unchanged by a homeomorphism. This pole choice puts C′ in the bounded component of R2∖C. Then C,C′⊂R2, and by [F3] and [F1] each curve has exactly two complementary components with the curve as their common boundary; write D and D′ for the bounded components of R2∖C and R2∖C′. By [F2] the closed regions D‾ and D′‾ are closed 2-disks, hence D and D′ are open disks.

2.1F1F3step 1.1algebra

The pole gives actual nesting. By the pole choice in step 1.1, C′⊂D. The exterior of D‾ is connected and unbounded, misses C′, and thus lies in the unbounded complementary component of C′. Since the two compact circles are disjoint, a collar of C also misses C′, so C lies in that same unbounded component. Hence the closed bounded region D′‾ misses C and its exterior and is contained in D. This uses the specified pole; two side-by-side curves in a previously fixed planar chart would not have this nesting.

3.1F3F4step 1.1step 2.1construct

Topological polygonal reduction in disjoint collars. It suffices to prove the topological annulus assertion for polygonal curves. A smooth embedded circle has a smooth tubular collar by [F4]. A sufficiently fine inscribed polygon in that collar projects one-to-one onto the circle: on every sufficiently short regular arc its tangent has positive component in the original tangent direction, so the projection is locally monotone; compact separation of distant arcs excludes other intersections. Thus the polygon is a continuous normal graph r=u(s) with arbitrarily small displacement. In collar coordinates, choose a smooth cutoff χ(r) equal to 1 near r=0 and 0 near the collar boundary, and make the approximation so small that ∥χ′∥∞∥u∥∞<1. The map Ht(s,r)=(s,r+tχ(r)u(s)) is a homeomorphism: on each normal fibre it is strictly increasing, fixes the ends, and has continuous inverse, since its increase is bounded below by (1−∥χ′∥∞∥u∥∞) times the fibre increment. It is identity off the collar and sends the original curve to its polygon at t=1. Choose the two collars disjoint and apply these maps simultaneously. This is a topological ambient isotopy, sufficient for complementary homeomorphism types; no smooth isotopy extension is applied to a polygonal endpoint. The two nested polygonal curves retain their disks and middle region under this homeomorphism.

4.1F6step 2.1step 3.1construct

A crosscut and its finite polygonal collar. Now the curves are polygonal and D′‾⊂D. Put M=D∖D′‾. This open region is nonempty and connected: join any two of its points by a polygonal path in the connected disk D using [F6], perturb its segments to avoid vertices of C′ and cross its edges transversely, and replace every run through D′‾ by a path in a thin exterior polygonal collar of C′. Compact nesting keeps that collar inside D; its side strips and vertex sectors connect around the whole polygon. Finitely many detours give a path in M, so [F6] makes it polygonally connected. Choose p∈C,q∈C′ in edge interiors and short straight access segments entering M. Join their other ends in M, perturb to make all intersections finite, subdivide there and erase cycles in the resulting finite edge walk. Trim in the endpoint collars to obtain a simple polygonal crosscut P with interior in M. Construct a companion P1 on one fixed side of P: choose disjoint small disks at its bends, disjoint thin rectangles on its truncated straight segments, and endpoint rectangles meeting only the appropriate edge interiors of C,C′. Finitely many nonincident pieces have positive separation; choose all widths smaller than it. Offset each segment in those rectangles and connect its offsets in the intervening bend sectors by short bevels. Since each bend sector joins just its two consecutive rectangles and the neighbourhoods miss all nonincident pieces, this is an embedded polygonal P1, disjoint from P, ending at nearby p1∈C,q1∈C′. The rectangles and bend sectors between the two arcs, with the short boundary intervals α1⊂C,α2⊂C′, form a closed thin strip S⊂M‾. Its boundary is precisely P∪P1∪α1∪α2, with no unintended intersection or boundary portion.

5.1F1F2F8step 4.1algebra

The four faces, not all graph cycles. Use the endpoints p,p1∈C, q,q1∈C′ from step 4.1, and divide each boundary into two arcs α1,α1′ and α2,α2′, with α1,α2 bounding the thin strip together with P,P1. The embedded graph G=C∪C′∪P∪P1 has four vertices, six edges and is 2-connected: after deleting any vertex, the surviving bridge and the remaining boundary arcs still connect it. Subdivide each of its six edges once to get a simple 2-connected graph, with ten vertices and twelve edges and the same faces. By [F8] it has 2−10+12=4 complementary faces, each bounded by a graph cycle. The exterior of C is one face and the interior of C′ another; their boundary cycles are C,C′. The thin-strip interior is a third face, with boundary γ=α1∪P1∪α2∪P. Following the other side of either bridge, the local cyclic order of its three incident edges forces the fourth face walk to use α1′,P1,α2′,P, so its boundary is γ1=α1′∪P1∪α2′∪P. This also follows by exhaustively following the two directed sides of the six edges: the two boundary faces, the strip and this last walk use every edge side once. There are other mixed bridge cycles in G; they are not face boundaries. Let Z be the closure of the strip face and Z1 the closure of the fourth open face. By [F2] these Jordan-bounded closures are closed disks. They satisfy M‾=Z∪Z1 and Z∩Z1=P∪P1: the four-face enumeration exhausts the middle region, and only the two bridges border both middle faces. In particular Z1 is the closure of its open face, not a set obtained by removing a strip interior and retaining unrelated boundary arcs.

6.1F2step 5.1algebra

The middle region is an annulus. By [F2] each of Z and Z1 is a closed 2-disk, and its boundary is divided by the four points p,p1,q,q1 into the four arcs α1,α2,P,P1, respectively α1′,α2′,P,P1, in the cyclic order P,α1,P1,α2 (respectively P,α1′,P1,α2′). A closed disk whose boundary is split by four points is homeomorphic to the square [0,1]×[0,1] with the four boundary arcs corresponding to the four sides; transporting the splitting of Z and of Z1 through such homeomorphisms describes the gluing of step 5.1 as the identification of the left edges of two squares with each other and of the right edges with each other, which is the standard description of S1×[0,1]: the free boundary consists of the two circles α1∪α1′=C and α2∪α2′=C′. Hence M‾≅S1×[0,1] and M≅S1×(0,1) is an open annulus with boundary C∪C′.

7.1step 3.1step 5.1step 6.1∎

Conclusion. In the polygonal case the complement of C∪C′ has exactly the three components D′, (R2∖D‾)∪{∞} and M by steps 1.1, 2.1 and 4.1: the first two are open disks and the third is an open annulus whose boundary is C∪C′. The annulus cobounded by C and C′ is unique, because the interior of any such closed cobounding annulus is the connected complementary component adjacent to both circles, and exactly one component does; and the complementary disks D′ and (R2∖D‾)∪{∞} are bounded by C′ and by C respectively. step 3.1 transfers this conclusion back to the given pair of disjoint smoothly embedded circles in S2: the collar homeomorphism constructed there carries each complementary component of the polygonal pair onto a complementary component of the original pair preserving the homeomorphism types, and it carries the annulus onto the annulus cobounded by C and C′. This proves every claim.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Oriented Reidemeister moves

Definition

An oriented Reidemeister move is one of the following local replacements of a regular oriented link diagram (Regular oriented link diagrams), performed inside a small closed disk while the diagram outside the disk is left unchanged, and understood up to planar isotopy outside the disk (Planar isotopy of link diagrams). All sign and orientation variants of each local picture are included.

(R1) Kink creation and deletion. A single strand is replaced by a small kink: two arcs crossing once and otherwise disjoint from the strand, with the over/under datum of the new crossing as in the picture. The move is allowed in both directions (create or delete the kink) and with both signs of the kink, the sign being the sign of the crossing created. The orientations of the two strands of the kink are the two possible consistent orientations through the crossing, and all of them occur in the move list.

(R2) Slide of two strands. Two local arcs belonging to two strands are replaced so that the number of crossings between those strands changes by two of opposite signs: the standard picture slides one strand over or under the other, creating or deleting the pair of crossings, with the two crossings being the two possible signs in either order. All four orientation patterns of the two strands at the two crossings are included.

(R3) Sliding a strand past a crossing. One strand is moved across an existing crossing of two other strands, so that the crossing of the two other strands passes from one side of the moving strand to the other; the three crossings of the local picture keep their signs and the over/under data are preserved. The three strands must have a consistent strict height order: one is above both others, one below both, and the remaining strand is between them. The cyclic over/under pattern is excluded. Every orientation pattern and all six height orders of this standard local picture are included.

A local R2 or R3 picture is braid-like when, after a local coordinate change and deformation, all participating strands run in the same positive longitudinal direction. This is a local condition and does not require a globally chosen braid axis. R1 is kept as a separate kink move.

The strands keep their orientations throughout every move. The union of the three move types, together with planar isotopy, generates the equivalence relation on diagrams used in the Reidemeister equivalence theorem below; the moves are sometimes called R1, R2, R3 as above. Since every move changes the diagram only inside a disk, it has a type and a finite list of local orientation/sign variants, and each variant is realized by an ambient isotopy as the next items show. The complete list of variants is required because the Markov and Alexander constructions below act on oriented diagrams and count signed crossings.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Each oriented Reidemeister move is realized by an ambient isotopy

Statement

Let D,D′ be regular oriented diagrams of links L,L′. If the diagrams differ by one oriented Reidemeister move or a planar isotopy, the links are equivalent. The local Reidemeister replacement has a realization supported in a ball over its move disk, after choosing lifts that agree outside that ball. Comparing arbitrary realizing links may additionally require changing their heights away from the move disk.

Facts & Assumptions

Given: The regular oriented diagrams and their realizing links, as in Oriented links in the three-sphere and ambient isotopy.

[F1]

The three standard local moves have both signs and all consistent orientation and height-order variants (Oriented Reidemeister moves).

[F2]

A planar isotopy is a smooth ambient isotopy transporting the decorated diagram (Planar isotopy of link diagrams).

[F3]

A compactly supported smooth time-dependent vector field has global smooth evolution, with inverses given by reverse evolution (Compactly supported time-dependent vector fields have global evolution on a compact time interval).

Proof

technique · direct
1.1givenconstruct

Heights over a fixed regular diagram. Use the source identification determined by the oriented branches. Two lifts of the same diagram have the same planar map and smooth height functions h0,h1. Their linear interpolation preserves the strict height difference at every double point, so remains injective; it remains immersive because its projected derivative is nonzero. Compactness then makes every slice an embedding. This gives a smooth isotopy between any two realizing lifts, after a stationary time reparametrization. It generally moves portions outside a specified local disk.

2.1F1step 1.1construct

The local embedding families. Choose standard lifts of the move pictures, with the same boundary germs. For R1 the local spatial family (v2,v3+δv,v) remains embedded for all small δ, since the third coordinate is strictly monotone; its projections on the two sides have respectively one curl crossing and none. Fixed endpoint collars and a cutoff inside a slightly larger disk join this model to the unchanged arc. Changing the sign of the third coordinate gives the opposite crossing. For R2 use two graph arcs (v,0,−η) and (v,v2+δ,+η) on the moving central portions, with fixed collars: one sign of δ gives two crossings, the other none, and their heights keep them disjoint. For R3 use three transverse planar arcs with fixed distinct height levels, and slide one projected arc across the intersection of the other two, with the motion cut off before its boundary collars. Distinct height levels prevent any spatial collision. Permuting those levels covers all six consistent height orders, including a moving strand between the others. Reversing the parameter gives the inverse moves, and orienting each arc as prescribed covers the oriented variants. These are families of embedded arcs, not straight-line interpolations of arbitrary kinks. Standard representatives can be arranged to agree with the unchanged diagram and its chosen heights outside the ball.

3.1F3step 1.1step 2.1construct

Ambient extension with the stated support. For any of these compact smooth embedding families, extend time constantly past its stationary end collars. The graph (t,x)↦(t,Ft(x)) is an embedded submanifold of the time-space product. In a slice chart its vertical velocity (0,∂tF) extends by keeping its coordinate coefficients constant in the normal directions; retain only the spatial component of this extension. Finitely many such charts cover the compact moving trace. Euclidean bump functions, positive on smaller charts and normalized by their finite sum near the trace, combine these extensions into a smooth time-dependent spatial vector field agreeing with the velocity. Multiply by a further cutoff supported in the prescribed open ball, or in a relatively compact neighbourhood of the full trace for step 1.1. Integrate by [F3]. Uniqueness makes the flow follow Ft on the link and fix points outside its support. The construction is finite and uses no choice axiom.

4.1F1F2F3step 1.1step 2.1step 3.1∎

Planar isotopy and conclusion. A planar isotopy lifts by (x,y,z)↦(Φt(x,y),z). Its action on the compact link trace can be cut off in a large spatial ball, by the velocity construction in step 3.1, to extend smoothly over ∞. Step 1.1 adjusts its final heights to the chosen lift of D′. For a Reidemeister move, first adjust to the standard lift, apply the ball-supported family of step 2.1 extended by step 3.1, then adjust to L′. All the families preserve the component orientations. Thus arbitrary realizing links are equivalent, while support in the move ball is asserted precisely for the local replacement with fixed outside lift.

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Reidemeister's theorem for oriented diagrams

Statement

Assume ACω. Let D and D′ be regular oriented link diagrams. Then D and D′ represent equivalent oriented links if and only if D′ can be obtained from D by a finite sequence of planar isotopies and oriented Reidemeister moves R1, R2, R3.

Facts & Assumptions

Given: ACω and regular oriented diagrams D,D′.

[F1]

Each oriented Reidemeister move and each planar isotopy of diagrams is realized by an ambient isotopy of S3 carrying the represented oriented link to the represented oriented link, with ball support for a local replacement between lifts agreeing outside that ball (Each oriented Reidemeister move is realized by an ambient isotopy).

[F2]

Equivalence of oriented links is an equivalence relation, so a finite composition of realized moves produces an equivalence (Oriented links in the three-sphere and ambient isotopy).

[F3]

A regular diagram already has a smooth realizing link by its definition. Realizing lifts with the same decorated diagram are equivalent by the height interpolation in Each oriented Reidemeister move is realized by an ambient isotopy.

[F4]

Assume ACω: a smooth isotopy of links, constant near the ends, can be perturbed, relative to the ends, into general position, with the ordinary cusp, quadratic tangency and transverse triple-event properties of General-position isotopies of links.

[F5]

Under the general-position hypotheses, each exceptional time produces exactly one oriented Reidemeister move up to planar isotopy and the intervals between exceptional times are planar isotopies (Generic isotopies have only Reidemeister singular times).

[F6]

Under countable choice, Sard makes the image of a smooth two-dimensional manifold in S3 null: every differential has rank at most two, less than the target dimension (Morse-Sard for smooth manifolds). Compactly supported smooth vector fields give ambient point motions (Compactly supported time-dependent vector fields have global evolution on a compact time interval).

Proof

technique · direct
1.1F1F2given

The easy direction and empty case. An empty diagram represents the empty link, and the moves preserve component count. When both diagrams are empty they are already identical; neither equivalence nor a move sequence can relate an empty diagram to a nonempty one. If D′ is obtained from D by a finite sequence of planar isotopies and oriented Reidemeister moves, then by [F1] each step is realized by an ambient isotopy of S3 carrying the represented link to the next one, and [F2] makes the composition of the finitely many ambient isotopies again an equivalence; hence D and D′ represent equivalent oriented links.

1.2F2F3F4F6givenconstruct

Keep the whole track off infinity. Let L0,L1⊂R3 realize D,D′; by the assumed equivalence there is an ambient isotopy H of S3 carrying their oriented images to one another. Reparametrize time to make its restricted link isotopy stationary near both ends. Extend that track constantly to C×R; its image is null by [F6]. Choose a small coordinate ball about ∞ disjoint from L0∪L1, and a point q in that ball outside the track. A compactly supported vector field in the ball, equal to the velocity of a short coordinate path from q to ∞, gives a diffeomorphism K with K(q)=∞ and fixing both endpoint links, by [F6]. Then Ft=K∘Ht∣L0 is a smooth link isotopy avoiding ∞ at every time, with exactly the original endpoint diagrams. Its compact trace is contained in a large ball in R3. The regular endpoint projections satisfy the hypothesis of [F4].

2.1F4F5step 1.2

General position and the finite sequence. Apply [F4] to F with a strong neighbourhood small enough to keep the end projections fixed: there is a smooth isotopy F′ in general position whose end projections are D and D′. By [F5] the movement from t=0 to t=1 of the projections of F′ is a finite sequence of planar isotopies and oriented Reidemeister moves taking the projection of F0′ to that of F1′, that is, taking D to D′ up to planar isotopy.

3.1F1F2step 1.1step 2.1∎

Conclusion. Steps 1.1 and 2.1 prove the two implications, so D and D′ represent equivalent oriented links if and only if they are connected by planar isotopies and finitely many oriented Reidemeister moves. The axiom of countable choice is used exactly in [F6] (track avoidance by Sard) and [F4] (parametric transversality); the realization direction is choice-free.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

Coherence of Seifert circles and the height of a diagram

Definition

Assume the Axiom of Choice. Let C,C′ be two disjoint oriented circles in S2 (Seifert smoothing and Seifert circles of an oriented link diagram) and let A be the annulus they cobound, which exists and has the two circles as its boundary circles by Two disjoint circles in the two-sphere cobound an annulus. Orient A once, a choice of one of its two orientations (Oriented smooth manifolds and oriented charts); this orientation induces a boundary orientation on each of the two boundary circles C and C′, and reversing the orientation of A reverses both induced orientations (Induced boundary orientation).

Coherence. The circles C and C′ are coherent when, with respect to one, equivalently any, orientation of A, the given orientations of C and C′ agree with the two induced boundary orientations in opposite senses: one of the given orientations agrees and the other disagrees. Equivalently, the two given oriented circles represent the same element of H1(A;Z), which is the formulation used in the survey. The two formulations are the classical description of one-dimensional coherent orientation of an annulus, and the equivalence of reversing the orientation of A is immediate because both induced boundary orientations flip, so the relation "opposite senses of agreement" is independent of the chosen orientation. When the given orientations agree with the induced boundary orientations in the same sense (both agree or both disagree), the circles are incoherent.

Height. For an oriented diagram D with Seifert picture S and Seifert circles C1,…,Cm (Seifert smoothing and Seifert circles of an oriented link diagram), the height of D is the number of distinct unordered pairs {i,j} of indices such that Ci and Cj are incoherent:

h(D):=#{{i,j}:1≤i<j≤m, Ci and Cj incoherent}.

Coherence is defined for every pair of Seifert circles of the picture, not only for pairs joined by a signed arc; the signed arcs only record the crossings of the original diagram. In particular h(D)=0 means that all pairs of Seifert circles of D are coherent. The axiom of choice is consumed exactly through the annulus lemma, which supplies the annulus A and its two boundary circles; the rest of the definition, including the invariance under reversing the orientation of A, is choice-free. Distinct Seifert circles are disjoint, so the definition applies to every one of the finitely many pairs, including pairs with no signed arc between them. The empty and one-circle pictures have height zero.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

Defect regions, reducing arcs and the Yamada-Vogel reducing move

Definition

Assume the Axiom of Choice. Let D be an oriented diagram with Seifert picture S, consisting of the Seifert circles C1,…,Cm and the finite set of signed arcs recording the crossings (Seifert smoothing and Seifert circles of an oriented link diagram), and let coherence of pairs of Seifert circles be as in Coherence of Seifert circles and the height of a diagram.

A region of S is a connected component of S2∖S, where S includes both the circles and the signed arcs. Thus a region contains no signed arc in its interior. Its boundary can contain signed arcs and portions of Seifert circles. A region is a defect region if two Seifert circles Ci≠Cj that are incoherent both occur in the boundary of that region. The existence of a defect region when the height is positive is the content of the defect-region lemma below.

A reducing arc is a simple arc α contained in a defect region together with its endpoints, joining an incoherent pair Ci,Cj of Seifert circles and meeting the union of the Seifert circles exactly in its two endpoints; the arc may be taken polygonal inside the region.

The Yamada-Vogel reducing move performed along α slides one of the two circles, say Ci, over the other along α: in the diagram this is a Reidemeister II move of the original diagram in which a neighbourhood of the arc is replaced by the standard band picture of two crossings of opposite signs, so that in the new Seifert picture the incoherent pair Ci,Cj is replaced by two coherent Seifert circles Ca,Cz joined by two signed arcs of opposite signs, all other Seifert circles unchanged, Ca bounds a disk containing no other new Seifert circle, and Cz bounds a disk containing all Seifert circles that were contained in the annulus cobounded by Ci and Cj. The inverse of a reducing move is also allowed. A diagram is reducible when it admits a reducing arc; the move is defined for every defect region and the resulting picture is again a Seifert picture of an oriented diagram of the same link.

The choice axiom is used exactly where coherence is used, through the annulus lemma of Two disjoint circles in the two-sphere cobound an annulus; the local band picture of the move itself is an explicit Reidemeister II replacement of two oppositely signed crossings.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

A positive-height diagram has a defect region

Statement

Assume AC. If h(D)>0, the Seifert picture S, including its signed arcs, has a complementary region exposing two incoherent circles. That region supports a reducing arc disjoint from all signed arcs.

Facts & Assumptions

Given: AC and a diagram D of positive height with its finite Seifert picture.

[F1]

Distinct Seifert circles separate S2 into planar regions; two circles have their common annulus and two complementary disks (Two disjoint circles in the two-sphere cobound an annulus).

[F2]

A signed arc joins two distinct coherent circles, and the finite arcs have disjoint interiors (Seifert smoothing and Seifert circles of an oriented link diagram).

[F3]

A defect region is a component of S2∖S exposing an incoherent pair. An embedded arc in it with one endpoint on each exposed circle is a reducing arc (Defect regions, reducing arcs and the Yamada-Vogel reducing move).

[F4]

Coherence means opposite agreement signs relative to the annulus boundary orientations; height counts incoherent pairs (Coherence of Seifert circles and the height of a diagram).

Proof

technique · direct
1.1F1F4givenconstruct

Ignore the signed arcs temporarily. The complementary dual graph of the circles alone has one vertex per region and one edge per circle. Adding a disjoint Jordan circle splits a region in two, so this graph is a finite tree by [F1]. At a vertex with three or more boundary circles, give each its sign relative to the oriented region boundary. For any pair these are also their signs relative to their common annulus boundary. Two equal signs give an incoherent pair by [F4]. Thus three exposed circles always contain such a pair. If every vertex has degree at most two, the tree is a path, the circles form one chain, and coherence of every consecutive pair makes all pairs coherent by transporting their angular orientations along the chain. Positive height therefore implies either a vertex with at least three boundary circles, or a two-boundary annular region whose two circles are incoherent.

2.1F2F3F4step 1.1algebra

Restore the signed arcs. In the two-boundary incoherent case no signed arc can lie in that annulus: by [F2] it would have to join its incoherent boundary circles. Thus it is already a defect region. In a region with k≥3 boundary circles, collapse its capped boundary circles to k vertices on a sphere. The signed arcs give a finite plane multigraph with no loops. Each edge joins circles with opposite boundary signs by [F2], so the graph is bipartite. Some complementary face exposes at least three vertices: otherwise every nontrivial face walk could use only a pair of vertices. Remove successively empty digons between parallel edges. Such removals preserve the assertion that faces expose at most two vertices, since the merged digon uses that same pair. After these removals, a face using only two vertices must be the walk along a single bridge and back, or the boundary of a region between parallel edges containing another component; the latter exposes a vertex of that component as well. A connected component with two adjacent edges to different vertices has a face walk exposing those three vertices. Hence every remaining connected component has at most two vertices; if more than one component remains, their common exterior face exposes all their vertices, including isolated vertices. Since the graph has k≥3 vertices, this contradicts the assumed face property. Choose a face exposing at least three circles and restore the collapsed disks. Two of its circle boundary portions have the same sign, hence are incoherent by [F4], so that face is a defect region of the full picture.

3.1F1F3step 1.1step 2.1construct∎

A reducing arc. Choose points in the relative interiors of the two exposed circle boundary portions, avoiding the finitely many signed-arc endpoints. Short inward access arcs enter the open face. Its connected open planar interior is polygonally connected; join those access arcs inside it, perturb finitely many segments to make intersections finite, and erase loops in the resulting finite path. This gives an embedded arc meeting the entire Seifert picture only at its endpoints. It is the required reducing arc by [F3]. AC is inherited from [F1] and the coherence definition.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

A reducing move lowers the height by one

Statement

Assume the Axiom of Choice. If a diagram D′ is obtained from an oriented diagram D by a Yamada-Vogel reducing move, then h(D′)=h(D)−1. Consequently every sequence of reducing moves starting at D has length at most h(D).

Facts & Assumptions

Given: AC, an oriented diagram D with Seifert circles C1,…,Cm, a reducing arc α joining an incoherent pair Ci,Cj, and the diagram D′ obtained by the reducing move along α (Defect regions, reducing arcs and the Yamada-Vogel reducing move).

[F1]

The pair Ci,Cj is incoherent; in the new Seifert picture the two circles are replaced by two coherent circles Ca,Cz joined by two signed arcs of opposite signs, all other circles are unchanged, Ca bounds a disk Da containing no other Seifert circle of the new picture, and Cz bounds a disk Dz containing all Seifert circles that were contained in the annulus cobounded by Ci and Cj (Defect regions, reducing arcs and the Yamada-Vogel reducing move).

[F2]

Coherence of every pair of Seifert circles is defined through the annulus they cobound; the height is the number of incoherent unordered pairs (Coherence of Seifert circles and the height of a diagram).

[F3]

Two disjoint circles cobound an annulus, whose two complementary regions are the two disks bounded by the circles; the three regions determine which third circles lie in the annulus and which in the two disks (Two disjoint circles in the two-sphere cobound an annulus). AC is inherited from this lemma.

Proof

technique · direct
1.1F1F3givenconstruct

Partition the unchanged circles. Let A be the common annulus and let Di,Dj be its complementary open disks. Each unchanged circle lies entirely in exactly one of these three regions. The reducing strip lies in A and misses every other circle and signed arc. One new boundary surrounds the small empty strip disk Da; the other surrounds the old middle region A after the strip surgery, giving Dz. In particular no unchanged circle is in Da.

2.1F1F2F3step 1.1algebra

Comparing coherences for a third circle. For p∉{i,j} with Cp⊂A, Cp cannot be essential in A, since it would separate the endpoints of the reducing arc. Reading the boundary orientations before and after the strip surgery therefore gives (Cp,Cz)=(Cp,Ca)=(Cp,Ci)=(Cp,Cj): the two new circles are coherent with Cp exactly when the old pair was coherent with Cp. For Cp⊂Di one has (Cp,Cz)=(Cp,Ci) and (Cp,Ca)=(Cp,Cj), and for Cp⊂Dj the two roles are interchanged. These identities follow from the descriptions of Da and Dz in [F1] and the annulus decomposition of [F3], coherence being the same relation read in the regions of the new picture.

3.1F1F2step 2.1algebra

Counting the incoherent pairs. By step 2.1 pairs of two unchanged circles keep their coherence. For each unchanged Cp, step 2.1 preserves the number of incoherent pairs involving Cp and one of the two replaced circles; the moved pair itself is incoherent in D by [F1] and the new pair Ca,Cz is coherent in D′ by [F1]. Hence the number of incoherent unordered pairs drops by exactly one: h(D′)=h(D)−1.

4.1F2step 3.1∎

Conclusion. Since each reducing move lowers the height by one and the height is a nonnegative integer, a sequence of k reducing moves from D satisfies 0≤h(D)−k, so k≤h(D) and the sequence terminates after at most h(D) moves. AC is inherited exactly from [F3].

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

A height-zero diagram represents a closed braid

Statement

Assume AC. If h(D)=0 for an oriented diagram D, its oriented link is equivalent to the closure of an explicitly read-off braid. For a nonempty diagram, an isotopy of its Seifert picture on S2 and a choice of planar chart put its Seifert circles in one nested chain about a point p, with compatible orientations. Reading its signed crossing strips in angular order from a cut ray at p gives a finite Artin word whose closure is the link of D. The empty diagram gives the empty word in B0 and the empty closure.

Facts & Assumptions

Given: AC, a finite oriented diagram D with height zero and its smooth Seifert circles and signed crossing strips.

[F1]

Height zero means every pair of circles is coherent: their orientations agree with the boundary orientations of their common annulus in opposite senses (Coherence of Seifert circles and the height of a diagram).

[F2]

Oriented Seifert smoothing replaces a crossing by two parallel, equally directed local arcs; signed strips have disjoint interiors in the complement and recover every original crossing with its sign (Seifert smoothing and Seifert circles of an oriented link diagram, Defect regions, reducing arcs and the Yamada-Vogel reducing move).

[F3]

Jordan separation gives two complementary components and Schoenflies identifies their closed regions with disks; two disjoint circles on S2 have a unique common annulus (Jordan–Brouwer separation, Jordan–Schönflies extension for plane curves, Two disjoint circles in the two-sphere cobound an annulus).

[F4]

Smooth compact curves have tubular collars, and a smooth isotopy through embeddings extends to an ambient isotopy; AC supplies the countable choice assumed for extension (The Euclidean tubular neighbourhood theorem, A smooth isotopy of a compact manifold extends to an ambient isotopy, AC implies DC implies countable choice).

[F5]

A finite Artin word gives a geometric braid by the choice-free homomorphism, with each letter represented by the fixed signed half twist. The fixed-framing closure glues its strands cyclically; B0,B1 have empty presentations (The Artin presentation surjects onto the geometric braid group, The elementary geometric half twist, its support disc, and its opposite, The closure of a geometric braid, The braid group by Artin presentation).

Proof

technique · direct
1.1F1F3construct

The complementary dual tree. Make one vertex for each component of the complement of the n circles in S2 and one edge for each circle, joining its two adjacent components. With no circles this is one vertex. Inserting a new disjoint Jordan circle splits one existing region into two, by [F3], and replaces that vertex by two vertices joined by one edge. Induction therefore gives a tree with n edges. Suppose a vertex has three incident circles. Orient its common region and let ϵi∈{+1,−1} record each given circle orientation relative to the induced boundary orientation on that region. For any pair of those boundary circles the annulus between them includes the common region, so its boundary orientation there is the same; coherence [F1] requires ϵi=−ϵj. Three such signs cannot all be pairwise opposite. Thus every vertex has degree at most two and the tree is a path. Choose a pole in either end disk and use stereographic projection from that pole. The circles are now nested in the remaining plane. Their annulus orientations show that all are oriented in the same angular sense. This argument never assumes the original diagram or Seifert graph is connected.

2.1F1F2F3step 1.1construct

Crossing strips occur only between successive circles. At an oriented smoothing the two arcs bounding its strip are parallel and point in the same direction by [F2]. If they belonged to the same Jordan circle, the connected strip interior would lie on the same complementary side of that circle at both arcs. But equally directed arcs on opposite sides of this strip have that interior on opposite local sides; the inside of an oriented Jordan circle lies consistently on one local side. This contradicts separation [F3]. Hence a strip joins distinct circles. Its connected interior lies in one complementary component, so in the path of step 1.1 it joins precisely two consecutive circles across their common annulus. Within one such annulus, disjoint strips have the same cyclic order of their endpoints on the two coherently oriented boundaries: cutting along one strip makes a disk, and two further strips with interlaced endpoints would cross by Jordan separation.

3.1F3F4step 1.1step 2.1construct

Smooth straightening, rather than merely a topological change of coordinates. We give the local construction needed to straighten the finite picture. A smooth simple curve has a tubular collar [F4]. A sufficiently fine inscribed polygon, with its corners rounded in pairwise disjoint small disks, is a smooth normal graph over that curve: on a compact regular arc its tangent varies uniformly little on a sufficiently fine partition, so normal projection is locally strictly monotone; the collar and compact separation of nonadjacent arcs make it globally one-to-one. Multiplying its smooth normal displacement by a parameter is an isotopy through smooth normal graphs, extended by [F4]. A polygonal disk can be triangulated by induction: at a convex vertex, if its adjacent-vertex triangle contains no other vertex use that diagonal; otherwise choose a vertex in that triangle farthest towards the convex corner from its opposite edge and use the resulting interior diagonal. A vertex or edge blocking this latter diagonal would contradict the maximal choice; separation keeps its interior in the disk. Each diagonal splits the polygon into smaller polygons, so the induction terminates. To straighten a curve around a fixed interior point, root this triangulation at the triangle containing that point, after a sufficiently small generic perturbation of the polygon vertices so no vertex-pair line passes through it. Such perturbations avoid finitely many proper line conditions, preserve simplicity and keep the point inside by compact separation. Remove leaf triangles in reverse order, leaving the root triangle. Each removal replaces a two-edge boundary arc by its diagonal in the empty triangle. In affine coordinates the two-edge ear is a graph. Choose its rounded version and the rounded diagonal with identical smooth endpoint germs; interpolate their graph functions smoothly, retaining these germs. Every parameter slice is therefore smooth, including the initial and final slices; the literal polygon is only the combinatorial guide. To realize a small graph displacement u(s) ambiently, use the smooth collar map (s,r)↦(s,r+χ(r)u(s)) with ∥χ′∥∞∥u∥∞<1. Split a larger displacement into finitely many small ones inside the clear local graph neighbourhood. These are explicit smooth diffeomorphisms, fixing the collar boundary and the unchanged germs. At every time it is an embedded smooth arc, and no other portion of the curve meets that neighbourhood. The fixed interior point is untouched. The remaining rounded triangle is star-shaped about that point and is taken to a small concentric circle by positive radial interpolation, again with support away from the point. By [F4] these finitely many smooth curve isotopies are ambient isotopies. Apply them first to the outermost nested circle, transporting its entire interior picture, and then successively inside the disk bounded by each already fixed round circle. Choose the centre p inside the innermost disk; the construction fixes p throughout. This puts all circles concentrically about p. The same triangle-arc operation straightens finitely many disjoint crossing strips in an annulus: cut along a spanning strip, leave small neighbourhoods of the strip endpoints and their smooth incident circle/arc germs fixed. Approximate only the remaining compact smooth arc pieces in their disjoint normal collars by rounded polygonal graphs, as above. Each remaining proper strip core cuts the disk into two disks by Jordan separation and Schoenflies. Cutting successively along their disjoint cores gives finitely many disk faces, each triangulated by the preceding diagonal construction; remove the resulting triangles between each core and its target, fixing endpoint collars and already fixed cores. Once a core is straight its thin strip is straightened in its normal collar. All supports miss the other fixed cores. The relative disk-face boundary pieces retain their original smooth germs, and the rounded graph interpolation and explicit collar diffeomorphisms apply to the moving interiors; neither a polygonal endpoint nor a cornered whole face is used as the source of smooth isotopy extension. Round corners and use fixed collars as above. An integral winding of its first spanning strip is removed before this construction by (r,θ)↦(r,θ−2πkf(r)), where f is smooth, equals 1 at the inner boundary and 0 at the outer boundary. Its isotopy uses parameter multiples of f and rotates the entire inner disk at the same time, so is a sphere isotopy; it does not require a boundary-fixed annulus isotopy of a nonzero Dehn twist. Thus every straightening used here is realized through smooth embeddings and [F4], not inferred from the homeomorphism assertion of Schoenflies.

4.1F2F4step 2.1step 3.1construct

Choose compatible crossing-event angles. Starting with the outermost circle, choose distinct angular positions for its crossing-strip endpoints in their cyclic order. Pass inward through the circle chain. The endpoints shared with the preceding annulus already have angles, and by step 2.1 their order agrees on both boundaries. Insert the new events for the next annulus into the appropriate open gaps in that finite circular order, avoiding all previously chosen event angles. If there are no shared events, choose any initial angles in the specified cyclic order. An orientation-preserving smooth reparametrization of each circle realizes these finite assignments: on the intervening open arcs choose positive smooth derivatives with the prescribed integrals, using matching endpoint collars. The annulus strip construction in step 3.1 then takes each crossing strip to a small radial strip at its assigned angle, while the remaining annulus pieces match the two boundary reparametrizations. To interpolate prescribed boundary circle coordinates across a strip-free annulus, take increasing lifts f0,f1 with fi(θ+2π)=fi(θ)+2π and use the angular coordinate (1−χ(r))f0(θ)+χ(r)f1(θ), where χ is smooth and constant near both ends; its angular derivative is positive. The strip-relative construction gives the same boundary collars, so the finitely many constructions glue smoothly. No connectedness assumption is needed; an annulus with no strips can use any interpolation between its boundary coordinates.

5.1F2F4F5step 1.1step 4.1construct

Recover the closed braid and read its word. In each radial crossing strip put back the original signed crossing, using the model of two strands with increasing angular parameter and one signed half twist in the radial-depth disk. Outside the strips use the concentric oriented circles. Step 1.1 makes their angular directions agree, and a chart of the opposite orientation, if necessary, chooses that direction as increasing height. When reversing the planar orientation, reverse the depth coordinate as well. The combined three-dimensional coordinate change preserves ambient orientation and represents the same oriented link; read its over/under information in these new depth coordinates. The finitely many strips have distinct event angles by step 4.1. Cut at an unused angle and read their half twists in angular order; reverse the chronological list when using the rightmost-first geometric product convention. This is an explicit finite Artin word in Bn. Its geometric representative [F5] and the recovered diagram have exactly the same oriented strands and signed strip models. To transfer the sphere isotopies to link equivalence, place a lift of the diagram in a thin normal collar of the projection sphere, keeping the higher branch higher in each crossing strip. Compressing each normal fibre by a positive factor preserves these strict orders and is an isotopy of the link. A smooth sphere isotopy lifts in that collar by its tangential velocity field, extended constantly along sufficiently short normal fibres and cut off farther away; equivalently apply [F4] to the projection sphere and choose this collar extension. It transports every crossing strip without interchanging its depths. Its ambient extension therefore carries the original oriented link to this closed braid. The remaining local strip lifts with the same projected arcs and strict depth orders are joined by linear interpolation of the depths, which preserves embeddings. Thus the standard closures agree up to ambient isotopy. Each open strand period advances once around the pages; a closed component may advance several periods according to its permutation cycle, as [F5] specifies. There is no assertion that every component has degree one.

6.1F1F3F4F5step 1.1step 2.1step 3.1step 5.1∎

Boundary cases and conclusion. If n=0, the diagram and strip set are empty and [F5] gives the empty word and empty closure. If n=1, step 2.1 rules out every strip, and step 3.1 takes its sole circle to a round circle; the empty word in B1 closes to that unknot. For every n>1, steps 1.1-5.1 give the sphere isotopy, the chosen planar chart and the explicit word with the original oriented link as closure. All constructions are finite; AC is inherited only from [F1], [F3], [F4]. The preliminary change of chart is allowed on S2, and is not asserted to be a planar isotopy of the original fixed chart.

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Alexander's theorem: every link is a closed braid

Statement

Assume the Axiom of Choice. Every oriented link L⊂R3⊂S3 is equivalent to the oriented closure β^ of a geometric braid β on some number n≥0 of strands, with n≥1 when L is nonempty.

Facts & Assumptions

Given: AC, an oriented link L⊂R3, and the Yamada-Vogel reducing algorithm (Defect regions, reducing arcs and the Yamada-Vogel reducing move, Coherence of Seifert circles and the height of a diagram).

[F1]

Assume AC. Every oriented link has a regular projection; AC yields ACω, which is used in the existence proof (Existence of regular projections, AC implies DC implies countable choice).

[F2]

If the height of a diagram is positive, the Seifert picture contains a defect region and hence a reducing arc (A positive-height diagram has a defect region).

[F3]

A reducing move lowers the height by one, so no sequence of reducing moves starting at D has more than h(D) terms (A reducing move lowers the height by one).

[F4]

A diagram of height zero represents the closure of an explicitly read-off braid: a sphere isotopy and a choice of planar chart give a nested coherent chain, and reading its signed crossing strips in angular order from a cut ray gives the braid word. The empty diagram gives the empty braid in B0 (A height-zero diagram represents a closed braid).

Proof

technique · direct
1.1F1F2F3F4given

Choosing a diagram and a first reduction. If L is empty, its empty diagram is read by [F4] as the empty braid in B0, proving the assertion. For a nonempty L, by [F1] fix a regular projection D0 of L with its over/under and orientation data, and let h0:=h(D0)∈N. If h0=0, [F4] already presents L as the closure of a braid. If h0>0, then by [F2] the Seifert picture of D0 contains a defect region and a reducing arc; performing the reducing move produces a diagram D1 of the same oriented link with h(D1)=h0−1 by [F3].

2.1F2F3step 1.1

Termination of the algorithm. Iterate step 1.1. The sequence of heights is a strictly decreasing sequence of nonnegative integers, because each reducing move is a Reidemeister II move of the diagram, which does not change the represented oriented link, and lowers the height by exactly one; hence after exactly h0 steps the algorithm stops at a diagram Dh0 with h(Dh0)=0 representing L.

3.1F4step 2.1

Reading the braid. By [F4] the height-zero diagram Dh0 is put in closed-braid form by a sphere isotopy and a choice of planar chart; the braid word read from the nested chain has a closure equivalent to the link of Dh0, which is L. Hence L is equivalent to the oriented closure β^ of an explicit geometric braid β on n≥1 strands.

4.1F1F2F3F4step 3.1∎

Conclusion. Steps 1.1-3.1 give the required braid. AC is used in [F1] (regular projections, through the bridge from AC to ACω) and in the AC-stated height and reducing-move chain [F2], [F3], which rests on the annulus lemma.

DefinitionDefinition: AI-adaptedProof: Not applicableOpen item page →

Markov conjugation and stabilization moves

Definition

Let (Bn)n≥0 be the braid groups of the Artin presentation (The braid group by Artin presentation), with generators σ1,…,σn−1 and the two Artin relations, and let

ιn ⁣:Bn⟶Bn+1,ιn(σi):=σi(1≤i≤n−1),

be the standard inclusion homomorphism. It is well defined: the defining relators of Bn are among the defining relators of Bn+1 under the assignment σi↦σi, so Von Dyck's theorem: maps of generators that satisfy the relators extend uniquely from a presented group extends the assignment to a homomorphism. Its image is generated by σ1,…,σn−1, and we write ιn(β) for it; whether this homomorphism is injective is not needed anywhere on this page, and no subgroup identification is made. For the geometric description, use the base configuration Q(n+1) of The elementary geometric half twist, its support disc, and its opposite. Express β as an Artin word and replace its letters by the half twists with the same indices at Q(n+1). The letters with i<n fix the rightmost point, since their support discs exclude it. This gives a representative of ιn(β) under The Artin presentation surjects onto the geometric braid group; it is independent of the word as a geometric braid class. Thus adding the strand on the right includes expressing the old braid at the first n target basepoints. Those points differ from Q(n), so the original parametrized strands are not literally left untouched. The final half twist is the target σn.

On the disjoint union ⨆n≥0Bn define two kinds of elementary Markov moves.

Conjugation. For β,γ∈Bn, replace β by γβγ−1, an element of the same group Bn. The number of strands does not change. When β is a geometric braid and γ the class of a braid in Gn, this is realized on closures by reading γ around the axis, as the closure-preservation lemma below shows.

Stabilization and destabilization. For n≥1 and β∈Bn the positive stabilization is βσn:=ιn(β)⋅σn∈Bn+1 and the negative stabilization is βσn−1:=ιn(β)⋅σn−1∈Bn+1; the inverse operations, from Bn+1 to Bn, are called destabilizations. Both signs occur, and the sign is the sign of the exponent of the new generator. By The Artin presentation surjects onto the geometric braid group and The elementary geometric half twist, its support disc, and its opposite the product βσn±1 is the class of the target-basepoint geometric word just described, composed with the target half twist σn±1 in its fixed product position; that half twist fixes the first n−1 target points. Inserting the generator inside a factorization need not produce a conjugate of that fixed stabilization. It nevertheless gives a Markov sequence: if β=uv, old conjugation by v gives vu, its right stabilization is (vu)σn±1, and conjugation by v−1 gives uσn±1v. Only moving a factor past the whole word is cyclic conjugacy. At n=0 no σ0 exists, so the empty braid has only conjugation moves and is an isolated Markov class.

Markov equivalence. Two braids β,β′ (possibly with different numbers of strands) are Markov equivalent when they can be connected by a finite sequence of moves each of which is a conjugation, a positive or negative stabilization, or a destabilization. Markov equivalence is an equivalence relation on ⨆nBn: it is reflexive, symmetric by allowing inverses of the listed moves, and transitive by concatenating sequences.

Convention on orientation and side. All braids are read with the product and side conventions fixed on the pages geometric-braids-and-artin-generators and artin-presentation-completeness-and-braid-combing: the rightmost factor is traversed first in geometric time, and stabilization always adds the new strand on the right. The definition is choice-free: no choice principle is used in the Artin presentation, in the inclusion, or in the identification of σn with the elementary geometric half twist.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Markov moves preserve the oriented closure up to isotopy

Statement

Assume ACω. If β′ is obtained from β by a conjugation or by a stabilization or destabilization in the sense of Markov conjugation and stabilization moves, then β′^ and β^ are equivalent oriented links.

Facts & Assumptions

Given: ACω, braids β∈Bn and β′ obtained from β by one Markov move, and the closure construction of The closure of a geometric braid with axis A.

[F1]

Conjugation replaces β by γβγ−1∈Bn, and stabilization replaces β by the product of the standard inclusion of β with σn±1∈Bn+1, the new strand being added on the right; σn is the elementary geometric half twist of the last two strands (Markov conjugation and stabilization moves, The elementary geometric half twist, its support disc, and its opposite).

[F2]

The closure of a braid based at Q meets every page in exactly n points; its components are the cycles of the endpoint permutation and its orientation runs from height 0 to height 1 along the strands (The closure of a geometric braid).

[F3]

Every geometric braid is braid-isotopic to a smooth braid, and the isotopy may be taken arbitrarily close to it and fixing the endpoints; ACω is assumed there (Every geometric braid is braid-isotopic to a smooth braid).

[F4]

Each oriented Reidemeister move, in particular the elimination of a kink, is realized by an ambient isotopy supported in a small ball (Each oriented Reidemeister move is realized by an ambient isotopy).

[F5]

Under countable choice a smooth isotopy of a compact boundaryless manifold through embeddings extends to ambient isotopy, with support in a prescribed neighbourhood of the compact trace (A smooth isotopy of a compact manifold extends to an ambient isotopy).

[F6]

Under countable choice changing a representative by endpoint-fixed braid isotopy leaves its oriented closure equivalent; smooth families with endpoint collars glue on one circle per permutation cycle (The closure depends only on the braid isotopy class).

[F7]

A smooth Euclidean map with nonsingular derivative has a smooth local inverse without choice; a previously constructed global inverse is therefore smooth wherever the derivative is nonsingular (Choice-free smooth inverse function theorem in Euclidean space).

Proof

technique · direct
1.1F1F2F3F5F6construct

An explicit conjugation homotopy of configuration loops. Use smooth representatives of β,γ with flat endpoint collars by [F3], [F6]. For 0≤u≤1, let cu be the initial path of γ from its starting configuration Q to γ(u), parametrized smoothly as γ(uψ(t)), where ψ is increasing and constant on endpoint collars. Form the free configuration loop, in chronological order, cu−1⋅β⋅cu, with the three pieces each occupying one third of the page parameter. It is based at the moving configuration γ(u) and is smooth jointly in u and page parameter, including seams, by these collars. Label the moving basepoints by the original strands of γ. Then a point labelled j follows cu−1 to qj, follows β to qπβ(j), and follows cu to the moving point labelled πβ(j). Thus the permutation cycles are the fixed cycles of πβ in these transported labels. Concatenate each cycle on R/kZ and apply the fixed φ of [F2]. This gives a smooth isotopy of the compact union of cycle circles through embedded oriented closed n-braids: points are distinct at every page, and the page parameter has positive derivative. At u=0 it is β with constant pauses; at u=1 its chronological pieces are γ−1,β,γ, so its actual product is γβγ−1. The pauses and different positive piece durations are endpoint-fixed reparametrizations covered by [F6]. Make the u parameter constant near its ends and apply [F5] to the compact cycle embedding family. Its trace lies in S3∖A, so choose support away from A. This proves conjugation invariance by an actual ambient isotopy. For n=0 it is the identity of the empty link. No continuity in a discrete word variable is asserted.

1.2F1F2F3F4F5F6F7construct

Put the stabilization in a fixed-framing local ball. Here n≥1 by [F1]. Choose a finite elementary-word representative of β in which all old strands lie in a central disk cluster, with a short empty word interval at the cutting page; [F6] permits its endpoint collars. Before adding a point, match the original n-point base configuration to the first n points of the (n+1)-point configuration. The increasing real affine map x↦n+1n+2x−14(n+2) does so: substituting qj(n)=(2j−n−1)/(4(n+1)) gives qj(n+1)=(2j−n−2)/(4(n+2)). Extend its isotopy on the central cluster to the disk by finitely many small affine changes multiplied by a smooth cutoff equal to one on the cluster and zero outside a larger disk. Choose each increment with derivative norm of its displacement below one, just as in the point-motion construction below. This preserves order, carries each old half twist to its corresponding first-n half twist, and introduces no rotation or framing twist. The resulting cycle family gives an ambient closure isotopy by [F5]. It therefore justifies the geometric realization of the standard inclusion, although the two canonical basepoint tuples differ. The added last point is initially to the right of this cluster. Move it along the real corridor beyond the cluster to w=1−ε, with 0<ε small, keeping all old strands fixed. Such point motions and the following collar motions are smooth disk isotopies: subdivide a compact collision-free point path into finitely many small displacements v and use x↦x+χ(x)v, with support missing the other points and ∥Dχ∥∞∣v∣<1. Its inverse exists uniquely by contraction of x↦y−χ(x)v, The explicit iterates have geometrically decreasing successive differences, so their limit exists uniquely. Its Jacobian determinant is 1+Dχ⋅v>0; [F7] makes that inverse smooth. Thus it is an orientation-preserving diffeomorphism equal to identity off its disk. Apply these motions independently of page angle for the added point, and periodically in the empty cutting-page interval for the selected last old strand. Move that old strand's short collar to w=1−2ε, retaining its connections to the old braid outside the interval along a narrow real corridor disjoint from the other strands. Transport the elementary last-pair half twist with this motion. Its connecting arc remains the real interval between these two points; an arc-fixing transverse compression in its disk collar makes its supporting strip arbitrarily thin. These are isotopies from the identity, not an arbitrary page-preserving change of framing, and their closed cycle families extend ambiently by [F5]; their endpoints therefore have the same oriented closures as the original fixed-product stabilization and old braid. The new point's whole constant closed strand and the relocated old collar now lie near a0=(0,1)∈A. Indeed the fixed formula [F2] is z=1−∣w∣2e2πit: if ∣w−1∣<3ε, then ∣z∣2<6ε. Choose the half-twist strip inside this region and a short cutting-page interval, so a small ball at a0 contains the entire added meridian circle, the last-pair crossing and only this one old-strand collar. All other old strands stay in the remote cluster. In local coordinates (Re⁡z,Im⁡z,Im⁡w) with Re⁡w=1−∣z∣2−(Im⁡w)2, the new constant point gives the planar circle ∣z∣=2ε−ε2; the old collar is a short outer arc at radius 4ε−4ε2. The one transported half twist joins this circle to that arc by one signed crossing strip. Recovering the crossing from this local Seifert picture is exactly the oriented Reidemeister-I curl, positive or negative according to σn±1. Unwind it by [F4], leaving the old arc and the rest of the link. This local isotopy is in S3; it may pass through the fixed axis inside this ball, as required when page count changes. The fixed disk formula introduces no extra full twist. This proves stabilization invariance for both signs and does not assume the original new basepoint was near the axis.

2.1F3F4F5F6step 1.1step 1.2∎

Inverses and finite sequences. Run step 1.2 backwards for destabilization and step 1.1 backwards for inverse conjugation. A finite composition of these ambient isotopies preserves the oriented link. Countable choice is used only by the smoothing and isotopy-extension suppliers [F3], [F5], [F6]; the local Reidemeister model is [F4].

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Braid-isotopic closed braids are conjugate

Statement

Assume the Axiom of Choice. Let β,β′ be n-braids and suppose β^ and β′^ are isotopic through closed n-braids about the standard axis (an isotopy of the closed braids through braids in the complement of the axis). Then β′ is conjugate to β in Bn.

Facts & Assumptions

Given: AC, two n-braids β,β′ based at Q, and an isotopy of their closures through closed n-braids about the standard axis A.

[F1]

Every closed n-braid about A meets each page in exactly n points, and the closure of a braid based at Q is obtained from its strand images by the fixed diffeomorphism φ of the closure construction (The closure of a geometric braid).

[F2]

The unordered configuration space Cn(D∘) of n points in the open disk is the quotient of the ordered configuration space by the label permutations, with basepoint the orbit [Q] of the base configuration; its points are n-element subsets (Unordered configuration spaces Cn(X)).

[F3]

Raw slicing induces a bijection S from the geometric braid group Gn to π1(Cn(D∘),[Q]), and the map Φ ⁣:Gn→π1(Cn(D2),[Q]), [β]↦(ι∗C[S(β)])−1, is a group isomorphism (Geometric braid classes and the unordered configuration fundamental group).

[F4]

Assume AC: the published surjection BnArtin→Gn is an isomorphism, so the Artin braid group Bn of The braid group by Artin presentation is identified with the geometric braid group (The Artin presentation is complete for geometric braids).

[F5]

Two loops in a path-connected space are freely homotopic if and only if their classes in the fundamental group are conjugate; free homotopy classes correspond bijectively to conjugacy classes (Free homotopy classes of loops are conjugacy classes, Based loops and the fundamental group).

[F6]

AC implies countable choice, so the general continuous braid's chosen smooth closure model is available under the Given hypothesis (AC implies DC implies countable choice).

Proof

technique · direct
1.1F1F2F3F4F6

The loop of a closed braid. Fix a page Pθ of the closure construction and identify it with the open disk D∘; a closed n-braid about A meets each page in exactly n points by [F1], and as the page turns once about the axis these points trace a continuous loop c ⁣:S1→Cn(D∘) in the unordered configuration space of [F2]. For a general continuous braid use the selected smooth model of [F1], available by [F6]. It is endpoint-fixed braid-isotopic to the given braid, so raw slicing of [F3] sends its geometric class to the class of this closure loop. The group isomorphism in [F3] is instead Φ(β)=(ι∗C[c])−1. Apply that fixed induced map and inversion when transporting conjugacy; both carry conjugate classes to conjugate classes. Composing Φ with [F4] gives the fixed group isomorphism used below.

1.2F1F2given

Isotopy of closed braids gives a homotopy of the loops. An isotopy of β^ to β′^ through closed n-braids about A gives, by reading the n intersection points with the turning page at each stage, a homotopy of the loops c and c′ in Cn(D∘): the intersection points depend continuously on the isotopy parameter because the isotopy is continuous and the page meets every intermediate closed braid transversely in exactly n points. Hence c and c′ are freely homotopic loops.

2.1F4F5F6step 1.1step 1.2∎

Conclusion. By [F5] the free homotopy of step 1.2 makes the classes of c and c′ conjugate in π1(Cn(D∘),[Q]); first applying ι∗C and inversion, and then transporting through the fixed isomorphisms of step 1.1 gives that [β] and [β′] are conjugate in Bn, so β′ is conjugate to β in Bn. The axiom of choice supplies [F4], through Artin-presentation completeness, and the countable choice for the smooth closure model in [F6].

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Braid-like Reidemeister moves on closed braids are braid isotopies

Statement

Let D,D′ be closed braid diagrams whose Seifert pictures are nested chains, and suppose D′ is obtained from D by a braid-like Reidemeister move of type II or III, by a planar isotopy, or by a sequence of such moves. Then the corresponding closed braids are braid-isotopic in the sense of the conjugacy bridge of Braid-isotopic closed braids are conjugate, and the braids read from D and D′ are conjugate in Bn. Consequently every braid isotopy appearing in a Markov sequence may be replaced by a conjugation move.

Facts & Assumptions

Given: The closed braid diagrams and finite sequence of moves in the Statement; no choice axiom is assumed.

[F1]

The braid-like II and III pictures have all participating local strands oriented in the same braid direction (Oriented Reidemeister moves).

[F2]

The Artin presentation has inverse cancellation, far commutation, and the adjacent relation σiσi+1σi=σi+1σiσi+1 (The braid group by Artin presentation).

[F3]

There is a choice-free homomorphism from the Artin presentation to geometric braids sending the generators to the fixed geometric half twists; the far and adjacent relations are realized by endpoint-fixed geometric braid isotopies (The Artin presentation surjects onto the geometric braid group, Far commutativity of elementary geometric half twists, The geometric three strand braid relation).

[F4]

A braid isotopy is an endpoint-fixed homotopy of distinct disk points, and the finite-word models used here are smooth with endpoint collars and their literal closures use the fixed standard disk framing (Braid isotopy relative to the top and bottom endpoints, The closure of a geometric braid).

Proof

technique · direct
1.1F1F2F3construct

Read a local move at a fixed cut. Choose a cut outside the small disk of the move, and straighten its local angular foliation to a braid rectangle. The untouched part gives two fixed word contexts. A braid-like II pair contributes σiσi−1 or σi−1σi, so removal is inverse cancellation. For III put s=σi, t=σi+1. The positive and negative cases are sts=tst and its inverse. The four remaining possible over/under depth orders give sts−1=t−1st,s−1ts=tst−1,st−1s−1=t−1s−1t,s−1t−1s=ts−1t−1. The first two follow by multiplying sts=tst on the appropriate left and right by t−1 or s−1, and the last two are their inverses. These exhaust the six orders of three distinct strand depths; the two omitted sign triples would require a cyclic strict depth order. Thus the complete words agree by [F2]. By the homomorphism and actual geometric relations of [F3], each replacement is an endpoint-fixed geometric braid isotopy, also in either fixed word context. This does not require injectivity of that homomorphism.

1.2F2F3F4construct

Planar isotopy, reading order and cyclic cut. Transport the nested circles, crossing strips and a cut with the planar isotopy. This transports their oriented cyclic orders; no crossing is created and no over/under sign changes. The read word depends only on these orders. More explicitly, cut the transported annular picture and lift its circles to parallel oriented intervals. Every crossing strip is an event involving two consecutive intervals. Events meeting a common interval have their order fixed by that interval's oriented order. Any two total readings extending these finite orders differ by successive exchanges of adjacent incomparable events: move the first event of one reading to its position in the other, noting that every event passed must be incomparable, and induct on the remaining finite list. Incomparable crossing strips use disjoint pairs of intervals, so their generator indices differ by at least two; exchanging them is precisely far commutation [F2], realized geometrically by [F3]. A cut passing one or more events changes a product uv to vu=u−1(uv)u, an explicit conjugation, and records the same closed braid with a different starting page. Consequently the end readings of a transported picture, and readings using any other admissible cut, differ only by far commutations and conjugations. The transported angular foliation supplies a family of closed braids; at the end its reading agrees with the usual nested-chain reading by the same finite-order argument. There is no assertion that a fixed page gives fixed based endpoints throughout an arbitrary planar isotopy.

2.1F2F3F4step 1.1step 1.2∎

Closed isotopy and conjugacy. Apply step 1.1 to every local II or III replacement and step 1.2 to the planar pieces and changes of cut. Equal-word replacements yield based geometric isotopies by [F3], hence closed isotopies by [F4]. A cyclic cut change is realized by moving the starting page around the same closed braid, so also gives an isotopy through closed braids. The finite concatenation is therefore a braid isotopy of the closed braids. Algebraically the equal-word replacements leave the element unchanged and each cyclic change conjugates it; their finite composition is a conjugation in Bn. This proves both conclusions directly and makes the replacement by conjugation in a Markov sequence explicit. No choice-stated converse identification of geometric and Artin braids is used.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Non-braid-like Reidemeister moves are generated by braid-like moves and reductions

Statement

Assume the Axiom of Choice. Every non-braid-like Reidemeister move of type II acting on one Seifert circle, and every non-braid-like Reidemeister move of type III, is a composition of Reidemeister I moves, braid-like Reidemeister II and III moves, Yamada-Vogel reducing moves, and their inverses. Hence any two diagrams of the same link can be connected by a sequence of moves of the four types I, braid-like II, braid-like III and reducing (and inverses).

Facts & Assumptions

Given: The Axiom of Choice, an oriented link diagram, its Seifert picture and coherence data (Coherence of Seifert circles and the height of a diagram, Defect regions, reducing arcs and the Yamada-Vogel reducing move).

[F1]

Coherence of two Seifert circles and the height of a diagram are defined through the annulus cobounded by the two circles; AC is consumed there (Coherence of Seifert circles and the height of a diagram, The Axiom of Choice).

[F2]

A reducing move along an arc joining an incoherent pair replaces the pair by two coherent circles joined by two oppositely signed arcs, leaving all other circles unchanged (Defect regions, reducing arcs and the Yamada-Vogel reducing move).

[F3]

Braid-like II and III moves locally have their strands in one common braid direction; type I is the separately permitted move adding or deleting a kink (Oriented Reidemeister moves).

Proof

technique · direct
1.1F1F2F3givenconstruct

The non-braid-like R2 on one circle. Read the five panels in the top row of Traczyk's Figure 3, printed p. 411. The upper arc points left and the lower arc right. Name their outside ports UL,UR,LL,LR; the outgoing ports are UL,LR. Since both smoothed arcs belong to one circle C, the untouched outside paths join UL to LL and LR to UR. The first arrow adds one R1 curl to the lower arc, producing a small circle C1 coherent with C. The second arrow adds a second, oppositely signed R1 curl on that curl. Oriented smoothing now gives the old continuation, the middle diamond circle C1 and the upper loop circle C2. Work on the side of C containing the move disk, choosing the pole on its other side only to read the annulus orientations: C1,C2 are side by side there, C1 has the boundary direction of C, and C2 the opposite direction. Thus C1 is coherent with C, C2 with C1, and C2 is incoherent with C. The third arrow pulls the upper loop through the upper horizontal arc, which is over both new crossings in the displayed picture. This antiparallel R2 is a reducing move on C,C2, not on the first coherent pair: smoothing gives its empty small circle Ca above the horizontal arc and the other new circle Cz surrounding the unchanged C1. The last arrow cancels the two lower self-crossings by R2. Its inverse is a reducing move: smoothing after cancellation gives two side-by-side circles of the same orientation, hence an incoherent pair, while smoothing before cancellation gives the coherent pair C1,Cz; Ca is unchanged. The two reducing strips are different. The crossing counts are 0,1,2,4,2, and the last panel retains exactly the two desired crossings between the original arcs and all four outside germs. The sequence is therefore R1, R1, reducing R2, inverse reducing R2. The two R1 signs are opposite, as are the crossings in each R2 pair. Reflecting the spatial height makes the upper arc under both target crossings; reversing the two arc orientations and running the sequence backwards supplies the remaining sign, orientation and inverse variants. The outside paths and all other Seifert circles stay fixed throughout.

1.2F1F2F3construct

The antiparallel R2 on distinct circles. Before creation of its two crossings the opposite local directions on the exposed boundaries make the two circles incoherent in their common annulus. The move disk consequently supplies a reducing arc, and its R2 is precisely a reducing move, with either over/under choice. Running this creation backwards gives the inverse reducing move. A coherent pair is not made incoherent merely by reversing crossing signs.

2.1F2F3step 1.1step 1.2construct

The non-braid-like R3. Choose the arc with orientation opposite to the other two as the detouring arc. In the bottom square of Figure 3, pull a short part of that arc past one branch by an antiparallel R2, perform the now braid-like R3 at the other end of the detour, and remove the detour by the reverse antiparallel R2. Tracking the three original pairwise crossings leaves exactly the target R3 picture. If a detour R2 acts on one Seifert circle, replace it by step 1.1; if it acts on distinct circles use step 1.2. The consistent total height orders choose which detour is over or under. Choosing the odd-oriented arc in each orientation pattern, and reversing or reflecting the pictured square as needed, covers all non-braid-like variants.

3.1F1F2F3step 1.1step 1.2step 2.1∎

Conclusion. Steps 1.1, 1.2 and 2.1 show that every non-braid-like R3 and every non-braid-like R2 is generated by the four allowed move types, while type I moves are separately allowed and braid-like II and III moves are already in the move list. Since by Reidemeister's theorem for oriented diagrams any two diagrams of the same oriented link are connected by the oriented moves R1, R2, R3, replacing each non-braid-like R2 and R3 by the compositions above and keeping the remaining moves gives the required sequence through the four move types. AC is inherited from [F1] and supplies the countable choice used by Reidemeister equivalence (AC implies DC implies countable choice).

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Braid-like moves can be moved to height zero

Statement

Assume the Axiom of Choice. Any two diagrams of the same oriented link can be connected, allowing sphere isotopies of the decorated Seifert picture and changes of planar chart, by a sequence of moves of the following kinds: braid isotopies of closed braid diagrams, Markov stabilizations and destabilizations, and Yamada-Vogel reducing moves and their inverses grouped into sequences with positive-height intermediate diagrams; height-zero endpoints of these sequences are allowed. Moreover, every type I, braid-like type II and braid-like type III move occurring in such a sequence may be assumed to be performed at height zero after this normalization, that is, as a braid isotopy or as a stabilization. Sphere isotopies transport the crossing strips and their depth data; they are not asserted to be planar isotopies of the original chart.

Facts & Assumptions

[F1]

Every non-braid-like R2 and R3 is generated by type I moves, braid-like R2 and R3 moves, reducing moves and their inverses, with type I retained as a separate allowed kink move (Non-braid-like Reidemeister moves are generated by braid-like moves and reductions).

[F2]

Braid-like R2 and R3 moves of closed braid diagrams and planar isotopies are braid isotopies of the underlying closed braids (Braid-like Reidemeister moves on closed braids are braid isotopies).

[F3]

A reducing move lowers the height by exactly one and is a Reidemeister II move of the diagram; the height is a nonnegative integer (A reducing move lowers the height by one).

[F4]

Braid isotopies of closed braids correspond to conjugation moves, and stabilizations and destabilizations are the Markov strand-changing moves (Markov conjugation and stabilization moves).

[F5]

A positive-height diagram has a defect region supporting a reducing arc (A positive-height diagram has a defect region).

[F6]

A height-zero Seifert picture can be put in a concentric nested chain by sphere isotopy and a choice of chart, transporting its signed crossing strips and representing the same oriented link (A height-zero diagram represents a closed braid).

Proof

technique · direct
1.1F5givenconstruct

A reduction away from a local move disk. For a type I move adding a kink, choose a disk U meeting the initial Seifert picture in one proper arc. For a braid-like II or III move, smoothing inside its standard move disk gives parallel, coherently oriented proper arcs; signed crossing arcs lie between adjacent arcs. Each inside component of a complementary region is consequently a cap or a terminated strip meeting one outside component, or an uncut strip between two coherent circles E,F which can join two outside components. Take a defect region R supplied by [F5]. If a component of R∖U exposes an incoherent pair, an arc inside that component is the required reduction away from U. Suppose none does. Every outside component then exposes at most two distinct circles, and if there are two they are coherent: indeed three exposed circles have two with the same orientation as boundary components of the region, by the two possible boundary orientations, and those two are incoherent in their cobounding annulus. Any outside component attached to an uncut strip exposes its pair E,F and therefore cannot expose a further circle. Thus along every chain of uncut strips all outside components expose the same coherent pair. Caps and terminated strips cannot join different outside components or introduce a further circle; components wholly inside U also expose at most a coherent pair. Since R is connected, it would expose at most this coherent pair (or a single circle if there is no joining strip), contradicting that it is a defect region. A reduction away from U therefore exists. This establishes avoidance only for the specified local move disks, with deletion of a kink handled by reversal.

2.1F3step 1.1construct

Pushing the move to height zero. Let a type I move adding a kink, a braid-like II move or a braid-like III move t start at positive-height Y. By step 1.1 choose a reducing move r disjoint from its local disk. The supports are disjoint, so r and t commute. Replace Y→ttY by Y→rrY→ttrY→r−1tY. The copy of t starts at height one lower by [F3]. Repeat at each positive starting height; after finitely many reductions this copy of t starts at height zero, surrounded by the reductions and their inverses. For a type I deletion, reverse the corresponding adding construction.

3.1F1F2F3F4F6step 2.1

Height-zero moves and the type I return path. Height zero alone does not make a diagram a closed braid in its original planar chart. First normalize its Seifert picture by [F6], transporting the local move disk as well. For II and III the two smoothed local pictures agree up to isotopy of their coherently directed proper arcs. In the straightening construction of [F6], retain these arcs as marked pieces and put their crossing strips in one small angular interval; outside that interval the two pictures use the same nested circles and strips. The local parallel arcs can be straightened in a rectangle with fixed boundary germs, so both recovered diagrams are closed braids and the transported move is braid-like. Now [F2] applies, giving a braid isotopy, hence a conjugation by [F4]. For type I retain the marked arc and the side on which its kink is to be added. In the circle chain, choose the end disk lying on that side of its supporting circle, and normalize with the axis in that end disk; either end disk is allowed by [F6]. Thus an outward kink on the innermost circle in the old chart uses the opposite end disk, beyond all the enclosing circles. Let k be the number of circles between the marked arc and this chosen end disk. Carry a short part of the marked strand beneath those k strands by braid-like II moves, add the kink on the end-disk side of the now nearest strand, and return it under the same strands. The end-disk-side kink smooths to a new circle in that end disk, coherent with every old circle, so both diagrams at this stage have height zero; with the end strand indexed last and a cut at the kink, its reading is the signed Markov stabilization of [F4], up to cyclic conjugation. Crossing sign determines the sign of stabilization independently of the selected side. The return across each strand consists of a braid-like III move followed by an inverse reducing move, as in the survey's printed p. 20 prescription and the last two arrows of Traczyk's Figure 4: the new kink circle and the crossed old circle are distinct, and the final antiparallel II cancels the temporary transport crossings. The marked outside germs and the prescribed kink side are restored. For k=0 this is a stabilization directly; for k>0 the inverse reductions can create positive height, even when the original marked strand was nearest the old axis. Nearest-strand position alone therefore does not suffice to identify a type I move with stabilization. Apply step 2.1 to each braid-like II or III move introduced in this finite return path, and reverse the construction for an inverse type I move.

4.1F1F2F3F6step 2.1step 3.1

The replacement sequence. Replace every non-braid-like II or III move using [F1], then replace the resulting type I and braid-like II and III moves by steps 2.1 and 3.1. The braid isotopies and Markov moves now occur in the normalized height-zero pictures. Undo each normalization before resuming the original reducing path. Sphere isotopy preserves coherence and transports reducing arcs, so these insertions do not change any height or reducing endpoint. Cut each remaining finite sequence of reducing moves and inverses at every height-zero diagram. Every resulting portion has positive-height intermediate diagrams; its endpoints can have height zero, including the last reduction from height one to zero and the first inverse reduction from zero to one. This gives the stated grouping of reducing content with the height-zero moves.

5.1F1F3step 3.1step 4.1∎

Conclusion. Steps 1.1-4.1 exhibit the required sequence and show that all braid-like content may be assumed at height zero. AC is inherited from the coherence and reducing-move apparatus of the height function.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Ordinary exchange moves are Markov sequences

Statement

Let n≥2, P,Q∈Bn−1, placed on the first n−1 strands of Bn, and t=σn−1. The ordinary exchange β=PtQt−1⟼β′=Pt−1Qt is realized by conjugations, one ordinary negative stabilization into Bn+1, and one ordinary negative destabilization back to Bn. It can also be realized with one positive stabilization and one positive destabilization. These are the ordinary moves of Markov conjugation and stabilization moves; conjugations are counted separately. The boxes P,Q are arbitrary on their specified strands.

Facts & Assumptions

Given: n≥2, P,Q∈Bn−1 and their specified strand placements.

[F1]

The Artin relations are σiσi+1σi=σi+1σiσi+1 and far commutation for index difference greater than one (The braid group by Artin presentation).

[F2]

Conjugations in a fixed braid group and u↔uσk±1 between Bk and Bk+1 are ordinary Markov moves (Markov conjugation and stabilization moves).

Proof

technique · direct
1.1F1givenalgebra

Supports and mixed relations. Append strand n+1 and write s=σn. Every letter of P,Q has index at most n−2, so both boxes commute with s and s−1, including when n=2 and the boxes are trivial in B1. From sts=tst obtain t−1st=sts−1, hence ts−1t−1s=t2s−1t−1. Also st−1s−1=t−1s−1t by inversion and rearrangement of the same relation, and therefore s2t−1s−1=t−1s−1t2: indeed s2t−1s−1=st−1s−1t=t−1s−1t2, where the first equality follows from st−1s−1=t−1s−1t. These identities hold without moving either box across t.

2.1F1F2step 1.1constructalgebra

Stabilization and the first weaving. Put A=Pt2 and B=t−1Qt−1 in Bn. Then BA=A−1βA. Conjugate β to BA, negatively stabilize to BAs−1, and conjugate by B−1 to As−1B. The latter equals E1=Pts−1t−1sQt−1 by step 1.1. Thus E1 is reached by exactly one negative ordinary stabilization and conjugations. With Figure 11's ports numbered from outermost to innermost, the old innermost strand is n and the added inner strand is n+1; the two boxes occupy the first n−1 old ports. The successive crossing letters of the right-hand weaving are t,s−1,t−1,s, followed after Q by t−1, giving precisely the chronological record E1. For these ordinary strands the indicated full twist on a single added strand is 1.

3.1F1F2step 1.1step 2.1algebra

The exchange calculation and the last weaving. Define E5=Pt−1Qst−1s−1t. Its successive left-hand weaving letters read chronologically, with the same ports and cut, are s,t−1,s−1,t, after the initial t−1 and Q. Since P,Q commute with s, use step 1.1 to compute sE1s−1=Pt−1Qs2t−1s−1=Pt−1Qt−1s−1t2=E5. This gives an explicit conjugation from the first weaving to the last; it proves the comparison without presuming any unverified intermediate diagram arrow. Conjugating E5 by t2 now gives (t2Pt−1Qt−1)s−1=(t2β′t−2)s−1. Its parenthesis belongs to Bn, so an ordinary negative right destabilization deletes s−1; conjugating by t−2 yields β′. No later Garside or Markov theorem is used.

4.1F1F2step 2.1step 3.1algebra∎

Signs and endpoints. Steps 2.1-3.1 use exactly one negative stabilization and one negative destabilization. The map σi↦σi−1 preserves both Artin relations and all strand placements, so it is an involutive automorphism. Apply the negative sequence to the mirrored boxes and then mirror every word and move: it gives a positive sequence from β′ to β. Reversing that sequence gives the asserted positive sequence from β to β′. All groups and conjugators are explicit finite words, so the proof is choice-free. At n=2 the same formulas hold with P=Q=1; no σ0 is used.

Remarks

The source weaving labels above are chronological records. The geometric product runs its rightmost factor first, so its actual element is the reversed record. Word reversal preserves the Artin relations, carries conjugations to conjugations by the reversed inverse conjugator, and carries a right stabilization to a left one, which cyclic conjugation makes a right stabilization of the same sign. Thus it preserves ordinary Markov sequences. The abstract word equations and move sequence proved here remain exactly as displayed.

This proves the ordinary exchange on the displayed supports. Identifying a multiple-reduction ambiguity with a finite succession of these ordinary exchanges requires a separate strand-by-strand argument; cabling this calculation alone does not supply that argument.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Reducing-move peaks can be lowered to the four-band case

Statement

Assume AC. Consider a finite reducing sequence with height-zero endpoints and strictly positive interior diagrams. An empty reducing portion needs no replacement. Every nonempty such portion may be replaced by a path of strictly smaller maximum height, or reduced to irreducible interior peaks supported by at most four bands of parallel Seifert circles, arranged as in Figures 7 and 8 of the source. The replacement may include finite height-zero ordinary Markov exchange portions; its remaining reducing content is split at height-zero diagrams.

Facts & Assumptions

Given: AC, a finite sequence whose endpoints have height zero, whose interior diagrams have strictly positive height, and whose consecutive diagrams are related by a reducing move or its inverse. For a nonempty reducing portion let H be its maximum height; the empty portion requires no replacement.

[F1]

The height is a nonnegative integer, a reducing move lowers it by exactly one, its inverse raises it by one, and a height-zero diagram can be put in closed-braid form by sphere isotopy and a choice of planar chart (A reducing move lowers the height by one, Coherence of Seifert circles and the height of a diagram, A height-zero diagram represents a closed braid).

[F2]

A reducing move is realized by a reducing arc lying in a defect region and joining an incoherent pair of Seifert circles; a local maximum (a peak) of the sequence is a triple Y(r),Y^,Y(s) of consecutive diagrams with two reducing arcs αr,αs of Y^, which may be assumed transverse and meeting minimally (Defect regions, reducing arcs and the Yamada-Vogel reducing move).

[F3]

Two disjoint reducing arcs whose moves involve three or four distinct Seifert circles commute: the moves may be performed in either order with the same result. If the two arcs involve the same two Seifert circles, the moves form a non-commuting pair, and after one is performed the other is no longer a reducing move (Defect regions, reducing arcs and the Yamada-Vogel reducing move).

[F4]

A region of the Seifert picture with at least three exposed Seifert circles is a defect region, and therefore supports a reducing arc (A positive-height diagram has a defect region).

[F5]

Type I moves and braid-like II and III moves can be replaced by height-zero braid isotopies or Markov moves together with reducing content. Between height-zero portions, split the reducing content into sequences whose intermediate diagrams have positive height and whose endpoints have height zero (Braid-like moves can be moved to height zero).

[F6]

If P,Q∈Bn−1 and t=σn−1, PtQt−1 and Pt−1Qt differ by an explicit finite sequence of ordinary signed Markov moves and conjugations (Ordinary exchange moves are Markov sequences, Markov conjugation and stabilization moves).

[F7]

The fixed real base configuration is symmetric; positive generators are the anticlockwise geometric half twists. The geometric product runs its rightmost factor first. The Artin map to geometric braids is choice-free and surjective and, under AC, an isomorphism (Geometric braids in the disc with setwise endpoints, The elementary geometric half twist, its support disc, and its opposite, Stacking of geometric braids is a well-defined associative operation on isotopy classes, The Artin presentation surjects onto the geometric braid group, The Artin presentation is complete for geometric braids).

Proof

technique · direct
1.1F1F2algebra

Intersection numbers of the two arcs at a peak. Let Y(r),Y^,Y(s) be a peak with arcs αr,αs, assumed transverse and meeting minimally, and let n=∣αr∩αs∣. If n≥2, then smoothing out one or more of the intersection points produces a reducing arc αr′ with the same endpoints as αr that is disjoint from αr and meets αs in fewer than n points; inserting r′ at the peak replaces it by the two peaks Y(r),Y^,Y(r′) and Y(r′),Y^,Y(s), both of the same height h(Y^). Repeating this finitely many times we may assume that the two arcs of every peak meet in at most one point, the maximum height being unchanged because insertion only replaces one peak by two peaks of equal height.

1.2F1F2F6F7construct

Height-one peaks with their actual crossing choices. Exactly one pair C1,C2 is incoherent, and all other e=n−2 circles are coherent with both and with each other. In a chart where C1,C2 bound their common annulus, their orientations are opposite. An additional circle cannot lie in either end disk: coherence with that end circle and with the other would impose opposite orientation requirements. It also cannot be essential in the common annulus, since its two neighbouring annuli would likewise impose opposite requirements. Thus its bounded disk lies in the common annulus and misses C1,C2. Coherence with either fixes its orientation, so two such additional circles cannot be side by side: they would have equal planar orientations and be incoherent. Consequently the additional circles form one coherent nested band, possibly empty. This proves the inside/outside-band description used by the survey's height-one lemma, printed p.24, without a connectedness assumption. Cut its outside region along the two disjoint reducing arcs. On each side the signed crossing strips involve the coherent band and one of C1,C2 only: a strip cannot join the incoherent pair. Reading the two sides gives arbitrary boxes P,Q on the core e strands and one selected strand each. After either reduction choose the cut before the P side in the frame where the two selected single-circle ports are the first two inner ports and the coherent e-band is outside them. Thus n=e+2, both arbitrary contexts act on the last n−1 ports, and the two new crossings are σ1 and its inverse. The physical records are Pσ1Qσ1−1 and Pσ1−1Qσ1. To put their supports in [F6], use an actual old-strand conjugation. The distinct paths eπiuqj, 0≤u≤1, form a geometric braid δ because the base configuration in [F7] is symmetric. For any braid path γ, the square eπiuγ(t) has the same δ along its two endpoint edges and gives Rπ(γ)=δγδ−1 in the rightmost-first convention. It sends the positive ith half twist to the positive (n−i)th, since the disk rotation preserves orientation. By the surjection and completeness in [F7], an Artin word for δ therefore satisfies δσiδ−1=σn−i. Hence its old-strand conjugation takes the two last-supported boxes to first-n−1-supported boxes and σ1 to t=σn−1. Rename these conjugated contexts P,Q. The two resulting records are exactly PtQt−1 and Pt−1Qt, with signs retained. If the other reducing arc starts on the Q side, its record is QtϵPt−ϵ, cyclically conjugate to Pt−ϵQtϵ. These formulas retain every original signed strip in its whole core-plus-one box, rather than fixing the peak's given crossing choices. Isotopic arcs bounding an empty strip give the same pair of records with one context empty. If records are taken chronologically, reverse them for the rightmost-first product: the two actual records are cyclically conjugate to rev⁡(P)t−1rev⁡(Q)t and rev⁡(P)trev⁡(Q)t−1, so [F6] applies in reverse direction with the same supports. Therefore the height-one peak is replaced at height zero by conjugations and, when the choices differ, the ordinary exchange sequence [F6]. At e=0, P=Q=1 and both records are 1, so no strand change is needed. This is the source's equivalence of the resulting diagrams, not literal equality or an appeal to the later four-band or Markov theorem.

2.1F2F4step 1.1algebra

Peaks whose arcs meet once. Now let ∣αr∩αs∣=1. If some reducing arc αt is disjoint from both αr and αs, then inserting t at the peak replaces it by two peaks, each carrying a disjoint pair of arcs. Suppose no third arc of the defect region supporting αr and αs is disjoint from both; then the region has the single arrangement exhibited in the source (survey, printed pp. 21-22) and the two arcs act on four distinct Seifert circles. Consider the region R outside the circles of the defect region and the signed arcs joining them. If R contains a Seifert circle, then some region of the picture has three exposed Seifert circles, hence is a defect region by [F4], and it supports a reducing arc disjoint from both αr and αs, a contradiction. If R contains no Seifert circle, then it contains no signed arc between its four exposed circles either, and a pair of diagonally opposed circles can be joined by a reducing arc inside R, again disjoint from both. Hence in every case a third disjoint arc exists, insertion is possible, and the peak may be replaced by two peaks with disjoint pairs of arcs; the height is unchanged.

3.1F1F2F3step 2.1algebra

Commuting pairs, insertions and peaks of height one. Let a peak have disjoint arcs αr,αs. If their moves involve three or four distinct Seifert circles they form a commuting pair: performing the two reducing moves in either order gives the same diagram Y′ with h(Y′)=h(Y^)−2, so the peak is replaced by the valley Y(r),Y′,Y(s) and is removed, and any new peaks created in this replacement lie at levels strictly below h(Y^). If the arcs involve the same two Seifert circles, the pair is non-commuting; call the peak irreducible if in addition no reducing arc disjoint from both is available that involves a Seifert circle other than those two. If such an arc αt exists, insertion at the peak replaces it by two peaks whose pairs of arcs involve at least three circles, hence are commuting pairs, and each is removed by a valley, so the peak is replaced by content of strictly smaller height. The height-one case is verified separately; its two diagrams need not be literally equal. Summarizing: after finitely many replacements, either the maximum height of the sequence strictly decreases, or every remaining peak is irreducible.

4.1F2F3step 3.1algebra

Irreducible peaks have at most four bands. Let Y(r),Y^,Y(s) be an irreducible peak of height at least two, let C1,C2 be its two Seifert circles, and for i=1,2 let Di be the disk bounded by Ci that contains neither αr nor αs. The complement of D1∪D2∪αr∪αs in the sphere has two components; by irreducibility neither component contains a defect region, since such a region would support a reducing arc disjoint from αr,αs and involving a further circle. Hence the Seifert circles in either component, if any, form a band of mutually coherent parallel circles oriented oppositely to C1 and C2, and the same argument applied inside D1 and D2 shows that each Di contains no defect region either, so Ci is the outer circle of a band. Therefore the diagram consists of at most four bands of parallel Seifert circles, joined by braids on the sums of the band multiplicities, with bands of multiplicity zero allowed, exactly the configuration classified by the source (survey, printed pp. 23-24; Traczyk's Figures 7 and 9).

5.1F1F5step 1.1step 1.2step 2.1step 3.1step 4.1∎

Conclusion on the source's reducing portions. If the reducing portion is empty, nothing is required. Otherwise its height-zero endpoints force each maximal positive height to occur at interior peaks. The finite arc-intersection insertions, commuting valleys and height-one exchange comparisons of the preceding steps either remove all peaks at the current maximum height or leave only irreducible peaks of the four-band shape classified in step 4.1. Height-zero exchange portions are ordinary Markov paths by step 1.2; split the remaining reducing content at their height-zero diagrams, as in [F5]. Thus the maximum height decreases when no irreducible peak remains, and otherwise the stated special configuration is reached. All replacement paths retain the original endpoints and the given first reducing choices. AC is inherited from the coherence/reducing and geometric-completeness suppliers.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Block interchanges transport arbitrary braid boxes

Statement

For p,q≥0, use the strand-placement homomorphisms from Bp to the first p strands and from Bq to the last q strands of Bp+q, using The braid group by Artin presentation. Write α⊗β for the product of these images and set Qp,q:=∏j=1q(σp+j−1σp+j−2⋯σj), where rows are multiplied in increasing j and a descending row with upper index smaller than its lower index is empty. For arbitrary α∈Bp, β∈Bq, (α⊗β)Qp,q=Qp,q(β⊗α). On the right, β occupies the first q strands and α the last p strands. The inverse interchange satisfies (β⊗α)Qp,q−1=Qp,q−1(α⊗β). In particular, either uniform overcrossing or uniform undercrossing of whole blocks transports arbitrary internal braid boxes; it does not require those boxes to commute with a twist on only part of their strands.

Facts & Assumptions

Given: nonnegative integers p,q, the presented braid groups, and the specified row order for Qp,q.

[F1]

The Artin relations are σiσi+1σi=σi+1σiσi+1 and σiσj=σjσi for ∣i−j∣>1; B0 and B1 are trivial (The braid group by Artin presentation).

Proof

technique · direct
1.1F1given

Strand placements and empty blocks. The assignments σi↦σi from Bp and σj↦σp+j from Bq preserve every defining relation, so they define homomorphisms into Bp+q. Their images commute: the closest possible generator indices are p−1 and p+1, whose difference is two. This defines α⊗β independently of word representatives. If p=0 or q=0, every row of Qp,q is empty or there are no rows; Qp,q=1, and both identities reduce to the same braid on the nonempty block. Hence assume p,q≥1.

1.2F1algebra

The descending-row identity. Put Dn,l=σnσn−1⋯σl. For l≤k<n, commute a leading σk past σn,…,σk+2, replace σkσk+1σk by σk+1σkσk+1, and commute the last σk+1 past σk−1,…,σl. Every latter index differs from k+1 by at least two. The resulting word is Dn,lσk+1. Thus σkDn,l=Dn,lσk+1. When n=k+1 the initial commuting segment is empty; when k=l the final commuting segment is empty. Both endpoint cases therefore use the same braid relation.

2.1step 1.2algebra

Generators of the first block. For 1≤i<p, push σi through the rows of Qp,q using step 1.2. At row j its index is k=i+j−1, with j≤k<p+j−1, exactly the required range for Dp+j−1,j; after that row the index is i+j. After all q rows, σiQp,q=Qp,qσq+i. Multiplying this equality by the appropriate inverses gives σi−1Qp,q=Qp,qσq+i−1 as well. For p=1 there are no first-block generators to check.

3.1F1step 2.1algebra

Generators of the second block. In BN, N=p+q, index reflection σi↦σN−i preserves the Artin relations. It sends Qp,q to Qq,p: the reflected word is the product of the grid entries σq−i+j first in increasing i=1,…,q, then increasing j=1,…,p, whereas Qq,p orders the same grid first by j, then by i. To transpose these orders, only pairs with i<i′ and j>j′ must change order. Their indices differ by (i′−i)+(j−j′)≥2, so every such swap is a far commutation. Apply step 2.1 to Qq,p and its first-block generator σq−j, 1≤j<q, then reflect back: this gives σp+jQp,q=Qp,qσj, and the same formula for inverse generators. For q=1 this verification is vacuous.

4.1step 1.1step 2.1step 3.1algebra∎

Arbitrary boxes and inverse crossings. Apply steps 2.1 and 3.1 successively to any words for α and β, including inverse letters. They give (α⊗β)Qp,q=Qp,q(β⊗α) with the indicated shifted embeddings. Multiplying by Qp,q−1 on both sides gives the asserted inverse identity. The displayed equations use algebraic word order: the rightmost factor runs first geometrically. Thus Qp,q physically takes ordered input blocks (q,p) to output blocks (p,q), and Qp,q−1 takes physical input (p,q) to output (q,p). Each strand of one block crosses each of the other once with uniform sign, and order inside either block is preserved. A chronological record is obtained by reversing the actual word; when chronological input is written (p,q), the negative chronological interchange is Qq,p−1, whose actual geometric word is its reversal. These are distinct reading conventions, not a reflection or change of generator sign. The two algebraic identities already proved transport the boxes at their specified input/output frames. Thus both signs transport arbitrary boxes, with all zero-width and one-width cases covered in steps 1.1, 2.1 and 3.1. No closure equivalence or Markov theorem is used.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Compensated band kinks decompose into ordinary Markov moves

Statement

Assume AC. Let n≥m≥0 and α∈Bn. Set W0=1; for m≥1 put Cm=Qm,m,Tm=(σ1⋯σm−1)m,Wm=CmTm−1, using the block interchange of Block interchanges transport arbitrary braid boxes. Place Wm on the last m original strands and m newly appended strands: write ιk(σi)=σk+i for a consecutive strand placement and put k=n−m. Then α ιk(Wm)∈Bn+m is related to α∈Bn by conjugations and exactly m positive ordinary destabilizations of Markov conjugation and stabilization moves, decreasing the strand number from n+m to n. Replacing every generator of Wm by its inverse gives the analogous packet with m negative destabilizations. Reversing these sequences gives the corresponding m stabilizations. The surrounding braid α is arbitrary and may mix the selected original strands with every other original strand.

There is also a packet with the compensation in the reverse word order: set V0=1 and, for m≥1, Vm=TmQm,m−1=Qm,m−1(1m⊗Tm). The first Tm is on the selected original strands, and the last is on the new strands. For arbitrary α as above, αιk(Vm) is related to α by conjugations and exactly m negative ordinary destabilizations; its generator mirror has m positive destabilizations. Reversing these sequences gives the corresponding stabilizations. Thus the original packet claims and these reverse-order packet claims hold with the same arbitrary-box hypothesis.

Facts & Assumptions

Given: AC, n≥m≥0, α∈Bn, and the specified word and strand-placement conventions. For m≥1 define Am=Qm,m−1,Bm=σ2m−2⋯σm,hm=σm−1⋯σ1,gm=BmTm−1hm. Empty rows and products are 1. All the latter words lie in B2m−1, except Tm,hm, which already lie on the first m strands. Products are concatenations; the Artin action satisfies ρ(uv)=ρ(u)∘ρ(v). In substitution calculations, u(x) abbreviates ρ(u)(x).

[F1]

Under AC the Artin representation is faithful (The Artin representation is faithful).

[F2]

The representation of The Artin representation on a free group uses ρ(σi)(xi)=xixi+1xi−1, ρ(σi)(xi+1)=xi and fixes the remaining free generators (Artin automorphisms of the free group).

[F3]

Conjugation in a fixed Br and the deletion of a final σr−1±1 from a word on the first r−1 strands followed by that generator are ordinary Markov moves (Markov conjugation and stabilization moves).

[F4]

Qp,q is the increasing product of its descending rows; strand placements preserve the Artin relations and are homomorphisms, and the whole-block transport identities hold (Block interchanges transport arbitrary braid boxes, The braid group by Artin presentation).

Proof

technique · direct
1.1F3F4given

Empty and single-strand packets. For m=0, W0=1 and no move is needed, including when n=0. For m=1, C1=σ1, T1=1 and αιn−1(W1)=ασn; one positive destabilization gives α. Here n≥1, so no nonexistent σ0 is used. Henceforth m≥2.

1.2F2F4construct

The substitutions needed for the packet identity. Work first in B2m−1. Put r=m−1, a=x1⋯xr, t=xm, yj=xm+j for 1≤j≤r, w=y1⋯yr, and P=at. For 0≤q≤m, temporarily use the free group on m+q generators for Qm,q; its substitution sends xi to Pxm+iP−1 for i≤q, sends xq+j to xj for j≤m, and fixes higher generators. Prove this by induction on q, beginning with the empty word and adjoining one fixed free generator at each rank increase. The next descending row on indices q+1,…,m+q+1 sends xq+1 to Lxm+q+1L−1, L=xq+1⋯xm+q, and sends xq+1+j to xq+j for j≤m. The preceding Qm,q sends L to P by induction and fixes xm+q+1; it retains the previously established images of the first q generators. This proves the formula at q+1. Specializing back to the ambient F2m−1 at q=r gives Am(xj)=PyjP−1 for j≤r and Am(xr+j)=xj for j≤m. The descending word hm sends x1 to ata−1 and xj to xj−1 for 2≤j≤m; it fixes the yj. Finally Tm conjugates all x1,…,xm by P and fixes the yj. To check the last formula, δ=σ1⋯σm−1 sends xj to x1xj+1x1−1 for j<m and xm to x1. Put Pℓ=x1⋯xℓ. For 0≤ℓ≤m, induction gives δℓ(xj)=Pℓx1+((j+ℓ−1) mod m)Pℓ−1. Indeed δ(Pℓ)=Pℓ+1x1−1 when ℓ<m, and δ(xk)=x1x1+(k mod m)x1−1, including k=m; the two x1 factors cancel in the conjugation. At ℓ=m the image is PxjP−1. All statements concern the frozen homomorphism convention of [F2].

2.1F1F2step 1.2algebra

A general braid identity. We prove AmBmTm−1=hmι1(Wr)hm−1. Let K=ρ(AmBm). The descending Bm sends t to t(y1⋯yr−1)yr(y1⋯yr−1)−1t−1 and yj to xm+j−1. Step 1.2 consequently gives K(xj)=PyjP−1 for j≤r, K(t)=ata−1, K(yj)=xj, and K(P)=Pwa−1. Applying Tm−1 first therefore gives ρ(AmBmTm−1)(xj)=aw−1yjwa−1, ρ(AmBmTm−1)(t)=aw−1twa−1 and ρ(AmBmTm−1)(yj)=xj. To calculate the other side, put v=x2⋯xm. The shifted Wr fixes x1, sends xj+1 to vw−1yjwv−1 and sends yj to xj+1; these are the equal-block version of the row and full-twist substitutions in step 1.2. Also hm−1(xj)=xj+1 for j≤r, hm−1(t)=v−1x1v, hm(v)=a and hm(x1)=ata−1. Substitution now gives exactly the same three displayed images on all 2m−1 generators. By [F1] the two braid words are equal.

3.1F3F4step 2.1constructalgebra

One destabilization with an arbitrary surrounding braid. In Bn+m shift the local words by k=n−m, suppressing this shift in the calculation. Since Cm=Amσ2m−1Bm, cyclically conjugating αAmσn+m−1BmTm−1 past its initial segment αAmσn+m−1 gives (BmTm−1αAm)σn+m−1. The parenthesized word R uses only indices at most n+m−2: the surrounding α uses at most n−1, and the local Am,Bm,Tm have at most n+m−2. Thus [F3] deletes its final generator and gives R∈Bn+m−1. Conjugate this whole word by gm−1. Direct cancellation, without commuting any factor through α, gives gm−1Rgm=hm−1α(AmBmTm−1)hm=(hm−1αhm)ι1(Wm−1) by step 2.1. Restoring shifts, the new surrounding braid α1=hm−1αhm lies in the original Bn, because hm uses only its last m strands. The remaining packet occupies the last m−1 original strands and m−1 new strands. This is the required one-step reduction of packet width.

4.1F1F3F4step 1.1step 3.1algebra

Induction, signs and conjugation bookkeeping. Repeat step 3.1 with packet widths m,m−1,…,2, then use step 1.1 at width one. At every stage the surrounding braid is an arbitrary element of the same original Bn, conjugated by a word supported on its selected original strands; the active packet width and the number of new strands both decrease by one. The result has exactly n strands after exactly m positive destabilizations. The resulting original braid is H−1αH, where H is the ordered product in Bn of the shifted hj used in those stages; one final conjugation gives α. The generator-inversion map σi↦σi−1 preserves both Artin relations, hence is an involutive automorphism compatible with all strand placements. Apply it to the entire sequence for the surrounding braid obtained by applying that same automorphism to α. This gives the mirrored packet with exactly m negative destabilizations and endpoint α. Its compensation is the opposite full twist: the hm substitution of step 1.2 fixes P and shifts xj to xj−1 until its first wrap, when it becomes PxmP−1. Thus hmm also conjugates each xj by P, so faithfulness gives hmm=Tm. Mirroring this equality gives mirror⁡(Tm)=(σ1⋯σm−1)−m=Tm−1, since the mirror of hm is the inverse of that ascending product. Consequently the positive packet has a negative full-twist compensation and its mirror a positive one. Reversing either sequence gives the stabilization statement. Conjugations are counted separately from these m strand-changing moves. AC is used only through faithful action [F1]; no closure-isotopy-to-Markov implication is invoked.

5.1F1F3F4step 1.1step 4.1algebra∎

Reverse-order compensated packets. Word reversal rev⁡ fixes each signed generator and reverses multiplication. The Artin braid relations are palindromic, and reversing a far commutation gives the same relation, so reversal is an involutive anti-automorphism compatible with strand placements. It carries a conjugation by g to one by rev⁡(g)−1, and carries a right stabilization to a left stabilization, which cyclic conjugation turns into a right stabilization of the same sign; the inverse applies to destabilizations. Its use therefore preserves the number and signs of ordinary strand-changing moves. To verify rev⁡(Qm,m)=Qm,m directly, label a row cell by (i,j), 1≤i,j≤m, with generator σm+j−i. After reversal set I=m−i+1,J=m−j+1; its index becomes m+I−J. The reversed order lists first J then I, whereas the rows of Qm,m list first I then J. Every pair that changes order has I>I′ and J<J′, so the generator indices differ by (I−I′)+(J′−J)≥2 and far commutation suffices. Reversal of Tm is hmm=Tm by step 4.1; m=0,1 are the empty or trivial cases. Consequently the mirrored packet of step 4.1 is Qm,m−1Tm, and its word reversal is Vm=TmQm,m−1. Apply the negative sequence of step 4.1 to rev⁡(α), reverse every word and move, and cyclically conjugate its starting word Vmrev⁡(rev⁡(α))=Vmα to αVm. This gives exactly m negative ordinary destabilizations and conjugations to α, with no commutation of a twist through α. Whole-block naturality in [F4] gives TmQm,m−1=Qm,m−1(1m⊗Tm) on the local old/new strands. Mirroring the whole argument gives the positive variant, and reversing gives stabilizations. This proves every additional assertion of the Statement without changing the original packet sequence.

Remarks

The abstract packet theorem permits arbitrary internal boxes in α. Applying it to a pictured geometric band move still requires identifying its cut, strand placements and full-twist compensation with Wm or its mirrored packet. The statement alone does not identify Traczyk's Figure 10 or certify the remaining Figure 8 and Figure 11 comparisons.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Band exchanges decompose into ordinary Markov moves

Statement

Assume AC. Let c,p,q≥0, P∈Bc+p and Q∈Bc+q, each placed on the first indicated strands. Put n=c+p+q, and place t=Qp,q and u=Qq,p after the first c strands, with the block-interchange conventions of Block interchanges transport arbitrary braid boxes. Then the typed band exchange β=PtQt−1⟼β′=Pu−1Qu is a finite sequence of ordinary Markov moves. If p,q>0, put m=max⁡(p,q) and δ=∣p−q∣. One explicit sequence uses δ positive stabilizations and δ positive destabilizations, together with m negative stabilizations and m negative destabilizations; conjugations are counted separately. If p=0 or q=0, the endpoints coincide and no move is needed. The boxes P,Q are arbitrary on their supports and may mix the core strands with the relevant band.

Facts & Assumptions

Given: AC, the nonnegative widths c,p,q and arbitrary boxes on the stated strand placements.

[F1]

The Artin braid relation and far commutation define these braid groups, with B0,B1 trivial (The braid group by Artin presentation).

[F2]

A compensated packet Qm,mTm−1, Tm=(σ1⋯σm−1)m, with arbitrary surrounding braid has a sequence of m positive ordinary destabilizations; its mirror has m negative destabilizations, and reversing gives stabilizations (Compensated band kinks decompose into ordinary Markov moves).

[F3]

Whole-block interchanges transport arbitrary boxes, including full twists, through the positive and negative uniform crossing, with the prescribed changed strand placements (Block interchanges transport arbitrary braid boxes).

[F4]

Under AC the frozen Artin representation is faithful; its positive generator sends xi to xixi+1xi−1 and xi+1 to xi, and ρ(vw)=ρ(v)∘ρ(w) (The Artin representation is faithful, Artin automorphisms of the free group, The Artin representation on a free group).

[F5]

For single-strand exchanged bands, ordinary exchange with arbitrary boxes on the first n−1 strands has one negative stabilization and one negative destabilization (Ordinary exchange moves are Markov sequences).

[F6]

Conjugations and v↔vσk±1 between Bk and Bk+1 are ordinary Markov moves (Markov conjugation and stabilization moves).

Proof

technique · direct
1.1F1F3given

Empty bands and conventions. If p=0 or q=0, both interchanges are 1 by [F3], and both displayed endpoints are literally PQ in the same group; no padding of a nonexistent band is attempted. Assume henceforth p,q>0. All products below are braid-group products with the action convention in [F4]. Core strands are fixed by the local words, whose indices are shifted by c. Descending and ascending products of length zero are 1.

1.2F2F4algebra

The equal-width braid relation. First suppose p=q=m. In Bc+3m put t=Qm,m on blocks 1,2 and s=Qm,m on blocks 2,3. Let their free generators be xj,yj,zj, 1≤j≤m, and write a=x1⋯xm, b=y1⋯ym. The rectangular row induction from the packet proof gives ρ(t)(xj)=ayja−1, ρ(t)(yj)=xj and fixes zj; for s it gives ρ(s)(yj)=bzjb−1, ρ(s)(zj)=yj and fixes xj. To recall that induction explicitly, for Qm,k on m+k strands put a=x1⋯xm; its images are axm+ia−1 for 1≤i≤k and xj for the input generator xk+j, 1≤j≤m. At k=0 this is identity. Adjoin a fixed generator and multiply by the next descending row: that row sends xk+1 to (xk+1⋯xk+m)xk+m+1(xk+1⋯xk+m)−1 and xk+1+j to xk+j for 1≤j≤m; the previous substitution sends the parenthesized product to a. This proves the induction and the displayed equal-block formulas at k=m. Now both tst and sts send xj to abzjb−1a−1, yj to ayja−1 and zj to xj; they fix the core generators. For example t(a)=aba−1 and t(b)=a, giving the first image in tst; the other substitutions follow directly. Faithfulness in [F4] proves tst=sts for every m≥1, including m=1.

1.3F1F3F6algebra

Padding one strand is an ordinary stabilization. Suppose p<q, and allow either crossing sign: set v+=Qp,q or v−=Qq,p−1, shifted by c. At the cyclic cut after the crossing, PvϵQvϵ−1 is conjugate to Qvϵ−1Pvϵ. Add one strand after the p-band, set P+=Pσc+p and replace the crossing by v++=Qp+1,q or v−+=Qq,p+1−1. Then vϵ+=Rϵvϵ, where R+=σc+p+1⋯σc+p+q and R−=σc+p+1−1⋯σc+p+q−1. For the positive word, interleave the jth letter of R+ before the jth row of Qp,q: it commutes past every earlier old-row letter, whose maximum index is c+p+j−2, at distance at least two. This makes precisely the rows of Qp+1,q. For the negative word, split the last descending row of Qq,p+1 and invert; its inverse is exactly R−. The largest index in P is c+p−1, so P commutes with either Rϵ. Whole-block naturality [F3] gives σc+pvϵ+=vϵ+σn, because this generator acts within the enlarged first block of width p+1. Consequently Q(vϵ+)−1P+vϵ+=Qvϵ−1Pvϵσn. The left is the padded braid at its cyclic cut, and the right is exactly one positive ordinary stabilization of the original braid at its cyclic cut. The box Q remains on the first c+q old strands. Thus padding changes no component by a tensor identity; it is an actual Markov stabilization with explicit word equality.

2.1F2F3F5F6step 1.2algebra

A compensated equal-width exchange. For m=1, [F5] supplies the sequence directly, with the trivial full twist T1=1. The following calculation covers m≥2. Write zi for the positive full twist Tm on block i, i=1,2,3. The boxes P,Q∈Bc+m commute with s, since their largest generator is c+m−1 and the smallest generator of s is c+m+1. They also commute with z2,z3. By [F3], sz3=z2s and t2z2=z2t2; these transport twists on entire blocks and do not pass a partial twist through a box. Put A=Pt2, B=t−1Qt−1 and X=t2β′t−2. Then BA=A−1βA. Rearranging sts=tst from step 1.2 gives t−1st=sts−1 and st−1s−1=t−1s−1t; consequently ts−1t−1s=t2s−1t−1 and s2t−1s−1=t−1s−1t2. These identities and the box commutations give E1=As−1B=Pts−1t−1sQt−1 and t2sE1s−1t−2=Xs−1; explicitly sE1s−1=Pt−1Qt−1s−1t2 using ts−1t−1s=t2s−1t−1 and s2t−1s−1=t−1s−1t2. Conjugate β to BA and apply the reverse mirrored packet of [F2], producing BAs−1z2=BAz3s−1 by m negative ordinary stabilizations. Conjugate by B−1 to Az3s−1B. Since A is an old-strand braid, it commutes with z3. Conjugation by t2s therefore yields z2Xs−1: move z3 left past A, use sz3=z2s and t2z2=z2t2, and use the untwisted equality above. Conjugate next by z2−1 to Xs−1z2, exactly a mirrored compensated packet with arbitrary surrounding braid X. Its m negative ordinary destabilizations give X, and conjugation by t−2 gives β′. This completes the equal-width exchange with every box retained. Reversing this sequence gives the reverse exchange with the same numbers and signs.

3.1F2F4F6step 1.1step 2.1step 1.3construct∎

Unequal widths and move counts. For 0<p<q, repeat step 1.3 exactly δ=q−p times. At stage j, the enlarged box lies in Bc+p+j, the other box stays in Bc+q, and the ordinary added generator is σc+p+j+q; hence every use satisfies the same support inequalities. The two padded endpoints have equal band width m=q. Step 2.1 relates them with m negative stabilizations and m negative destabilizations; apply step 1.3 backwards for the negative-crossing endpoint to remove the δ positive kinks. For p>q>0, cyclically start with Q instead: the first endpoint becomes Qt−1Pt on block order (q,p), and the second becomes QuPu−1. These are respectively the negative and positive endpoints with the smaller first band q; pad that first band by step 1.3 and use the reverse of step 2.1 before removing the padding. For p=q no padding is needed. Thus for every positive pair of widths the stated δ positive and m negative stabilizations and matching destabilizations suffice, with conjugations accounted separately. Zero widths were settled in step 1.1, so all permitted cases are covered. AC is used only through the faithful-action and packet suppliers; no Markov closure-equivalence theorem is invoked.

Remarks

This is an abstract typed band-exchange calculation. The algebraic support convention places the first box on core plus p strands. In actual geometric time, Qp,q and Qq,p−1 take input blocks (q,p) to output blocks (p,q), positively and negatively respectively; their chronological records are their word reversals. Thus geometric input/output labels must be distinguished from algebraic word order. Applying it to a reducing-move ambiguity still requires identification of the source ports, boxes and frames with these two endpoints.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

The first four-band comparison is a compensated band stabilization

Statement

Assume AC. Let a,b≥1, c,d≥0, n=a+b+c+d, and choose arbitrary A∈Ba+c, B∈Bb+c, C∈Ba+d, D∈Bb+d. Write ιs(σi)=σs+i, use the row interchanges Qp,q of Block interchanges transport arbitrary braid boxes, and set T0=T1=1, Tr=(σ1⋯σr−1)r for r≥2. Define the following words in Bn: A0=ιb(A),B0=ιa(B),C0=ιb(C),D0=ιa(D),X=Qb,a,Y=ιa+b(Qc,d), α=C0Y−1A0XB0YD0X−1. In Bn+d put A1=ιb+d(A),B1=ιa+d(B),C1=ιb+d(C),D1=ιa+d(D),Y1=ιa+b+d(Qc,d), H=Qb,d+aQd,a,U=Qb+d,a+d−1,β=C1Y1−1A1H ιa(Td) B1Y1D1U. Each factor is placed on the first strands unless a shift is displayed. Finally set ℓ=a+d+c and R=ιb(Qℓ,d),K=ιb(Tℓ−1) ιa+b(Qd,c)∈Bn,V=ιn−d(TdQd,d−1). Then RβR−1=αKVK−1. Consequently α is related to β by conjugations and exactly d negative ordinary stabilizations, from n to n+d strands. At d=0 the two words coincide and the count is zero. These are statements about the displayed algebraic words; identifying a diagram's chronological record with its actual geometric word is a separate convention.

Facts & Assumptions

Given: AC, the stated widths and arbitrary boxes, with every strand placement as in the Statement.

[F1]

Artin braid relations and far commutations define these groups; strand placements are homomorphisms (The braid group by Artin presentation).

[F2]

Uniform whole-block interchanges transport arbitrary internal words, including inverses, in the precise shifted placements (Block interchanges transport arbitrary braid boxes).

[F3]

Under AC the frozen Artin representation is faithful; ρ(uv)=ρ(u)∘ρ(v), ρ(σi)(xi)=xixi+1xi−1 and ρ(σi)(xi+1)=xi (The Artin representation is faithful, The Artin representation on a free group, Artin automorphisms of the free group).

[F4]

With arbitrary surrounding braid, the reverse-order packet TdQd,d−1 on the last d old strands and d new strands has exactly d negative ordinary destabilizations and conjugations to that surrounding braid; its reversed sequence gives stabilizations (Compensated band kinks decompose into ordinary Markov moves).

[F5]

Conjugations and the ordinary signed stabilization/destabilization are Markov moves (Markov conjugation and stabilization moves).

Proof

technique · direct
1.1F1F2given

Zero widths and complete embeddings. If d=0, Y=Y1=1, R=V=1, H=X, U=X−1, and all primed boxes are their unprimed counterparts; therefore β=α and the asserted equality and count hold. Now assume d≥1. All words are defined: A0,C0 fit inside the last ℓ old strands after the first b, B0,D0 fit after the first a, their primed versions are shifted by d, H uses the first a+b+d strands, and U uses the first a+b+2d strands. The unshifted part of K uses only the ℓ old strands and Qd,c uses the last c+d old strands, so K∈Bn. When c=0, Y=Y1=1 and Qd,c=1, with no padding or generator indexed zero; all subsequent products on the c-block are empty and all corresponding individual-generator rows are absent.

1.2F3construct

Substitution rules used below. For Qp,q on p+q generators write P=x1⋯xp. Its images are Pxp+iP−1 for input xi, 1≤i≤q, and xj for input xq+j, 1≤j≤p. Induct on the number of rows q, adjoining a fixed generator each time: the next descending row sends xq+1 to (xq+1⋯xq+p)xq+p+1(xq+1⋯xq+p)−1 and xq+1+j to xq+j; the preceding substitution sends that parenthesized product to P. This proves every individual-generator formula, including p=0 or q=0. Equivalently, in the ordered input basis (f1,…,fq,g1,…,gp) and output names (x1,…,xp,y1,…,yq), the positive interchange sends fi to PyiP−1 and gj to xj; its inverse sends each xj to gj and each yi to G−1fiG, G=g1⋯gp. The names on each side designate the corresponding consecutive indices of the same free basis, rather than a change of generator sign. For r≥1, Tr conjugates each of its r free generators by their product. Indeed for δ=σ1⋯σr−1, δ(xj)=x1x1+(j mod r)x1−1; setting Ph=x1⋯xh, induction gives δh(xj)=Phx1+((j+h−1) mod r)Ph−1 for 0≤h≤r, since δ(Ph)=Ph+1x1−1. At h=r this is conjugation by the full product. Every Artin generator fixes the full boundary product, so all local words do too. These rules follow solely from [F3], and apply after any displayed shift.

2.1F3step 1.2algebra

The upper routing identity on all generators. Put RB=ιa(Qb+c+d,d) and Z=Qa,dQd,aιa(Td). We prove RHιa(Td)=XRBZ. Within this substitution calculation, capitals denote products of free generators, not the braid boxes. For this comparison label the input consecutive blocks by sizes (a,d,b,c,d) and label the output free generators by (b,a,c,d,d) as bi,aj,ch,dj,ej, with products B,A,C,D,E and P=ACD. The domain rows consist respectively of xj, xa+j, xa+d+i, xa+d+b+h and xn+j in their specified ranges. These are explicit output index assignments: for example output aj=xb+j, ch=xb+a+h, dj=xb+a+c+j, and ej=xn+j. Both sides send an input generator in the first a-block to BPEP−1ajPE−1P−1B−1; a generator in the next d-block to BPEejE−1P−1B−1; a generator in the following b-block to bi; and the final c,d blocks to ch,dj respectively. Here is the full substitution verification. By step 1.2, H sends the first a generators to BFajF−1B−1, the next d to BfjB−1, and the next b to bi, in the intermediate order (b,d,a,c,d) with middle product F; it leaves the last two blocks in place. The rightmost Td first conjugates that input d-block by its own product, and R then sends fj to PejP−1 and the remaining ℓ intermediate generators to the corresponding a,c,d generators, giving the stated five rows. On the other side the monodromy Qa,dQd,a sends aj to AFA−1ajAF−1A−1 and fj to AfjA−1; consequently Z sends fj to AFfjF−1A−1 while its a-row is unchanged. The word RB sends fj to SejS−1, S=BCD, and the remaining (b,c,d) generators to those same output blocks. Finally X sends each current aj to BajB−1 and each current bi to bi in the output order (b,a), fixing c,d,e; in particular X(AS)=BACD=BP. Substituting gives exactly the same five rows above, including the full internal d-block conjugation. Faithfulness gives the asserted routing identity.

3.1F3step 1.2step 2.1algebra

The compensated lower routing identity on all generators. We prove RHιa(Td)UR−1=KVK−1. Capitals again denote free-generator products in this calculation. Use this time the ordered free basis (bi,aj,dj,ch,ej) of sizes (b,a,d,c,d), with products B,A,D,C,E, and set P=ADC, M=D−1PE, N=MP−1. Both sides fix every bi,ch, send aj to NajN−1, send dj to MejM−1 and send ej to M−1djM. For the left side, apply the words from right to left using step 1.2. After R−1 the intermediate order is (b,d,a,d,c): each old a,d,c generator is the corresponding generator in the shifted last ℓ slots, and a new ej is L−1fjL, L the product of those last ℓ slots. The inverse interchange U moves the (b,d) group past (a,d); its rule conjugates the latter by the inverse product of the relocated (b,d) group. The next Td conjugates the middle d-block by its product. After applying H, in the intermediate order (b,d,a,d,c) with products (B,G,A,F,C), the old aj image is F−1GajG−1F, the old dj image is F−1GgjG−1F, the old ch is ch, and the new image is L1−1fjL1, L1=F−1GAFC; the b-row is fixed. These follow by cancellation of the conjugating (B,F) product against the leading B in the H substitutions of step 2.1. The final R sends gj to PejP−1, each fj to dj, and the remaining a,c generators to themselves. It sends L1 to D−1(PEP−1)ADC=D−1PE=M. Substitution gives precisely the five rows asserted, since R(F−1G)=D−1PEP−1=N. For the right side, split the old slots temporarily as (b,a,c,d) with last two products F,G. By step 1.2, K=ιb(Tℓ−1)ιa+b(Qd,c) sends aj to P−1ajP, each generator of the c-block to P−1DchD−1P, and each last-d generator to P−1djP; it fixes the new ej and preserves P. Thus K−1(aj)=PajP−1, K−1(dj)=PgjP−1, and K−1(ch)=PG−1fhGP−1. The packet V=TdQd,d−1 fixes a,f, sends gj to ej, and sends ej to E−1GgjG−1E. In particular K(V(P))=K(AFE)=D−1PE=M. Applying K to these substituted images gives NajN−1, MejM−1, ch and E−1P−1DdjD−1PE=M−1djM, with the b-row fixed. All individual generators have now been checked, so faithfulness proves the identity.

4.1F1F2step 2.1step 3.1algebra

Strip arbitrary boxes by whole-block naturality. The word R interchanges the whole old ℓ-block with the new d-block. Hence [F2] gives RC1Y1−1A1=C0Y−1A0R, because the three old words lie wholly in that ℓ-block and their primed embeddings are shifted by d. Insert the upper routing identity of step 2.1 into RβR−1. The factor Z acts only on the first a+d strands, whereas B1,D1 start after that block and Y1 is farther out; their smallest possible generator is at least a+d+1, while Z has largest at most a+d−1. Thus every letter of Z commutes with every letter, including inverse letters, of B1Y1D1. Next RB interchanges the entire old (b+c+d) block after the first a with the new d block, so [F2] gives RBB1Y1D1=B0YD0RB. The calculation is therefore RβR−1=C0Y−1A0XB0YD0[RBZUR−1]. By steps 2.1 and 3.1, the bracket equals X−1KVK−1. This yields exactly αKVK−1. Every box is arbitrary on its stated whole block; no full twist on only part of a box is commuted through it.

5.1F3F4F5step 1.1step 4.1construct∎

The ordinary move sequence and all boundaries. Conjugate α in Bn by K−1. By the reverse of [F4], append V to this arbitrary surrounding braid by exactly d negative ordinary stabilizations: K−1αK↦(K−1αK)V. Conjugate by K and then by R−1 to obtain β using step 4.1. The strand number rises from n to n+d; K remains in the original Bn and every intermediate use of [F4] has its selected last d old strands and d new strands. At d=1 all full twists on the single copied band are 1, but the necessary old full twist in K on ℓ strands is retained. At c=0, the c rows and cap words disappear, and every substitution and naturality argument above still holds with C=1 as a free-product name; this does not set the arbitrary braid box C to identity. At d=0 use step 1.1, with no move and no nonexistent generator. These distinctions cover all stated widths and trivial or nontrivial boxes. AC is inherited only from the faithful-action and packet suppliers; no closure-equivalence or Markov theorem is used.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

The second four-band comparison is a compensated band destabilization

Statement

Assume AC. Let a,b≥1, c,d≥0, n=a+b+c+d, and choose arbitrary A∈Ba+c, B∈Bb+c, C∈Ba+d, D∈Bb+d. Use ιs(σi)=σs+i and the words Qp,q of Block interchanges transport arbitrary braid boxes. Set T0=T1=1 and Tr=(σ1⋯σr−1)r for r≥2. In Bn put A1=ιb+d(A),B1=ιa+d(B),Cin=C,Din=D, H=Qb,d+aQd,a,F=Hιa(Td),α=DinA1FB1CinF−1. In Bn+d define Y1=ιa+b+d(Qc,d),U=Qb+d,a+d−1,β=DinY1−1A1FB1Y1CinU. Finally set K=ιb(Qd,a) ιa+b(Qc,d−1)∈Bn,V=ιn−d(TdQd,d−1). Then FY1UY1−1=KVK−1 and Y1βY1−1=αKVK−1. Consequently β is related to α by conjugations and exactly d negative ordinary destabilizations, from n+d to n strands. At d=0 the words coincide. These are algebraic word assertions; source chronology is interpreted separately.

Facts & Assumptions

Given: AC, the stated widths, arbitrary boxes and displayed placements.

[F1]

Artin relations define the groups and their shifted strand embeddings (The braid group by Artin presentation).

[F2]

Whole-block interchanges transport arbitrary internal braid words and their inverses (Block interchanges transport arbitrary braid boxes).

[F3]

Under AC the Artin representation is faithful, with ρ(uv)=ρ(u)∘ρ(v) and σi:xi↦xixi+1xi−1, xi+1↦xi (The Artin representation is faithful, The Artin representation on a free group, Artin automorphisms of the free group).

[F4]

An arbitrary surrounding braid followed by the last-band packet TdQd,d−1 reduces by d negative ordinary destabilizations and conjugations, including d=0,1 (Compensated band kinks decompose into ordinary Markov moves).

[F5]

Conjugations and signed ordinary stabilization/destabilization are Markov moves (Markov conjugation and stabilization moves).

Proof

technique · direct
1.1F1F2given

Placements and zero widths. If d=0, Y1=V=1, F=Qb,a, U=F−1, K=1, and α=β. Assume d≥1. The boxes A1,B1 fit in Bn, while Cin,Din use the first a+d,b+d strands. Their largest generator is at most a+d−1,b+d−1, respectively, whereas Y1 starts at generator a+b+d+1; hence both commute with Y1. The word F uses the first a+b+d strands, and U the first a+b+2d strands of Bn+d. Both factors of K fit in Bn, and V occupies the last d old and d new strands. If c=0, Y1 and the second factor of K are identity; all c-rows below are absent.

1.2F3construct

Complete substitution rules. On ordered input blocks (f1,…,fq,g1,…,gp), the action of Qp,q has output blocks (x1,…,xp,y1,…,yq) and images fi↦XyiX−1, gj↦xj, where X=x1⋯xp. Its inverse sends xj↦gj, yi↦G−1fiG, G=g1⋯gp. To verify these rules, adjoin the descending row for the next input fi: that row conjugates its last generator by the intervening consecutive product and shifts the other generators one place; induction on q sends that product to X. This checks each individual input generator, with empty blocks giving identity. The action of Tr conjugates all r generators by their ordered product. For δ=σ1⋯σr−1 and Ph=x1⋯xh, direct substitution gives δ(Ph)=Ph+1x1−1; induction gives δh(xj)=Phx1+((j+h−1) mod r)Ph−1 for 0≤h≤r. At h=r this proves the assertion. Each local word fixes the product of all its generators.

2.1F3step 1.2algebra

The routing identity on every generator. We prove J:=FY1UY1−1=KVK−1. Within this calculation capitals denote products of free generators, not braid boxes. Use the consecutive numeric basis (bi,dj,aj,ch,ej) of sizes (b,d,a,c,d), and write its products B,D,A,C,E. Thus bi=xi, dj=xb+j, aj=xb+d+j, ch=xb+d+a+h and ej=xn+j. Put N=CE−1C−1D. Both sides fix bi,ch and have the following remaining individual images: dj⟼CejC−1,aj⟼NajN−1,ej⟼E−1C−1DdjD−1CE. Here is the full left-side substitution. Apply the factors from right to left by step 1.2. After Y1−1 the ordered intermediate blocks are (B,D,A,G,C); the c-row is unchanged and each new ej has image C−1gjC. The next U gives order (A,G,B,D,C): bi,dj move without conjugation, while aj maps to (BD)−1aj(BD) and the new row to C−1(BD)−1gj(BD)C. The factor Y1 interchanges the last (D,C) blocks, giving order (A,G,B,C,D), sends dj to CdjC−1 and fixes ch, so the intervening product BD becomes BCDC−1. Finally F=Hιa(Td) maps the first blocks (A,G,B) to (B,D,A): first Td conjugates each gj by G, and H sends aj to BGajG−1B−1, gj to BgjB−1 and bi to bi. The final output names are (B,D,A,C,E), so the intermediate last D is now numeric E, and the intermediate G is numeric D. Substitute into all five rows. The old dj becomes CejC−1; the aj conjugator reduces to CE−1C−1D=N; the new row reduces to E−1C−1DdjD−1CE; bi,ch stay fixed.

3.1F3step 1.2step 2.1algebra

The other side on the same indexed basis. Temporarily name the old input slots of K by (B,A,C,G), with final d-block product G, and keep the new block E. Step 1.2 gives K(aj)=DajD−1, K(ch)=ch, K(gj)=C−1djC, with the first b and new e rows fixed. Its inverse therefore sends numeric dj to CgjC−1, numeric aj to CG−1C−1ajCGC−1, and numeric ch to ch. The packet V sends gj to ej, sends ej to E−1GgjG−1E, and fixes aj,ch,bi. Applying K last gives dj↦CejC−1, aj↦(CE−1C−1D)aj(CE−1C−1D)−1, and ej↦E−1C−1DdjD−1CE, with the other two rows fixed. These are all images of the same numeric basis used in step 2.1, not merely block products or permutations. Faithfulness now proves J=KVK−1.

4.1F1F3F4F5step 1.1step 3.1algebra∎

Arbitrary boxes and the ordinary sequence. By step 1.1, Y1 commutes with Din,Cin. Direct cancellation gives Y1βY1−1=DinA1FB1CinY1UY1−1=α(FY1UY1−1)=αKVK−1. This uses no commutation with A1,B1 and no partial-band assumption about any box. Conjugate β by Y1 and then by the old-strand word K−1; the result is (K−1αK)V. Apply [F4] to this arbitrary surrounding braid, removing exactly d last new strands by negative ordinary destabilizations and conjugations, and conjugate by K to obtain α. The strand number decreases from n+d to n. For d=1 only T1=1 is used, with K retained. For c=0 the free-product name C is the empty product, although the arbitrary box C∈Ba+d is retained throughout the displayed braid equality. At d=0 use step 1.1. AC enters solely through [F3], [F4]; no Markov equivalence theorem is a supplier.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

The four-band case is a Markov sequence

Statement

Assume the Axiom of Choice. In the irreducible four-band d-pair configuration of the preceding item, the two ways of reducing the peak to height zero produce closed braids X′,Z′ whose braid words differ by a finite sequence of braid isotopies and ordinary Markov stabilizations and destabilizations. The source comparison has two comparison blocks, its right and left columns, each with one multiple-strand stabilization and one multiple-strand destabilization; each is expanded into finitely many ordinary moves. The uniform over/under ambiguity in a multiple reduction is resolved by a band exchange, which also has a finite ordinary Markov sequence. Consequently the peak can be cancelled by Markov moves.

Facts & Assumptions

Given: AC, the irreducible four-band d-pair configuration of Reducing-move peaks can be lowered to the four-band case with its two reducing arcs, occurring as an interior peak of a reducing portion with height-zero endpoints, and the two multiple reductions of its peak to height zero.

[F1]

The irreducible configuration consists of at most four bands of mutually coherent parallel Seifert circles joined by braids; a multiple reduction slides all strands of one band over, or all under, the strands of the other in one band slide, which is a sequence of ordinary reducing moves, and the two ways of reducing the peak use the arc pairs αr,αp in one order and αs,αu in the other (Reducing-move peaks can be lowered to the four-band case).

[F2]

The standard closure uses its fixed disk framing, and has exactly n points in each page; components correspond to endpoint-permutation cycles (The closure of a geometric braid).

[F3]

For arbitrary P,Q∈Bn−1, n≥2, the ordinary exchange Pσn−1Qσn−1−1↦Pσn−1−1Qσn−1 has an explicit sequence of conjugations, one ordinary stabilization and one ordinary destabilization, with either sign (Ordinary exchange moves are Markov sequences).

[F4]

Braid-like II and III moves of closed braid diagrams together with planar isotopies are braid isotopies and give conjugate read words; signed ordinary stabilizations and destabilizations are the strand-changing Markov moves (Braid-like Reidemeister moves on closed braids are braid isotopies, Markov conjugation and stabilization moves).

[F5]

For arbitrary braid boxes on entire blocks of widths p,q, the uniform block interchange Qp,q satisfies (α⊗β)Qp,q=Qp,q(β⊗α), and its inverse satisfies the reversed identity (Block interchanges transport arbitrary braid boxes).

[F6]

Assume AC. For an arbitrary surrounding braid α∈Bn, a compensated packet Wm=Qm,mTm−1 on the last m original strands and m new strands is realized by m ordinary positive destabilizations and conjugations, from n+m strands to n; its mirror uses m negative destabilizations, and reversal gives the corresponding stabilizations. The reverse-order negative packet Vm=TmQm,m−1=Qm,m−1(1m⊗Tm) has the same m negative ordinary destabilizations with arbitrary surrounding braid (Compensated band kinks decompose into ordinary Markov moves).

[F7]

Braids acting on disjoint consecutive strand blocks commute: their generator indices differ by at least two, so the far-commutation relation applies to every pair of letters, including inverses (The braid group by Artin presentation).

[F8]

For arbitrary core-plus-band boxes P∈Bc+p and Q∈Bc+q, the typed band exchange PQp,qQQp,q−1↦PQq,p−1QQq,p has a full ordinary Markov sequence: pad the smaller band by actual positive stabilizations at a cyclic cut, use compensated equal-width exchange and remove the padding. Zero-width endpoints coincide (Band exchanges decompose into ordinary Markov moves).

[F10]

The frozen positive geometric generator is an anticlockwise half rotation: its first indexed point goes through negative second coordinate; this is part of the fixed oriented transverse-disc convention (The elementary geometric half twist, its support disc, and its opposite).

[F11]

The geometric product [γ][β]=[γ⋆β] runs β during the first half of the height interval and γ during the second (Stacking of geometric braids is a well-defined associative operation on isotopy classes).

[F12]

With arbitrary four boxes on the stated original and shifted blocks, the first comparison words satisfy RβR−1=αKVK−1, with K an old-strand full-twist/placement word and V the reverse-order negative packet. Therefore they differ by exactly d negative ordinary stabilizations and conjugations, including the c=0,d=0 cases (The first four-band comparison is a compensated band stabilization).

[F13]

With the displayed arbitrary old boxes and whole-block routings, the second comparison satisfies Y1βY1−1=αKVK−1, with K in the old Bn and V the reverse-order negative packet. It therefore consists of exactly d negative ordinary destabilizations and conjugations (The second four-band comparison is a compensated band destabilization).

[F14]

The Artin generators map to the fixed geometric half twists by a choice-free surjective homomorphism; under AC this map is an isomorphism. Geometric braids are paths of distinct disk points based at the fixed symmetric real configuration, with isotopy given by endpoint-fixed homotopies (The Artin presentation surjects onto the geometric braid group, The Artin presentation is complete for geometric braids, Geometric braids in the disc with setwise endpoints).

Proof

technique · direct
1.1F1given

The two multiple reductions. The specified peak is X←Y→Z, with the two first ordinary reducing moves fixed as part of the given data. In Traczyk's Figure 7, one uniform multiple reduction begins with the specified move along αr to X and is completed by the uniform reduction along αp; the other begins with the specified move along αs to Z and is completed along αu. These are distinct arc pairs, rather than merely the same two moves in reversed order. They give the two height-zero diagrams X′,Z′ shown in the bottom of Traczyk's Figure 7 and in Birman--Brendle's Figure 12, printed p. 26. We must compare their braid words, preserving the source's uniform over/under convention and arbitrary internal boxes.

1.2F4F5F7F10F11algebra

Chronological records and oriented frames. The geometric product runs its rightmost factor first. In the source picture we may list crossings and boxes chronologically along the oriented braid; call this list its chronological record. The actual braid-group element is its word reversal rev⁡, with each individual generator's sign unchanged. Reversal preserves the Artin relations, which are palindromic or far commutations, so it defines an involutive anti-automorphism. It takes a conjugation by g to one by rev⁡(g)−1, and takes a right stabilization to a left stabilization; cyclic conjugation converts the latter to a right stabilization of the same sign. Hence any ordinary Markov sequence for chronological records gives one for the actual elements with the same strand-changing counts. Moreover rev⁡(Qp,q)=Qq,p: label a cell of Qp,q by row j=1,…,q and entry i=1,…,p, with index p+j−i. After reversal put I=p−i+1, J=q−j+1; the entry is σq+I−J, ordered first by increasing J then I. The rows of Qq,p order this same grid first by I then J. The pairs that change order have I>I′ and J<J′, with index difference (I−I′)+(J′−J)≥2, so only far commutations are needed. Empty widths are included. At a Figure 8 cut we use increasing radial coordinate and positive depth as the transverse frame (er,eZ), so ports are numbered inner to outer. The clockwise braid tangent is −eθ, and (er,eZ,−eθ) preserves ambient orientation. At a Figure 11 cut we instead use (−er,−eZ), so ports are numbered outer to inner; the two transverse frames differ by an orientation-preserving half rotation, not a reflection. Positive anticlockwise half twists therefore retain the frozen generator convention: the first inner port passes through negative physical depth in the former frame, while the first outer port passes through positive physical depth in the latter. All chronological formulas below are read in their stated frame and are explicitly reversed to interpret them as actual braid-group words.

1.3F3F4algebra

The ordinary exchange comparison. If the ambiguity has the ordinary exchange form of [F3], put t=σn−1 and s=σn. Its first and last weaving words are E1=Pts−1t−1sQt−1 and E5=Pt−1Qst−1s−1t. The supplier proves sE1s−1=E5 using only the adjacent Artin relation and the commutation of P,Q with s. It also writes E1=As−1B with A=Pt2, B=t−1Qt−1 and BA=A−1(PtQt−1)A, and t2E5t−2=(t2(Pt−1Qt)t−2)s−1. These expose an ordinary stabilization and destabilization with both arbitrary boxes retained. This verifies the ordinary Figure 11 exchange. The multiple-band case is supplied by [F8], with the source endpoint trace below.

1.4F5F7F12construct

The right-column arbitrary-box slide. Retain the first comparison's placements and put F=Hιa(Td), Cout=C1, Dout=D1, Cin=C on the first a+d strands, Din=D on the first b+d strands, and G=Y1−1A1FB1Y1. The second panel is βR2=CoutGDoutU and the third panel is βR3=DinGCinU. The inverse whole-block naturality [F5] gives UCout=CinU and DoutU=UDin. Also Dout and Cin are on disjoint blocks, so they commute by [F7]. Thus conjugating by Cout−1 gives GDoutUCout=GCinUDin, and conjugating this by Din gives βR3. These are two explicit conjugations with arbitrary internal boxes; there is no strand change and no twist is commuted through part of a box.

1.5F4F7algebra

Inversion transfers signed comparisons. Word inversion reverses product order and inverts every letter. It takes conjugate braids to conjugate braids. If a chronological right stabilization is γ↦γσnϵ, its inverse is γ−1↦σn−ϵγ−1, and cyclic conjugation turns the latter into γ−1σn−ϵ. Thus inversion transfers any ordinary Markov sequence, reversing its crossing signs and keeping its strand-changing counts; reversal of the sequence additionally interchanges stabilization and destabilization. This is an assertion about words and move sequences, not an identification of a link with the closure of its inverse braid. Universal identities with arbitrary boxes may be applied to inverse boxes first and then inverted.

2.1F8F5F3algebra

The Figure 11 band-exchange endpoints. Cut immediately before the upper box P in the initial and final panels of Traczyk's Figure 11, and number the ports from outermost to innermost. Write their widths as (c,p,q): the outer core has width c, the middle exchanged band width p, and the inner exchanged band width q. The upper box acts on the first c+p ports. Read the right-hand crossing in the braid direction: it uniformly interchanges (p,q) with (q,p), so the lower box Q acts on the first c+q ports; the left-hand crossing restores (p,q). With the first crossing positive in this outer-first frame, the initial chronological record is PQp,qQQp,q−1. In the final panel the two crossing signs are reversed, while the two boxes remain in the same positions with the same oriented endpoint frames. The negative first interchange with input (p,q) is Qq,p−1, so the final chronological record is PQq,p−1QQq,p. By step 1.2, their actual words, after a cyclic cut, are respectively rev⁡(P)Qq,p−1rev⁡(Q)Qq,p and rev⁡(P)Qp,qrev⁡(Q)Qp,q−1; [F8] applies in reverse direction, with the same core-plus-band supports. If the projection convention reads the first crossing negative, reverse the same comparison. By [F8] these endpoints have a full ordinary Markov sequence for arbitrary boxes and unequal widths: the after-crossing cyclic-cut identity Q(vϵ+)−1(Pσc+p)vϵ+=(Qvϵ−1Pvϵ)σc+p+q makes padding an actual stabilization, and the equal-width computation retains the compensation through its weaving conjugation and restores it before destabilizing. Zero widths give identical endpoints by [F8]. This proves the source's complete Figure 11 exchange by an independently reproducible computation, without requiring acceptance of its intermediate arrows. The relevant reducing-choice endpoint factorization is given below.

2.2F2F6F12step 1.2algebra

The first right-column strand-changing arrow. Cut the initial Figure 8 right-column diagram immediately before the striped box C, with inner-to-outer ports (b,a,d,c) in the frame of step 1.2. Put n=a+b+c+d, X=Qb,a and Y=ιa+b(Qc,d). The chronological record is αC=C0Y−1A0XB0YD0X−1, exactly α of [F12]; it is the cyclic record before C of the initial before-A record. At the corresponding cut in the second right-column panel the ports are (b,dnew,a,dold,c), with total n+d. The striped and black boxes now begin after the first b+d ports, and the grid and textured boxes after the first a+d ports. The left cap remains the negative interchange Y1−1 of the last (d,c) blocks, the upper weave is the positive reversal of (b,dnew,a) given by H=Qb,d+aQd,a, its + square is the full Td on the middle copied band after H, the right cap is Y1, and the lower weave is the negative uniform interchange of the two whole blocks (a+d,b+d) given by U=Qb+d,a+d−1. Thus the complete chronological record is βR2=C1Y1−1A1H ιa(Td) B1Y1D1U, with every shifted embedding specified in [F12]. The old and copied d port labels interchange at the end of this record; they are not asserted to remain the same physical strands inside arbitrary boxes. By the full general identity of [F12], RβR2R−1=αCKVK−1, where ℓ=a+d+c, R=ιb(Qℓ,d), K=ιb(Tℓ−1)ιa+b(Qd,c) and V=ιn−d(TdQd,d−1). In particular the old full twist in K and the reverse compensation order in V are retained. Conjugating αC by K−1, adding this packet by [F6], and conjugating by K and R−1 gives the second panel with exactly d negative ordinary stabilizations. Step 1.2 reverses this complete sequence to interpret the actual geometric words, preserving those counts and signs. For d=0 the words coincide; for c=0 the caps disappear but the same general identity applies. This verifies the first strand-changing source arrow with all boxes and frame compensation present.

2.3F13step 1.2algebra

The second right-column strand-changing arrow. At the third panel cut before Din, the inner-to-outer ports are (b,dold,a,dnew,c). The first inner box is Din; the left cap Y1−1 moves (dnew,c) to (c,dnew), and the outer black box A1 acts on the now consecutive (a,c) block. After the upper routing and the grid box, Y1 returns (c,dnew) to (dnew,c), so Cin acts on (a,dold) and U closes the record with the two d blocks interchanged. The cap is the inverse of a (c,d) interchange and therefore requires input (d,c), also when the widths differ. The upper positive routing H takes (b,dold,a) to (a,dold,b) and its positive square is ιa(Td). The grid box is B1, the right cap is Y1, the inner striped box is Cin, and the lower uniform negative routing is U. Hence its complete chronological record is βR3=DinY1−1A1FB1Y1CinU. In the bottom-right panel the same cut has only (b,d,a,c): the cap crossings have disappeared and the lower three-band routing with its negative square is precisely F−1. Its record is αR4=DinA1FB1CinF−1. These are exactly [F13], whose old-strand conjugator is K2=ιb(Qd,a)ιa+b(Qc,d−1). Therefore the third to fourth right-panel arrow is d negative ordinary destabilizations and conjugations. The source's copied and old d ports interchange when this closed record returns to its cut; the boxes have the specified supports irrespective of internal permutations. Step 1.2 interprets the actual group words without changing signs or counts. At c=0 there are no caps; at d=0 the panels coincide, as [F13] proves.

3.1F12F13step 1.2step 1.4step 1.5step 2.2step 2.3algebra

Both left-column strand-changing arrows. Apply the universal right-column computations to A−1,B−1,C−1,D−1 and invert their complete words, then use a cyclic cut. The old bottom-left record before A1 is αL4=A1DinFCinB1F−1; the third left-panel record is βL3=A1Y1DinU−1CinY1−1B1F−1. These are the cyclic inverses of αR4,βR3 with inverse box inputs. Their source ports at the cut before A1 are (b,dold,a,c,dnew). The left positive cap gives (b,dold,a,dnew,c); the upper positive routing U−1 then gives (a,dnew,b,dold,c); the right negative cap gives (a,dnew,b,c,dold). The lower negative routing and negative square are F−1 and return (b,dnew,a,c,dold). The boxes encountered are respectively outer A1, inner Din, inner Cin, outer B1, as the displayed record specifies. By steps 1.5 and 2.3, the source arrow L4→L3 is d positive ordinary stabilizations and conjugations. Next the second left-panel record before C1 is βL2=C1U−1D1Y1−1B1F−1A1Y1, while the initial left-panel record before C0 is αL1=C0XD0Y−1B0X−1A0Y. These are the cyclic inverses of βR2,αC with inverse boxes. For βL2 start with ports (b,dold,a,dnew,c); the upper U−1 gives (a,dnew,b,dold,c), the right Y1−1 gives (a,dnew,b,c,dold), the lower F−1 gives (b,dnew,a,c,dold), and the left Y1 gives (b,dnew,a,dold,c). Thus the outer striped and textured boxes have exactly C1,D1 placements and the lower black/grid boxes exactly A1,B1 placements; all crossing signs and the negative square are those of the inverse records. By steps 1.5 and 2.2, L2→L1 is d positive ordinary destabilizations and conjugations. The L3→L2 slide is the inverse-box/inversion transfer of step 1.4, so consists only of conjugations. The zero-width cases transfer as well. Finally step 1.2 turns every chronological comparison into the actual word comparison.

3.2F1F5F8step 1.1step 2.1algebra

Classify the commissioned reducing choices. Use the source's band convention in [F1]: every multiple reduction has all its individual slides over, or all under, the other band. The given peak specifies the first ordinary reducing move to X or Z, including its crossing choice. Therefore the sign of the initial uniform multiple reduction r or s is fixed by that first move; the remaining copies in that uniform slide cannot independently change signs. The free choices are the final uniform reductions p and u. Write the four original band widths as a,b,c,d, with the left and right bands a,b>0, and top/bottom c,d≥0, and retain A,B,C,D for the black, grid, striped and textured boxes. For the right-column Figure 8 endpoint arising from p, cut before A with inner-to-outer ports (b,a,c,d). The upper central crossing interchanges b,a, leaving A on (a,c) and B on (b,c); thus Pp=AQb,aB acts on the first k+c strands, k=a+b. After the outer right cap interchanges c,d, the lower portion has D on (b,d), its central inverse interchange returns the core order, and C acts on (a,d); hence Qp=DQb,a−1C acts on the first k+d strands. In the inner-first frame of step 1.2, the outer band passes over the inner at the right cap, so that cap is positive Qc,d; the left cap is its negative inverse. The endpoint's chronological record is therefore PpQc,dQpQc,d−1, with the cap words shifted by k. Reversing the uniform over/under choice of p reverses precisely these two cap crossing signs; the four boxes and the central crossings from the fixed first reduction are unchanged in their oriented frames. The other chronological record is PpQd,c−1QpQd,c, exactly the typed exchange of [F8] with core k and widths (c,d). For the left-column endpoint arising from u, cut before C with ports (b,a,d,c). The upper portion is Pu=CQb,aD∈Bk+d; the lower portion after the right cap is Qu=BQb,a−1A∈Bk+c. Its displayed cap signs in that same inner-first frame give the chronological record PuQc,d−1QuQc,d, and changing the uniform u choice gives PuQd,cQuQd,c−1, the typed exchange with widths (d,c). All these chronological products use the consecutive strand placements fixed at their respective cuts; word reversal from step 1.2 converts them to actual elements and reverses the exchange direction, preserving each box's support. Thus [F8] resolves every free choice under the commissioned uniform convention for both final reductions, with arbitrary internal boxes and no arbitrary-suffix or mixed-choice assertion. If c=0 or d=0, the corresponding last multiple reduction has no individual operations and the two cap words are 1, so there is no such ambiguity.

4.1F1F5F7F8F10F11F14step 1.2step 3.2algebra

The initial uniform choices and the old frame rotation. The actual first reducing moves r,s in the given peak may have the opposite crossing choices to the conveniently drawn source reference. We compare the completed braids, without changing those first moves. First justify changing the old n-strand frame. Let Rv(x)=eπivx for 0≤v≤1. Because the fixed real base configuration is symmetric, the distinct paths Rv(qj) give a geometric n-braid δ returning to the same unordered configuration. For any braid path γ(t), the square (v,t)↦Rv(γ(t)) is a configuration homotopy whose two endpoint edges are this same path δ. Its boundary identity, with the rightmost-first stacking convention [F11], is R1(γ)=δγδ−1. Rotation preserves transverse orientation and carries the fixed positive half twist on adjacent points i,i+1 to the positive half twist on points n−i,n−i+1. By [F14] choose an Artin word for δ; completeness transfers the square identity to the Artin group, giving δσiδ−1=σn−i. The assertion extends to inverse letters and arbitrary words. Thus index reversal Jn(σi)=σn−i is an old-strand conjugation, not a mirror or a Markov strand change. It carries a box on the last r strands to one on the first r. On a rectangular interchange it gives Jp+q(Qp,q)=Qq,p: reflecting generator indices turns cell (j,i) into index q−j+i; exchange the rectangular row and column order, moving only incomparable cells with index difference at least two, just as in step 1.2. Now cut the right completed diagram before the grid box B. Its two central uniform interchanges are the ones created by the initial reduction r, and its remaining record factors as PX−1QX, with P=BYD acting on the last b+c+d ports, Q=CY−1A on the last a+c+d ports, and X=Qb,a on the inner a+b ports. At this cut the fixed final reduction is contained wholly in the two outside cap routings Y,Y−1. Change to the outer-first frame using Jn. With core width c+d, the boxes Jn(P),Jn(Q) act on the first c+d+b,c+d+a ports, and the central inverse interchange is ιc+d(Qa,b−1). The endpoint is therefore the negative endpoint of the typed exchange [F8] with active widths (b,a). Reversing the initial uniform choice gives its positive endpoint with ιc+d(Qb,a) and inverse return; the box braids, outside cap choices and their oriented frames are retained. For the left completed diagram the cut before D gives the same factorization with P=DY−1B and Q=AYC, with the same last-block supports and the same outside core. Hence [F8] also compares its actual initial s choice with the reference choice. Conjugation by the old δ and step 1.2 transfer these comparisons back to the actual braid words. Together with step 3.2 this compares every commissioned uniform initial and final choice to the reference, without modifying the specified peak, selecting independent signs inside a uniform band slide, or assuming partial-band box commutation. Empty outside bands merely reduce the core width; a,b≥1 ensure both active initial reductions are present.

4.2F5F7step 2.3step 3.1algebra

The middle comparison. In the bottom-right and bottom-left panels, all four boxes act on old n strands. The black and textured boxes A1,Din are disjoint; likewise the grid and striped boxes B1,Cin are disjoint, as their stated placements show. At the cut before A1, the right record is A1FB1CinF−1Din and the left record is A1DinFCinB1F−1. The routing F is the full positive three-band weave including its middle-band positive full twist; the lower routing reverses that entire weave and twist and is F−1. Moving the routed twist on its own whole band uses [F5]. By [F7], conjugation by Din changes the right record into DinA1FB1CinF−1=A1DinFCinB1F−1, the left record. This retains every arbitrary box, makes no partial-band commutation, and has no strand change. Step 1.2 transfers the conjugation to the actual words.

5.1F1F8F12F13step 1.1step 1.4step 2.2step 2.3step 3.1step 3.2step 4.1step 4.2∎

Assembly of the source comparison. The right comparison block comprises the first arrow of step 2.2, the box slide of step 1.4, and the second arrow of step 2.3: one multiple negative stabilization and one multiple negative destabilization, each expanded into d ordinary strand changes. Step 4.2 joins it to the left block, whose arrows and box slide are verified in step 3.1: one multiple positive stabilization and one multiple positive destabilization, each likewise expanded into d ordinary strand changes. These are finite sequences even when a band's width is greater than one; at d=0 the strand-changing parts are empty. The reducing-choice exchanges of steps 3.2 and 4.1 contribute additional finite ordinary Markov sequences, so no fixed bound of two individual stabilizations and two individual destabilizations is asserted. All diagrams therefore have the claimed finite comparison, and the specified peak can be cancelled by Markov moves. AC is inherited from the peak-lowering, faithful-action and compensated-packet suppliers.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Reidemeister moves between closed braid diagrams factor through Markov moves

Statement

Assume the Axiom of Choice. Let B and C be closed braid diagrams representing the same oriented link. Then B can be transformed into C by a sequence of braid isotopies of closed braid diagrams, Markov stabilizations and destabilizations, and their inverses; equivalently, the braids read from B and C are Markov equivalent.

Facts & Assumptions

Given: AC, closed braid diagrams B,C for the same oriented link, and the Reidemeister equivalence theorem (Reidemeister's theorem for oriented diagrams).

[F1]

Every non-braid-like R2 and R3 is generated by type I moves, braid-like R2 and R3 moves, reducing moves and inverses; this is the generating lemma (Non-braid-like Reidemeister moves are generated by braid-like moves and reductions).

[F2]

Type I and braid-like moves may be moved to height zero; sphere isotopy and a chart choice then put the pictures in closed-braid form, where the replacement moves are braid isotopies or stabilizations (Braid-like moves can be moved to height zero).

[F3]

Braid-like R2 and R3 moves on closed braids and planar isotopies are braid isotopies, hence conjugations in the braid group (Braid-like Reidemeister moves on closed braids are braid isotopies).

[F4]

Reducing portions with height-zero endpoints can be lowered or reduced to irreducible four-band peaks. The supplier's proof 1.1 and 2.1 insert reductions at the same peak diagram; proof 3.1 replaces compatible pairs by descending valleys, and proof 1.2 removes height-one peaks by height-zero ordinary exchanges (Reducing-move peaks can be lowered to the four-band case).

[F7]

At an irreducible four-band peak the two neighbours admit descending reductions to height-zero braids whose words are Markov equivalent. Proof 4.1 also constructs the old-strand conjugation reversing strand indices, without changing signs (The four-band case is a Markov sequence).

[F5]

The regular-projection existence used in the hypothesis is ACω-stated and is discharged from AC by the choice-implication bridge (AC implies DC implies countable choice).

[F6]

A reducing move lowers height by exactly one and preserves the oriented link; its inverse raises height by one (A reducing move lowers the height by one).

Proof

technique · direct
1.1F1F5given

Reidemeister sequence and phase one. By the oriented Reidemeister equivalence theorem there is a finite sequence of planar isotopies and oriented R1, R2, R3 moves from B to C; by [F5] this uses ACω, discharged from AC. Apply [F1] to every non-braid-like R2 and R3 in the sequence: the sequence is replaced by one using only type I moves, braid-like R2 and R3 moves, and reducing moves and their inverses.

2.1F2F3F7step 1.1

Phase two: braid-like content at height zero. Apply [F2] to replace each type I and braid-like II or III move. Use the sphere/chart normalizations in [F2] before applying [F3]; height zero in the original chart alone is insufficient. Keep the coherent oriented circle order and a transported cut in each normalized picture. The finite crossing-order argument for planar readings in [F3] compares compatible straightenings by far commutations and cyclic conjugations; if the two end disks are exchanged, strand-index reversal is the old-strand conjugation of [F7]. Thus normalization choices change only conjugacy of the readings. Take the endpoint normalizations to be the given closed-braid forms of B,C. The II and III moves now occur on closed braid diagrams and are braid isotopies by [F3], hence conjugations. The type I replacement has a stabilization or destabilization at height zero and, when the kink must return across other strands, includes braid-like III moves and inverse reductions as specified by [F2]. Group the resulting transformation into height-zero Markov portions interleaved with reducing sequences. Split each reducing sequence at its height-zero diagrams, so its intermediate diagrams have strictly positive height and its endpoints have height zero.

3.1F4F6F7step 2.1construct

Phase three: track a fixed maximum height. Consider a reducing portion from step 2.1, with maximum height H. If it is empty there is nothing to eliminate. Otherwise H>0 and each occurrence of H is an interior peak by [F6]. Track the constructions of [F4], rather than assuming a height bound in its alternative conclusion. At a peak X←Y→Z, inserting a reduction t replaces it by X←Y→Y(t)←Y→Z; Y still has height H and every new neighbour has height H−1. The arc surgeries in [F4] first decrease intersection numbers of each resulting pair until they are at most one, then replace one-intersection pairs by disjoint pairs. These are finite refinements at this same Y, never insertions above H. A compatible disjoint pair is replaced by its common double reduction of height H−2. An available third compatible arc gives two such valleys. For H=1, the height-one exchange construction of [F4] removes the peak at height zero. The remaining disjoint pairs at H≥2 are irreducible four-band peaks. For each use [F7]: descend from both neighbours, which start at H−1, to height zero, compare there by its finite Markov sequence, and reverse the second descent. By [F6] that entire replacement has height at most H−1. After the finitely many refinements and replacements at all original height-H occurrences, no height-H diagram remains. Split at the inserted height-zero Markov portions. Every remaining reducing portion has smaller maximum, so induction on the nonnegative integer H eliminates it. This yields a finite Markov sequence between the normalized endpoints.

4.1F1F2F4F7step 3.1∎

Conclusion. Steps 1.1-3.1 factor the chosen Reidemeister sequence between B and C through Markov moves, so the braids read from B and C are Markov equivalent. AC is inherited through the AC-stated generating, height and four-band items.

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Markov's theorem for braid closures

Statement

Assume the Axiom of Choice. Let β,β′ be braids. Then β^ and β′^ are equivalent oriented links if and only if β and β′ are Markov equivalent, that is, related by a finite sequence of conjugations in a fixed braid group, stabilizations β↦βσn±1 into the next braid group, and their inverses.

Facts & Assumptions

Given: AC, braids β,β′ and the Markov moves of Markov conjugation and stabilization moves.

[F1]

Assume ACω: each single conjugation, stabilization or destabilization preserves the oriented closure up to ambient isotopy (Markov moves preserve the oriented closure up to isotopy); AC yields ACω (AC implies DC implies countable choice).

[F2]

If two closed braids are isotopic through closed braids about the axis, the braids read from them are conjugate (Braid-isotopic closed braids are conjugate).

[F3]

Assume AC: two closed braid diagrams representing the same oriented link are related by braid isotopies of closed braid diagrams, stabilizations and destabilizations, and inverses, so the braids read from them are Markov equivalent (Reidemeister moves between closed braid diagrams factor through Markov moves).

[F4]

Assume ACω: two regular diagrams represent equivalent oriented links if and only if they are related by finitely many planar isotopies and oriented Reidemeister moves (Reidemeister's theorem for oriented diagrams).

[F5]

The closure of a braid is an oriented link whose components and orientation are those of the closure construction (The closure of a geometric braid).

Proof

technique · direct
1.1F1F5given

Markov equivalence implies isotopy of closures. Let β and β′ be Markov equivalent, so there is a finite sequence of braids β=β0,β1,…,βk=β′ in which each consecutive pair is related by a conjugation, a stabilization or a destabilization. By [F1] each single step preserves the oriented closure up to ambient isotopy; composing the finitely many ambient isotopies gives an equivalence of β^ and β′^.

1.2F2F3F5

Isotopy of closures implies Markov equivalence. Conversely, assume β^ and β′^ are equivalent oriented links. If either is empty, both are empty and [F5] forces both braid strand counts to be zero; they are the unique element of B0 and hence Markov equivalent. Otherwise both strand counts are positive. Use the closed braid diagrams obtained from the closure construction applied to β and β′, and read each diagram at its original cutting ray. The braids read this way are β and β′ themselves, up to braid isotopy. Apply [F3] to these two diagrams to obtain their Markov equivalence. If a different cutting ray is used, [F2] gives a conjugate braid, which lies in the same Markov class. Thus β and β′ are Markov equivalent.

2.1F1F2F3F4step 1.1step 1.2∎

Conclusion. Steps 1.1 and 1.2 give the equivalence. AC discharges the countable-choice hypothesis of [F1] and supplies the AC hypotheses of the factorization result [F3] and the conjugacy bridge [F2]. The Reidemeister theorem [F4] is used inside [F3], rather than to identify a braid read at its original cutting ray.

5 · Examples, counterexamples and false statements

None yet.

Sources