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Noncrossing Partition Lattices and Kreweras Complements

1 · Prerequisites

2 · Summary

For a finite-type Coxeter system and a chosen Coxeter element c, the noncrossing poset is the absolute-order interval [1,c]. The page proves its finite lattice structure from the ordered positive-root complex, then transports that structure between Coxeter elements and identifies the type-A set-partition model.

Definitions and conventions

Coxeter elements, the noncrossing interval [1,c], and the Kreweras map w ↦ w⁻¹c defines the Coxeter-element convention, the interval NC⁡(W,c), the Kreweras map K(w)=w−1c, and the componentwise convention for reducible systems.

Root geometry and conjugacy

Moved space of a reversed reflection product with independent normals proves the moved-space identity for a reversed product of independent reflection normals. Intersection of root subcomplexes and purity under convexity proves the common-face intersection and purity facts used by the lattice argument. Coxeter elements of tree type are conjugate by source and sink firings proves conjugacy of Coxeter elements when the finite diagram components are trees.

Lattice and complement

Finite noncrossing intervals are lattices, independently of the Coxeter element proves meets, joins, reducible product structure, and independence of the lattice isomorphism type from the chosen finite-type Coxeter element. The Kreweras complement of [1,c], and the type-A model by noncrossing set partitions proves the group-theoretic Kreweras identities for every finite type and the cycle and noncrossing-partition model in type A. The type-A criterion is proved in both directions; no general Catalan-count product is asserted.

The earlier braided-and-symmetric-monoidal-categories supplies the symmetric-group Coxeter presentation used to identify the type-A model.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-10-08Open item page →

Coxeter elements, the noncrossing interval [1,c], and the Kreweras map w ↦ w⁻¹c

Definition

Let (W,S) be a Coxeter system with S finite, Coxeter diagram Γ, word length ℓ, and presented group W (Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups, Coxeter diagrams: edges, labels, components and finite type). Let V=RS have its Coxeter form B and canonical reflection homomorphism ρ:W→GL(V); its reflection set is T={wsw−1:w∈W, s∈S} (The real Coxeter form, its radical, reflections, and form-preserving maps, The canonical reflection homomorphism, roots, reflections, and the positive cone). Extend the finite-type formulas of Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator (1),(2) to this possibly infinite group by defining ℓT(w):=min⁡{k∈N:w=t1⋯tk, ti∈T},u≤Tv  ⟺  ℓT(v)=ℓT(u)+ℓT(u−1v). The minimum exists because S⊆T generates W; the empty product is 1.

(1) Coxeter elements. Put n=∣S∣ and choose a bijection σ:{1,…,n}→S. The product cσ=σ(1)⋯σ(n) is the Coxeter element for that ordering; a Coxeter element of (W,S) is any such product, with each simple reflection used exactly once. For n=0 the unique empty ordering has empty product 1; for n=1 the product is the sole simple reflection. If Γ is connected and W is of finite type, the bipartite construction of The bipartite Coxeter element, its ordered prefix roots, and the conditional vector map mu(a) = -2(c-1)^{-1}a (1) gives one particular Coxeter element. This definition makes no claim that distinct orderings give conjugate elements.

(2) The noncrossing interval. For an irreducible system and a Coxeter element c, define NC⁡(W,c):=[1,c]≤T={w∈W:1≤Tw≤Tc}, with the order induced by ≤T. Since T generates W, ℓT is a word length: it is subadditive, vanishes only at 1, and ℓT(1)=0. Thus u≤Tu; if u≤Tv and v≤Tu, adding the two defining equalities gives ℓT(u−1v)=ℓT(v−1u)=0, so u=v. If u≤Tv≤Tw, subadditivity gives ℓT(w)≤ℓT(u)+ℓT(u−1w)≤ℓT(u)+ℓT(u−1v)+ℓT(v−1w)=ℓT(w), so equality holds throughout and u≤Tw. Hence ≤T is a partial order, and 1 is the least element of the interval. The chosen c is part of the definition; independence up to isomorphism for finite type is proved in Finite noncrossing intervals are lattices, independently of the Coxeter element ↗ (4), not assumed here.

(3) Reducible systems. If the connected components of Γ have vertex sets S1,…,Sk, then W≅WS1×⋯×WSk by Disconnected diagrams, direct products, and comparison of invariant forms (1), where WSi=⟨Si⟩ (The subgroup ⟨S⟩ generated by a subset, the cyclic subgroup ⟨g⟩, and cyclic groups). A Coxeter element c has coordinates ci, each a Coxeter element for (WSi,Si). Define NC⁡(W,c):=∏i=1kNC⁡(WSi,ci) with componentwise order; for k=0 this is the one-element empty product. This agrees with the ambient interval [1,c]≤T: conjugates of a simple generator stay in its component, so T is the disjoint union of the component reflection sets Ti in their respective factors. Any reflection factorization of (wi) projects to one in each factor, giving ℓT((wi))≥∑iℓTi(wi); concatenating shortest factorizations in the factors gives the reverse inequality. Hence reflection length is the sum of the component lengths, and the absolute-order relation is componentwise. For k=0, W={1} and both the interval and product are singletons.

(4) The Kreweras map. Define K:NC⁡(W,c)→W,K(w):=w−1c. This is well-defined as a map to W by the group operations. It is not defined here as a map into NC⁡(W,c), and no bijectivity or order-reversal is asserted; those properties are proved in The Kreweras complement of [1,c], and the type-A model by noncrossing set partitions ↗ (1).

(5) Abstentions. This definition asserts no finiteness, lattice property, conjugacy of Coxeter elements, independence from c, or Kreweras-complement property beyond the definitions above. In the reducible case the product definition in (3) is justified locally as the ambient absolute interval; the lattice theorem remains a separate finite-type result. No form of the Axiom of Choice is used.

Remarks

  • The set of Coxeter elements need not be a union of W-conjugacy classes. Take the presentation with generators s1,s2,s3 and only the relations si2=1 (the free product C2∗C2∗C2). On the set X of words with no equal adjacent letters, let si delete the first letter when it is si, and otherwise prepend si. Each operation is an involution in the permutation group Sym⁡(X) (The symmetric group Sym⁡(X): the bijections of a set X under composition, Sym⁡(X) is a group under composition, and it is non-abelian whenever X has at least three distinct elements), so the presentation's universal property gives a homomorphism to that group. A word with no equal adjacent letters sends the empty word to its own letter string, whereas a product of k generators sends it to a string of length at most k. Therefore s2s1s2s3s2 has word length five. It equals s2(s1s2s3)s2, but cannot be a once-each product of three generators.
LemmaStatement: AI-adaptedProof: AI-adaptedaudited 2026-10-08Open item page →

Moved space of a reversed reflection product with independent normals

Statement

Let (V,⟨⋅,⋅⟩) be a finite-dimensional real inner-product space, let k≥0, and let σ1,…,σk∈V be linearly independent unit vectors (Real and complex inner-product spaces and their induced length, Linear independence: a finite list v:n→V is independent when ∑i<nλivi=0V forces every λi=0F, and a subset S⊆V is independent when every injective finite list into S is independent, For a subspace W of a finite-dimensional inner product space, V=W⊕W⊥). For a unit vector v define the orthogonal reflection R(v)x:=x−2⟨x,v⟩v. For a linear map A:V→V write M(A):=im⁡(A−idV) (Linear map between vector spaces over the same field, Kernel and image of a linear map). Then:

(1) The moved space. M(R(σk)R(σk−1)⋯R(σ1))=span⁡(σ1,…,σk), and its dimension is k. When k=0, the product is the identity, the span of the empty set is {0}, and the moved space is {0}.

(2) Reflection length. For a finite-type Coxeter system, specialize the inner-product space of (1) to V=RS with Coxeter form B (The real Coxeter form, its radical, reflections, and form-preserving maps); this is positive definite by Finiteness criterion: W is finite exactly when the Coxeter form is positive definite. Let ρ be the canonical reflection homomorphism and T the reflection set (The canonical reflection homomorphism, roots, reflections, and the positive cone, Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator). If σ1,…,σk are roots, choose any ri∈T with ρ(ri)=R(σi); such reflections exist by Descent of the reflection representation, unit root norms, and conjugation of reflections (3),(4). Then ℓT(rkrk−1⋯r1)=dim⁡M(R(σk)⋯R(σ1))=k. The product order is the reverse of the root list, as in The Brady-Watt ordered root complex X(c), its subcomplexes X(sigma) and X(sigma,rho), and their positive-cone realizations (1),(2): the reflection with normal σ1 acts first on vectors when applying the product.

(3) Limits. Clause (1) needs only linear independence and unit norms; the normals need not lie in a common open half-space and no Coxeter complex is needed. Clause (2) uses finite type so that the Coxeter form is a positive definite inner product. No crystallographic assumption or Choice is used.

Facts & Assumptions

Given: A finite-dimensional real inner-product space and a finite list of linearly independent unit vectors; for clause (2), a finite-type Coxeter system, its canonical reflection representation and roots.

[F1]

In a finite-dimensional inner-product space, V=U⊕U⊥ for every subspace U (For a subspace W of a finite-dimensional inner product space, V=W⊕W⊥). If a linear map sends U into itself and is injective on finite-dimensional U, it is onto U by rank-nullity (Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T).

[F2]

For a unit vector v, the displayed formula gives R(v)x−x=−2⟨x,v⟩v, so R(v)−id has image span⁡(v) (take x=v) and R(v) fixes v⊥. Also ⟨R(v)x,v⟩=−⟨x,v⟩, whence R(v)2=id, and expanding ⟨R(v)x,R(v)y⟩ gives ⟨x,y⟩ because ⟨v,v⟩=1. Thus it is an orthogonal reflection.

[F3]

For finite-type W, the Coxeter form is positive definite and ρ(W) preserves it. Every root has unit norm, and for every t∈T its operator ρ(t) is the reflection R(α) for a root α (Finiteness criterion: W is finite exactly when the Coxeter form is positive definite, Descent of the reflection representation, unit root norms, and conjugation of reflections (2)–(4)).

[F4]

The reflection length ℓT(g) is the least number of factors from T in a factorization of g (Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator (1)).

Proof

technique · show that the product fixes exactly the orthogonal complement of the span, then use a rank bound for products of reflections

Given: The data in the Statement. For clause (1), put U=span⁡(σ1,…,σk) and A=R(σk)⋯R(σ1).

1.1F1F2F5algebra

(Moved space of the product.) If k=0, then A=idV and M(A)={0}=U. Otherwise each R(σi) sends U into U and fixes U⊥ pointwise, so A(U)⊆U, A fixes U⊥, and M(A)⊆U. To prove the reverse inclusion, let x∈U satisfy Ax=x, set x0=x, and for i=1,…,k set xi=R(σi)xi−1. Then xk=Ax=x0, so 0=xk−x0=∑i=1k(xi−xi−1)=−2∑i=1k⟨xi−1,σi⟩σi. Linear independence forces every coefficient to vanish. Thus xi=xi−1 for every i, and each reflection fixes x; hence x⊥σi for every i. Since x∈U, this gives x∈U∩U⊥={0}. Therefore (A−id)∣U:U→U is injective, and rank-nullity makes it surjective. Thus U⊆M(A), so M(A)=U and dim⁡M(A)=k by [F5].

2.1F3F4step 1.1algebra∎

(Reflection-length rank bound.) Assume the finite-type hypotheses of clause (2) and let g=rk⋯r1, so ρ(g)=A by [F3]. For any two invertible linear maps X,Y, XY−id=(X−id)+X(Y−id), hence M(XY)⊆M(X)+XM(Y) and dim⁡M(XY)≤dim⁡M(X)+dim⁡M(Y) because X is invertible. Iterating this inequality, any factorization of g into m elements of T gives dim⁡M(ρ(g))≤m, since each image under ρ is an orthogonal reflection with one-dimensional moved space by [F3]. Step 1.1 gives dim⁡M(ρ(g))=k, so every reflection factorization has at least k factors. The displayed factorization g=rk⋯r1 has exactly k, and therefore ℓT(g)=k, including the empty-product case.

LemmaStatement: AI-adaptedProof: AI-adaptedaudited 2026-10-08Open item page →

Intersection of root subcomplexes and purity under convexity

Statement

Let (W,S) be an irreducible finite-type Coxeter system, and let c be the bipartite Coxeter element with positive-root order, ordered root complex X(c), cones c[F], c[Y], and realizations ∣Y∣=c[Y]∩Sn−1⊂V=RS of The Brady-Watt ordered root complex X(c), its subcomplexes X(sigma) and X(sigma,rho), and their positive-cone realizations (1)–(3), The Coxeter plane, ordered-root enumeration, and invertibility of rho(c) - id (3), and Real and complex inner-product spaces and their induced length. The vertices of every face are linearly independent unit positive roots, all in a common open half-space (The factorization criterion, linear independence of the faces, and the geometric simplicial structure of X(sigma) (2)); all spans are taken in V (Linear subspace of a vector space). Set c[∅]={0}, and let Y,Z be subcomplexes of X(c) (An abstract simplicial complex, The geometric realization of an abstract simplicial complex). Then:

(1) Realization of an intersection. If Y∩Z is the subcomplex consisting of simplices common to both, then c[Y∩Z]=c[Y]∩c[Z],∣Y∩Z∣=∣Y∣∩∣Z∣. If Y and Z have no common vertex, this reads c[Y]∩c[Z]={0} and ∣Y∩Z∣=∅.

(2) Purity under convexity. Suppose Y∩Z has at least one vertex. Put C=c[Y]∩c[Z] and K=∣Y∣∩∣Z∣. If C is convex, then every maximal simplex F of Y∩Z satisfies span⁡(F)=span⁡(C). Under the common-open-half-space condition above, convexity of C is equivalent to geodesic convexity of K: for any two points of K, the shorter great-circle arc between them lies in K. Consequently all maximal simplices have dimension dim⁡span⁡(C)−1, so Y∩Z is pure (all maximal simplices have the same dimension).

(3) Small cases. If Y∩Z consists of one vertex v, its unique maximal simplex is {v} and its span is span⁡(C)=span⁡(v). If Y∩Z has no vertex, then c[Y∩Z]={0}, ∣Y∩Z∣=∅, and (2) is vacuous.

(4) Limits. The result does not identify span⁡(c[Y]∩c[Z]) with span⁡(c[Y])∩span⁡(c[Z]). In particular, it does not determine M(α)∩M(β) for Y=X(α) and Z=X(β). No Choice is used.

Facts & Assumptions

Given: The bipartite ordered root complex X(c) of an irreducible finite-type Coxeter system, and two of its subcomplexes Y,Z.

[F1]

The vertex set Φ+ is finite. Every face has linearly independent unit vertices, these vertices lie in a common open half-space, and for any two faces F,F′ one has c[F]∩c[F′]=c[F∩F′] (The Brady-Watt ordered root complex X(c), its subcomplexes X(sigma) and X(sigma,rho), and their positive-cone realizations, The Coxeter plane, ordered-root enumeration, and invertibility of rho(c) - id, The factorization criterion, linear independence of the faces, and the geometric simplicial structure of X(sigma) (2),(4)).

[F2]

The empty face is a simplex, subcomplexes are closed under taking faces, and their realizations are the sphere sections of their positive-cone unions (An abstract simplicial complex, The geometric realization of an abstract simplicial complex, The Brady-Watt ordered root complex X(c), its subcomplexes X(sigma) and X(sigma,rho), and their positive-cone realizations (3)).

[F3]

A finite-dimensional subspace of a normed space is closed (A finite-dimensional normed subspace is closed). For each fixed v∈V, x↦⟨x,v⟩ is continuous by Cauchy–Schwarz (Cauchy–Schwarz: ∣⟨u,v⟩∣≤∥u∥∥v∥, with equality exactly for linearly dependent vectors).

Proof

technique · use unique simplex carriers for the cone identity, then use a limit point and finiteness of the face set to prove purity

Given: The finite root complex and subcomplexes Y,Z above. Write C=c[Y]∩c[Z] and K=∣Y∣∩∣Z∣.

1.1F1F2

(Cone and realization intersections.) If x∈c[Y]∩c[Z], then x∈c[F] for some face F∈Y and x∈c[F′] for some face F′∈Z. By [F1], x∈c[F∩F′], and F∩F′ is a common face, so x∈c[Y∩Z]. The reverse inclusion follows from Y∩Z⊆Y,Z. Intersecting this cone equality with Sn−1 gives the realization equality. If there is no common vertex, every common face is empty, so the cone intersection is c[∅]={0} and its sphere section is empty.

1.2F1F3F4algebra

(Closed face cones.) The empty-face cone is {0} and is closed. Let F={v1,…,vm} be a nonempty face, U=span⁡(F), and let G=(⟨vi,vj⟩)i,j be its Gram matrix. For any nonzero coefficient vector a, aTGa=∥∑iaivi∥2>0 by independence, so G is invertible. For x∈U, its unique coordinate vector in the basis F is G−1(⟨x,vj⟩)j, whose coordinates are continuous by [F3]. Hence c[F] is the intersection of the closed subspace U with the inverse images of the closed half-line [0,∞) under these coordinate maps; it is closed in V. Since X(c) has finitely many faces, c[Y], c[Z], and C are finite unions or intersections of closed face cones and are closed.

1.3F1algebra

(Cone and spherical convexity.) Every nonzero vector of C is a nonnegative combination of positive-root vertices, so the common open-half-space functional in [F1] is strictly positive on it; in particular K contains no antipodal pair. If C is convex, the segment between any u,v∈K lies in C and avoids 0; normalizing that segment gives the shorter great-circle arc, so K is geodesically convex. Conversely, suppose K is geodesically convex. For nonzero x,y∈C, write x=ru, y=sv with r,s>0 and u,v∈K. If u=v, then x+y∈C. Otherwise the normalized positive combination (ru+sv)/∥ru+sv∥ lies on the shorter arc from u to v, hence in K, so again x+y∈C. Thus C is closed under addition and nonnegative scaling, and is convex.

2.1F1step 1.1step 1.2algebra

(Full span of each maximal simplex.) Let L=span⁡(C). Since Y∩Z has a vertex, L≠{0}. Choose a maximal simplex F of Y∩Z; it is nonempty. Suppose U:=span⁡(F) is a proper subspace of L. The point x:=∑v∈Fv has strictly positive coordinates in the independent list F. The common vertices span L by step 1.1, so some common vertex q lies outside U. For 0<t≤1, convexity gives xt=(1−t)x+tq∈C, and xt∉U. Take tj=1/j for j≥2. There are finitely many faces of Y∩Z, so one face F′ has xtj∈c[F′] for infinitely many j. Along that subsequence xtj→x, and step 1.2 makes c[F′] closed; hence x∈c[F′]. Since x∈c[F], [F1] gives x∈c[F∩F′]. The coordinates of x in the independent family F are all strictly positive, so uniqueness of those coordinates forces F⊆F′. Maximality gives F=F′, contradicting xtj∉U. Therefore span⁡(F)=L.

3.1F2F4step 1.1step 2.1

(Empty and one-vertex cases; dimensions.) If Y∩Z has no vertex, its sole face is ∅, step 1.1 gives the empty realization, and the nonempty hypothesis of (2) fails. If it has exactly one vertex v, its only nonempty face is {v}; then C=c[{v}] is a ray, its span is span⁡(v), and the unique maximal simplex spans it. In the general nonempty case, step 2.1 gives span⁡(F)=L for every maximal simplex. By [F4], ∣F∣=dim⁡L and dim⁡F=dim⁡L−1; hence all maximal simplices have the same dimension, as claimed.

4.1F1step 3.1∎

The span of K equals L: every nonzero point of C is a positive scalar multiple of its normalization in K, and K⊆C. This also verifies the span formulation for the single-vertex case and completes (2)–(3).

LemmaStatement: Literature-sourcedProof: AI-adaptedaudited 2026-10-08Open item page →

Coxeter elements of tree type are conjugate by source and sink firings

Statement

Let (W,S) be a Coxeter system with S finite, ∣S∣=n, Coxeter diagram Γ, and Coxeter elements c,c′ defined as once-each products (Coxeter diagrams: edges, labels, components and finite type, Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups, Coxeter elements, the noncrossing interval [1,c], and the Kreweras map w ↦ w⁻¹c (1)). Suppose W is of finite type. Then every connected component of Γ is a tree: finiteness of W makes the Coxeter form B positive definite, and its restriction to each component is positive definite (Finiteness criterion: W is finite exactly when the Coxeter form is positive definite (1)); the positive-definite diagram exclusions show each component has no cycle (Exclusions for positive definite diagrams: trees, valency, labels, chains and arms (2)); hence each nonempty connected component is a tree (Trees, forests, leaves and isolated vertices).

(1) Orientation moves on a tree. Let T be a finite tree. A source in an orientation is a vertex whose incident arrows all point away from it; a sink is one whose incident arrows all point towards it. Firing a source or sink reverses all its incident arrows. Every orientation of T is acyclic, and any two orientations are connected by a finite sequence of firings.

(2) Orderings and orientations. An ordering of S orients each edge {s,t} of Γ from its earlier vertex to its later vertex. This orientation is acyclic, and every acyclic orientation is obtained from some ordering. Its product is independent of the chosen ordering that realizes the orientation, so an acyclic orientation determines a Coxeter element. If a source or sink s is fired, the new Coxeter element is scs=scs−1.

(3) Conjugacy. If every connected component of Γ is a tree, then any two Coxeter elements are conjugate by a product of simple reflections. In particular this holds in finite type, since then every connected component is a tree. If S=∅, then W={1} and the unique Coxeter element is conjugate to itself.

(4) Limits. Finite type is sufficient, not necessary, for the conjugacy assertion: its proof only needs each component to be a tree. No conjugacy claim is made for diagrams with cycles. No finite classification or geometric realization is used, and no Choice is needed.

Facts & Assumptions

Given: A Coxeter system (W,S) with finite S, its labelled diagram Γ, Coxeter form B, and Coxeter elements defined by once-each orderings. In clauses (1)–(3), finite type means W is finite.

[F1]

The diagram has a finite vertex set S, an edge exactly when m(s,t)≥3, and connected components that partition S; a connected component is nonempty. Finite type means that W is finite (Coxeter diagrams: edges, labels, components and finite type).

[F2]

If W is finite, then B is positive definite. Its restriction to the span of a component's simple roots is positive definite; a connected positive-definite Coxeter diagram has no cycle. A nonempty connected acyclic finite graph is a tree (Finiteness criterion: W is finite exactly when the Coxeter form is positive definite (1), Exclusions for positive definite diagrams: trees, valency, labels, chains and arms (2), Trees, forests, leaves and isolated vertices).

[F3]

Every Coxeter element is a product of the simple generators in some ordering, with each used once. Under the component decomposition, it has the corresponding component Coxeter elements as coordinates (Coxeter elements, the noncrossing interval [1,c], and the Kreweras map w ↦ w⁻¹c (1),(3)).

[F4]

In the Coxeter presentation, s2=1 for every s∈S; if m(s,t)=2 then (st)2=1 and therefore st=ts. The edge set of Γ is exactly the pairs with m(s,t)≥3 (Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups, Coxeter diagrams: edges, labels, components and finite type).

[F5]

A partial order is reflexive, antisymmetric, and transitive (Partial order and partially ordered set).

[F6]

The multiplication map from the product of the standard subgroups of the connected components to W is an isomorphism, and the component coordinates of a Coxeter element are the products in the restricted orderings (Coxeter elements, the noncrossing interval [1,c], and the Kreweras map w ↦ w⁻¹c (3), The subgroup ⟨S⟩ generated by a subset, the cyclic subgroup ⟨g⟩, and cyclic groups).

Proof

technique · finite induction on the tree's vertices, followed by translating firings into conjugations

Given: The data above; for a finite tree T, two orientations ω,ω′; and, for the conjugacy clauses, two once-each orderings defining c,c′.

1.1F1F2construct

(Firings connect tree orientations.) Every orientation of a tree is acyclic because a directed cycle would be an undirected cycle. Prove firing connectivity by induction on the number of vertices. For one vertex there is only one orientation; for two vertices a single firing of either endpoint reverses the only edge. For a tree with at least three vertices, an endpoint v of a longest simple path is a leaf: a neighbour outside the path would extend it, and a second neighbour on the path would create a cycle. Let u be its neighbour and put T′=T∖{v}. This is a smaller tree, since a path between remaining vertices cannot use a leaf internally and deleting a vertex creates no cycle. By induction, a finite firing sequence changes ω∣T′ to ω′∣T′. Lift each firing at x≠u directly, since its incident edges are unchanged. Immediately before a firing at u, its incident arrows in T′ all point in one direction; if the edge uv points the other way, fire the leaf v first, which is always legal and flips only uv. Now fire u. The restriction to T′ follows the inductive sequence. At the end, if uv has the wrong direction, fire v once more. This reaches ω′ and proves (1).

1.2F1F4F5choosealgebra

(Orderings encode acyclic orientations.) An ordering gives no directed cycle because the position strictly increases along each oriented edge. Conversely, in a finite acyclic orientation there is a source: otherwise repeatedly following an incoming edge would revisit a vertex and give a directed cycle. Remove a source and repeat to obtain an ordering realizing every edge direction. If two such orderings realize the same orientation, they are linear extensions of the partial order generated by its directed paths. To connect the extensions, move the first vertex of one extension left through the preceding vertices of the other; each crossed vertex is incomparable with it, and induction repeats this on the remaining vertices. Incomparable vertices cannot be joined by an edge, so their generators commute by [F4]. Thus all these orderings give the same product, proving the orientation-to-element assertion.

2.1F4step 1.2algebra

(One firing is conjugation.) If s is a source, choose a realizing ordering that starts with s and write c=sz. After firing s, the ordering with s moved to the end realizes the new orientation, so its product is zs=s(sz)s−1 by s2=1. If s is a sink, choose a realizing ordering ending in s, write c=zs, and move s to the beginning after firing; the new product is sz=s(zs)s−1. This also covers an isolated vertex: it is both source and sink, firing changes no edge, and it commutes with all other generators. Hence each firing conjugates by its simple reflection.

3.1F2F3F6step 1.1step 2.1constructalgebra∎

(Componentwise conjugacy.) If S=∅, both products are 1. Otherwise suppose every connected component of Γ is a tree; finite type guarantees this by [F2]. Restrict the orderings defining c,c′ to each component. By step 1.1 a finite firing sequence connects the resulting orientations, and by step 2.1 each firing conjugates the corresponding component product by a simple reflection. Thus each component pair is conjugate by some wi∈WSi. The component decomposition in [F6] combines these into w=(wi)∈W with c′=wcw−1. If Γ is connected, this conjugator is a product of simple reflections, as each firing uses one. This proves (3).

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-10-08Open item page →

Finite noncrossing intervals are lattices, independently of the Coxeter element

Statement

Let (W,S) be a Coxeter system of finite type with S finite, reflection set T, reflection length ℓT, absolute order ≤T, and noncrossing interval NC⁡(W,c)=[1,c]≤T (The real Coxeter form, its radical, reflections, and form-preserving maps, The canonical reflection homomorphism, roots, reflections, and the positive cone, Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator, Carter's reflection-length formula, the absolute order on a finite Coxeter group, and moved-space rigidity under a common upper bound, Coxeter elements, the noncrossing interval [1,c], and the Kreweras map w ↦ w⁻¹c (2)). For a connected system, denote by γ the designated bipartite Coxeter element and use its ordered root complex X(γ), subcomplexes X(σ), and root sets Pσ={α∈Φ+:tα≤Tσ} (The Brady-Watt ordered root complex X(c), its subcomplexes X(sigma) and X(sigma,rho), and their positive-cone realizations, The mu-dot-root identities, the cone separation, and the canonical simple systems of the subintervals [1, sigma] (4), The bipartite Coxeter element, its ordered prefix roots, and the conditional vector map mu(a) = -2(c-1)^{-1}a (1)). Then:

(1) Binary meets in the bipartite interval. For all a,b∈[1,γ], their common lower bounds have a greatest element a∧b. If a=1, b=1, or Pa∩Pb=∅, then a∧b=1. In the last case, X(a)∩X(b) has no vertices (though it contains the empty face), and its realization is empty. This case occurs in rank two: for the Coxeter system m(s,t)=3, take γ=st, a=s, b=t; then a,b≤Tγ and their distinct singleton root sets are disjoint.

Otherwise choose a maximal simplex F={v1<⋯<vr} of X(a)∩X(b) and put σ:=R(vr)R(vr−1)⋯R(v1). Then σ≤Tγ, M(σ)=span⁡(F)=span⁡(∣X(a)∣∩∣X(b)∣), and σ=a∧b. In every case, M(a∧b)=span⁡(∣X(a)∣∩∣X(b)∣),Pa∧b=Pa∩Pb, where span⁡(∅)={0}.

(2) Joins and the lattice property. The common upper bounds of any a,b∈[1,γ] form a nonempty finite set and have a least element a∨b. Thus [1,γ] is a finite lattice with least element 1 and greatest element γ (Lattices, distributive lattices, and order ideals). The meet of any nonempty finite subset is obtained by iterating the binary meet of (1).

(3) Reducible systems. If the connected components of Γ have vertex sets S1,…,Sk, write W=WS1×⋯×WSk and c=(c1,…,ck) under the component decomposition (Coxeter diagrams: edges, labels, components and finite type, Disconnected diagrams, direct products, and comparison of invariant forms, Coxeter elements, the noncrossing interval [1,c], and the Kreweras map w ↦ w⁻¹c (3)); each ci is a Coxeter element of WSi. Let Ti be the reflection set of (WSi,Si). Then [1,c]≤T=∏i=1k[1,ci]≤Ti=∏i=1kNC⁡(WSi,ci) as posets, where Ti is the reflection set of (WSi,Si) and each factor uses its own absolute order. The empty product when S=∅ is a singleton. Consequently every finite-type noncrossing interval is a finite lattice.

(4) Independence of the Coxeter element. Any two Coxeter elements c,c′ of a finite-type W are conjugate: choose w∈W with c′=wcw−1 (Coxeter elements of tree type are conjugate by source and sink firings (3)). Then Ad⁡w:[1,c]≤T⟶[1,c′]≤T,x⟼wxw−1, is a lattice isomorphism. It preserves reflection length and satisfies M(wxw−1)=ρ(w)M(x); hence the isomorphism type of NC⁡(W,c) is independent of c.

(5) Limits. No assertion is made about whether the whole absolute order Abs⁡(W) is a lattice, about intervals [1,w] when w is not a Coxeter element, or about non-finite types. No finite classification, crystallographic hypothesis, or Axiom of Choice is used; the finite noncrystallographic types are included.

Facts & Assumptions

Given: The finite-type Coxeter system and its absolute order, the bipartite root complex for the connected case, and a,b∈[1,γ].

[F1]

Carter's formula gives ℓT(w)=dim⁡M(w); ≤T is a partial order; it is invariant under conjugation; and for u,v≤Tδ, u≤Tv if and only if M(u)⊆M(v) (Carter's reflection-length formula, the absolute order on a finite Coxeter group, and moved-space rigidity under a common upper bound (1)–(3)).

[F2]

For connected rank at least two, Pσ=Φ+∩M(σ), it spans M(σ), and Pσ is the positive root set of the reflection subgroup with a simple system spanning M(σ) (The mu-dot-root identities, the cone separation, and the canonical simple systems of the subintervals [1, sigma] (4)(i)). In particular P1=∅, M(1)={0}, and X(1) has empty realization.

[F3]

For connected rank at least two, a face of the bipartite root complex is an increasing root tuple whose reverse product lies below γ with length the tuple size (The factorization criterion, linear independence of the faces, and the geometric simplicial structure of X(sigma) (1)–(2)); every positive root reflection lies below γ (The mu-dot-root identities, the cone separation, and the canonical simple systems of the subintervals [1, sigma] (4)(i)).

[F4]

The common-face cone and realization identities hold for subcomplexes (Intersection of root subcomplexes and purity under convexity (1)). For each σ≤Tγ, c[X(σ)] is the positive cone on Pσ and ∣X(σ)∣ is its sphere section (The separating-root lemma, the exact facet halfspaces of the added cones, and the spherical convexity of |X(sigma)| (2)–(3)).

[F5]

The moved space of the reversed reflection product on an independent face is its linear span (Moved space of a reversed reflection product with independent normals (1)).

[F6]

The component decomposition identifies W with ∏iWSi; the component product in Coxeter elements, the noncrossing interval [1,c], and the Kreweras map w ↦ w⁻¹c (3) agrees with the ambient absolute interval.

[F7]

For rank one, the simple root es is the unique positive root, the simple reflection sends it to −es, and its reflecting involution is s (The canonical reflection homomorphism, roots, reflections, and the positive cone, Root sign coherence and the action of simple reflections on positive roots (2)). Clause (1) of A2 gives M(s)=Res (Moved space of a reversed reflection product with independent normals (1)).

[F8]

In rank one the presentation has generator s and relation s2=1; any involution assigned to s extends to a homomorphism from W (Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups).

[F9]

In the rank-two Coxeter system m(s,t)=3, the Coxeter form has B(es,es)=B(et,et)=1 and B(es,et)=−1/2, and the canonical homomorphism sends s,t to rs,rt (The real Coxeter form, its radical, reflections, and form-preserving maps, The canonical reflection homomorphism, roots, reflections, and the positive cone). Thus ρ(st)es=et and ρ(st)et=−es−et, so in the basis (es,et) the matrix of ρ(st) is (0−11−1). Its cube is the identity matrix, while the matrix itself is nonidentity. The presentation imposes (st)3=1 (Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups), hence st has order exactly 3.

[F10]

In the ambient connected rank-at-least-two case, [F3] gives s≤Tγ and t≤Tγ because es,et∈Φ+; [F2] then gives Ps=Φ+∩M(s) and Pt=Φ+∩M(t). Clause (1) of Moved space of a reversed reflection product with independent normals gives M(s)=Res and M(t)=Ret. Every root has B-norm one, and es,et∈Φ+ (Root sign coherence and the action of simple reflections on positive roots (2)); their reflecting involutions are s,t (The canonical reflection homomorphism, roots, reflections, and the positive cone). Since B(es,es)=B(et,et)=1 (The real Coxeter form, its radical, reflections, and form-preserving maps), any positive root in either of these lines is the corresponding simple root. Thus Ps={es} and Pt={et}, which are distinct because the simple roots are linearly independent.

[F11]

In finite type, any two Coxeter elements are conjugate (Coxeter elements of tree type are conjugate by source and sink firings (3)).

Proof

technique · construct the meet from a maximal common face, obtain joins as meets of common upper bounds, then transfer and reduce componentwise

Given: The data above. In the connected case the root complex and root order are those for the bipartite element γ.

1.1F7F8constructalgebra

(Rank one.) Suppose S={s}. By [F8], W has at most two elements; the map s↦−1 to the group {1,−1} satisfies the presentation, so W={1,s}. The group is abelian, so T={s}, and the bipartite element is γ=s. By [F7], Φ={es,−es}, Φ+={es}, the reflection with normal es is s, and M(s)=Res. Thus P1=∅, Ps={es}, and X(s) has one vertex es, while X(1) has empty realization. If either a=1 or b=1, then a∧b=1 and both identities hold. Otherwise a=b=s, whose only common lower bounds are 1,s, so a∧b=s. Its unique maximal simplex is F={es} and its reverse reflection product is R(es)=s. Thus M(s)=span⁡(F)=span⁡(∣X(s)∣) and Ps=Ps∩Ps. This proves every clause of (1) in rank one.

1.2F1F2F9F10algebra

(Identity and empty intersections in rank at least two.) Assume ∣S∣≥2. If a=1 or b=1, the only element below 1 is 1, so a∧b=1; by [F2], M(1)={0}, P1=∅, and ∣X(1)∣=∅, so both displayed identities hold. Now suppose a,b≠1 and Pa∩Pb=∅. Any common lower bound τ has Pτ⊆Pa∩Pb=∅ by transitivity. By [F2], M(τ)=span⁡(Pτ)={0}, so ℓT(τ)=0 by [F1] and τ=1. Thus 1 is the greatest common lower bound and both identities again hold. To see that this case occurs, take m(s,t)=3, γ=st, a=s, b=t. By [F9], st has order 3, so it is neither the identity nor a reflection, since every reflection is conjugate to a simple involution. As it is a product of two reflections, ℓT(γ)=2. Also s−1γ=t and t−1γ=tst are reflections, so s,t≤Tγ. By [F10], Ps={es} and Pt={et}, which are distinct because the simple roots are linearly independent. Thus Ps∩Pt=∅.

1.3F1F2F3F4F5step 1.2algebra

(The nonempty common face in rank at least two.) Assume ∣S∣≥2 and Pa∩Pb≠∅, set Y=X(a), Z=X(b), and let C=c[Y]∩c[Z]. By [F4], c[X(a)]=c[Pa] and c[X(b)]=c[Pb]; each is a positive cone, so C is convex. The common-root set gives a common vertex, and [F4] identifies ∣Y∩Z∣=∣Y∣∩∣Z∣. Choose a maximal simplex F={v1<⋯<vr} of Y∩Z. It is a simplex of X(γ), so [F3] gives σ=R(vr)⋯R(v1)≤Tγ and ℓT(σ)=r. By [F5] and the purity conclusion of the preceding item, M(σ)=span⁡(F)=L:=span⁡(C). Since C⊆c[X(a)]=c[Pa] and span⁡(Pa)=M(a) by [F2], one has M(σ)⊆M(a); similarly M(σ)⊆M(b). With σ,a,b≤Tγ, rigidity [F1] gives σ≤Ta,b. Now Pσ⊆Pa∩Pb by transitivity. Conversely, each α∈Pa∩Pb is a common vertex of Y and Z, hence belongs to ∣Y∩Z∣⊆C and to L=M(σ). Thus M(tα)=span⁡(α)⊆M(σ); since tα,σ≤Tγ, rigidity gives tα≤Tσ, so α∈Pσ. Hence Pσ=Pa∩Pb. If τ≤Ta,b, then Pτ⊆Pσ, so M(τ)=span⁡(Pτ)⊆M(σ) by [F2]; rigidity gives τ≤Tσ. Therefore σ=a∧b. Finally, span⁡(∣Y∣∩∣Z∣)=span⁡(C) because every nonzero point of the cone normalizes into its sphere section. This proves all nonempty-case identities.

2.1F1step 1.1step 1.2step 1.3algebra

(Joins.) Let U={x∈[1,γ]:a≤Tx, b≤Tx}. It is nonempty because γ∈U, and finite because W is finite. Iterating the binary meet established in steps 1.1–1.3 gives the greatest lower bound m of U. Since a and b are lower bounds of every member of U, they satisfy a,b≤Tm; and m≤Tx for every common upper bound x. Thus m is the least common upper bound, a∨b. By induction on cardinality, the iterated binary meet of any nonempty finite subset is its greatest lower bound: this is immediate for a singleton, and adjoining one element replaces the existing meet m by m∧x. This proves (2).

3.1F1F11step 2.1algebra

(Conjugacy and independence of c.) For any finite-type W and Coxeter elements c,c′, [F11] gives c′=wcw−1 for some w∈W. Conjugation maps T bijectively to itself, so it preserves ℓT and ≤T; its inverse is conjugation by w−1. Hence it is an order isomorphism of the two intervals. Also ρ(wxw−1)−id=ρ(w)(ρ(x)−id)ρ(w)−1, so M(wxw−1)=ρ(w)M(x). An order isomorphism preserves greatest lower bounds and least upper bounds by their defining universal properties, and therefore is a lattice isomorphism once the bipartite interval is known to be a lattice. This proves (4) and transfers (1)–(2) to every Coxeter element in the connected case.

4.1F1F6step 1.1step 1.2step 1.3step 2.1step 3.1algebra∎

(Reducible systems.) If S=∅, then W={1}, the interval and the empty product are both one-element lattices. Otherwise use the component decomposition and interval identity [F6]. Each WSi is a connected finite-type Coxeter group, so its noncrossing interval is a finite lattice by steps 1.1–1.3, 2.1, and 3.1. Componentwise meets and joins make the finite product a lattice. This proves (3) and completes the theorem.

TheoremStatement: Literature-sourcedProof: AI-adaptedaudited 2026-10-08Open item page →

The Kreweras complement of [1,c], and the type-A model by noncrossing set partitions

Statement

(1) The general Kreweras complement. Let (W,S) be a Coxeter system of finite type with S finite, let n=∣S∣, let T be its reflection set, and let ℓT and ≤T be reflection length and absolute order (Coxeter diagrams: edges, labels, components and finite type, Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator). For a Coxeter element c, let NC⁡(W,c)=[1,c]≤T and K(w)=w−1c (Coxeter elements, the noncrossing interval [1,c], and the Kreweras map w ↦ w⁻¹c). The length of c is n: apply Moved space of a reversed reflection product with independent normals (2) to the independent unit simple-root normals in a once-each expression for c. Then K maps NC⁡(W,c) bijectively to itself and, for every w in this interval, K(K(w))=c−1wc,ℓT(K(w))=n−ℓT(w),wK(w)=c. It reverses order: if u≤Tv in the interval, then K(v)≤TK(u). Since NC⁡(W,c) is a finite lattice (Finite noncrossing intervals are lattices, independently of the Coxeter element (2)–(4)), K is a lattice anti-automorphism.

(2) Type A: reflection length. Let N≥1 and realize the Coxeter system of type AN−1 as SN on {1,…,N}, with si=(i i+1) for 1≤i<N and c=s1s2⋯sN−1=(1 2 ⋯ N) (The finite symmetric group Sn, one-line notation, and cycle notation, Sym⁡(X) is a group under composition, and it is non-abelian whenever X has at least three distinct elements, The symmetric group has the Coxeter presentation, Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups). Its reflection set T consists of all transpositions. For every w∈SN, ℓT(w)=N−#{cycles of w}, where fixed points count as one-cycles (Every permutation of a finite set is a product of pairwise disjoint cycles, uniquely up to reordering and cyclic rotation, Support, fixed points, disjoint cycles, cycle length, disjoint-cycle decompositions, and cycle type).

(3) Type A: the noncrossing criterion. For w∈SN, let π(w) be the partition of {1,…,N} into the supports of all cycles of w, including fixed points. Place the labels at equally spaced points on a circle in cyclic order 1,2,…,N,1, including one point for N=1 and two antipodal points for N=2. A partition is noncrossing when the convex hulls of distinct blocks are disjoint. A cycle is cyclically increasing when its entries, read in the direction of the cycle, advance in that cyclic order. Then w≤Tc⟺π(w) is noncrossing and every cycle of w is cyclically increasing.

(4) Type A: the partition model and its complement. Let NC⁡(N) be the noncrossing partitions of {1,…,N} ordered by refinement. The map w↦π(w) is a poset isomorphism ([1,c]≤T,≤T)  ⟶  (NC⁡(N), refinement); its inverse sends each block to the cycle that lists its elements in cyclically increasing order and multiplies those disjoint cycles. Put black vertices b1,…,bN and white vertices d1,…,dN alternately at equally spaced points on a circle. For π∈NC⁡(N), its classical Kreweras complement Kcl(π) is the coarsest partition Q of the white labels whose interleaving with π is noncrossing. Under the isomorphism of (4), π(K(w))=Kcl(π(w)),wπ wKcl(π)=c, where wπ is the inverse image of π. The complement is an order-reversing bijection, Kcl2 rotates labels by i↦i−1 (indices modulo N), and ∣Kcl(π)∣=N+1−∣π∣,π∧Kcl(π)=0^,π∨Kcl(π)=1^.

(5) Limits. No noncrossing set-partition model, crossing criterion, or Catalan count is asserted for finite Coxeter types other than type A. No Lie-theoretic root system is used in (2)–(4). The statements include N=1 and rank-zero finite Coxeter systems; no Choice is used.

Facts & Assumptions

Given: A finite-type Coxeter system and a Coxeter element c; in type A, the symmetric group SN with the indicated simple reflections and cyclic order.

[F1]

The simple-root normals form a linearly independent unit list. For any once-each product of the corresponding simple reflections, Moved space of a reversed reflection product with independent normals (2) gives ℓT(c)=∣S∣; for the empty list both sides are zero.

[F2]

ℓT is the word length in the conjugation-invariant set T, and u≤Tv means ℓT(v)=ℓT(u)+ℓT(u−1v) (Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator (1)–(2)).

[F3]

Adjacent transpositions give the Coxeter presentation of SN and their ordered product is the long cycle c=(1 2 ⋯ N) (The symmetric group has the Coxeter presentation, Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups). Cycle notation composes from right to left (The finite symmetric group Sn, one-line notation, and cycle notation).

[F4]

Every permutation has a unique disjoint-cycle decomposition up to reordering and cyclic rotation, with fixed points added as one-cycles when counting (Every permutation of a finite set is a product of pairwise disjoint cycles, uniquely up to reordering and cyclic rotation, Support, fixed points, disjoint cycles, cycle length, disjoint-cycle decompositions, and cycle type).

[F5]

The finite noncrossing interval [1,c]≤T is a lattice (Finite noncrossing intervals are lattices, independently of the Coxeter element (2)–(4)).

Proof

technique · prove the interval complement from length additivity, derive the type-A length and interval criteria by transposition moves, then identify the cycle-support order and trace the complementary regions

Given: The data above.

1.1F1

(Rank of a Coxeter element.) Write c=si1⋯sin, where each simple reflection occurs once. Its simple-root normals, in the reverse list, are independent unit roots. Clause (2) of [F1] applied to this reversed list gives ℓT(c)=n. If n=0, c=1 and the same equality is immediate.

1.2F1F2algebra

(The group-theoretic complement maps the interval to itself.) The set T is invariant under conjugation: conjugation permutes its defining conjugates of simple reflections. Conjugating a shortest reflection factorization and then conjugating back shows ℓT(gwg−1)=ℓT(w) for all g,w∈W. If w≤Tc, then ℓT(c)=ℓT(w)+ℓT(w−1c), so ℓT(K(w))=n−ℓT(w). Also K(w)−1c=c−1wc, whose reflection length is ℓT(w). Therefore ℓT(K(w))+ℓT(K(w)−1c)=n=ℓT(c) and K(w)≤Tc. Thus K(w) is in the interval. The map is injective, and the interval is finite because W is finite, so it is onto. Direct multiplication gives K(K(w))=c−1wc and wK(w)=c.

1.3F3F4algebra

(The reflection set in type A.) For N=1 there are no simple reflections and no transpositions. For N≥2, conjugating any simple reflection si=(i i+1) by g∈SN gives (g(i) g(i+1)) (Conjugating a cycle relabels each entry: g(a1 … ak)g−1=(g(a1) … g(ak))), so every reflection is a transposition. Conversely, for any transposition (a b) choose a permutation g with g(1)=a and g(2)=b; then gs1g−1=(a b), so every transposition is a reflection. A right multiplication by a transposition changes the number of cycles by exactly one: if its two labels lie in one cycle, it cuts that cycle at those labels into two; if they lie in different cycles, it joins the cycles. Thus any expression of w as r transpositions must have r≥N−#{cycles of w}, since reaching the identity requires increasing the cycle count to N one step at a time. Conversely each cycle (a1 a2 ⋯ am) is (a1 am)(a1 am−1)⋯(a1 a2), a product of m−1 transpositions. Multiplying these expressions over the disjoint cycles gives the matching upper bound and proves the formula. It also covers N=1, where the identity is the empty product.

1.4givenconstruct

(An interval block exists.) Every noncrossing partition with at least two blocks has a block consisting of consecutive vertices in the original cyclic order. A singleton block suffices. Otherwise choose a block B minimizing max⁡B−min⁡B in the linear order 1<⋯<N. If B is not a linear interval, there are successive elements b<b′ of B and a label x with b<x<b′. Let D be the block containing x. Any y∈D outside (b,b′) would make the chords bb′ and xy have alternating endpoints and therefore cross, contradicting disjointness of the block hulls. Hence D⊆(b,b′), so max⁡D−min⁡D<b′−b≤max⁡B−min⁡B, contrary to minimality. Thus B is a linear interval, and hence consecutive in the original cyclic order.

2.1F1F2F5step 1.2algebra

(Order reversal and lattice duality.) If u≤Tv, then ℓT(v)=ℓT(u)+ℓT(u−1v). The conjugation invariance just proved and v(u−1v)v−1=vu−1 give ℓT(K(v)−1K(u))=ℓT(c−1vu−1c)=ℓT(u−1v)=ℓT(v)−ℓT(u)=ℓT(K(u))−ℓT(K(v)). Hence K(v)≤TK(u). An order-reversing bijection of a lattice carries every least upper bound to a greatest lower bound and vice versa, by the defining universal properties. The lattice hypothesis is supplied by [F5].

2.2F2F3step 1.3induction

(Noncrossing increasing cycles lie below c: singleton removal.) Define wπ to be the product of the cyclically increasing cycles on the blocks of π. We prove wπ≤Tc by induction on N. The one-block partition gives wπ=c. If π has a singleton block {a}, remove it to obtain a noncrossing partition π′ on the remaining cyclically ordered set Y of size N−1, with long cycle cY and permutation w′. By induction, ℓY(w′)+ℓY(w′−1cY)=N−2. Regard these permutations as fixing a in SN, and let p be the predecessor of a in the cyclic order. For τ=(p a), direct evaluation on the labels gives c=cYτ. Since w′ fixes a, q=w′−1cY also fixes a; multiplying q on the right by τ joins the singleton cycle {a} to the cycle containing p. Hence ℓN(wπ)=ℓY(w′) and ℓN(wπ−1c)=ℓY(w′−1cY)+1. Their sum is N−1=ℓN(c), proving wπ≤Tc. This includes the discrete partition and the cases N≤2.

2.3F2step 1.3induction

(Elements below c have noncrossing increasing cycles.) Induct on d=N−1−ℓT(w) for w≤Tc. If d=0, then w=c. If d>0, take a shortest transposition factorization w−1c=t1⋯td. A shortest factorization of w followed by this one is a shortest factorization of c, so its prefix x=wt1 satisfies w≤Tx≤Tc and ℓT(x)=ℓT(w)+1. By induction, π(x) is noncrossing and its cycles are cyclically increasing. Since right multiplication by t1 lowers reflection length by one, step 1.3 shows that t1 splits one cycle of x. If that cycle is (b1 b2 ⋯ br) in cyclic order and t1=(bi bj) with i<j, the two resulting cycles have supports and cyclic orders (bi,bj+1,…,br,b1,…,bi−1)and(bj,bi+1,…,bj−1), with singleton cycles interpreted as fixed points. Both are cyclically increasing. Their convex hulls lie on opposite sides of the chord bibj and meet its line only at distinct endpoints, so are disjoint. Every other block hull was disjoint from the old block hull and remains disjoint from its two sub-hulls. Thus π(w) is noncrossing and every cycle is cyclically increasing.

3.1F2F3F4step 1.3step 1.4step 2.2induction

(Noncrossing increasing cycles lie below c: interval-block contraction.) Now suppose π has no singleton blocks and at least two blocks. By step 1.4 it has a consecutive block B={a,a+1,…,a+m−1} in cyclic order, with m≥2. Contract B to a single label a to obtain a cyclically ordered set Y of size N−m+1 and a noncrossing partition π′′ whose block at a is the singleton {a}. Let c′′ be the long cycle on Y and w′′=wπ′′, so w′′ fixes a. Write γB=(a a+1 ⋯ a+m−1). Extending permutations of Y to fix the deleted labels gives wπ=γBw′′ and c=c′′γB, with w′′ commuting with γB. By induction, ℓY(w′′)+ℓY(w′′−1c′′)=∣Y∣−1=N−m. The disjoint-cycle formula gives ℓN(wπ)=(m−1)+ℓY(w′′), while wπ−1c=w′′−1γB−1c′′γB=γB−1(w′′−1c′′)γB, so conjugation invariance and the cycle formula give ℓN(wπ−1c)=ℓY(w′′−1c′′). The two lengths sum to (m−1)+(N−m)=N−1=ℓN(c). Therefore wπ≤Tc. The one-block case was handled in step 2.2; these cases exhaust all partitions.

4.1F2step 1.3step 2.2step 3.1step 2.3algebra

(The partition map is a bijection and preserves order.) Steps 2.2, 2.3 and 3.1 show that each noncrossing partition has an interval element wπ and every interval element arises this way. Its cycle supports determine each of its cyclically increasing cycles, so this correspondence is bijective. If u≤Tv, choose a shortest transposition factorization of u−1v and append it to a shortest factorization of u. Every prefix is shortest, giving a chain in [1,c] from u to v whose steps multiply on the right by a transposition and raise length by one. By step 1.3 each step joins two cycles, so π(u) refines π(v). Conversely suppose π(u) refines π(v). For each block B of π(v), restrict v to its cyclically increasing cycle vB and let uB be the product of the cycles of u supported in B. The induced partition π(u)∣B is noncrossing, and its cycles remain cyclically increasing in the induced cyclic order on B. By steps 2.2, 2.3 and 3.1 applied to the labels in B, uB≤TvB. The blocks B are disjoint, and the cycle formula in step 1.3 gives additivity of reflection length across these supports for u, v, and u−1v. Summing ℓB(vB)=ℓB(uB)+ℓB(uB−1vB) over all B gives ℓT(v)=ℓT(u)+ℓT(u−1v), hence u≤Tv. Thus the bijection is an order isomorphism.

5.1F2step 1.2step 2.1step 1.3step 2.2step 3.1step 2.3step 4.1algebra

(The region partition is the classical complement.) For N=1, the sole black and white blocks are singletons, c=K(1)=1, and all assertions in (4) hold, with 0^=1^. Assume N≥2. Draw the convex hull edges of each black block of π in the alternating 2N-gon. These noncrossing chords cut the disk into polygonal regions. Group white vertices lying in the same region. Each region is a convex polygonal cell of the dissection by noncrossing chords, so grouping its white vertices gives a noncrossing partition. Any compatible white block must lie in one region, since a segment joining vertices in different regions crosses a black block edge. Hence this region partition is the coarsest interleaving partner, namely Kcl(π). Let α=wπ, and label di as the white vertex in the gap after bi. Tracing the boundary of the region at di to the next white vertex passes black vertex bi+1 and then follows the boundary edge of its black block back to its predecessor bj, where j=α−1(i+1). Thus the successor permutation of white vertices within their regions is q(i)=α−1(i+1)=α−1c(i). Its cycles are exactly the white blocks of Kcl(π), so q=wKcl(π)=α−1c and αq=c. Since K(w)=w−1c, this proves π(K(w))=Kcl(π(w)). By step 2.1 and the order isomorphism, Kcl is an order-reversing bijection, and its square is relabeling by c−1, namely i↦i−1. From steps 1.2 and 1.3, ℓT(K(w))=(N−1)−(N−∣π(w)∣)=∣π(w)∣−1. Applying the cycle formula to K(w) gives ∣Kcl(π)∣=N+1−∣π∣. If two labels belonged to a common block of both π and Kcl(π), the corresponding black and white chords would have alternating endpoints and cross; hence their only common lower bound in refinement order is 0^, giving π∧Kcl(π)=0^. Each cycle of wπ and wKcl(π) stays inside a class of the equivalence relation generated by their block memberships. Thus each permutation preserves every equivalence class setwise, so their product c also preserves each class setwise. Since c is transitive, the only such class is the whole label set. Every common upper bound is consequently 1^, so π∨Kcl(π)=1^.

6.1F1F2step 1.2step 2.1step 1.3step 1.4step 2.2step 3.1step 2.3step 4.1step 5.1∎

The general complement proof uses only finite reflection length, conjugation invariance of its defining set, and the lattice property of the interval. The type-A model uses the Coxeter presentation and permutations only; it invokes no Lie-theoretic root system, crystallographic hypothesis, finite classification, or Choice. No enumeration of the general finite-type interval is asserted.

5 · Examples, counterexamples and false statements

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