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Coxeter Presentations, Exchange, and Reduced Word Theorems

1 · Prerequisites

2 · Summary

The Hecke branch begins again with generators, words and length; it does not assume the exchange or reduced-word theorem. A presentation alone gives a group, while its reflection geometry is what supplies the word control needed to define T_w independently of a reduced expression.

The page is authored as six draft items, in dependency order. Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups fixes the conventions: a finite Coxeter matrix, the presented group via the free group on S and the normal closure of the relators, its universal property, reduced words and the length ℓ, and the standard parabolic subgroups WJ — with the explicit convention that no finiteness, faithfulness or completeness is asserted there. The geometric representation on the simple-root basis over a common splitting field, and the root set fixes one common characteristic-zero splitting field, one primitive 2m-th root per finite edge and the constants cst, and defines the involutions σs on the simple-root basis together with the root set as their orbit; no positivity, integrality or faithfulness is claimed.

The word control is built in The rank-two block computation, exact dihedral orders, the signed reflection action, and ambient reducedness: the rank-two block σsσt has trace c2−2 and determinant 1, its order is exactly m (including m=2 and m=∞), the assignment s↦σs descends to a representation W→GLK(E) with distinct generators and exact dihedral orders, the signed right action on {±1}×T is well defined, prefix reflections delete two letters, the sign-change set Φ(w) of a reduced word is expression-independent with #Φ(w)=ℓ(w), and alternating dihedral words are reduced in the ambient group. Length parity, exchange, two-letter deletion, and faithfulness of the signed reflection action then supplies the sign character and the parity laws ℓ(sw)=ℓ(w)±1, exchange on both sides, Tits two-letter deletion and faithfulness of the signed action. Matsumoto's theorem: braid connectivity of reduced expressions, with singleton detection in dihedral subgroups proves braid connectivity of the reduced expressions of an element, the equivalence between reduced and M-reduced words, and the singleton claim S∩⟨s,t⟩={s,t}. Finally Support, intrinsic parabolic presentations, minimal coset representatives and length additivity, with the type-A identification proves the support description WJ={w∈W:S(w)⊆J}, the intrinsic parabolic presentation with ℓJ=ℓ and WJ∩S=J, the unique minimal coset representatives with the descent characterisation and length additivity, and the type-A identification ℓ=inv⁡.

Reading and applications

Prerequisite pages: tensor-coherence-and-algebraic-descent, symmetric-groups-and-the-sign-homomorphism, splitting-fields, finite-fields-and-cyclotomic-extensions, group-homomorphisms-and-the-isomorphism-theorems. The companion coxeter-presentations-exchange-and-reduced-word-theorems-examples develops the calculations and failures needed to test these constructions. All six items are draft, and the two definitions record their later-derived features as justifiers rather than as part of their assertions.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups

Definition

Let S be a finite set and let m:S×S→{1,2,3,… }∪{∞} satisfy m(s,s)=1 for all s∈S and m(s,t)=m(t,s)∈{2,3,… }∪{∞} for all s≠t. Such an m is a Coxeter matrix on S.

The presented group. Let R⊆F(S) be the set of relators R:={s2:s∈S}∪{(st)m(s,t):s,t∈S, m(s,t)<∞}, where F(S) is the free group on S (Free group on a set of generators, Reduced words form the free group on an alphabet) and (st)k is the k-fold product (Powers gn: natural exponents in a monoid and integer exponents in a group, with g0=e), and let N:=⟨ ⁣⟨R⟩ ⁣⟩F(S) be the normal closure of R in F(S) (The normal closure of a subset of a group, The normal closure of R is the set of finite products of conjugates of elements of R and their inverses). Define W:=F(S)/N, and write s (as well as sN) for the image of s∈S in W (The quotient group G/N and coset product (gN)(hN)=ghN, Group and abelian group).

Universal property. For every group G and every map f:S→G with f(s)2=1 for all s∈S and (f(s)f(t))m(s,t)=1 in G for all s≠t with m(s,t)<∞, there is a unique group homomorphism φ:W→G with φ(s)=f(s) for all s∈S (Monoid homomorphism and group homomorphism, A homomorphism that kills a normal subgroup factors uniquely through the quotient group, Von Dyck's theorem: maps of generators that satisfy the relators extend uniquely from a presented group, Group presentation by generators and relations, Relators and relations; finitely generated, finitely related, and finite presentations, In ⟨X∣R⟩, the words u and v represent the same element if and only if u−1v∈⟨ ⁣⟨R⟩ ⁣⟩).

Length and reduced words. For w∈W put ℓ(w):=min⁡{k∈N: there exist s1,…,sk∈S with w=s1⋯sk}. The minimum exists: every element of the free group F(S) is represented by a finite word in S∪S−1 (Words in an alphabet with formal inverses, elementary cancellation, and reduced words), and s−1=s in W because s2∈R, so every element of W is the value of a finite word in S; the set of admissible k∈N is therefore nonempty and has a least element by the well-ordering principle (The well-ordering principle, The natural numbers N (von Neumann)). A word (s1,…,sk) in S is a reduced expression of w when w=s1⋯sk and k=ℓ(w), and nonreduced otherwise. The empty word (  ) is a word in S of length 0, and ℓ(1)=0; it is the reduced expression of 1.

Terminology. The pair (W,S) is the Coxeter system presented by the Coxeter matrix (S,m); when s↦s is understood, we also say that W is the Coxeter group presented by (S,m).

Standard parabolic subgroups. For J⊆S put WJ:=⟨{s:s∈J}⟩≤W (The subgroup ⟨S⟩ generated by a subset, the cyclic subgroup ⟨g⟩, and cyclic groups).

Conventions. m(s,t) is declared to be the order of st in W; a value m(s,t)=∞ imposes no relator on s,t. This definition asserts no finiteness, faithfulness, or completeness property of the presentation: the statements that the elements s∈S are pairwise distinct in W (so that S may be viewed as a labelled subset of W), that ℓ(s)=1, that a word is reduced exactly when it admits no two-letter deletion, and that (WJ,J) is the Coxeter system presented by the restricted matrix m∣J with intrinsic length equal to the ambient length are recorded with their justifiers below and are not used before those results.

Remarks

The properties announced in the Conventions paragraph are supplied later in this pair: pairwise distinctness of the simple generators, ℓ(s)=1, and the deletion characterisation of reduced words in Length parity, exchange, two-letter deletion, and faithfulness of the signed reflection action ↗; the intrinsic presentation and length of the standard parabolic subgroups in Support, intrinsic parabolic presentations, minimal coset representatives and length additivity, with the type-A identification ↗. Until those results are available, W and ℓ are to be read exactly as constructed above.

No form of choice is used in the construction: W is the quotient of the free group on the finite set S by an explicitly generated normal closure, and the only minimisation in the definition is over a nonempty set of natural numbers.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

The geometric representation on the simple-root basis over a common splitting field, and the root set

Definition

Let (S,m) be a finite Coxeter matrix and let W be the group presented by it as in Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups.

A common splitting field. Put h:=∏ (X2m(s,t)−1)∈Q[X], the product over all unordered pairs {s,t}⊆S with s≠t and m(s,t)<∞ (an empty product when #S≤1). Let K be a splitting field of h over Q (The rationals as equivalence classes of pairs of integers, The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution, Over an integral domain, degrees add under multiplication of nonzero polynomials, For every field F, F[x] is a unique factorisation domain, Every nonzero polynomial over a field has a splitting field, Every finite family of nonzero polynomials has a splitting field, obtained from their product, Polynomials that split and splitting fields of a polynomial or a family of polynomials). The prime subfield of K is Q (A field's prime subfield is isomorphic to Q in characteristic zero and to Fp in characteristic p, The characteristic of a ring: the least n≥1 with n⋅1R=0 when one exists, and 0 otherwise, The rationals form a field), so char⁡K=0 (The characteristic of a field is zero or a prime number) and Xn−1 is separable over K for every n≥1 (tn−1 is separable over K exactly when the characteristic does not divide n, and then a splitting field carries n distinct n-th roots of unity). For each finite edge, X2m(s,t)−1 is a factor of h (Over an integral domain, degrees add under multiplication of nonzero polynomials), so all its roots lie in K and, being separable of degree 2m(s,t), it has exactly 2m(s,t) distinct roots there (tn−1 is separable over K exactly when the characteristic does not divide n, and then a splitting field carries n distinct n-th roots of unity, Polynomials that split and splitting fields of a polynomial or a family of polynomials); the group μ2m(s,t)(K) of such roots is therefore finite of order 2m(s,t) and hence cyclic, so it contains a primitive 2m(s,t)-th root of unity (μn(K) is cyclic of order dividing n, and has a primitive n-th root of unity exactly when its order is n, The group μn(K) of n-th roots of unity in a field, and primitive n-th roots of unity). Fix one primitive 2m(s,t)-th root of unity ζst∈K for each unordered finite edge (a finite selection, since S is finite) and put cst:=cts:=ζst+ζst−1∈K(m(s,t)<∞),cst:=cts:=2(m(s,t)=∞).

The representation on the simple-root basis. Let E be the K-vector space with basis (αs)s∈S (Vector space over a field, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis). For each s∈S let σs:E→E be the unique K-linear map with σs(αs)=−αs,σs(αt)=αt+cstαs(t≠s). Directly from the definition each σs is an involution: σs2(αs)=αs and, for t≠s, σs2(αt)=σs(αt+cstαs)=αt+cstαs−cstαs=αt. Hence every σs is invertible with σs−1=σs, and the group these maps generate acts on E (Linear map between vector spaces over the same field, Invertible linear maps, linear isomorphisms, and inverse linear maps, Identity maps and composites of linear maps are linear).

The root set. The root set of the construction is Φ:={σs1σs2⋯σsk(αt):k≥0, s1,…,sk,t∈S}⊆E, the orbit of the simple roots under the group generated by the σs (each generator is an involution). Writing σ:W→GLK(E) for the representation supplied by The rank-two block computation, exact dihedral orders, the signed reflection action, and ambient reducedness ↗, one has Φ={σw(αs):w∈W, s∈S}, the W-orbit of the simple roots.

Conventions and limits. (i) cst=cts; cst=0 exactly when m(s,t)=2; cst=2 when m(s,t)=∞; replacing ζst by ζst−1 leaves cst unchanged, so the construction depends on the chosen primitive roots only through the numbers cst. (ii) No positivity of cst, no integrality of roots over Z, no identification of roots with reflections, and no faithfulness or definiteness of any form is asserted; those are separate matters, not part of this construction. (iii) S is finite throughout, so the basis (αs)s∈S is finite and every linear map is specified by finitely many values.

Remarks

The single recorded justifier of this definition is The rank-two block computation, exact dihedral orders, the signed reflection action, and ambient reducedness ↗, which proves σs2=id, the exact order of σsσt, and the induced homomorphism σ:W→GLK(E); only after that lemma is Φ literally the W-orbit of the simple roots.

No choice is used. The splitting field and the maps σs are constructions, and one primitive 2m-th root is fixed for each of finitely many edges; the convention cst=2 for m(s,t)=∞ introduces no root of unity at all.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

The rank-two block computation, exact dihedral orders, the signed reflection action, and ambient reducedness

Statement

Let (S,m) be a finite Coxeter matrix, and let K, E, (αs)s∈S, the constants cst and the linear maps σs be as in The geometric representation on the simple-root basis over a common splitting field, and the root set; let W and ℓ be as in Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups. The involution, representation, signed-action and word assertions below hold for every finite S, including S=∅ and #S=1. Only the rank-two assertions require distinct generators.

  1. Involutions. σs2=idE for every s∈S; each σs is invertible, and ℓ(s)=1.
  2. The rank-two block. For any distinct s,t∈S, put m:=m(s,t), c:=cst, P:=span⁡(αs,αt)⊆E, A:=σsσt and B:=A∣P. Then A(P)⊆P, and in the basis (αs,αt), B(αs)=(c2−1)αs+c αt,B(αt)=−c αs−αt;tr⁡(B)=c2−2,det⁡(B)=1. Moreover A(v)−v∈P for every v∈E. Consequently E=P⊕Q with Q:=span⁡{αu:u∈S, u∉{s,t}}, and for every v=p+q with p∈P, q∈Q and every k≥1, Ak(p+q)=Bkp+(idP+B+⋯+Bk−1)(A(q)−q)+q.
  3. Exact order of the rank-two product. For any distinct s,t and the notation of (2), when m<∞ put ζ:=ζst, of order 2m, so c=ζ+ζ−1; when m=∞ use c=2, with no ζ stipulated. Then: (a) if 2<m<∞, the characteristic polynomial of B is X2−(c2−2)X+1=(X−ζ2)(X−ζ−2) with distinct roots; B is diagonalisable, Bm=idP, idP+B+⋯+Bm−1=0, and Bk≠idP for 0<k<m; (b) if m=2, then c=0 and B=−idP; (c) if m=∞, then c=2, B≠idP, (B−idP)2=0, and Bk≠idP for every k≥1. Hence σsσt=A has order exactly m when m<∞, and infinite order when m=∞; in particular Am=idE and Ak≠idE for 0<k<m in the finite case.
  4. The representation and the exact dihedral orders. The assignment s↦σs respects every defining relator of W and induces a unique homomorphism σ:W→GLK(E) with σ(s)=σs. Consequently, for any distinct s,t∈S, one has s≠t in W and st has order exactly m(s,t) in W (infinite when m(s,t)=∞).
  5. The signed reflection action. Let T:={wsw−1:w∈W, s∈S}⊆W. For s∈S define Us:{±1}×T→{±1}×T by Us(ε,r):=(ε⋅(−1)δ(s,r), srs), where δ(s,r)=1 if r=s and δ(s,r)=0 otherwise. Then Us2=id for every s, and the assignment (ε,r)⋅s:=Us(ε,r) extends to a well-defined right action of W on {±1}×T (so (ε,r)⋅1=(ε,r) and (ε,r)⋅(wv)=((ε,r)⋅w)⋅v). Indeed the only relations to check are s2=1 and, for distinct s,t with m=m(s,t)<∞, (st)m=1: Us2=id because for r=s the two signs cancel while for r≠s one has srs≠s and Us(ε,srs)=(ε,r); and the alternating word s t s t⋯ of length 2m acts trivially because its 2m prefix reflections ri:=wi−1siwi−1−1 (with wj:=s1⋯sj) satisfy ri+m=ri for 1≤i≤m and r1,…,rm are pairwise distinct, so each reflection of the dihedral subgroup ⟨s,t⟩ occurs exactly twice and every accumulated sign is even while the conjugating coordinate returns to r because (st)m=1.
  6. Prefix reflections, deletion and expression independence. Let (s1,…,sk) be a word in S with value w=s1⋯sk and prefix reflections ri=s1⋯si−1sisi−1⋯s1, and let n(r):=#{i:ri=r}. (a) If ri=rj for some i<j, then s1⋯si^⋯sj^⋯sk=w: the two letters si,sj can be deleted. (b) (−1)n(r)=:η(r,w) depends only on w and r, and the right action satisfies (ε,r)⋅w=(ε η(r,w), w−1rw). (c) If the word is reduced (k=ℓ(w)), then n(r)∈{0,1} for every r, the map i↦ri is injective, and the set Φ(w):={r1,…,rk}={r∈T:η(r,w)=−1} is independent of the reduced expression chosen, with #Φ(w)=ℓ(w).
  7. Ambient reducedness of dihedral words. Let s≠t, m:=m(s,t), and for q≥1 let wq be the value of the alternating word of length q beginning with s. If m<∞ and q≤m, or if m=∞ and q≥1, then ℓ(wq)=q, every word in S representing wq has length at least q, and w1,…,wq are pairwise distinct. Thus dihedral alternating words are reduced in the ambient group W, not merely inside ⟨s,t⟩.

Facts & Assumptions

Given: A finite Coxeter matrix (S,m), the construction K,E,(αs),σs of The geometric representation on the simple-root basis over a common splitting field, and the root set, the group W of Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups, with no assumption on #S; distinct s,t∈S are fixed only for the rank-two assertions.

[F1]

Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups: W=F(S)/N is the group presented by the Coxeter matrix, with N the normal closure of R={s2:s∈S}∪{(st)m(s,t):s≠t, m(s,t)<∞}; for every group G and every map f:S→G with f(s)2=1 for all s∈S and (f(s)f(t))m(s,t)=1 for all s≠t with m(s,t)<∞, there is a unique homomorphism φ:W→G with φ(s)=f(s) for all s∈S. The length ℓ(w) is the least k with w=s1⋯sk for some s1,…,sk∈S, and ℓ(1)=0.

[F2]

The geometric representation on the simple-root basis over a common splitting field, and the root set: K is a field of characteristic 0; E is the K-vector space with basis (αs)s∈S; the map σs is the unique K-linear map with σs(αs)=−αs and σs(αt)=αt+cstαs for t≠s; and for distinct s,t with m(s,t)<∞ the constant is cst=ζst+ζst−1 with ζst a primitive 2m(s,t)-th root of unity, while cst=2 when m(s,t)=∞.

[F3]

Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis, Linear map between vector spaces over the same field: the basis vectors are linearly independent and span E; linear maps with the same values on that basis agree, so the primitive formulas in [F2] may be used to check every identity below on the basis.

[F4]

For the distinct pair and its subspaces P,Q in (2), Internal direct sum V=⨁i<nUi: the sum is everything and each summand meets the sum of the others only in 0V: if a vector space is the sum of subspaces meeting only in 0 it is their internal direct sum; here P=span⁡(αs,αt) and Q=span⁡{αu:u∉{s,t}} complement each other because the basis (αu)u∈S splits into the two parts.

[F6]

Eigenvalues, eigenvectors, eigenspaces Eλ(T)=ker⁡(T−λI), and the spectrum σF(T) of an endomorphism, A characteristic polynomial that splits into distinct linear factors forces diagonalisability: if the characteristic polynomial splits into distinct linear factors then the endomorphism is diagonalisable, and a polynomial in B vanishes on B as soon as it vanishes at every eigenvalue and B is diagonalisable.

[F7]

The order ∣G∣ of a finite group and the order ord⁡(g) of an element, with ord⁡(g)=∞ when no positive power of g is the identity: for g in a group, ord⁡(g) is the least k≥1 with gk=1 when such k exist, gord⁡(g)=1 then, and g has infinite order when gk≠1 for all k≥1.

[F8]

The group μn(K) of n-th roots of unity in a field, and primitive n-th roots of unity: a primitive n-th root of unity is an element of order exactly n; in particular, for a distinct pair with finite m=m(s,t), the fixed element ζ=ζst satisfies ζ2m=1 and ζj≠1 for 0<j<2m.

Proof

Given: The data of the statement: the finite Coxeter matrix (S,m), the field K, the space E with basis (αs), the maps σs and constants cst, the group W, with no assumption on #S; a distinct pair is fixed only in steps 1.2–2.3 and the rank-two portions of steps 3.1, 4.1 and 7.1.

Proof technique: direct; all identities for A and B are checked on the basis (αs,αt) of P and the complementary basis of Q.

1.1F2F3algebra

Involutions and nonidentity for every generator. Fix any s∈S. The primitive formulas of [F2] give σs2(αs)=σs(−αs)=αs and, for u≠s, σs2(αu)=σs(αu+csuαs)=αu+csuαs−csuαs=αu. By [F3] the square is the identity on E, so σs is invertible with inverse itself. Also σs(αs)=−αs≠αs, since the basis vector αs is nonzero and char⁡K=0; hence σs≠idE. These computations require no second generator; when S=∅, the assertions indexed by s are vacuous.

1.2F2F3algebra

The rank-two block. Fix any distinct s,t∈S and use the notation of (2). Since σt(αs)=αs+c αt and σt(αt)=−αt with c=cst, linearity of σs gives A(αs)=σs(αs+c αt)=(c2−1)αs+c αt and A(αt)=σs(−αt)=−c αs−αt, so A(P)⊆P and the matrix of B=A∣P in the basis (αs,αt) is (c2−1−cc−1), whose trace is (c2−1)+(−1)=c2−2 and whose determinant is (c2−1)(−1)−(−c)c=1. For u∉{s,t} one has A(αu)=σs(αu+ctuαt)=αu+(csu+ctuc)αs+ctuαt, so A(αu)−αu∈P; by linearity A(v)−v∈P for every v∈E.

2.1F3F4step 1.2algebra

The two-block recursion. The subspace Q=span⁡{αu:u∉{s,t}} satisfies P∩Q={0} and P+Q=E because the basis of E splits into the basis (αs,αt) of P and the basis of Q; hence E=P⊕Q, and every v∈E has unique coordinates v=p+q with p∈P, q∈Q. I claim that for every k≥1 and all such p,q one has Ak(p+q)=Bkp+Sk(A(q)−q)+q, where Sk:=idP+B+⋯+Bk−1. For k=1 this reads A(p)+A(q)=Bp+(A(q)−q)+q, which holds because A∣P=B. If it holds for k, then Bkp+Sk(A(q)−q)∈P and A(q)−q∈P by step 1.2, so applying A and using A∣P=B gives Ak+1(p+q)=B(Bkp+Sk(A(q)−q))+A(q)=Bk+1p+(BSk+idP)(A(q)−q)+q, and BSk+idP=B+B2+⋯+Bk+idP=Sk+1, which completes the induction.

2.2F2step 1.2algebra

The cases m=2 and m=∞. If m=2, then ζ has order 4, so ζ2=−1 and ζ−1=−ζ, whence c=ζ+ζ−1=0; substituting into the matrix of step 1.2 leaves B=(−100−1)=−idP, so B2=idP and idP+B=0. If m=∞, then c=2 by the convention in [F2], and the matrix of step 1.2 gives B=idP+M with M=B−idP=(2−22−2)≠0 and M2=0. The binomial expansion in the commutative algebra of endomorphisms then gives Bk=(idP+M)k=idP+kM for every k≥1, and kM≠0 because M≠0 and k⋅1K≠0 in the characteristic-zero field K; hence Bk≠idP for all k≥1, and in particular B≠idP.

2.3F5F6F7F8step 1.2algebra

The finite case 2<m<∞. By step 1.2 and [F5] the characteristic polynomial of B is χB(x)=x2−(c2−2)x+1; since c=ζ+ζ−1 satisfies c2−2=ζ2+ζ−2 and ζ2ζ−2=1, one has χB(x)=x2−(ζ2+ζ−2)x+1=(x−ζ2)(x−ζ−2). The two roots are distinct: ζ2=ζ−2 would give ζ4=1, impossible because ζ has order 2m>4. Hence the characteristic polynomial splits into distinct linear factors, and [F6] makes B diagonalisable, with eigenvalues ζ2 and ζ−2. Consequently Bm=idP, because on an eigenvector for λ∈{ζ2,ζ−2} the operator Bm scales it by λm=ζ±2m=1 by [F7] and [F8]; the operator Sm=idP+B+⋯+Bm−1 is zero, because the polynomial 1+x+⋯+xm−1 vanishes at both eigenvalues, where ζ±2m=1 and ζ±2≠1 since 0<2<2m; and Bk≠idP for 0<k<m, because on the eigenvector for ζ2 it scales by ζ2k≠1, again since 0<2k<2m and 2m is the least positive exponent killing ζ.

3.1F1F2F7step 1.1step 2.1step 2.2step 2.3

Order of A, the relators, and the exact dihedral orders. For any distinct pair s,t with finite m=m(s,t), step 2.1 gives Am(p+q)=Bmp+Sm(A(q)−q)+q=p+q for all p,q, using Bm=idP and Sm=0 from steps 2.2 and 2.3; hence Am=idE. If 0<k<m, choose p∈P with Bkp≠p, which exists because Bk≠idP; then Ak(p)=Bkp≠p, so Ak≠idE and A has order exactly m. For m=∞, Bk≠idP for every k≥1 gives Ak≠idE for every k≥1, so A has infinite order. Since this verifies each arbitrary distinct finite pair and step 1.1 proves σs2=idE for every generator, the map f:S→GLK(E), f(s)=σs, satisfies the hypotheses of the universal property in [F1], so there is a unique homomorphism σ:W→GLK(E) with σ(s)=σs for all s. For #S≤1 there are no distinct-pair relators, so the same universal property applies using only step 1.1; for S=∅ it gives the unique homomorphism from the trivial group. For every s∈S, step 1.1 now gives σ(s)≠idE, hence s≠1 and ℓ(s)≥1 by [F1]; the one-letter expression gives ℓ(s)≤1, so ℓ(s)=1. If s≠t then σs≠σt, because σs(αs)=−αs while σt(αs)=αs+cstαt and −2αs−cstαt≠0 by linear independence of αs,αt and 2≠0 in K; hence s≠t in W. Finally, st has order exactly m when m<∞: writing n for the order of st in W, one has n≤m because (st)m=1 in W and n is least, and m≤n because An=σ((st)n)=idE and m is the least positive exponent killing A; hence n=m. When m=∞, a positive exponent k with (st)k=1 would give Ak=idE, so no such exponent exists and st has infinite order.

4.1F1F7step 3.1

The signed action. First, Us is a bijection of {±1}×T with Us2=id: for r=s one has Us(ε,s)=(−ε,s) and applying Us again gives (+ε,s), while for r≠s one has srs≠s — if srs=s then multiplying on the left by s gives rs=1 and hence r=s−1=s using s2=1 in W — so δ(s,srs)=0 and the two applications of Us return (ε,r) from Us(ε,srs)=(ε,r). Next, fix a word (s1,…,sk) in S with wj:=s1⋯sj and prefix reflections ri:=wi−1siwi−1−1; applying Us1,Us2,…,Usk successively to (ε,r) returns (ε(−1)n(r), wk−1rwk) with n(r):=#{i:ri=r}, by induction on k: the i-th step multiplies the sign by (−1)δ(si, wi−1−1rwi−1), whose exponent is 1 exactly when r=wi−1siwi−1−1=ri, and it conjugates the coordinate to siwi−1−1rwi−1si=(wi−1si)−1r(wi−1si). Now fix any distinct a,b∈S with m=m(a,b)<∞; consider the alternating word (a,b,a,b,… ) of length 2m and write ri=(ab)i−1a for 1≤i≤2m for its prefix reflections. The closed form is proved by induction on i: the conjugating prefix is wi−1=(ab)(i−1)/2 for odd i and wi−1=(ab)(i−2)/2a for even i, and the identities (ab)−j=(ba)j and a(ba)j=(ab)ja rewrite wi−1siwi−1−1 as (ab)i−1a in both cases. Hence ri+m=(ab)i−1(ab)ma=(ab)i−1a=ri for 1≤i≤m, while r1,…,rm are pairwise distinct: an equality ri=rj with i<j≤m gives (ab)j−i=1 after right multiplication by a, contradicting 0<j−i<m and the fact from step 3.1 that ab has order exactly m. Therefore the sequence of 2m prefix reflections is r1,…,rm,r1,…,rm, so n(r)∈{0,2} for every r∈T and every accumulated sign (−1)n(r) is 1; the conjugating coordinate returns to r because w2m=(ab)m=1. So the permutation effected by each alternating word of length 2m is the identity, which verifies (UaUb)m=id and (UbUa)m=id for this arbitrary finite pair. There are no distinct-pair relators when #S≤1, so the same verification applies in those cases using only Us2=id. By [F1]'s universal property applied to f:S→Sym⁡({±1}×T), f(s)=Us, there is then a homomorphism ρ:W→Sym⁡({±1}×T) with ρ(s)=Us for all s. Define (ε,r)⋅w:=ρ(w−1)(ε,r); this is a right action, because (wv)−1=v−1w−1 gives (ε,r)⋅(wv)=ρ(v−1)(ρ(w−1)(ε,r))=((ε,r)⋅w)⋅v, and (ε,r)⋅s=ρ(s)−1(ε,r)=Us(ε,r).

5.1step 4.1algebra

Deletion. Suppose ri=rj for some i<j, that is wi−1siwi−1−1=wj−1sjwj−1−1. Multiplying on the left by wi−1−1 and on the right by wj−1 and using wi−1−1wj−1=sisi+1⋯sj−1 gives si sisi+1⋯sj−1=sisi+1⋯sj−1sj, hence sisi+1⋯sj=si+1⋯sj−1. Substituting this into the word (s1,…,sk) shows that deleting the two letters si and sj leaves the value unchanged: s1⋯si^⋯sj^⋯sk=w.

6.1F1step 4.1step 5.1

Expression independence and reduced words. For a word of w the sign accumulated in step 4.1 is (−1)n(r), and the right action (ε,r)⋅w is independent of the word; comparing with the displayed formula for two words of w shows that (−1)n(r)=η(r,w) depends only on w and r, and that (ε,r)⋅w=(ε η(r,w), w−1rw) for every word of w. If the word (s1,…,sk) is reduced, then n(r)≤1 for every r: if n(r)≥2, two indices would have ri=rj with i<j, and step 5.1 would produce a word of length k−2 for w, contradicting k=ℓ(w). Consequently the map i↦ri is injective, so the set {r1,…,rk}={r∈T:η(r,w)=−1} has exactly k=ℓ(w) elements; by the independence of η from the word, this set is independent of the reduced expression chosen.

7.1F1F7step 3.1step 4.1step 6.1∎

Ambient reducedness of alternating words. Let s≠t, m:=m(s,t), and let wq be the value of the alternating word (s,t,s,… ) of length q beginning with s, where q≤m if m<∞ and q≥1 is arbitrary if m=∞. Its prefix reflections are ri=(st)i−1s by the closed form of step 4.1. They are pairwise distinct: an equality ri=rj with i<j≤q gives (st)j−i=1 after right multiplication by s, which is impossible when m<∞ because then 0<j−i<q≤m and m is the least positive exponent killing st by step 3.1 and [F7], and equally impossible when m=∞ because by step 3.1 no positive power of st is 1. Hence exactly q elements r satisfy η(r,wq)=−1, namely the ri; on the other hand, for any word of length k in S representing wq [F1], step 6.1 gives #{r:η(r,wq)=−1}=#{r:n′(r) odd}≤∑rn′(r)=k, where n′(r) counts the occurrences of r among that word's prefix reflections. Therefore k≥q for every word for wq, so ℓ(wq)=q; in particular the values w1,…,wq are pairwise distinct, having distinct lengths.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

Length parity, exchange, two-letter deletion, and faithfulness of the signed reflection action

Statement

Let (S,m) be a finite Coxeter matrix, W the presented group, ℓ its length function, and let T, η, Φ and the right action of W on {±1}×T be as in The rank-two block computation, exact dihedral orders, the signed reflection action, and ambient reducedness.

  1. Sign character and parity. There is a unique homomorphism sgn⁡:W→{±1} with sgn⁡(s)=−1 for all s∈S, and sgn⁡(w)=(−1)ℓ(w) for all w∈W. Consequently, for all w∈W and s∈S, ℓ(sw)=ℓ(w)±1,ℓ(ws)=ℓ(w)±1, with ℓ(sw)≡ℓ(w)+1(mod2) and ℓ(ws)≡ℓ(w)+1(mod2).
  2. Exchange. Let w=s1⋯sk be a reduced expression and let s∈S satisfy ℓ(sw)=k−1. Then sw=s1⋯si^⋯sk for some i∈{1,…,k}; equivalently, w has a reduced expression beginning with s, and s is a prefix reflection of any reduced expression of w. Right-handed form: if ℓ(ws)=k−1 then ws=s1⋯si^⋯sk for some i.
  3. Deletion. If the word (s1,…,sk) in S is not reduced, then there are i<j with s1⋯si^⋯sj^⋯sk=s1⋯sk. Hence repeated deletion of two letters transforms every word into a reduced expression for the same element, and a word is reduced if and only if it cannot be shortened by deleting two letters.
  4. Faithfulness of the signed action. The right action of W on {±1}×T is faithful, so W embeds in Sym⁡({±1}×T); in particular distinct simple generators are distinct in W.

Facts & Assumptions

Given: A finite Coxeter matrix (S,m), the presented group W with its length ℓ, the reflection set T and the right action of W on {±1}×T of The rank-two block computation, exact dihedral orders, the signed reflection action, and ambient reducedness, and a word (s1,…,sk) in S in each claim below.

[F1]

Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups: W is presented by (S,m) with relators s2 and (st)m(s,t); for every group G and every map f:S→G with f(s)2=1 and (f(s)f(t))m(s,t)=1 whenever m(s,t)<∞, there is a unique homomorphism W→G with s↦f(s). The length ℓ(w) is the least k with w=s1⋯sk, and ℓ(1)=0, the empty word being the only word of length 0.

[F2]

The rank-two block computation, exact dihedral orders, the signed reflection action, and ambient reducedness: the right action of W on {±1}×T defined by Us(ε,r)=(ε(−1)δ(s,r),srs) satisfies (ε,r)⋅w=(ε η(r,w),w−1rw) with η(r,w)=(−1)n(r) depending only on w and r; T={wsw−1:w∈W, s∈S}; for a reduced expression w=s1⋯sk with prefix reflections ri=s1⋯si−1sisi−1⋯s1 one has n(r)∈{0,1} for every r, the map i↦ri is injective, and Φ(w):={r1,…,rk}={r∈T:η(r,w)=−1} is independent of the reduced expression, with #Φ(w)=ℓ(w); also ℓ(s)=1 for every s∈S.

[F3]

Monoid homomorphism and group homomorphism: a group homomorphism satisfies φ(uv)=φ(u)φ(v) and φ(1)=1, so for sgn⁡ one has sgn⁡(sw)=sgn⁡(s)sgn⁡(w).

Proof

Given: A finite Coxeter matrix (S,m), the group W and length ℓ, and the right action of W on {±1}×T with its function η and sets Φ(w).

1.1F1F3algebra

The sign character and the parity laws. The map f:S→{±1}, f(s)=−1, satisfies f(s)2=1 and (f(s)f(t))m(s,t)=(−1)2m(s,t)=1 for every finite edge, so the universal property in [F1] gives a unique homomorphism sgn⁡:W→{±1} with sgn⁡(s)=−1. For any word w=s1⋯sk this gives sgn⁡(w)=(−1)k; taking a word of length ℓ(w) shows sgn⁡(w)=(−1)ℓ(w), so every word for w has length congruent to ℓ(w) modulo 2. Next, ℓ(sw)≤ℓ(w)+1: a word of length ℓ(w) for w prefixed by s is a word of length ℓ(w)+1 for sw, and ℓ is a minimum; symmetrically ℓ(w)=ℓ(s⋅sw)≤ℓ(sw)+1. Since sgn⁡(sw)=sgn⁡(s)sgn⁡(w)=−sgn⁡(w) by [F3] and (−1)ℓ(sw)=sgn⁡(sw), the parities of ℓ(sw) and ℓ(w) are opposite, so ℓ(sw)≠ℓ(w); with the two inequalities this forces ℓ(sw)=ℓ(w)±1, and the congruence ℓ(sw)≡ℓ(w)+1 records the parity. The same argument with ws in place of sw, using ℓ(ws)≤ℓ(w)+1 and ℓ(w)=ℓ(ws⋅s)≤ℓ(ws)+1, gives ℓ(ws)=ℓ(w)±1 and ℓ(ws)≡ℓ(w)+1 modulo 2.

1.2F1F2

Faithfulness. Let w≠1. Then ℓ(w)≠0 because the only word of length 0 is the empty word with value 1 by [F1], so ℓ(w)≥1; choose a reduced expression w=s1⋯sk with k=ℓ(w)≥1. By [F2] the set Φ(w)={r1,…,rk} has #Φ(w)=ℓ(w)≥1, so pick r∈Φ(w), that is η(r,w)=−1. The action formula of [F2] then gives (1,r)⋅w=(η(r,w)⋅1, w−1rw)=(−1,w−1rw)≠(1,w−1rw), so w acts nontrivially on {±1}×T; hence the action is faithful and W embeds in Sym⁡({±1}×T). In particular, for s≠t in S the elements s,t have distinct images under this embedding because Us(1,s)=(−1,s) while Ut(1,s)=(1,tst), and the first coordinates −1 and 1 differ.

1.3F1F2algebra

Exchange. Let w=s1⋯sk be reduced and let ℓ(sw)=k−1. Choose a reduced expression sw=t1⋯tk−1; then w=s t1⋯tk−1 is a word of length k=ℓ(w), hence a reduced expression of w whose first prefix reflection is r1=s, so η(s,w)=−1 and s∈Φ(w) by [F2]. By the expression-independence of Φ(w) in [F2] applied to the reduced expression w=s1⋯sk, there is i∈{1,…,k} with s=ri=wi−1siwi−1−1, where wi−1=s1⋯si−1; multiplying this identity on the right by w=wi−1sisi+1⋯sk gives sw=wi−1si+1⋯sk=s1⋯si^⋯sk, which is the asserted deletion. For the right-handed form, note first that ℓ(u−1)=ℓ(u) for every u: reversing a reduced word for u gives a word of the same length for u−1, so ℓ(u−1)≤ℓ(u), and applying this to u−1 gives equality. If now ℓ(ws)=k−1, then w−1=sk⋯s1 is a reduced expression of w−1 and ℓ(s w−1)=ℓ((ws)−1)=ℓ(ws)=k−1, so the left-handed form applied to w−1 writes s w−1=sk⋯sj^⋯s1 for some j; inverting both sides gives ws=(s w−1)−1=s1⋯sj^⋯sk.

2.1F1step 1.1step 1.3algebra∎

Deletion. Let (s1,…,sk) be a word that is not reduced, and let j be the least index such that the prefix (s1,…,sj) is not reduced; such j exists and j≥2, and (s1,…,sj−1) is reduced with value u:=s1⋯sj−1. By the minimality of j the element usj has ℓ(usj)<j=ℓ(u)+1, so ℓ(usj)=ℓ(u)−1=j−2 by step 1.1, and the right-handed exchange of step 1.3 applied to the reduced expression (s1,…,sj−1) and the letter sj gives usj=s1⋯si^⋯sj−1 for some i≤j−1. Therefore s1⋯si^⋯sj^⋯sk=(s1⋯si^⋯sj−1) sj+1⋯sk=usjsj+1⋯sk=s1⋯sk, so deleting the letters at positions i and j leaves the value unchanged. Iterating, the length strictly decreases by 2 at each deletion and stops at length ℓ(w), leaving a reduced expression of the same element. Conversely, a reduced word cannot be shortened by deleting two letters, since the deleted word is a strictly shorter word for the same element.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

Matsumoto's theorem: braid connectivity of reduced expressions, with singleton detection in dihedral subgroups

Statement

Let (S,m) be a finite Coxeter matrix; let W and ℓ be the presented group and its length and let σ:W→GLK(E) be the geometric representation, as in Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups, The geometric representation on the simple-root basis over a common splitting field, and the root set and The rank-two block computation, exact dihedral orders, the signed reflection action, and ambient reducedness, with reflection set T and sign-change sets Φ(w) as in that lemma. Parts (1) and (2) hold for every finite S, including the empty and singleton cases; a distinct pair is required only for part (3).

  1. Braid moves. Call two words in S braid-equivalent when one is obtained from the other by finitely many replacements of an alternating subword s t s t⋯ of length m(s,t)<∞ by the alternating word t s t s⋯ of the same length. Then any two reduced expressions of the same element w∈W are braid-equivalent.
  2. M-reducedness. A word is reduced if and only if no sequence of braid moves followed by cancellation of a consecutive equal pair can shorten it (M-reduced in the sense of [Davis, Definition 3.4.1]).
  3. Singleton detection in dihedral subgroups. Fix distinct s,t∈S and put m:=m(s,t). The subgroup ⟨s,t⟩≤W is dihedral of order 2m when m<∞ and infinite dihedral when m=∞; every element of ⟨s,t⟩ acts on the quotient E/span⁡(αs,αt) as the identity, while for every u∈S∖{s,t} one has σu(αu)−αu=−2αu∉span⁡(αs,αt) (no induced quotient map for σu is assumed). Consequently S∩⟨s,t⟩={s,t}, and every element of ⟨s,t⟩ has a reduced expression with all letters in {s,t} (so the alternating words of length q≤m, and any q when m=∞, are reduced in W).

Facts & Assumptions

Given: A finite Coxeter matrix (S,m), the group W with length ℓ, the representation σ:W→GLK(E) on the space with basis (αs)s∈S, the reflection set T and the right action of W on {±1}×T with its function η and sets Φ(w); distinct s,t∈S and m:=m(s,t) are fixed only for the rank-two assertions.

[F1]

Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups: W is presented by (S,m) with relators s2 and (st)m(s,t); for every group G and every map f:S→G with f(s)2=1 and (f(s)f(t))m(s,t)=1 whenever m(s,t)<∞, there is a unique homomorphism W→G with s↦f(s). The length ℓ(w) is the least k with w=s1⋯sk, and a word of length ℓ(w) for w is a reduced expression of w.

[F2]

The geometric representation on the simple-root basis over a common splitting field, and the root set: K is a field of characteristic 0; E has basis (αs)s∈S; σs is the unique K-linear map with σs(αs)=−αs and σs(αu)=αu+csuαs for u≠s, where csu=ζsu+ζsu−1 for m(s,u)<∞ and csu=2 for m(s,u)=∞.

[F3]

The rank-two block computation, exact dihedral orders, the signed reflection action, and ambient reducedness: the assignment s↦σs induces a homomorphism σ:W→GLK(E); st has order exactly m(s,t) in W (infinite when m(s,t)=∞); the right action satisfies (ε,r)⋅w=(εη(r,w),w−1rw) with η(r,w) depending only on w and r; for a reduced word with prefix reflections ri=s1⋯si−1sisi−1⋯s1 one has n(r)∈{0,1} for all r and Φ(w)={r1,…,rk}={r:η(r,w)=−1} independent of the reduced expression, of cardinality ℓ(w); the prefix reflections of the alternating word (a,b,a,b,… ) with a=s1,b=s2 are ri=(ab)i−1a; and an alternating word of length q≤m, or of any length when m=∞, is reduced in W, its value having length q.

[F4]

Length parity, exchange, two-letter deletion, and faithfulness of the signed reflection action: for all w∈W and s∈S one has ℓ(sw)=ℓ(w)±1 and ℓ(ws)=ℓ(w)±1; if w=s1⋯sk is reduced and ℓ(sw)=k−1 then sw=s1⋯si^⋯sk for some i, and if ℓ(ws)=k−1 then ws=s1⋯si^⋯sk for some i; and a word is reduced if and only if it cannot be shortened by deleting two letters.

Proof

Given: The data of the statement; a distinct pair is fixed only in the rank-two arguments of steps 1.1 and 1.2.

Proof technique: strong induction on q=ℓ(w) for part (1), with the exchange condition; part (3) is a direct computation with the dihedral subgroup, and part (2) is an induction on the word length using part (1).

1.1F2F3algebra

The dihedral subgroup and the singleton claim. Fix any distinct s,t∈S, put m:=m(s,t) and r:=st. By [F3], r has order exactly m when m<∞ and infinite order when m=∞, and srs=s(st)s=(ss)(ts)=ts=r−1; hence srk=r−ks for every k by induction on k. It follows that D:=⟨s,t⟩ is exactly the set D0:={rk:k∈Z}∪{rks:k∈Z}: indeed D0 contains s=r0s and t=r−1s (as r−1s=(ts)s=t), and D0 is closed under inverses (the inverse of rk is r−k and (rks)2=1) and under multiplication, since ra⋅rb=ra+b, ra⋅rbs=ra+bs, ras⋅rb=ra(srb)=rar−bs=ra−bs and ras⋅rbs=ra(srb)s=rar−bss=ra−b; so D0 is a subgroup containing s and t, while every product of copies of s,t lies in D0 by this closure, whence D=D0. The elements rk and rks with 0≤k<m (all k∈Z when m=∞) are pairwise distinct: rk=rl with 0≤k<l<m gives rl−k=1, impossible by minimality of the order m, while for m=∞ the powers rk with k∈Z are distinct because r has infinite order; and rk=rls gives rk−l=s, so s∈⟨r⟩, which cannot happen because s∈⟨r⟩ would make s commute with r, giving t=sr=rs=sts, hence st=ts and r=r−1; then r2=1, so the order of r divides 2, an impossibility unless m=2, and in the case m=2 one has ⟨r⟩={1,r} with s≠1 and s≠r=st (the latter would give t=1). Hence D has exactly 2m elements and is dihedral when m<∞, and is infinite dihedral when m=∞. Next, P:=span⁡(αs,αt) is preserved by σs and σt, and σs(v)−v∈P and σt(v)−v∈P for every v∈E: on the basis, σs(αs)−αs=−2αs, σs(αt)−αt=cstαs and σs(αu)−αu=csuαs for u∉{s,t}, and symmetrically for σt. So for every product g of copies of σs,σt one has g(v)−v∈P for all v by induction on the number of factors, and therefore every element of D=⟨s,t⟩ acts as the identity on E/P. If u∈S∖{s,t}, then αu∉P because the basis is linearly independent, while σu(αu)=−αu, so σu(αu)−αu=−2αu∉P. Thus σu fails the congruence σu(v)−v∈P satisfied by every element of D, without requiring σu to preserve P or induce a quotient map; since σ is a homomorphism, u∉D. Hence S∩D={s,t}. Finally, every element of D has a reduced expression with letters in {s,t}: for 0≤k<m the element rk equals both (st)k and (ts)m−k=rk−m, alternating words of lengths 2k and 2(m−k) whose minimum is at most m, and the element rks equals both (st)ks and (ts)m−k−1t (for k=0 the first is the single letter s), alternating words of lengths 2k+1 and 2(m−k)−1 whose minimum is at most m; an alternating word of length at most m is reduced in W by [F3], so the shorter of the two is a reduced expression of the element in the letters {s,t}. When m=∞ every alternating word is reduced by [F3], and rk with k<0 is written as (ts)−k and rks with k<0 as (ts)−k−1t.

1.2F1F3algebra

Braid moves preserve the value. Fix distinct s,t∈S with m=m(s,t)<∞ and let Lm and Lm′ be the two alternating words of length m in {s,t}, beginning with s and with t. With r:=st of order m by [F3], one has Lm=(st)m/2=rm/2 and Lm′=(ts)m/2=r−m/2 when m is even, so their values coincide; and Lm=(st)(m−1)/2s=r(m−1)/2s while Lm′=t(st)(m−1)/2=(ts)(m−1)/2t=r−(m−1)/2t=r−(m−1)/2r−1s=r−(m+1)/2s=r(m−1)/2s when m is odd, using ts=r−1, t=r−1s and rm=1. Hence replacing an alternating subword of length m by the other alternating word of the same length leaves the value in W unchanged, and so does a finite sequence of such replacements.

1.3F1F3F4

The reduction step. Let q≥2 and let a=(a1,…,aq) and b=(b1,…,bq) be reduced words for the same element w∈W, with a1≠b1; write a:=a1, b:=b1, and assume as induction hypothesis that part (1) holds for all elements of length at most q−1. Since b is reduced, bw=b2⋯bq has length q−1, so the exchange condition of [F4] applied to the reduced word a and the letter b gives bw=a1⋯ai^⋯aq for some i∈{1,…,q}, that is w=b a1⋯ai^⋯aq. Multiplying the equality bw=a1⋯ai^⋯aq on the right by aqaq−1⋯ai+1 and then by ai shows that it is equivalent to b a1⋯ai−1=a1⋯ai; in particular i≠1, since i=1 would give b=a1=a, contrary to a≠b. Hence i≥2, and the word (b,a1,…,ai^,…,aq) is a reduced expression of w. Its tail (a1,…,ai^,…,aq) and the suffix (b2,…,bq) are reduced expressions of the same element bw of length q−1, so by the induction hypothesis they are braid-equivalent, and prepending the letter b to both words gives the braid-equivalence of (b,a1,…,ai^,…,aq) with (b,b2,…,bq)=b. If i<q, multiplying the equality b a1⋯ai−1=a1⋯ai on the right by ai+1⋯aq−1 shows that the words (b,a1,…,ai^,…,aq−1) and (a1,…,aq−1) are reduced expressions of the same element a1⋯aq−1 of length q−1, so by the induction hypothesis they are braid-equivalent, and appending the letter aq to both gives the braid-equivalence of (b,a1,…,ai^,…,aq) with a; combining the two braid-equivalences, a is braid-equivalent to b. Thus either a and b are braid-equivalent, or i=q, in which case w=b a1⋯aq−1 and (b,a1,…,aq−1) is braid-equivalent to b.

2.1F3step 1.3

The iteration. Keep the setup of step 1.3 and assume now that the two words are not braid-equivalent. For j∈{1,…,q} let σ(j) be the q-letter word consisting of the alternating word of length q−j+1 ending in a followed by the letters a2,a3,…,aj. Thus σ(q)=(a1,…,aq)=a and σ(q−1)=(b,a,a2,…,aq−1); let Cj denote the braid class of σ(j) when σ(j) is reduced. I claim that for every p∈{0,…,q−2}: either a and b are braid-equivalent, or σ(j) is reduced, represents w, and Cj is the class of a when j−q is even and the class of b when j−q is odd, for every j∈{q−p−1,…,q}. For p=0 this is step 1.3: either the words are braid-equivalent, or w=b a1⋯aq−1 and the word σ(q−1)=(b,a1,…,aq−1) is braid-equivalent to b, while σ(q)=a, which is the assertion for j=q−1,q. For the induction step, suppose the claim known for p−1, so that σ(q−p) and σ(q−p+1) are reduced words for w; apply step 1.3 to this pair of words. If they are braid-equivalent, then their classes coincide and, since q−p and q−p+1 have opposite parity, the known class rule forces the classes of a and b to coincide, and we are done. Otherwise the second alternative of step 1.3 holds for the pair, which says that deleting the last letter of σ(q−p) and prepending the first letter of σ(q−p+1) yields a word that is reduced, represents w, and is braid-equivalent to σ(q−p+1). Now σ(q−p) is the alternating word of length p+1 ending in a followed by a2,…,aq−p, so deleting its last letter leaves the alternating word of length p+1 ending in a followed by a2,…,aq−p−1; the first letter of σ(q−p+1) is the first letter of the alternating word of length p ending in a, which differs from the first letter of the alternating word of length p+1 ending in a, so prepending it produces the alternating word of length p+2 ending in a followed by a2,…,aq−p−1, which is exactly σ(q−p−1). Hence σ(q−p−1) is reduced, represents w, and lies in the class of σ(q−p+1); since q−p−1 and q−p+1 have the same parity, this agrees with the claimed class rule and the claim holds for p.

3.1F3step 1.1step 2.1

The final phase. Assume the claim of step 2.1 with p=q−2, and assume that a and b are not braid-equivalent; then every σ(j) for j∈{1,…,q} is a reduced word for w whose class is the class of a when j−q is even and the class of b when j−q is odd. In particular σ(1) is the alternating word of length q ending in a, and σ(2) is the alternating word of length q−1 ending in a followed by the single letter a2. Write v for the value of the alternating word of length q−1 ending in a and ε for the letter with which the alternating word of length q ending in a begins; then the word σ(1) has value εv and σ(2) has value va2, and since both represent w one has εv=va2, that is a2=v−1εv. So a2 is conjugate to ε∈{a,b} by the element v, which lies in the dihedral subgroup ⟨a,b⟩; hence a2∈⟨a,b⟩, and since a2∈S, step 1.1 gives a2∈{a,b}. The letter a2 cannot be a: the word σ(2) would then end in two equal letters a,a, and deleting that pair would express w by a word shorter than q=ℓ(w), so σ(2) would not be reduced. Hence a2=b, so σ(2) is the alternating word of length q ending in b; of the two alternating words of length q in {a,b}, σ(1) ends in a and σ(2) ends in b. To see that this forces q=m, compare the values of σ(1) and σ(2) in the dihedral group ⟨a,b⟩, whose elements are the powers of r=ab; if q is even these values are (r−1)q/2 and rq/2, equal exactly when rq=1, and if q is odd they are r(q−1)/2a and r−(q−1)/2b, equal exactly when rq−1=r−1, that is rq=1. Since σ(1) and σ(2) represent the same element, rq=1; when m=∞ no positive power of r is 1 by [F3], so this alternative cannot occur and m<∞, and then by minimality of the order m one has q≥m. Conversely q≤m: by [F3] the prefix reflections of the reduced word σ(1) are pairwise distinct, while the closed form of [F3] for alternating words gives ri+m=ri for these reflections (here (ab)m=1 because m<∞), so q>m would make the reflections at positions 1 and 1+m coincide, a contradiction. Hence q=m. Finally, the two words σ(1) and σ(2) are the two alternating words of length m, so one is obtained from the other by the single replacement of the alternating block of length m by the other alternating word of the same length; they are braid-equivalent, hence in the same class, but by the class rule their classes are the classes of a and of b respectively, so a and b are braid-equivalent, against the assumption.

4.1F1F3F4step 1.2step 3.1∎

Conclusion. Part (1) follows by strong induction on q=ℓ(w): for q≤1 two reduced words for w coincide, and for q≥2, reduced words with the same first letter are handled by the induction hypothesis applied to their tails, while reduced words with different first letters are handled by steps 1.3, 2.1 and 3.1, which produce braid-equivalence in every case. For part (2), let c=(c1,…,ck) be a word with value w. If c is reduced, no sequence of braid moves and deletions of consecutive equal pairs can shorten it: braid moves preserve the value and the length by step 1.2, deleting a consecutive pair preserves the value, and a shorter word for w would contradict k=ℓ(w). Conversely, if c is not reduced, we show by induction on k that it can be shortened; for k≤1 the word is reduced, so k≥2, and if the suffix (c2,…,ck) can be shortened, the same sequence shortens the whole word. Otherwise the suffix is M-reduced, so by induction on k it is reduced, with value w′ and ℓ(w′)=k−1. Since w=c1w′ has length at most k−1<ℓ(w′)+1, the length laws of [F4] give ℓ(c1w′)=ℓ(w′)−1=k−2, and the exchange condition of [F4] applied to the reduced word (c2,…,ck) and the letter c1 gives c1w′=c2⋯cj^⋯ck for some j, so that w′=c1c2⋯cj^⋯ck is a reduced expression of w′ of length k−1 beginning with c1. By part (1) it is braid-equivalent to (c2,…,ck), and prepending c1 to both words exhibits a sequence of braid moves taking c=(c1,c2,…,ck) to (c1,c1,c2,…,cj^,…,ck), which the deletion of the consecutive pair c1,c1 shortens. Hence a word is reduced if and only if it is M-reduced.

Remarks

The argument for part (1) is the induction of [Lusztig, Theorem 1.9] with its intermediate statements rendered explicitly: step 1.3 is the one-step reduction, step 2.1 is the parameterized family of intermediate words, and step 3.1 is the final dihedral case, in which the two alternating words of length q in the letters a,b are reduced, coincide in value exactly when q=m, and differ by one braid move. Part (2) is [Davis, Theorem 3.4.2(i)] and [Davis, Definition 3.4.1]; the singleton claim S∩⟨s,t⟩={s,t} is the step used at the end of [Lusztig, Theorem 1.9]. Neither root positivity nor geometric faithfulness is used.

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Support, intrinsic parabolic presentations, minimal coset representatives and length additivity, with the type-A identification

Statement

Let (S,m) be a finite Coxeter matrix with presented group W, length ℓ and geometric representation σ:W→GLK(E) as in Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups, The geometric representation on the simple-root basis over a common splitting field, and the root set and The rank-two block computation, exact dihedral orders, the signed reflection action, and ambient reducedness, and let J⊆S.

  1. Support and reduction. For every w∈W the set S(w) of letters occurring in a reduced expression of w is independent of the reduced expression, and every word in S representing w can be transformed into a reduced expression by repeatedly deleting two letters (Tits reduction, using only letters already present). Consequently WJ=⟨J⟩={w∈W:S(w)⊆J}.
  2. Intrinsic parabolic presentation. Let WJ∗ be the group presented by the restricted Coxeter matrix (J,m∣J) in the sense of Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups. The canonical homomorphism WJ∗→WJ, s↦s, is an isomorphism. Hence (WJ,J) is a Coxeter system, its intrinsic length function ℓJ agrees with the ambient length ℓ on WJ, and WJ∩S=J.
  3. Minimal coset representatives. Every right coset WJa:={ua:u∈WJ} (a∈W, Left and right cosets gH and Hg of a subgroup) has a unique element d of minimal length; it is characterized by ℓ(sd)>ℓ(d) for all s∈J, and it satisfies ℓ(ud)=ℓ(u)+ℓ(d)for all u∈WJ. Equivalently, every w∈W has a unique factorization w=u d with u∈WJ and d the minimal representative of the right coset WJd, and then ℓ(w)=ℓ(u)+ℓ(d). By inversion (w↦w−1 preserves lengths and interchanges the two coset families {WJa} and {aWJ}), every left coset aWJ:={au:u∈WJ} has a unique minimal element d, characterized by ℓ(ds)>ℓ(d) for all s∈J and satisfying ℓ(du)=ℓ(d)+ℓ(u) for all u∈WJ.
  4. Type A. Let n≥2, S={s1,…,sn−1} and m(si,sj):=3 if ∣i−j∣=1, m(si,sj):=2 if ∣i−j∣>1 (a Coxeter matrix of type An−1). Then si↦(i i+1) extends to an isomorphism W→Sn (the letters 1,…,n carry the library's symmetric group by the order-preserving identification with {0,…,n−1}, under which (i i+1) is the adjacent transposition (i−1 i)), and for every w∈W, ℓ(w)=inv⁡(φ(w)), the inversion number of the corresponding permutation (Inversions, inversion number, the sign sgn⁡(σ)=(−1)inv⁡(σ), and even and odd permutations). In particular a word in the si is reduced if and only if its length equals the inversion number of its value.

Facts & Assumptions

Given: A finite Coxeter matrix (S,m), the group W with its length function ℓ, the geometric representation σ:W→GLK(E), and a subset J⊆S for parts (1)-(3); the type-A matrix and data of part (4).

[F1]

Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups: W=F(S)/N is the group presented by (S,m) with relators s2 (s∈S) and (st)m(s,t) (s≠t, m(s,t)<∞); for every group G and every map f:S→G with f(s)2=1 and (f(s)f(t))m(s,t)=1 whenever m(s,t)<∞ there is a unique homomorphism W→G with s↦f(s). The length ℓ(w) is the least length of a word in S representing w, and ℓ(1)=0; for J⊆S, WJ=⟨{s:s∈J}⟩.

[F2]

Length parity, exchange, two-letter deletion, and faithfulness of the signed reflection action: for all w∈W, s∈S one has ℓ(sw)=ℓ(w)±1 and ℓ(ws)=ℓ(w)±1; if w=s1⋯sk is reduced and ℓ(sw)=k−1 then sw=s1⋯si^⋯sk for some i; and a word is reduced if and only if it cannot be shortened by deleting two letters, that is, any non-reduced word s1⋯sk has s1⋯si^⋯sj^⋯sk=s1⋯sk for some i<j.

[F3]

Matsumoto's theorem: braid connectivity of reduced expressions, with singleton detection in dihedral subgroups: any two reduced expressions of the same element are braid-equivalent, where a braid move replaces an alternating subword of length m(s,t)<∞ by the alternating word of the same length with the two letters interchanged; and a word is reduced if and only if it is M-reduced, that is, cannot be shortened by a sequence of braid moves and cancellations of consecutive equal pairs.

[F4]

The finite symmetric group Sn, one-line notation, and cycle notation: Sn=Sym⁡({0,1,…,n−1}) with composition (στ)(i)=σ(τ(i)), one-line notation σ=[σ(0),…,σ(n−1)], and cycle notation; a 2-cycle (a b) is a transposition.

[F5]

Inversions, inversion number, the sign sgn⁡(σ)=(−1)inv⁡(σ), and even and odd permutations: Inv⁡(σ)={(i,j):i<j, σ(i)>σ(j)} and inv⁡(σ)=∣Inv⁡(σ)∣.

[F6]

Cycles with disjoint supports commute: transpositions with disjoint supports commute. Generation by the zero-indexed adjacent transpositions is proved directly in step 1.2.

[F7]

Left and right cosets gH and Hg of a subgroup: "For g∈G, the left coset and right coset of H represented by g are gH:={gh:h∈H},Hg:={hg:h∈H}.", so the sets WJa={ua:u∈WJ} and aWJ={au:u∈WJ} are the two coset families of the subgroup WJ.

[F8]

The well-ordering principle: every nonempty subset of N has a least element.

Proof

Given: The data of the statement.

1.1F1F2F3

Support and Tits reduction. Let w∈W. If s1⋯sk and s1′⋯sk′ are reduced expressions of w, then by [F3] they are braid-equivalent, and a braid move replaces an alternating block in two letters by the other alternating word of the same length, so it preserves the set of letters occurring; hence the set S(w) of letters of a reduced expression is independent of the chosen expression. Next, let t1⋯tr be a word in S with value w and length r>ℓ(w); by [F2] there are indices i<j with t1⋯ti^⋯tj^⋯tr=t1⋯tr=w, so deleting the two letters leaves the value and uses only letters already present. Iterating, the length drops by two at each step until it reaches ℓ(w), and the resulting reduced expression of w uses only letters of the original word. The values of finite words in J form a subgroup: the empty word gives 1, concatenation gives products, and reversal gives inverses because j−1=j for j∈J. This subgroup contains J and lies in every subgroup containing J, so it equals WJ by The subgroup ⟨S⟩ generated by a subset, the cyclic subgroup ⟨g⟩, and cyclic groups. Consequently w∈WJ if and only if S(w)⊆J: if S(w)⊆J, a reduced expression of w is a word in the letters of J, so w∈⟨J⟩=WJ; conversely, if w∈WJ, then w is a product of elements of J, hence the value of some word in the letters J, which reduces as above to a reduced expression whose letters lie in J, so S(w)⊆J.

1.2F1F4F5F6

Type A. Let n≥2 and let W now be the group presented by the type-A matrix on S={s1,…,sn−1}; we use the library's symmetric group Sn=Sym⁡({0,…,n−1}) and the transpositions τi=(i i+1) for 0≤i≤n−2, which are the elements written (i+1 i+2) in the statement under the order-preserving identification of {0,…,n−1} with {1,…,n}. The relators hold in Sn: τi2=id⁡; τiτj=τjτi when ∣i−j∣>1, because the two transpositions have disjoint supports and disjoint cycles commute by [F6]; and τiτi+1τi=τi+1τiτi+1 by direct computation on the three symbols i,i+1,i+2. By the universal property in [F1] there is a homomorphism φ:W→Sn with φ(si+1)=τi. It is onto by the following direct argument. If a permutation π is not the identity, its one-line entries are not increasing (the unique increasing bijection of {0,…,n−1} fixes every entry), so some i has π(i)>π(i+1). Right multiplication by τi swaps these neighbouring entries and lowers inv⁡(π) by one: the total contributions of pairs involving either position and a third position are unchanged, while the inversion (i,i+1) disappears. Repeating reaches inversion number zero and hence the identity, expressing π as a product of adjacent transpositions; this uses only [F4] and [F5]. We show ∣W∣≤n!. Let H:=⟨s1,…,sn−2⟩≤W (so H={1} when n=2), and for 1≤j≤n−1 put rj:=sn−1sn−2⋯sj, and put rn:=1. Right multiplication by a generator obeys: if i≤j−2 then rjsi=sirj, because si commutes with every factor of rj; if i=j−1 then rjsi=rj−1; if i=j then rjsi=rj+1, by sj2=1 (with rn−1sn−1=rn); and if i>j then rjsi=si−1rj, because moving the final si left past si−2,…,sj and using sisi−1si=si−1sisi−1 turns it into a leading si−1, which then commutes left past si+1,…,sn−1; for rn=1 the same rules read rnsi=sirn for i≤n−2 and rnsn−1=rn−1. In each case rjsi has the form h rk with h∈H: either h=si with i≤n−2, or h=1, or h=si−1 with i−1≤n−2. Hence the union T=⋃j=1nHrj contains 1=rn and is stable under right multiplication by every generator; since every generator is its own inverse, every word in the generators lies in T, so W=T and ∣W∣≤n ∣H∣. The relators of the type-A matrix on n−2 generators hold among s1,…,sn−2 in W, so by [F1] there is a surjection from the corresponding presented group onto H; by induction on n, whose base n=2 gives ∣W∣=2, this yields ∣H∣≤(n−1)! and hence ∣W∣≤n!. On the other hand ∣Sn∣=n!, since a bijection of the n-element set {0,…,n−1} is determined by choosing the image of 0 in n ways, then the image of 1 in n−1 ways, and so on. Therefore the surjection φ between the finite groups W and Sn is a bijection, hence an isomorphism. It remains to identify ℓ with the inversion number. Right multiplication by τi interchanges the entries at positions i and i+1 in the one-line notation by [F4], so it exchanges the inversion statuses of the pairs (p,i),(p,i+1) for p<i and of the pairs (i,q),(i+1,q) for q>i+1, while the pair (i,i+1) becomes an inversion exactly when it was not one; hence inv⁡(στi)=inv⁡(σ)±1, with inv⁡(στi)=inv⁡(σ)−1 exactly when σ(i)>σ(i+1) by [F5]. Therefore every word of length k in the generators τi representing σ has inv⁡(σ)≤k (each letter changes the inversion number by one), so inv⁡(σ)≤ℓA(σ) for the length function ℓA of Sn with respect to the generating set {τi}; and if inv⁡(σ)>0 then some i has σ(i)>σ(i+1) (otherwise σ would be increasing, hence the identity), so ℓA(σ)≤ℓA(στi)+1=inv⁡(στi)+1=inv⁡(σ) by induction on the inversion number, giving ℓA=inv⁡. Finally ℓ(w)=ℓA(φ(w)) for all w∈W: a word of length ℓ(w) in the si representing w maps to a word of the same length in the τi representing φ(w), so ℓA(φ(w))≤ℓ(w); conversely a word τi1⋯τik of length k=ℓA(φ(w)) for φ(w) lifts to the word si1+1⋯sik+1, which maps to φ(w), so it represents w by injectivity of φ and ℓ(w)≤k. Hence ℓ(w)=inv⁡(φ(w)), and by [F2] a word in the si is reduced exactly when its length equals the inversion number of its value.

2.1F1F2F3step 1.1

The intrinsic parabolic presentation. Let WJ∗ be the group presented by (J,m∣J) as in [F1]. The map f:J→WJ≤W, j↦j, satisfies the relator conditions in W, because every relator of the restricted matrix is a relator of (S,m); so [F1] gives a homomorphism f:WJ∗→WJ, which is onto because the elements of J generate WJ. For injectivity define f′:WJ→WJ∗ by choosing, for w∈WJ, a reduced expression w=j1⋯jq in W and setting f′(w):=j1⋯jq (product in WJ∗); by step 1.1 all letters jl lie in J. This is well defined: another reduced expression j1′⋯jq′ of w is braid-equivalent to it by [F3], and every braid move involved replaces an alternating block in two letters of J by the other alternating word, which is a defining relation of WJ∗, so the two products agree. To see that f′ is a homomorphism it suffices to show f′(jw)=j f′(w) for j∈J and w∈WJ. If ℓ(jw)=ℓ(w)+1, then prepending j to a reduced expression of w gives a reduced expression of jw with letters in J, and the claim is immediate. If ℓ(jw)=ℓ(w)−1, then [F2] applied to a reduced expression w=j1⋯jq gives jw=j1⋯ji^⋯jq for some i. The deleted word is reduced of length q−1, and prepending j to it gives another reduced expression of w of length q, with every letter in J. By [F3] these two reduced expressions of w are braid-equivalent using only letters in J, so their products agree in WJ∗: j1⋯jq=j j1⋯ji^⋯jq. Multiplying this equality by j in WJ∗ gives j f′(w)=j1⋯ji^⋯jq=f′(jw). Iterating f′(jw)=jf′(w) along a word for any u∈WJ, and using f′(1)=1, gives f′(uw)=f′(u)f′(w), so f′ is a homomorphism. By construction ff′(w)=w for all w∈WJ, and f′f fixes each generator of WJ∗ because f′(j)=j for j∈J; hence f′f=idWJ∗. Thus f and f′ are inverse isomorphisms, and (WJ,J) is a Coxeter system. If w∈WJ and j1⋯jq is a reduced expression of w in W, then all jl∈J by step 1.1, so the same word of length q is a word in the generators J of WJ∗, giving ℓJ(w)≤ℓ(w); conversely a word in the letters J representing w is a word in S representing w, so ℓ(w)≤ℓJ(w); hence ℓJ=ℓ on WJ. If s∈S∩WJ, then S(s)={s}⊆J by step 1.1, so s∈J and WJ∩S=J.

3.1F1F2F7F8step 1.1∎

Minimal coset representatives. Fix a∈W. The set {ℓ(x):x∈WJa} is a nonempty subset of N and so has a least element by [F8]; choose d∈WJa of minimal length. For s∈J one has sd∈WJa, so ℓ(sd)≥ℓ(d); by [F2] ℓ(sd)=ℓ(d)±1, and therefore ℓ(sd)=ℓ(d)+1>ℓ(d). Now let u∈WJ with a reduced expression u=u1⋯up and let d=d1⋯dq be reduced, so q=ℓ(d); by step 1.1 all ul lie in J. Applying the Tits reduction of step 1.1 to the concatenated word u1⋯upd1⋯dq produces a reduced expression of ud that is a subsequence of it, hence splits as u′ d′ with u′ a subsequence of u1⋯up and d′ a subsequence of d1⋯dq; write u′,d′ also for their values. If d′ is not the full word d1⋯dq, then d′=(u′)−1ud∈WJd=WJa, while d′ has fewer than q letters, so ℓ(d′)<q=ℓ(d), contradicting the minimality of d. Hence d′ is the full d-word, so u′d=ud and u′=u; since u′ is a subsequence of the reduced word u1⋯up of length p=ℓ(u) and represents u, it must use all p letters, so the concatenation is reduced and ℓ(ud)=p+q=ℓ(u)+ℓ(d). If d′′∈WJa also has minimal length, write d′′=ud with u∈WJ; then ℓ(d′′)=ℓ(u)+ℓ(d) with ℓ(d′′)=ℓ(d), so ℓ(u)=0, u=1 and d′′=d; this gives uniqueness. Conversely, let d satisfy ℓ(sd)>ℓ(d) for all s∈J and write d=u d0 with d0 the minimal representative of WJd; additivity gives ℓ(d)=ℓ(u)+ℓ(d0), and if u≠1 and u=s1⋯sp is a reduced expression with p≥1, then ℓ(s1d)=ℓ(s2⋯spd0)≤(p−1)+ℓ(d0)<ℓ(d), contradicting the hypothesis; hence u=1 and d=d0 is the minimal representative. The factorization w=ud of an arbitrary w∈W is now obtained by taking d minimal in WJw and u:=wd−1∈WJ, and its uniqueness follows from the uniqueness of d. For left cosets, note that ℓ(u−1)=ℓ(u) for every u: reversing a reduced word for u gives a word of the same length for u−1, so ℓ(u−1)≤ℓ(u), and applying this to u−1 gives equality. Inversion is an anti-automorphism interchanging right and left cosets and fixing lengths, so applying the right-coset statement to inverses gives unique minimal elements of the left cosets aWJ, characterized by ℓ(ds)>ℓ(d) for s∈J and satisfying ℓ(du)=ℓ(d)+ℓ(u).

5 · Examples, counterexamples and false statements

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