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Finite Reflection Length and Orthogonal Moved Spaces

1 · Prerequisites

2 · Summary

Reflection length counts arbitrary conjugate reflections rather than simple generators, and in a finite Coxeter group every element is a product of exactly $\dim M (w)$ of them, where $M (w) = im (ρ (w) - i d)$ is the moved space in the positive definite reflection representation. This page develops that equality, the absolute order it grades, and the geometry of moved spaces of orthogonal operators.

Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator fixes the conventions for a Coxeter system $(W, S)$ of finite type: the reflection length $ℓ_{T} (w)$ as the least number of elements of the conjugate reflection set $T$ whose product is $w$ (the minimum exists because $S \subseteq T$ and $S$ generates $W$ ), the absolute order $u \leq_{T} v$ defined by the length identity $ℓ_{T} (v) = ℓ_{T} (u) + ℓ_{T} (u^{- 1} v)$ , the moved and fixed spaces $M (A) = im (A - i d)$ , $F (A) = \ker (A - i d)$ of an arbitrary linear map on the inner product space $(V, B)$ , and the orthogonal relation $B \leq_{O} A$ defined by rank additivity. Clause (4) records explicitly that the definition asserts no order property, no rank-length equality and no prefix description; those are proved by the items below.

The Wall form of an orthogonal operator, subspace restriction, and the interval structure of the orthogonal reflection-length order is the finite-dimensional linear algebra. It proves $M (A) = F (A)^{⊥}$ with $V = M (A) \oplus F (A)$ , that the Wall form $χ_{A} (u, v) = B ((A - i d) ∣_{M (A)}^{- 1} u, v)$ satisfies $χ_{A} (u, v) + χ_{A} (v, u) = - B (u, v)$ and is nondegenerate with symmetric part $- \frac{1}{2} B$ , and it constructs for every subspace $U \subseteq M (A)$ the operator $A_{U} = i d + H_{U}^{- 1}$ on $U$ and $i d$ on $U^{⊥}$ , where $H_{U}$ is the operator with $B (H_{U} u, v) = χ_{A} (u, v)$ and $H_{U} + H_{U}^{*} = - {i d}_{U}$ . Its main theorem is that $U \mapsto A_{U}$ is an order isomorphism from the subspaces of $M (A)$ onto the set ${B \in O (V) : B \leq_{O} A}$ , with inverse $B \mapsto M (B)$ ; it also proves the rank-length equality and the prefix description of $\leq_{O}$ for products of reflections. The restriction $A_{U}$ is an element of $O (V)$ and need not lie in $ρ (W)$ ; the companion plane-rotation example exhibits this inside $I_{2} (4)$ .

Root normals inside the moved space, factorizations into reflections, and independent normals passes back inside $W$ . For $w \neq 1$ it produces a root $α$ with $F (w) \subseteq H_{α}$ and $α \in M (w)$ : a generic point of $F (w)$ avoids the finitely many intersections $H_{α} \cap F (w)$ , and its chamber stabiliser is a parabolic containing a conjugate simple reflection whose normal is a root. There follows $ρ (t_{α}) \leq_{O} ρ (w)$ and the rank drop $\dim M (w) = 1 + \dim M (t_{α} w)$ , and induction factors every $w$ into exactly $\dim M (w)$ elements of $T$ , giving $ℓ_{T} (w) = \dim M (w)$ . The telescoping identity $ρ (t_{1} \dots t_{m}) - i d = \sum_{i = 1}^{m} ρ (t_{1} \dots t_{i - 1}) (ρ (t_{i}) - i d)$ supplies the reverse inequality $\dim M (t_{1} \dots t_{m}) \leq m$ and shows that the transported normals $ρ (t_{1} \dots t_{i - 1}) α_{i}$ are linearly independent whenever $\dim M (t_{1} \dots t_{m}) = m$ . The argument uses the finite chamber tiling of the prerequisite page, and it is choice-free.

Carter's reflection-length formula, the absolute order on a finite Coxeter group, and moved-space rigidity under a common upper bound assembles the order theory. It proves Carter's formula $ℓ_{T} (w) = \dim M (w) = \dim V - \dim F (w)$ , that $\leq_{T}$ is a partial order of finite rank with the prefix description of $u \leq_{T} v$ and with $ℓ_{T}$ as rank function, the triangle inequality and the invariance of $ℓ_{T}$ under inversion and conjugation, the inclusions $M (u) \subseteq M (v)$ and $F (v) \subseteq F (u)$ for $u \leq_{T} v$ , and the moved-space rigidity: for $α, β \leq_{T} δ$ one has $α \leq_{T} β$ if and only if $M (α) \subseteq M (β)$ , so that $u \mapsto M (u)$ is an order isomorphism from $[1, δ]_{T}$ onto its image. The common upper bound $δ$ is used through the restriction of $ρ (δ)$ to $M (α)$ and is indispensable; the companion rotation example shows that the converse implication fails without it.

Earlier pages: finite-reflection-arrangements-and-spherical-coxeter-complexes supplies the finite reflection arrangement, the chamber system $w C$ , the open faces $w C_{I}$ and the point stabilisers ${Stab}_{W} (x) = w W_{I} w^{- 1}$ used by the shortening argument, together with the root-reflection dictionary and the positive definite Coxeter form. The companion finite-reflection-length-and-orthogonal-moved-spaces-examples tests the constructions in $S_{5}$ , $S_{4}$ and $I_{2} (4)$ . All four items and the companion computations are choice-free.

3 · Logical flowchart

4 · Definitions, theorems and proofs

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

Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator

Definition

Let $(W, S)$ be a Coxeter system of finite type with $S$ finite and length function $ℓ$ (Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups, Coxeter diagrams: edges, labels, components and finite type), with canonical reflection representation $ρ : W \to G L (V)$ on $V = R^{S}$ , Coxeter form $B$ , root system $Φ$ and reflection set $T = {w s w^{- 1} : w \in W, s \in S}$ (The real Coxeter form, its radical, reflections, and form-preserving maps, The canonical reflection homomorphism, roots, reflections, and the positive cone). Since $W$ is finite, $B$ is positive definite (Finiteness criterion: W is finite exactly when the Coxeter form is positive definite), so $(V, B)$ is a real inner product space (Real and complex inner-product spaces and their induced length), and $ρ (w)$ preserves $B$ for every $w \in W$ (Descent of the reflection representation, unit root norms, and conjugation of reflections (2)). Write $O (V)$ for the group of $B$ -preserving invertible linear maps $V \to V$ (Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces).

(1) Reflection length. For $w \in W$ put

$ℓ_{T} (w) : = \min {k \in N : there are t_{1}, \dots, t_{k} \in T with w = t_{1} t_{2} \dots t_{k}},$

where the empty product ( $k = 0$ ) is the identity. The minimum exists because $S \subseteq T$ and $S$ generates $W$ (Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups), so the admitted $k$ form a nonempty subset of $N$ , which has a least element (The well-ordering principle).

(2) Absolute order. For $u, v \in W$ define $u \leq_{T} v$ if and only if

$ℓ_{T} (v) = ℓ_{T} (u) + ℓ_{T} (u^{- 1} v) .$

(3) Moved and fixed spaces. For a linear map $A : V \to V$ (Linear map between vector spaces over the same field) define the moved space and the fixed space

$M (A) : = im (A - {i d}_{V}), F (A) : = \ker (A - {i d}_{V})$

(Kernel and image of a linear map). For $A, B \in O (V)$ define the relation $B \leq_{O} A$ if and only if

$\dim M (A) = \dim M (B) + \dim M (B^{- 1} A) .$

(4) Conventions and abstentions. For $u \in W$ write $M (u) : = M (ρ (u))$ and $F (u) : = F (ρ (u))$ . This definition asserts no property of $\leq_{T}$ and $\leq_{O}$ beyond the displayed formulas: it asserts neither that either relation is a partial order, nor that $ℓ_{T} (w) = \dim M (w)$ , nor that $B \leq_{O} A$ means that a shortest reflection factorization of $B$ is a prefix of one of $A$ . Those properties are proved in Carter's reflection-length formula, the absolute order on a finite Coxeter group, and moved-space rigidity under a common upper bound ↗, the recorded justifier of this definition, and by the restriction and factorization lemmas of this page. No Choice is used: $S$ , $W$ , $Φ$ and $T$ are finite and every object is finite-dimensional or set-theoretic.

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

The Wall form of an orthogonal operator, subspace restriction, and the interval structure of the orthogonal reflection-length order

Statement

Let $(V, B)$ be a finite-dimensional real inner product space (Real and complex inner-product spaces and their induced length); on this page $V = R^{S}$ with the positive definite Coxeter form $B$ of a Coxeter system of finite type (The real Coxeter form, its radical, reflections, and form-preserving maps, Finiteness criterion: W is finite exactly when the Coxeter form is positive definite), and $O (V)$ is the group of $B$ -preserving invertible linear maps (Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces). Let $M$ , $F$ and $\leq_{O}$ be as in Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator. Then:

(1) Basic identities. For every $A \in O (V)$ one has $F (A) = F (A^{- 1})$ , $M (A) = M (A^{- 1})$ , $M (A) = F (A)^{⊥}$ , $V = M (A) \oplus F (A)$ , $A (M (A)) = M (A)$ , $A (F (A)) = F (A)$ , and $(A - i d) ∣_{M (A)} : M (A) \to M (A)$ is a bijection (In finite dimension, $W^{⊥ ⊥} = W$ and $\dim W + \dim W^{⊥} = \dim V$ , Rank-nullity: $\dim_{F} V = nullity T + rank T$ ). For all $X, Y \in O (V)$ , $M (X Y) \subseteq M (X) + X M (Y)$ and hence $\dim M (X Y) \leq \dim M (X) + \dim M (Y)$ .

(2) The Wall form. For $A \in O (V)$ put

$χ_{A} (u, v) : = B ((A - i d) ∣_{M (A)}^{- 1} u, v) (u, v \in M (A)) .$

Then $χ_{A}$ is a bilinear form on $M (A)$ satisfying

$χ_{A} (u, v) + χ_{A} (v, u) = - B (u, v) (u, v \in M (A));$

in particular $χ_{A}$ is nondegenerate and its symmetric part is $- \frac{1}{2} B ∣_{M (A)}$ (The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ , Adjoints satisfy $(S + T)^{*} = S^{*} + T^{*}$ , $(λ T)^{*} = \overline{λ} T^{*}$ , $(S T)^{*} = T^{*} S^{*}$ , and $T^{* *} = T$ ).

(3) Subspace restriction. Let $A \in O (V)$ , let $U \subseteq M (A)$ be a subspace and let $Π_{U} : V \to U$ be the orthogonal projection (The orthogonal projection $P_{W} v$ is the $W$ -component in $V = W \oplus W^{⊥}$ , For a subspace $W$ of a finite-dimensional inner product space, $V = W \oplus W^{⊥}$ ). There is a unique $H_{U} \in End (U)$ with $B (H_{U} u, v) = χ_{A} (u, v)$ for all $u, v \in U$ , and it satisfies $H_{U} + H_{U}^{*} = - {i d}_{U}$ , so that $H_{U}$ is invertible. Define $A_{U} \in O (V)$ by $A_{U} u = u + H_{U}^{- 1} u$ for $u \in U$ and $A_{U} = i d$ on $U^{⊥}$ . Then $M (A_{U}) = U$ and

$(A_{U} - i d) ∣_{U}^{- 1} = H_{U}, H_{U} u = Π_{U} ((A - i d) ∣_{M (A)})^{- 1} u (u \in U) .$

An element $A \in O (V)$ is a reflection (that is, $\dim M (A) = 1$ ) if and only if $A = i d - 2 Π_{L}$ for the line $L = M (A)$ ; in particular every line $L \subseteq V$ is the moved space of exactly one reflection of $O (V)$ (For an endomorphism in finite dimension, preserving lengths, preserving inner products, carrying orthonormal bases to orthonormal bases, and $T^{*} T = I$ are equivalent).

(4) The restriction theorem. For every subspace $U \subseteq M (A)$ one has $A_{U} \leq_{O} A$ , that is $\dim M (A) = \dim U + \dim M (A_{U}^{- 1} A)$ ; and for $U \subseteq U^{'} \subseteq M (A)$ one has $(A_{U^{'}})_{U} = A_{U}$ , hence $A_{U} \leq_{O} A_{U^{'}}$ . Conversely every $B \in O (V)$ with $B \leq_{O} A$ satisfies $M (B) \subseteq M (A)$ , $χ_{A} ∣_{M (B)} = χ_{B}$ and $B = A_{M (B)}$ . Consequently the assignment $U \mapsto A_{U}$ is a bijection from the set of subspaces of $M (A)$ onto ${B \in O (V) : B \leq_{O} A}$ , with inverse $B \mapsto M (B)$ , and it is an order isomorphism for inclusion of subspaces and $\leq_{O}$ .

(5) Rank-length equality and prefixes. Let $A \in O (V)$ and let $r_{1}, \dots, r_{k} \in O (V)$ be reflections with $A = r_{1} r_{2} \dots r_{k}$ . Then $k \geq \dim M (A)$ , and $A$ is a product of exactly $\dim M (A)$ reflections. Moreover $B \leq_{O} A$ if and only if there are a shortest factorization $A = r_{1} \dots r_{m}$ (so $m = \dim M (A)$ ) and an index $k$ with $B = r_{1} \dots r_{k}$ .

Facts & Assumptions

Given: A finite-dimensional real inner product space $(V, B)$ with the positive definite Coxeter form of a finite-type Coxeter system, an element $A \in O (V)$ , and the moved space $M (A) = im (A - i d)$ , the fixed space $F (A) = \ker (A - i d)$ and the relation $\leq_{O}$ of Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator.

[F1]

For a linear map $T : V \to W$ with $V$ finite-dimensional, $\dim V = \dim \ker T + \dim im T$ ; in particular an injective endomorphism of a finite-dimensional vector space is bijective. Rank-nullity: $\dim_{F} V = nullity T + rank T$

[F2]

For a subspace $W$ of a finite-dimensional inner product space, $V = W \oplus W^{⊥}$ , $W^{⊥ ⊥} = W$ and $\dim W + \dim W^{⊥} = \dim V$ . For a subspace $W$ of a finite-dimensional inner product space, $V = W \oplus W^{⊥}$ In finite dimension, $W^{⊥ ⊥} = W$ and $\dim W + \dim W^{⊥} = \dim V$

[F3]

The orthogonal projection $P_{W} : V \to W$ sends $v = w + z$ with $w \in W$ , $z \in W^{⊥}$ to $P_{W} v = w$ ; equivalently $P_{W} v$ is the unique vector of $W$ with $v - P_{W} v \in W^{⊥}$ , and $P_{W} w = w$ for every $w \in W$ . The projection is linear: combining the unique decompositions of $x$ and $y$ gives $P_{W} (a x + b y) = a P_{W} x + b P_{W} y$ . The orthogonal projection $P_{W} v$ is the $W$ -component in $V = W \oplus W^{⊥}$

[F4]

$A \in O (V)$ means that $A$ is invertible and $B (A u, A v) = B (u, v)$ for all $u, v \in V$ ; equivalently $B (A u, v) = B (u, A^{- 1} v)$ for all $u, v \in V$ . Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces For an endomorphism in finite dimension, preserving lengths, preserving inner products, carrying orthonormal bases to orthonormal bases, and $T^{*} T = I$ are equivalent

[F5]

$B$ is symmetric and bilinear, and positive definite: $B (u, u) \geq 0$ with $B (u, u) = 0$ only for $u = 0$ ; consequently $B$ is nondegenerate, so $B (x, y) = 0$ for all $y \in V$ implies $x = 0$ , and the same holds for the restriction of $B$ to any subspace. Real and complex inner-product spaces and their induced length Finiteness criterion: W is finite exactly when the Coxeter form is positive definite

[F6]

$M (A) = im (A - i d)$ , $F (A) = \ker (A - i d)$ and $B \leq_{O} A$ holds if and only if $\dim M (A) = \dim M (B) + \dim M (B^{- 1} A)$ ; no order property of $\leq_{O}$ is asserted by the definition. Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator

Proof

technique · direct

1.1F2F4F5algebra

For $y \in V$ one has $B ((A - i d) x, y) = B (x, (A^{- 1} - i d) y)$ for all $x \in V$ by [F4], so $y \in M (A)^{⊥}$ holds exactly when $(A^{- 1} - i d) y = 0$ and hence exactly when $A^{- 1} y = y$ , that is, when $y \in F (A^{- 1})$ ; thus $M (A)^{⊥} = F (A^{- 1})$ . Since $A^{- 1} y = y$ holds exactly when $A y = y$ , one has $F (A^{- 1}) = F (A)$ , and $A^{- 1} - i d = - A^{- 1} (A - i d)$ shows $M (A^{- 1}) = M (A)$ . Consequently $M (A)^{⊥} = F (A)$ , and [F2] gives $M (A) = M (A)^{⊥ ⊥} = F (A)^{⊥}$ , $V = M (A) \oplus F (A)$ and $\dim M (A) + \dim F (A) = \dim V$ . Moreover $A (A - i d) = (A - i d) A$ , so $A (M (A)) = M (A)$ ; and $A (F (A)) \subseteq F (A)$ with equality because $A$ is injective.

1.2F1F4algebra

For $X, Y \in O (V)$ the identity $X Y - i d = (X - i d) + X (Y - i d)$ gives $M (X Y) \subseteq M (X) + X M (Y)$ ; the invertible $X$ restricts to an injective map $M (Y) \to X M (Y)$ , whose image therefore has dimension $\dim M (Y)$ by [F1]; consequently $\dim M (X Y) \leq \dim M (X) + \dim M (Y)$ .

2.1step 1.1F1

By step 1.1 the direct sum $V = M (A) \oplus F (A)$ gives $F (A) \cap M (A) = {0}$ , so the restriction $(A - i d) ∣_{M (A)} : M (A) \to M (A)$ , whose image lies in the $A$ -stable space $M (A)$ , has trivial kernel and is therefore injective; by [F1] it is bijective, and $S_{A} : = ((A - i d) ∣_{M (A)})^{- 1} \in G L (M (A))$ is defined.

2.2step 1.1F3F4F5algebra

Let $A \in O (V)$ with $\dim M (A) = 1$ , put $L : = M (A)$ and let $Π_{L}$ be the orthogonal projection onto $L$ [F3]. By step 1.1 the line $L$ and the space $F (A) = L^{⊥}$ are $A$ -stable, so $A$ fixes $L^{⊥}$ pointwise; for $0 \neq u \in L$ one has $A u = c u$ with $c \in R$ , and since $A$ preserves $B$ by [F4] and $B (u, u) \neq 0$ by [F5], $c^{2} B (u, u) = B (A u, A u) = B (u, u)$ forces $c^{2} = 1$ , while $c = 1$ would put $u \in L \cap F (A) = {0}$ ; hence $c = - 1$ and $A = i d - 2 Π_{L}$ . Conversely if $L$ is a line then $A^{'} : = i d - 2 Π_{L}$ acts as $- 1$ on $L$ and as the identity on $L^{⊥}$ , so it preserves $B$ and is invertible, and $M (A^{'}) = L$ has dimension one; if $A^{'} \in O (V)$ is a reflection with $M (A^{'}) = L$ , the first part applied to $A^{'}$ gives $A^{'} = i d - 2 Π_{L}$ . Hence every line $L \subseteq V$ is the moved space of exactly one reflection of $O (V)$ , namely $i d - 2 Π_{L}$ .

3.1step 1.1step 2.1F4F5algebra

Applying step 2.1 to the orthogonal element $A^{- 1}$ , whose moved space is $M (A)$ by step 1.1, gives the bijection $T_{A} : = ((A^{- 1} - i d) ∣_{M (A)})^{- 1} : M (A) \to M (A)$ . For $u, v \in M (A)$ put $w : = S_{A} u$ and $z : = T_{A} v$ , so that $(A - i d) w = u$ and $(A^{- 1} - i d) z = v$ ; then $B (S_{A} u, v) = B (w, (A^{- 1} - i d) z) = B (w, A^{- 1} z) - B (w, z) = B (A w, z) - B (w, z) = B ((A - i d) w, z) = B (u, T_{A} v)$ , using [F4] and symmetry of $B$ . Moreover $A^{- 1} - i d = - A^{- 1} (A - i d)$ on the $A$ -stable space $M (A)$ , so the inverse there is $T_{A} = - S_{A} A$ and $S_{A} + T_{A} = S_{A} (i d - A) = - ((A - i d) ∣_{M (A)})^{- 1} (A - i d) ∣_{M (A)} = - {i d}_{M (A)}$ .

4.1step 1.1step 3.1F4F5algebra

The Wall form $χ_{A} (u, v) = B (S_{A} u, v)$ is bilinear on $M (A)$ , and for $u, v \in M (A)$ the transpose identity of step 3.1 gives $B (S_{A} v, u) = B (v, T_{A} u) = B (T_{A} u, v)$ , so $χ_{A} (u, v) + χ_{A} (v, u) = B (S_{A} u, v) + B (T_{A} u, v) = B ((S_{A} + T_{A}) u, v) = - B (u, v)$ . If $χ_{A} (u, v) = 0$ for all $v \in M (A)$ , then $B (S_{A} u, \cdot)$ vanishes on $M (A)$ and, since $S_{A} u \in M (A) = F (A)^{⊥}$ , also on $F (A)$ , hence on $V$ ; by [F5] $S_{A} u = 0$ , and $S_{A}$ is injective, so $u = 0$ : the form $χ_{A}$ is nondegenerate, and its symmetric part is $\frac{1}{2} (χ_{A} (u, v) + χ_{A} (v, u)) = - \frac{1}{2} B (u, v)$ on $M (A)$ .

5.1step 3.1step 4.1F1F3F5algebra

Let $U \subseteq M (A)$ and define $H_{U} : = Π_{U} S_{A} ∣_{U} : U \to U$ . For $u, v \in U$ , [F3] gives $B (H_{U} u, v) = B (S_{A} u, v) = χ_{A} (u, v)$ ; if another operator has these pairings, its difference from $H_{U}$ pairs to zero with every $v \in U$ , so it equals $H_{U}$ by [F5]. Define $H_{U}^{t} : = Π_{U} T_{A} ∣_{U}$ , with $T_{A}$ from step 3.1. That step and [F3] give $B (H_{U}^{t} u, v) = B (T_{A} u, v) = B (u, S_{A} v) = B (u, H_{U} v)$ , so $H_{U}^{t}$ is the adjoint $H_{U}^{*}$ of The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ . Since $S_{A} + T_{A} = - {i d}_{M (A)}$ , compression to $U$ gives $H_{U} + H_{U}^{t} = - {i d}_{U}$ . If $H_{U} u = 0$ , then $0 = B (H_{U} u, u) = χ_{A} (u, u) = - \frac{1}{2} B (u, u)$ by step 4.1, hence $u = 0$ ; thus $H_{U}$ is injective and invertible by [F1], including when $U = 0$ .

6.1step 5.1F1F2F5algebra

Define $A_{U} \in End (V)$ by $A_{U} u = u + H_{U}^{- 1} u$ for $u \in U$ and $A_{U} = i d$ on $U^{⊥}$ ; this is well defined and linear because $V = U \oplus U^{⊥}$ [F2]. Then $A_{U} - i d = H_{U}^{- 1} Π_{U}$ has image $U$ , so $M (A_{U}) = U$ and $(A_{U} - i d) ∣_{U}^{- 1} = H_{U}$ . Put $T : = H_{U}^{- 1} \in G L (U)$ and $T^{t} : = (H_{U}^{t})^{- 1}$ ; inverting the transpose relation of step 5.1 shows that $T^{t}$ is the transpose of $T$ , that is, $B (T x, y) = B (x, T^{t} y)$ for all $x, y \in U$ . Multiplying $H_{U} + H_{U}^{t} = - {i d}_{U}$ on the left by $T$ and on the right by $T^{t}$ , and also on the left by $T^{t}$ and on the right by $T$ , gives $T + T^{t} = - T T^{t} = - T^{t} T$ , hence $(i d + T) (i d + T^{t}) = {i d}_{U} = (i d + T^{t}) (i d + T)$ . Therefore $B (A_{U} u, A_{U} v) = B ((i d + T^{t}) (i d + T) u, v) = B (u, v)$ for all $u, v \in U$ , the transpose of $i d + T$ being $i d + T^{t}$ ; on $U^{⊥}$ the operator $A_{U}$ is the identity, and $U ⊥ U^{⊥}$ , so $A_{U}$ preserves $B$ on $V$ . If $A_{U} x = 0$ , then $B (x, y) = B (A_{U} x, A_{U} y) = 0$ for every $y \in V$ , so $x = 0$ by [F5] and $A_{U}$ is injective, hence invertible by [F1]: thus $A_{U} \in O (V)$ .

7.1step 1.1step 5.1step 6.1F1F6algebra

Fix $U \subseteq M (A)$ and let $A_{U}$ be as in step 6.1. In the direct sum $V = M (A) \oplus F (A)$ of step 1.1 write $x = m + f$ ; then $(A - A_{U}) x = (A - i d) m - H_{U}^{- 1} Π_{U} m$ , so $x \in \ker (A - A_{U})$ exactly when $(A - i d) m = H_{U}^{- 1} Π_{U} m$ . With $u : = Π_{U} m \in U$ this equation reads $m = S_{A} H_{U}^{- 1} u$ , and it is consistent because $Π_{U} S_{A} Π_{U} = H_{U}$ is step 5.1: the solutions are exactly the $x = S_{A} H_{U}^{- 1} u + f$ with $u \in U$ and $f \in F (A)$ arbitrary. Hence $\ker (A - A_{U}) = S_{A} H_{U}^{- 1} (U) \oplus F (A)$ has dimension $\dim U + \dim F (A)$ , so $rank (A - A_{U}) = \dim M (A) - \dim U$ . Since $A_{U}^{- 1} A - i d = A_{U}^{- 1} (A - A_{U})$ , the space $M (A_{U}^{- 1} A)$ is the image of $A - A_{U}$ under $A_{U}^{- 1}$ and has the same dimension $rank (A - A_{U})$ ; therefore $\dim M (A) = \dim U + \dim M (A_{U}^{- 1} A)$ , that is, $A_{U} \leq_{O} A$ by [F6].

7.2step 1.1step 5.1step 6.1F6algebra

Let $B \leq_{O} A$ and put $C : = B^{- 1} A$ , so that $\dim M (A) = \dim M (B) + \dim M (C)$ by [F6]. Since $A - i d = (B - i d) C + (C - i d)$ , one has $M (A) \subseteq M (B) + M (C)$ , and comparing dimensions gives $M (A) = M (B) \oplus M (C)$ ; in particular $M (B) \subseteq M (A)$ . Fix $w \in U : = M (B)$ and write $m : = S_{A} w = m_{B} + m_{C}$ with $m_{B} \in M (B)$ and $m_{C} \in M (C)$ ; then $w = (A - i d) m = (B - i d) (C m) + (C - i d) m$ with $(B - i d) (C m) \in M (B)$ and $(C - i d) m \in M (C)$ , so comparing the two direct summands gives $C m = m$ and $(B - i d) m = w$ . Hence $m - S_{B} w \in F (B) = M (B)^{⊥} = U^{⊥}$ , where the first equality uses step 1.1 applied to $B$ ; therefore for all $u, v \in U$ one has $χ_{A} (u, v) = B (S_{A} u, v) = B (S_{B} u, v) = χ_{B} (u, v)$ , because $S_{A} u - S_{B} u \in U^{⊥}$ . Thus the operator defined by $χ_{A}$ on $U$ in step 5.1 is $H_{U} = S_{B}$ , and the construction of step 6.1 for $A$ and $U$ gives $A_{M (B)} = i d + S_{B}^{- 1}$ on $U$ and $i d$ on $U^{⊥}$ , while $B = i d + S_{B}^{- 1}$ on $U$ and $B = i d$ on $U^{⊥} = F (B)$ by step 1.1, so $B = A_{M (B)}$ .

8.1step 5.1step 6.1step 7.1algebra

Let $U \subseteq U^{'} \subseteq M (A)$ and apply step 6.1 to $A_{U^{'}}$ , whose moved space is $M (A_{U^{'}}) = U^{'}$ . Since $(A_{U^{'}} - i d) ∣_{U^{'}}^{- 1} = H_{U^{'}}$ , the Wall form of $A_{U^{'}}$ on $U^{'}$ is $χ_{A_{U^{'}}} (x, y) = B (H_{U^{'}} x, y)$ ; for $x, y \in U$ this equals $B (Π_{U^{'}} S_{A} Π_{U^{'}} x, y) = B (S_{A} x, y) = χ_{A} (x, y)$ by step 5.1, because $Π_{U^{'}} x = x$ and $y \in U \subseteq U^{'}$ . Hence the operator attached to $A_{U^{'}}$ and $U$ by step 5.1 is again $H_{U}$ , and the construction of step 6.1 gives $(A_{U^{'}})_{U} = A_{U}$ ; applying step 7.1 with $A_{U^{'}}$ in place of $A$ and the subspace $U \subseteq M (A_{U^{'}}) = U^{'}$ then gives $A_{U} \leq_{O} A_{U^{'}}$ .

8.2step 1.2step 2.2step 7.1

Let $r_{1}, \dots, r_{k} \in O (V)$ be reflections and $A = r_{1} r_{2} \dots r_{k}$ ; since $\dim M (r_{i}) = 1$ by step 2.2, the subadditivity of step 1.2 gives $\dim M (A) \leq k$ . Conversely every $A \in O (V)$ is a product of exactly $\dim M (A)$ reflections: if $M (A) = 0$ then $A = i d$ is the empty product, and otherwise one picks a line $L \subseteq M (A)$ and applies step 7.1 to $U : = L$ , obtaining $\dim M (A_{L}^{- 1} A) = \dim M (A) - 1$ , so by induction on $\dim M (A)$ the element $A_{L}^{- 1} A$ is a product of $\dim M (A) - 1$ reflections and $A = A_{L} \cdot (A_{L}^{- 1} A)$ is a product of $\dim M (A)$ of them.

9.1step 6.1step 7.2step 8.1

The assignment $U \mapsto A_{U}$ from subspaces of $M (A)$ to ${X \in O (V) : X \leq_{O} A}$ is injective, because $A_{U} = A_{U^{'}}$ forces $U = M (A_{U}) = M (A_{U^{'}}) = U^{'}$ by step 6.1, and surjective by step 7.2; with inverse $X \mapsto M (X)$ it is a bijection. It is an order isomorphism: if $U \subseteq U^{'}$ then $A_{U} \leq_{O} A_{U^{'}}$ by step 8.1, and conversely $A_{U} \leq_{O} A_{U^{'}}$ gives $U = M (A_{U}) \subseteq M (A_{U^{'}}) = U^{'}$ by the inclusion clause of step 7.2.

9.2step 1.2step 2.2step 8.2F6algebra

Suppose $A = r_{1} \dots r_{m}$ with $m = \dim M (A)$ and $B = r_{1} \dots r_{k}$ ; the $r_{i}$ are involutions by step 2.2, so $B^{- 1} A = r_{k + 1} \dots r_{m}$ , whence $\dim M (B) \leq k$ and $\dim M (B^{- 1} A) \leq m - k$ by step 8.2, while $\dim M (A) \leq \dim M (B) + \dim M (B^{- 1} A)$ by step 1.2; both inequalities are therefore equalities and $B \leq_{O} A$ by [F6].

10.1step 8.2F6∎

Conversely, if $B \leq_{O} A$ , write $B = r_{1} \dots r_{k}$ with $k = \dim M (B)$ and $B^{- 1} A = s_{1} \dots s_{l}$ with $l = \dim M (B^{- 1} A)$ , both by the factorization clause of step 8.2; then $A = r_{1} \dots r_{k} s_{1} \dots s_{l}$ is a product of $k + l = \dim M (A)$ reflections by [F6], that is, a shortest factorization of $A$ by step 8.2, of which $B$ is the prefix of length $k$ .

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

Root normals inside the moved space, factorizations into reflections, and independent normals

Statement

Let $(W, S)$ be a Coxeter system of finite type with $S$ finite, with $V = R^{S}$ , positive definite Coxeter form $B$ , canonical reflection representation $ρ$ , root system $Φ$ and reflection set $T$ (The real Coxeter form, its radical, reflections, and form-preserving maps, The canonical reflection homomorphism, roots, reflections, and the positive cone, Finiteness criterion: W is finite exactly when the Coxeter form is positive definite), with the chamber system $w C$ , the open faces $w C_{I}$ and the root hyperplanes $H_{α} = {x \in V : B (x, α) = 0}$ of the transferred dual action (The finite reflection arrangement, its chambers, the spherical chamber complex, and the coset face poset), and let $ℓ_{T}$ , $M$ , $F$ and $\leq_{O}$ be as in Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator; for $u \in W$ write $M (u) = M (ρ (u))$ and $F (u) = F (ρ (u))$ . Then:

(1) Root normals in the moved space. Let $w \in W$ with $w \neq 1$ . Then $\dim M (w) > 0$ and there is a root $α \in Φ$ with $F (w) \subseteq H_{α}$ . For every such root one has $α \in M (w)$ , and the reflection $t_{α} \in T$ with $ρ (t_{α}) = r_{α}$ (The inversion formula $∣ N (w) ∣ = ℓ (w)$ , the root-reflection dictionary and strong exchange (1), Reflections: involutivity, form invariance, fixed hyperplane, and exact rank-two order (2)) satisfies

$ρ (t_{α}) \leq_{O} ρ (w), \dim M (w) = 1 + \dim M (t_{α} w) .$

(2) Factorizations and Carter's formula. Every $w \in W$ is a product $w = t_{1} t_{2} \dots t_{k}$ of $k = \dim M (w)$ reflections $t_{i} \in T$ , and no product of fewer elements of $T$ represents $w$ ; equivalently

$ℓ_{T} (w) = \dim M (w) .$

(3) Independent normals. Let $t_{1}, \dots, t_{m} \in T$ and choose roots $α_{i} \in Φ$ with $ρ (t_{i}) = r_{α_{i}}$ . Then $\dim M (t_{1} \dots t_{m}) \leq m$ ; and if $\dim M (t_{1} \dots t_{m}) = m$ — in particular if $t_{1} \dots t_{m}$ is a shortest reflection factorization of its product — then the $m$ vectors

$α_{1}, ρ (t_{1}) α_{2}, ρ (t_{1} t_{2}) α_{3}, \dots, ρ (t_{1} t_{2} \dots t_{m - 1}) α_{m}$

are linearly independent (Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent) and span $M (t_{1} \dots t_{m})$ .

Facts & Assumptions

Given: The finite-type Coxeter datum $V = R^{S}$ , $B$ , $ρ$ , $Φ$ , $T$ , $W$ and the elements $w, t_{1}, \dots, t_{m}$ above; $M (u) = M (ρ (u))$ , $F (u) = F (ρ (u))$ and $ℓ_{T}$ are as in Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator.

[F1]

For $I \subseteq S$ the open face is $C_{I} = {v \in C : B (v, e_{s}) = 0 for s \in I, B (v, e_{s}) > 0 for s \notin I}$ , the root hyperplane is $H_{α} = {v \in V : B (v, α) = 0}$ , and $w C_{I} \subseteq w H_{e_{s}}$ for $s \in I$ ; moreover $V ∖ {0}$ is the disjoint union of the sets $w C_{I}$ over the left cosets $w W_{I}$ with $I ⊊ S$ , and ${Stab}_{W} (x) = w W_{I} w^{- 1}$ for every $x \in w C_{I}$ . The finite reflection arrangement, its chambers, the spherical chamber complex, and the coset face poset The finite chamber tiling, the face-stabiliser identification, and the spherical Coxeter complex as a triangulation of the sphere

[F2]

No finite family of proper subspaces of a finite-dimensional vector space over an infinite field covers the whole space. A finite-dimensional vector space over an infinite field is not a finite union of proper subspaces

[F3]

Every root $α \in Φ$ satisfies $B (α, α) = 1$ , and there is a unique $t_{α} \in T$ with $ρ (t_{α}) = r_{α}$ . The inversion formula $∣ N (w) ∣ = ℓ (w)$ , the root-reflection dictionary and strong exchange

[F4]

The representation $ρ$ is injective. The root-length criterion and faithfulness of the canonical reflection representation

[F5]

For $a \in V$ with $B (a, a) \neq 0$ the map $r_{a}$ is linear, preserves $B$ , fixes every $v$ with $B (v, a) = 0$ , and satisfies $r_{a} (a) = - a$ ; consequently $r_{a} (v) - v \in R a$ for every $v$ , so $M (r_{a}) = R a$ . Reflections: involutivity, form invariance, fixed hyperplane, and exact rank-two order Descent of the reflection representation, unit root norms, and conjugation of reflections

[F6]

On the positive definite space $(V, B)$ the Wall form lemma holds: (1) $\dim M (X Y) \leq \dim M (X) + \dim M (Y)$ for $X, Y \in O (V)$ and $M (A) = F (A)^{⊥}$ for $A \in O (V)$ ; (3) for every subspace $U \subseteq M (A)$ the operator $A_{U}$ of the lemma satisfies $M (A_{U}) = U$ , and every line $L \subseteq V$ is the moved space of exactly one reflection of $O (V)$ , namely $i d - 2 Π_{L}$ ; (4) $A_{U} \leq_{O} A$ and $\dim M (A) = \dim U + \dim M (A_{U}^{- 1} A)$ for every $U \subseteq M (A)$ . The Wall form of an orthogonal operator, subspace restriction, and the interval structure of the orthogonal reflection-length order

[F7]

$ρ$ is a group homomorphism, so $ρ (u v) = ρ (u) ρ (v)$ and $ρ (1) = {i d}_{V}$ (The canonical reflection homomorphism, roots, reflections, and the positive cone). $ℓ_{T} (w)$ is the least $k$ over products of $k$ elements of $T$ ; $M (u) = M (ρ (u))$ , $F (u) = F (ρ (u))$ ; and $T = {w s w^{- 1} : w \in W, s \in S}$ with $(W, S)$ presented as in Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups. Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator

[F8]

For a subspace $U$ of a finite-dimensional space $Z$ , $\dim U \leq \dim Z$ , with equality exactly when $U = Z$ . A finite-dimensional space has a basis, obtained as the extension of any linearly independent subset; a basis is an independent spanning set and its cardinality is the dimension of the space. If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis

[F9]

If a vector space has a spanning subset of cardinality $n$ , then every linearly independent subset has at most $n$ elements. If $V$ has a spanning set with $n$ elements, then every linearly independent subset of $V$ is finite with at most $n$ elements; in particular $V$ has no linearly independent subset equinumerous with $N$

[F10]

A list of vectors is linearly dependent exactly when some nontrivial linear relation holds, and dependence of $w_{1}, \dots, w_{m}$ lets one of the vectors be solved for as a combination of the others; the span of a set is the set of its finite linear combinations. Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$

Proof

technique · direct

1.1F1F2F4F5F6given

Let $w \neq 1$ . First $\dim M (w) > 0$ : otherwise $M (ρ (w)) = 0$ , so $ρ (w) - i d = 0$ and $ρ (w) = i d$ , whence $w = 1$ by [F4], a contradiction. Next, a root $α$ with $F (w) \subseteq H_{α}$ exists. If $F (w) = 0$ then F6 gives $M (w) = F (w)^{⊥} = V \neq 0$ , so $S \neq \emptyset$ and $α : = e_{s}$ for any $s \in S$ is a root with $F (w) = 0 \subseteq H_{α}$ . Suppose now that $F (w) \neq 0$ ; let $Φ^{'} \subseteq Φ$ be the finite set of roots with $F (w) \subseteq̸ H_{α}$ , so that each $H_{α} \cap F (w)$ with $α \in Φ^{'}$ is a proper subspace of the finite-dimensional real space $F (w)$ ; the finite family consisting of those subspaces and ${0}$ consists of proper subspaces of $F (w)$ , since $F (w) \neq 0$ ; by [F2] its union does not cover $F (w)$ , so there is $0 \neq x \in F (w)$ with $x \notin H_{α}$ for every $α \in Φ^{'}$ , including when $Φ^{'} = \emptyset$ ; for every root $β$ the implication $x \in H_{β} \Rightarrow F (w) \subseteq H_{β}$ holds, and $x$ is fixed by $w$ , so $w \in {Stab}_{W} (x)$ . By [F1] there are $w_{0} \in W$ and $I ⊊ S$ with $x \in w_{0} C_{I}$ and ${Stab}_{W} (x) = w_{0} W_{I} w_{0}^{- 1}$ ; since $w \neq 1$ lies in this stabiliser, $I \neq \emptyset$ , so for $s \in I$ one has $x \in w_{0} C_{I} \subseteq w_{0} H_{e_{s}} = H_{ρ (w_{0}) e_{s}}$ by [F1], the equality following from the $B$ -invariance of $ρ (w_{0})$ [F5]; the implication above with $β : = ρ (w_{0}) e_{s}$ gives $F (w) \subseteq H_{ρ (w_{0}) e_{s}}$ for the root $ρ (w_{0}) e_{s} \in Φ$ , so in this case a root with the required property exists as well. Finally, if $F (w) \subseteq H_{α}$ , then $B (f, α) = 0$ for all $f \in F (w)$ , so $α \in F (w)^{⊥} = M (w)$ by F6.

1.2F3F5

Let $t \in T$ and let $α \in Φ$ satisfy $ρ (t) = r_{α}$ , as supplied by [F3]. Then $r_{α} (v) - v = - 2 B (v, α) α$ for every $v$ by [F5], so $M (t) = M (r_{α}) = R α$ and $\dim M (t) = 1$ ; also $ρ (t)^{- 1} = ρ (t) = ρ (t^{- 1})$ because $t^{2} = 1$ .

1.3F6

For $X, Y \in O (V)$ one has $\dim M (X Y) \leq \dim M (X) + \dim M (Y)$ by F6.

1.4F7algebragiven

For $t_{1}, \dots, t_{m} \in T$ , the telescoping identity is $ρ (t_{1} \dots t_{m}) - i d = \sum_{i = 1}^{m} ρ (t_{1} \dots t_{i - 1}) (ρ (t_{i}) - i d)$ : each summand is $ρ (t_{1} \dots t_{i}) - ρ (t_{1} \dots t_{i - 1})$ , by the homomorphism property of $ρ$ . Thus $M (t_{1} \dots t_{m}) \subseteq \sum_{i = 1}^{m} ρ (t_{1} \dots t_{i - 1}) M (t_{i})$ .

1.5F8F9F10

Linear-algebra tool. Let $Z$ be a finite-dimensional vector space spanned by vectors $w_{1}, \dots, w_{m}$ . Then $\dim Z \leq m$ : by [F8] $Z$ has a basis $B$ with $∣ B ∣ = \dim Z$ , and $B$ is linearly independent while $Z$ is spanned by $w_{1}, \dots, w_{m}$ , so [F9] gives $\dim Z \leq m$ . If moreover $\dim Z = m$ , then $w_{1}, \dots, w_{m}$ are linearly independent: otherwise a nontrivial relation expresses some $w_{j}$ as a combination of the remaining $m - 1$ vectors by [F10], those remaining vectors still span $Z$ by [F10], and [F9] would give $\dim Z \leq m - 1$ , a contradiction.

2.1step 1.2step 1.3

A product of $k$ elements of $T$ has moved dimension at most $k$ : by step 1.2 each factor has moved dimension $1$ , and step 1.3 applied $k$ times bounds the moved dimension of the product by the sum $k$ .

2.2step 1.1step 1.2F3F6

Let $α \in Φ$ satisfy $F (w) \subseteq H_{α}$ and $α \in M (w)$ , as produced by step 1.1. Then $R α$ is a line contained in $M (ρ (w))$ , and by F6 and [F3] the unique reflection of $O (V)$ with moved space $R α$ is $i d - 2 Π_{R α} = r_{α} = ρ (t_{α})$ ; hence $ρ (t_{α}) = (ρ (w))_{R α}$ in the notation of [F6]. Applying F6 with $A = ρ (w)$ and $U = R α$ gives $ρ (t_{α}) \leq_{O} ρ (w)$ and $\dim M (w) = \dim R α + \dim M (ρ (t_{α})^{- 1} ρ (w)) = 1 + \dim M (t_{α} w)$ , the last equality because $ρ (t_{α})^{- 1} ρ (w) = ρ (t_{α}^{- 1} w) = ρ (t_{α} w)$ by steps 1.1 and 1.2.

2.3step 1.2step 1.4step 1.5F8

Let $t_{1}, \dots, t_{m} \in T$ and put $v_{i} : = ρ (t_{1} \dots t_{i - 1}) α_{i}$ and $Z : = span {v_{1}, \dots, v_{m}}$ . Steps 1.2 and 1.4 give $M (t_{1} \dots t_{m}) \subseteq Z$ . Since $Z$ is spanned by these $m$ vectors, step 1.5 gives $\dim Z \leq m$ ; [F8] applied to the subspace $M (t_{1} \dots t_{m})$ of $Z$ yields $\dim M (t_{1} \dots t_{m}) \leq \dim Z \leq m$ .

3.1step 2.1step 2.2F7

Carter's formula and factorization: every $w \in W$ satisfies $ℓ_{T} (w) = \dim M (w)$ , and $w$ is a product of exactly $\dim M (w)$ elements of $T$ . If $w = 1$ , then $\dim M (w) = 0$ and the empty product represents $1$ , giving both assertions. If $w \neq 1$ with $\dim M (w) = k \geq 1$ , step 2.2 supplies $t : = t_{α} \in T$ with $\dim M (t w) = k - 1$ ; by induction on $k$ (applied to $t w$ , whose moved dimension is $k - 1$ ) there are $t_{2}, \dots, t_{k} \in T$ with $t w = t_{2} \dots t_{k}$ , so $w = t \cdot (t w) = t t_{2} \dots t_{k}$ is a product of $k$ elements of $T$ . For the reverse inequality let $w = s_{1} \dots s_{j}$ with $s_{1}, \dots, s_{j} \in T$ ; then $k = \dim M (w) = \dim M (ρ (s_{1}) \dots ρ (s_{j})) \leq j$ by step 2.1, so no shorter product of elements of $T$ represents $w$ and $ℓ_{T} (w) = k$ by [F7].

4.1step 1.5step 2.3step 3.1F8given∎

Suppose $\dim M (t_{1} \dots t_{m}) = m$ and put $v_{i} : = ρ (t_{1} \dots t_{i - 1}) α_{i}$ as in step 2.3. Step 2.3 gives $M (t_{1} \dots t_{m}) \subseteq Z$ with $m = \dim M (t_{1} \dots t_{m}) \leq \dim Z \leq m$ , so $\dim Z = m$ and $M (t_{1} \dots t_{m}) = Z$ by [F8]. Thus the $v_{i}$ span $M (t_{1} \dots t_{m})$ , and by step 1.5 the vectors $v_{1}, \dots, v_{m}$ are linearly independent and hence form a basis of $M (t_{1} \dots t_{m})$ : this is the independence and spanning assertion of (3). Finally, if $t_{1} \dots t_{m}$ is a shortest reflection factorization of its product $u : = t_{1} \dots t_{m}$ , then $m = ℓ_{T} (u) = \dim M (u)$ by step 3.1, so the hypothesis $\dim M (t_{1} \dots t_{m}) = m$ holds and the same conclusion applies. This proves (1), (2) and (3).

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

Carter's reflection-length formula, the absolute order on a finite Coxeter group, and moved-space rigidity under a common upper bound

Statement

Let $(W, S)$ be a Coxeter system of finite type with $S$ finite, with $V = R^{S}$ , positive definite Coxeter form $B$ , canonical reflection representation $ρ$ , root system $Φ$ , reflection set $T$ and length function $ℓ$ (Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups, The real Coxeter form, its radical, reflections, and form-preserving maps, The canonical reflection homomorphism, roots, reflections, and the positive cone, Finiteness criterion: W is finite exactly when the Coxeter form is positive definite), and let $ℓ_{T}$ , $M$ , $F$ and $\leq_{T}$ be as in Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator. Then:

(1) Carter's formula. For every $w \in W$ ,

$ℓ_{T} (w) = \dim M (w) = \dim V - \dim F (w) .$

(2) The absolute order. $\leq_{T}$ is a partial order on $W$ (Partial order and partially ordered set), and:

(i) $u \leq_{T} v$ holds if and only if there are reflections $t_{1}, \dots, t_{m} \in T$ and an index $k \leq m$ such that $u = t_{1} \dots t_{k}$ and $v = t_{1} \dots t_{m}$ are shortest reflection factorizations, that is, $k = ℓ_{T} (u)$ and $m = ℓ_{T} (v)$ (a shortest reflection factorization of $u$ is a prefix of one of $v$ );

(ii) $ℓ_{T} (1) = 0$ , $u <_{T} v$ implies $ℓ_{T} (u) < ℓ_{T} (v)$ , and $ℓ_{T} (y) = ℓ_{T} (x) + 1$ whenever $y$ covers $x$ ; hence $ℓ_{T}$ is a rank function and every interval $[u, v]_{T} = {x \in W : u \leq_{T} x \leq_{T} v}$ is finite (Graded poset, rank function, and rank levels);

(iii) $∣ ℓ_{T} (u) - ℓ_{T} (v) ∣ \leq ℓ_{T} (u^{- 1} v)$ , $ℓ_{T} (u^{- 1}) = ℓ_{T} (u)$ and $ℓ_{T} (v u v^{- 1}) = ℓ_{T} (u)$ for all $u, v \in W$ ;

(iv) $u \leq_{T} v$ implies $M (u) \subseteq M (v)$ and $F (v) \subseteq F (u)$ .

(3) Moved-space rigidity under a common upper bound. Let $α, β, δ \in W$ with $α \leq_{T} δ$ and $β \leq_{T} δ$ . Then

$α \leq_{T} β ⟺ M (α) \subseteq M (β);$

in particular $M (α) = M (β)$ implies $α = β$ , and $u \mapsto M (u)$ is an order isomorphism from $[1, δ]_{T}$ onto its image ordered by inclusion. The proof of the converse uses the common upper bound $δ$ , through the restriction of $δ$ to the subspace $M (α)$ ; the converse is claimed only under this hypothesis (see the companion example of the $I_{2} (4)$ rotation, where the hypothesis fails).

Facts & Assumptions

Given: The finite-type Coxeter datum $W, S, V, B, ρ, Φ, T$ and the elements $α, β, δ \in W$ above; $ℓ_{T}$ , $M$ , $F$ and $\leq_{T}$ are as in Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator.

[F1]

Every $w \in W$ is a product of $k = \dim M (w)$ elements of $T$ , no product of fewer elements of $T$ represents $w$ , and $ℓ_{T} (w) = \dim M (w)$ . Root normals inside the moved space, factorizations into reflections, and independent normals

[F2]

The Wall form lemma holds on the positive definite space $(V, B)$ : (1) $M (A) = F (A)^{⊥}$ and $V = M (A) \oplus F (A)$ for $A \in O (V)$ ; (4) for $B \leq_{O} A$ one has $M (B) \subseteq M (A)$ and $B = A_{M (B)}$ , the assignment $U \mapsto A_{U}$ is a bijection from subspaces of $M (A)$ onto ${B \in O (V) : B \leq_{O} A}$ , and it is an order isomorphism for inclusion and $\leq_{O}$ , with $(A_{U^{'}})_{U} = A_{U}$ and $A_{U} \leq_{O} A_{U^{'}}$ for $U \subseteq U^{'}$ ; (5) an element of $O (V)$ is a product of exactly $\dim M$ reflections, and $B \leq_{O} A$ holds exactly when $B$ is a prefix of a shortest factorization of $A$ . The Wall form of an orthogonal operator, subspace restriction, and the interval structure of the orthogonal reflection-length order

[F3]

$u \leq_{T} v$ means $ℓ_{T} (v) = ℓ_{T} (u) + ℓ_{T} (u^{- 1} v)$ , $B \leq_{O} A$ means $\dim M (A) = \dim M (B) + \dim M (B^{- 1} A)$ , and $T = {w s w^{- 1} : w \in W, s \in S}$ is closed under inversion; for $u \in W$ one has $M (u) = M (ρ (u))$ and $F (u) = F (ρ (u))$ . Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator

[F4]

$\leq_{T}$ is a relation on the finite set $W$ ; $\leq_{T}$ is a partial order exactly when it is reflexive, antisymmetric and transitive, and a rank function on a finite poset is a map $ρ$ with $ρ (minimal) = 0$ and $ρ (y) = ρ (x) + 1$ across covers. Partial order and partially ordered set Graded poset, rank function, and rank levels

[F5]

The canonical reflection homomorphism, roots, reflections, and the positive cone (1): $ρ : W \to G L (V)$ is a group homomorphism into the group of invertible linear maps, so $ρ (v)^{- 1} = ρ (v^{- 1})$ and $ρ (v)^{- 1} (V) = V$ .

Proof

technique · direct

1.1F1F2

For every $w \in W$ the factorization lemma gives $ℓ_{T} (w) = \dim M (w)$ [F1], and the Wall form lemma gives $V = M (w) \oplus F (w)$ F2, so $\dim M (w) = \dim V - \dim F (w)$ ; this is Carter's formula (1).

1.2F1

For $x, y \in W$ one has $ℓ_{T} (x y) \leq ℓ_{T} (x) + ℓ_{T} (y)$ : shortest factorizations $x = t_{1} \dots t_{k}$ and $y = s_{1} \dots s_{l}$ with $k = ℓ_{T} (x)$ , $l = ℓ_{T} (y)$ exist by [F1] and concatenate to $x y = t_{1} \dots t_{k} s_{1} \dots s_{l}$ , a product of $k + l$ elements of $T$ . Also $ℓ_{T} (x^{- 1}) = ℓ_{T} (x)$ , because if $x = t_{1} \dots t_{k}$ then $x^{- 1} = t_{k} \dots t_{1}$ , giving $ℓ_{T} (x^{- 1}) \leq ℓ_{T} (x)$ , and applying this to $x^{- 1}$ gives equality.

1.3F1

$ℓ_{T} (x) = 0$ if and only if $x = 1$ : by [F1] an element of reflection length $0$ is a product of $0$ elements of $T$ , which is the identity, and conversely the empty product represents $1$ ; in particular $1$ is the only element of reflection length $0$ .

2.1step 1.1F3

For $u, v \in W$ one has $u \leq_{T} v$ if and only if $ρ (u) \leq_{O} ρ (v)$ : by [F3] the two relations read $ℓ_{T} (v) = ℓ_{T} (u) + ℓ_{T} (u^{- 1} v)$ and $\dim M (v) = \dim M (u) + \dim M (u^{- 1} v)$ , and $ℓ_{T} = \dim M$ is step 1.1.

2.2step 1.1F3F5algebra

Conjugation invariance: for $u, v \in W$ , [F5] gives $ρ (v u v^{- 1}) - i d = ρ (v) (ρ (u) - i d) ρ (v)^{- 1}$ . Since $ρ (v)^{- 1} (V) = V$ , taking images and using [F3] yields $M (v u v^{- 1}) = ρ (v) M (u)$ . The invertible map $ρ (v)$ preserves the dimension of this subspace, so $\dim M (v u v^{- 1}) = \dim M (u)$ and hence $ℓ_{T} (v u v^{- 1}) = ℓ_{T} (u)$ by step 1.1.

2.3step 1.2step 1.3

The relation $\leq_{T}$ is reflexive, antisymmetric and transitive, and the triangle inequality holds. Reflexive: $ℓ_{T} (u) = ℓ_{T} (u) + ℓ_{T} (u^{- 1} u)$ by step 1.3. Antisymmetric: if $u \leq_{T} v$ and $v \leq_{T} u$ , then $ℓ_{T} (u^{- 1} v) = ℓ_{T} (v) - ℓ_{T} (u)$ and $ℓ_{T} (v^{- 1} u) = ℓ_{T} (u) - ℓ_{T} (v)$ , while $ℓ_{T} (v^{- 1} u) = ℓ_{T} ((u^{- 1} v)^{- 1}) = ℓ_{T} (u^{- 1} v)$ by step 1.2, so $ℓ_{T} (u^{- 1} v) = 0$ and $u^{- 1} v = 1$ , that is $u = v$ , by step 1.3. Transitive: if $u \leq_{T} v \leq_{T} w$ , then $ℓ_{T} (w) = ℓ_{T} (u) + ℓ_{T} (u^{- 1} v) + ℓ_{T} (v^{- 1} w)$ , while $ℓ_{T} (u^{- 1} w) \leq ℓ_{T} (u^{- 1} v) + ℓ_{T} (v^{- 1} w)$ and $ℓ_{T} (w) \leq ℓ_{T} (u) + ℓ_{T} (u^{- 1} w)$ by step 1.2; all inequalities are therefore equalities and $ℓ_{T} (w) = ℓ_{T} (u) + ℓ_{T} (u^{- 1} w)$ , that is $u \leq_{T} w$ . Triangle inequality: $ℓ_{T} (v) \leq ℓ_{T} (u) + ℓ_{T} (u^{- 1} v)$ and $ℓ_{T} (u) \leq ℓ_{T} (v) + ℓ_{T} (v^{- 1} u) = ℓ_{T} (v) + ℓ_{T} (u^{- 1} v)$ by step 1.2, so $∣ ℓ_{T} (u) - ℓ_{T} (v) ∣ \leq ℓ_{T} (u^{- 1} v)$ .

2.4step 1.2F1

Prefix form: $u \leq_{T} v$ holds if and only if there are reflections $t_{1}, \dots, t_{m} \in T$ and an index $k \leq m$ such that $u = t_{1} \dots t_{k}$ and $v = t_{1} \dots t_{m}$ are shortest factorizations, that is $k = ℓ_{T} (u)$ and $m = ℓ_{T} (v)$ . If $u \leq_{T} v$ , then [F1] supplies shortest factorizations $u = t_{1} \dots t_{k}$ and $u^{- 1} v = s_{1} \dots s_{l}$ with $l = ℓ_{T} (u^{- 1} v)$ , and $v = u \cdot (u^{- 1} v) = t_{1} \dots t_{k} s_{1} \dots s_{l}$ has length $k + l = ℓ_{T} (v)$ , so it is shortest and exhibits the required prefix. Conversely, given such factorizations, $u^{- 1} v = (t_{1} \dots t_{k})^{- 1} t_{1} \dots t_{m} = t_{k} \dots t_{1} t_{1} \dots t_{m} = t_{k + 1} \dots t_{m}$ , so $ℓ_{T} (u^{- 1} v) \leq m - k$ and hence $ℓ_{T} (u) + ℓ_{T} (u^{- 1} v) \leq k + (m - k) = m = ℓ_{T} (v)$ , while $ℓ_{T} (v) \leq ℓ_{T} (u) + ℓ_{T} (u^{- 1} v)$ by step 1.2; thus equality holds and $u \leq_{T} v$ .

3.1step 1.3step 2.3step 2.4F4

Part (ii). First $ℓ_{T} (1) = 0$ by step 1.3. If $u <_{T} v$ , then $ℓ_{T} (v) = ℓ_{T} (u) + ℓ_{T} (u^{- 1} v)$ with $u^{- 1} v \neq 1$ , so $ℓ_{T} (u^{- 1} v) \geq 1$ by step 1.3 and $ℓ_{T} (u) < ℓ_{T} (v)$ . Suppose now that $y$ covers $x$ , so $x <_{T} y$ , and put $d : = ℓ_{T} (y) - ℓ_{T} (x) = ℓ_{T} (x^{- 1} y) \geq 1$ ; by step 2.4 there are $t_{1}, \dots, t_{m} \in T$ with $y = t_{1} \dots t_{m}$ and $x = t_{1} \dots t_{k}$ shortest, so $d = m - k$ . If $d \geq 2$ , put $z : = t_{1} \dots t_{k + 1}$ ; then $z^{- 1} y = t_{k + 2} \dots t_{m}$ is a product of $m - k - 1$ elements of $T$ , so $ℓ_{T} (z^{- 1} y) \leq m - k - 1$ and, by the triangle inequality of step 2.3, $ℓ_{T} (z) \geq ℓ_{T} (y) - ℓ_{T} (z^{- 1} y) \geq m - (m - k - 1) = k + 1$ , while $ℓ_{T} (z) \leq k + 1$ ; hence $ℓ_{T} (z) = k + 1$ and step 2.4 applied to the shortest factorizations $x = t_{1} \dots t_{k}$ and $z = t_{1} \dots t_{k + 1}$ gives $x <_{T} z$ , and applied to $z = t_{1} \dots t_{k + 1}$ and $y = t_{1} \dots t_{m}$ gives $z <_{T} y$ — contradicting that $y$ covers $x$ . Hence $d = 1$ . Every minimal element is $1$ : if $x$ is minimal and $x \neq 1$ , then $1 <_{T} x$ because $ℓ_{T} (x) = ℓ_{T} (1) + ℓ_{T} (x)$ by steps 1.3, a contradiction; and $ℓ_{T} (1) = 0$ . Consequently $ℓ_{T}$ is a rank function on the finite poset $(W, \leq_{T})$ [F4], and every interval $[u, v]_{T}$ is contained in the finite set $W$ .

3.2step 2.1F2F3

Part (iv): if $u \leq_{T} v$ , then $ρ (u) \leq_{O} ρ (v)$ by step 2.1, so F2 gives $M (u) \subseteq M (v)$ , and then $F (v) = M (v)^{⊥} \subseteq M (u)^{⊥} = F (u)$ by F2 and [F3].

4.1step 2.1step 2.3step 3.2F2∎

Let $α \leq_{T} δ$ and $β \leq_{T} δ$ . If $α \leq_{T} β$ , then $M (α) \subseteq M (β)$ by step 3.2. Conversely assume $M (α) \subseteq M (β)$ ; by step 2.1 one has $ρ (α) \leq_{O} ρ (δ)$ and $ρ (β) \leq_{O} ρ (δ)$ , so F2 gives $ρ (α) = (ρ (δ))_{M (α)}$ , $ρ (β) = (ρ (δ))_{M (β)}$ , and $M (β) \subseteq M (δ)$ ; since the restriction assignment of F2 is an order isomorphism and $M (α) \subseteq M (β)$ , one has $ρ (α) = ((ρ (δ))_{M (β)})_{M (α)} = (ρ (β))_{M (α)} \leq_{O} ρ (β)$ , and step 2.1 gives $α \leq_{T} β$ . Hence $α \leq_{T} β$ if and only if $M (α) \subseteq M (β)$ ; in particular $M (α) = M (β)$ yields both $α \leq_{T} β$ and $β \leq_{T} α$ , so $α = β$ by antisymmetry in step 2.3, and the map $u \mapsto M (u)$ is an order isomorphism from $[1, δ]_{T}$ onto its image ordered by inclusion, being order-preserving and order-reflecting by the equivalence just proved and injective by the equality statement. This proves (1), (2) and (3).

5 · Examples, counterexamples and false statements

None yet.

Finite Reflection Length and Orthogonal Moved Spaces

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator

Definition

The Wall form of an orthogonal operator, subspace restriction, and the interval structure of the orthogonal reflection-length order

Statement

Facts & Assumptions

Proof

Root normals inside the moved space, factorizations into reflections, and independent normals

Statement

Facts & Assumptions

Proof

Carter's reflection-length formula, the absolute order on a finite Coxeter group, and moved-space rigidity under a common upper bound

Statement

Facts & Assumptions

Proof

5 · Examples, counterexamples and false statements

Sources