LemmaStatement: AI-generatedProof: AI-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Common relative facets have one enlarged standard subset

Statement

In the finite-rank canonical Coxeter representation with simple-root basis $(α_{i})$ and dual space $Θ$ , fix $J$ . Let $(x, I), (y, K) \in C h a m (J)$ be algebraic chamber labels such that $x C_{I}$ and $y C_{K}$ are adjacent relative chambers in $Θ_{J}$ , with common closed wall $F$ . There are unique indices $a \notin I$ and $b \notin K$ describing this wall, and $F = \overline{x C_{I + a}} = \overline{y C_{K + b}}, I + a = K + b = : L .$ Here $I + a = I \cup {a}$ , and closures are taken in $Θ_{J}$ . The two chambers lie on opposite sides of the supporting relative hyperplane of $F$ .

Facts & Assumptions

[F1]

Coordinate faces and their disjoint translated stratification, including the descriptions of chamber walls and the equality $∣ I ∣ = ∣ J ∣$ for a chamber $x C_{I} \subseteq Θ_{J}$ , are supplied by Convexity, stratification and faces of the relative cone. Wall and adjacency use Chambers, walls and adjacency; in particular adjacent chambers are distinct connected components and are disjoint.

[F2]

The algebraic chamber-label set $C h a m (J)$ consists of pairs $(x, I)$ satisfying $W_{J} x = x W_{I}$ with $x$ of minimum length in $x W_{I}$ , as defined in Algebraic chamber-label set. The Statement additionally specifies that the corresponding face translates are adjacent relative chambers.

[F3]

Two points in the closed fundamental chamber in the same orbit are equal; the translating element fixes that point, by Chamber collisions, point stabilizers, and the intersection rule, part (3), with its basis vectors $e_{i}$ identified with $α_{i}$ .

Proof

Given: The finite-rank canonical Coxeter representation, root basis and dual action, with the notation and hypotheses in the Statement and the labels of [F2]. Goal: prove the stated assertions.

Proof technique: direct.

1.1F1given

Put $E = Θ_{J}$ . The linear span of the positive coordinate orthant $C_{I}$ is $Θ_{I}$ , so $x C_{I} \subseteq E$ implies $x Θ_{I} \subseteq E$ . By [F1], $∣ I ∣ = ∣ J ∣$ , and hence these subspaces have the same finite dimension and $x Θ_{I} = E$ . Likewise $y Θ_{K} = E$ . Thus both chamber closures are full-dimensional linear images of nonnegative orthants in $E$ .

2.1F1step 1.1

Pull the closed wall of $x C_{I}$ back by $x$ . The distinct facets of the nonnegative orthant in $Θ_{I}$ are obtained by setting exactly one of its coordinates indexed outside $I$ to zero. Consequently there is one and only one $a$ with $F = \overline{x C_{I + a}}$ . Applying the same argument to $y C_{K}$ gives a unique $b$ .

3.1F1F3step 2.1

The relative interior of a facet closure is its face with every other coordinate strictly positive. Equality of the closed walls therefore gives $x C_{I + a} = y C_{K + b}$ . Choose a point in this common face and write it $x f = y g$ with $f \in C_{I + a}$ and $g \in C_{K + b}$ . Both $f$ and $g$ lie in the closed fundamental chamber, and $y^{- 1} x f = g$ . By [F3], $f = g$ . Their zero-coordinate sets coincide, so $I + a = K + b$ .

4.1F1step 1.1step 2.1step 3.1

Choose $p \in x C_{I + a} = y C_{K + b}$ . On $E$ set $λ_{i} (q) = (x^{- 1} q) (α_{i})$ for $i \notin I$ and $μ_{j} (q) = (y^{- 1} q) (α_{j})$ for $j \notin K$ . By step 1.1 these are coordinate systems: $x C_{I}$ is given by all $λ_{i} > 0$ and $y C_{K}$ by all $μ_{j} > 0$ . At $p$ , only $λ_{a}$ and $μ_{b}$ vanish, while all the other listed coordinates are positive. Their common wall $F$ spans a hyperplane $H$ of $E$ , and $\ker λ_{a} = H = \ker μ_{b}$ , since each kernel contains $F$ and has codimension one.

5.1F1step 4.1∎

The finitely many other coordinate inequalities remain strict in some open ball $U$ about $p$ in $E$ . Thus $x C_{I} \cap U = {q \in U : λ_{a} (q) > 0}$ and $y C_{K} \cap U = {q \in U : μ_{b} (q) > 0}$ . Since the two nonzero functionals have the same kernel, $μ_{b} = c λ_{a}$ for a nonzero real scalar $c$ : choose $v \in E$ with $λ_{a} (v) = 1$ and write $q = λ_{a} (q) v + (q - λ_{a} (q) v)$ , whose second term lies in their common kernel. If $c > 0$ , then $p + t v$ for sufficiently small $t > 0$ belongs to both chambers, contrary to their disjointness in [F1]. Hence $c < 0$ . The whole first chamber has $λ_{a} > 0$ and the whole second has $λ_{a} < 0$ , proving the opposite-side assertion.

Depends on

Used by · 2

Dependency tree · 2 levels

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