RemarkRemark: Literature-sourcedProof: 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.

Distinct roots can define the same relative hyperplane

PDF p. 7: §3.1, paragraph after (3.3)

Statement

The map from positive $J$ -roots to the relative arrangement $H^{J}$ , taking $α$ to $H_{α} \cap Θ_{J}$ , need not be injective. In particular, distinct positive $J$ -roots can have the same restricted hyperplane.

Statement references

Tits cone intersections, relative roots and arrangements

Proof

Given: A finite Coxeter system with its real simple-root representation and the relative-root notation in the Statement.

[F1] For a Coxeter matrix with $m (s, t) = 3$ , the simple-root form has $B (e_{s}, e_{t}) = - 1 / 2$ , and the simple reflection acts by $s (v) = v - 2 B (v, e_{s}) e_{s}$ . The roots are the vectors $w e_{u}$ with $w \in W$ and $u$ a simple generator. Real Coxeter representation and contragredient notation

[F2] A positive root is a root whose coordinates in the simple-root basis are all nonnegative. Root signs, inversions and descent translation

[F3] For $J \subseteq S$ , the space $Θ_{J}$ consists of the linear functionals vanishing on $e_{j}$ for $j \in J$ . A $J$ -root is a root whose hyperplane does not contain $Θ_{J}$ , and $H^{J}$ consists of its restricted hyperplanes. Tits cone intersections, relative roots and arrangements

[F4] For a real vector space with finite basis $(e_{u})_{u \in S}$ and $J \subseteq S$ , the evaluation $f \mapsto f (γ)$ on $E = {f \in V^{*} : f (e_{j}) = 0 for every j \in J}$ is nonzero exactly when $γ \notin span {e_{j} : j \in J}$ . Annihilators, relative hyperplanes and complexification

Proof technique: direct.

Goal: Exhibit two distinct positive $J$ -roots with the same restricted hyperplane.

1.1F1F2choosealgebra

Take $S = {s, t}$ with $m (s, t) = 3$ and $J = {s}$ . Set $α = e_{t}$ and $β = s e_{t}$ . By the reflection formula, $β = e_{t} + e_{s}$ . Both are roots by [F1], both have nonnegative simple-root coordinates, and they are distinct because $e_{s}, e_{t}$ are basis vectors. Thus both are positive roots.

2.1F3F4step 1.1algebra∎

For every $f \in Θ_{J}$ , one has $f (e_{s}) = 0$ , so $f (β) = f (e_{t}) = f (α)$ . Consequently $H_{α} \cap Θ_{J} = H_{β} \cap Θ_{J}$ . Since $e_{s}, e_{t}$ are linearly independent, $α = e_{t} \notin span {e_{s}}$ . By [F4], the evaluation $f \mapsto f (α)$ is nonzero on $Θ_{J}$ ; the equality $f (β) = f (α)$ makes $f \mapsto f (β)$ nonzero there as well. Thus neither hyperplane contains $Θ_{J}$ . Hence both roots are positive $J$ -roots and their common intersection is a member of $H^{J}$ , proving the stated noninjectivity.

Source notes

The article is cited for attribution of the general observation. The displayed rank-two example proves the noninjectivity independently of that observation.

Depends on

Dependency tree · 2 levels

Each result is drawn at its shortest distance from this item. Arrows run from prerequisites to the items that use them; this item has a heavier outline. External source nodes link directly to the cited text. The view shows up to two dependency steps; open a prerequisite to continue.

Sources

Normalisers of parabolic subgroups of Artin–Tits groups and Tits cone intersections · PDF p. 7, §3.1, paragraph after (3.3)