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.

The relative root deletion set is the restricted-hyperplane union

Statement

Let $(W, S)$ be a finite-rank Coxeter system in its canonical real representation $V = R^{S}$ , with basis $(α_{s})$ , dual $Θ$ , root system $Φ$ , contragredient action, and Tits cone $U$ . For $I \subseteq S$ put $E_{I} = {f : f (α_{i}) = 0 (i \in I)}$ and $C_{I} = {f \in E_{I} : f (α_{s}) > 0 (s \notin I)}$ . Fix $J \subseteq S$ and put $E = E_{J}$ , $T = U \cap E$ , and $T^{\circ} = {int}_{E} T$ . Write $H_{α} = {f : f (α) = 0}$ , $Φ^{J} = {α \in Φ : E \subseteq̸ H_{α}}$ and $A = {E \cap H_{α} : α \in Φ^{J}}$ . Every member of $A$ is a hyperplane in $E$ , and $T ∖ ⋃_{α \in Φ^{J}} H_{α} = T ∖ ⋃_{K \in A} K$ . The analogous equality holds with $T^{\circ}$ in place of $T$ .

Statement references

Annihilators, relative hyperplanes and complexification

Tits cone geometry with its exact interior boundary

Facts & Assumptions

[F1]

The Coxeter representation, root hyperplanes and cone are those of Tits cone geometry with its exact interior boundary.

[F2]

Nonzero restricted evaluations have hyperplane kernels by Annihilators, relative hyperplanes and complexification.

Proof

Given: The objects and hypotheses of the statement. Goal: Establish its conclusion.

1.1F1F2given

The definition of $Φ^{J}$ says exactly that evaluation at $α$ is nonzero on $E$ . Its kernel there is $E \cap H_{α}$ , which is a hyperplane by [F2].

2.1step 1.1givenalgebra∎

Both $T$ and $T^{\circ}$ are subsets of $E$ . For any such subset $B$ , membership in $B \cap H_{α}$ equals membership in $B \cap (E \cap H_{α})$ . Taking unions over $Φ^{J}$ proves both equalities; repeated traces do not change a union. If $E = {0}$ , the index set is empty and the same argument applies.

Depends on

Used by · 2

Dependency tree · 2 levels

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