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Only finitely many hyperplanes separate two Tits-cone points

PDF p. 8: Lemma 3.5

Statement

Let (W,S) be a Coxeter system, with geometric roots Φ=Φ+⊔Φ− in the real vector space with simple-root basis (αi)i∈Γ0, and let Θ be its algebraic dual with the contragredient action. Put C‾={h∈Θ:h(αi)≥0 for every i∈Γ0} and Cone=⋃w∈WwC‾. For α∈Φ put Hα={h∈Θ:h(α)=0} and H={Hα:α∈Φ}. Say that Hα separates f,g∈Cone when f(α) and g(α) have opposite strict signs.

For f,g∈Cone, only finitely many hyperplanes in H separate f and g. The generating set may be infinite; in finite rank these conventions agree with the stated Tits-cone and separation definitions.

Facts & Assumptions

[F1]

Each root has exactly one of the two signs, Φ+=Φ∩∑iR≥0αi and Φ−=−Φ+. For every w∈W, the set {α∈Φ+:wα∈Φ−} has cardinality ℓ(w) and is therefore finite. Root-sign and inversion-length facts

[F2]

Every vector in the geometric representation is a finite linear combination of simple roots. Each root is the image of a simple-root basis vector under an invertible representation map, and (wh)(v)=h(w−1v) for the contragredient action. Coxeter-system, roots, length and weak-order conventions

[F3]

In finite rank, the nonnegative chamber, its translates, root hyperplanes and opposite-strict-sign separation are the conventions in the cited definitions. Reflecting hyperplanes, fundamental chamber and Tits cone Separation by a hyperplane

Proof

Given: The Coxeter system, root notation and points f,g∈Cone in the Statement.

Proof technique: direct.

Goal: Show that the set of separating root hyperplanes is finite.

1.1F2algebragiven

Independence of the root representative. Every root β is nonzero, since it is an invertible image of a basis vector by [F2]. A nonzero simple-root coefficient of β gives, by scaling the corresponding coordinate functional, an h0∈Θ with h0(β)=1. For every h∈Θ, the vector h−h(β)h0 belongs to Hβ; hence Θ=Rh0⊕Hβ and Hβ is a linear hyperplane. If another root γ has Hγ=Hβ, applying evaluation at γ to this vector gives h(γ)=ch(β) for all h∈Θ, where c=h0(γ). Evaluation at γ is nonzero by the same coordinate argument, so c≠0. Consequently f(γ)g(γ)=c2f(β)g(β), and opposite strict signs are independent of the root representative. This argument works for any rank and uses only coordinate functionals on the given basis.

1.2F1F2given

For any x∈Cone choose w∈W and h∈C‾ with x=wh. If α∈Φ+ and x(α)<0, then h(w−1α)<0 by [F2]. A positive root is a finite nonnegative linear combination of simple roots, so h evaluates it nonnegatively. By the sign decomposition in [F1], w−1α must therefore be negative. Hence {α∈Φ+:x(α)<0} is contained in the finite inversion set {α∈Φ+:w−1α∈Φ−}.

2.1F1F3step 1.1step 1.2given∎

For each separating hyperplane Hβ, one of β and −β is positive by [F1], and both define the same hyperplane. For that positive root α, opposite strict signs imply f(α)<0 or g(α)<0. Thus every separating hyperplane is the image, under α↦Hα, of a member of the union of the two finite sets obtained in 1.2 for x=f and x=g. That union and its image are finite, proving the claim. The identification in [F3] gives the finite-rank formulation in the article's geometric conventions.

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