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Adjacent relative chambers force a finite crossing component

Statement

In the finite-rank canonical Coxeter representation, fix J and let (x,I),(y,K) label adjacent relative chambers xCI,yCK⊆ΘJ. Let a∉I be the index of their common wall, and put L=I+a=I∪{a} and u=y−1x. Then ΦL+∖ΦI+⊆N(u),∣ΦL∖ΦI∣≤2ℓ(u). In particular WD is finite, where D=[I,a] is the vertex set of the connected component containing a in the Coxeter diagram induced on I+a.

Facts & Assumptions

[F1]

The common wall is the facet at a and the two chambers are on opposite sides of its supporting hyperplane, by Common relative facets have one enlarged standard subset.

[F2]

Roots have one sign, dual evaluation obeys (xf)(β)=f(x−1β), and ∣N(u)∣=ℓ(u), by Root signs, inversions and descent translation.

[F3]

The roots of a standard parabolic are exactly the ambient roots supported on its vertex subset, by Intersections of standard parabolics, the parabolic root subsystem, and global minimality of coset representatives, part (2).

[F4]

Finiteness of ΦI+a∖ΦI implies finiteness of W[I,a] by Finite root difference forces finite connected parabolic.

[F5]

For M⊆S, ΘM is the subspace annihilating the simple roots indexed by M, and CM consists of its points strictly positive on all remaining simple roots. A relative chamber xCI⊆ΘJ has ∣I∣=∣J∣, and its wall at a has supporting relative hyperplane Hxαa∩ΘJ, where Hγ={q:q(γ)=0}. These coordinate-face and wall descriptions are supplied by Convexity, stratification and faces of the relative cone.

[F6]

The notation [I,a] denotes the vertices of the connected component containing a in the Coxeter diagram induced on I+a, by Connected component notation [I,a].

Proof

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

Proof technique: direct.

1.1F2F3F5given

Let β∈ΦL+∖ΦI+. By [F2, F3], its simple-root coefficients are nonnegative, supported on I+a, and its coefficient ca at a is strictly positive. Thus f(β)=caf(αa) for every f∈ΘI. By [F5], the positive coordinate orthant CI spans ΘI, so xCI⊆ΘJ implies xΘI⊆ΘJ. Since ∣I∣=∣J∣, the subspaces have the same finite dimension and xΘI=ΘJ. Dual evaluation now gives q(xβ)=caq(xαa) for every q∈ΘJ.

2.1F1F2F5step 1.1

For q=xf with f∈CI, [F5] gives q(xαa)=f(αa)>0. The common wall's supporting hyperplane is exactly Hxαa∩ΘJ by [F5]. Consequently [F1] puts all of yCK in the half-space where q(xαa)<0. Since ca>0, step 1.1 shows that evaluation on xβ is positive on xCI and negative on yCK.

3.1F2F5step 2.1choose

Choose g∈CK, which is nonempty since one may prescribe value zero on the basis roots indexed by K and value one on the others, by [F5]. Then (yg)(xβ)=g(y−1xβ)<0 by step 2.1 and [F2]. The root uβ cannot be positive: a positive root has nonnegative simple-root coefficients, whereas g has nonnegative values on every simple root by [F5], so its evaluation would be nonnegative. Thus uβ is negative by [F2], and β∈N(u).

4.1F2F3F4F6step 3.1∎

The positive part ΦL+∖ΦI+ of the root difference has at most ℓ(u) members by [F2] and step 3.1. By [F2, F3], negation is a bijection from its positive part to its negative part, and the two parts are disjoint. Hence ∣ΦL∖ΦI∣≤2ℓ(u). Apply [F4] with the component D of [F6] to obtain finiteness of WD. This uses a finite inversion set, regardless of the size or type of WI or WL.

Depends on

Used by · 1

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