Chamber open sets used for the topological equivalence

PDF p. 25: Lemma 6.6, proof: definitions of XC and XC,Cprime

Let $(W,S)$ be a Coxeter system with finite $W$, and let $J\subseteq S$. Use the finite-dimensional relative space $E={\mathrm{\Theta }}_J$, its finite set ${\mathcal{H}}^J$ of distinct linear hyperplanes, and the relative complexified complement ${\mathcal{X}}_J^{\mathbb{C}}$ from Relative complexified hyperplane complement. Thus ${\mathcal{X}}_J^{\mathbb{C}}$ consists of $x+iy$ with $x,y\in E$ and no $H\in {\mathcal{H}}^J$ containing both $x$ and $y$. The arrangement has ambient domain $E$; its chambers are the connected components of $E\setminus {\bigcup }_{H\in {\mathcal{H}}^J}H$, and adjacent chambers are distinct chambers sharing a wall, as in Chambers, walls and adjacency.

For $H\in {\mathcal{H}}^J$ and a chamber $C$, let $D_H(C)$ be the open half-space bounded by $H$ that contains $C$: if $a:E\to \mathbb{R}$ is a nonzero linear functional with kernel $H$, it is either $\{v:a(v)>0\}$ or $\{v:a(v)<0\}$, according to which contains $C$. Define

$X_C=\{x+iy\in {\mathcal{X}}_J^{\mathbb{C}}:\text{ for every }H\in {\mathcal{H}}^J,x\in H\Rightarrow y\in D_H(C)\}\mathrm{.}$

Let $C,C^{\mathrm{\prime }}$ be adjacent, and let $H_0$ be their unique separating hyperplane. Define

$X_{C,C^{\mathrm{\prime }}}=\{x+iy\in {\mathcal{X}}_J^{\mathbb{C}}:x\in D_{H_0}(C),\text{ and for every }H\in {\mathcal{H}}^J,x\in H\Rightarrow y\in D_H(C)\cap D_H(C^{\mathrm{\prime }})\}\mathrm{.}$

Define $X_{C^{\mathrm{\prime }},C}$ by interchanging $C$ and $C^{\mathrm{\prime }}$ in this formula. These are subsets of ${\mathcal{X}}_J^{\mathbb{C}}$ with its subspace topology.

Relative complexified hyperplane complement

Chambers, walls and adjacency

Locally finite arrangements and relative face terminology

Sign cells and local finite-arrangement reductions

Finite galleries and separating hyperplanes

Given: A finite Coxeter group $W$, its simple generators $S$, a subset $J\subseteq S$, and the relative arrangement and complement in the Definition.

Goal: Justify the finiteness, defining half-spaces, adjacent-chamber convention and openness used by the Definition.

1.1 Since the simple generators form a subset $S\subseteq W$ and $W$ is finite, $S$ and its index set ${\mathrm{\Gamma }}_0$ are finite. By [F1], $\mathrm{\Phi }$ is the image of the finite set $W\times {\mathrm{\Gamma }}_0$ under $(w,i)\mapsto w{\alpha }_i$, so $\mathrm{\Phi }$ is finite. By [F2], ${\mathcal{H}}^J$ is an image of a subset of $\mathrm{\Phi }$, with coincident hyperplanes counted once, and is therefore finite. The simple-root space, its dual and the subspace $E$ are finite-dimensional. With ambient domain $T=E$, which is open and convex in itself, this is a locally finite arrangement by [F4]; every member meets $E$ at zero. If $E$ has dimension zero there are no proper hyperplanes, so the arrangement is empty and these conclusions remain valid. [F1, F2, F4, given]

2.1 Fix $H\in {\mathcal{H}}^J$. By [F2], choose a root $\alpha$ with $H=H_{\alpha }\cap E$ and $E\subseteq ̸H_{\alpha }$. Define $a_H:E\to \mathbb{R}$ by $a_H(f)=f(\alpha )$. Evaluation is linear by [F1], its kernel is exactly $H$, and the noncontainment supplies $f\in E$ with $a_H(f)\mathrm{\ne }0$, so it is nonzero. The choices can be made for all $H$ by finite choice, since step 1.1 gives a finite family. For any linear functional $a$ on $E$, in the coordinates $c:E\to {\mathbb{R}}^n$ of [F4] there are coefficients $b_i$ with $a(c^{-1}(z))={\sum }_{i=1}^n{b}_i{z}_i$. Thus $\mathrm{\mid }a(u)-a(v)\mathrm{\mid }\le ({\sum }_i\mathrm{\mid }{b}_i\mathrm{\mid })\mathrm{\parallel }c(u)-c(v){\mathrm{\parallel }}_2$, which proves continuity directly: for a desired positive tolerance $\eta$ use coordinate distance less than $\eta \mathrm{/}(1+{\sum }_i\mathrm{\mid }{b}_i\mathrm{\mid })$. Hence $H$ is closed and the two strict half-spaces are open in $E$. [F1, F2, F4, step 1.1, construct]

3.1 The hypotheses of [F5] now hold by steps 1.1 and 2.1. Therefore every chamber lies entirely in exactly one of the two strict half-spaces for each $H$. To check independence of the defining functional, let $a,b$ be nonzero linear functionals with the same kernel $H$, and choose $e\in E$ with $a(e)=1$. For any $v\in E$, $v-a(v)e\in H$, so $b(v)=a(v)b(e)$. Here $b(e)\mathrm{\ne }0$, since otherwise $b$ would vanish identically. Thus replacing $a$ by $b$ either preserves or exchanges its two strict half-spaces; the one containing $C$ is unchanged. This proves that $D_H(C)$ is well-defined. The complement and complex notation are those of [F3]. [F3, F5, step 1.1, step 2.1, algebra]

4.1 For adjacent distinct chambers $C,C^{\mathrm{\prime }}$, their one-edge gallery is shortest, since a gallery of length zero has equal endpoints. By [F6] it crosses exactly their separating hyperplanes, so there is exactly one, denoted $H_0$. For $H\mathrm{\ne }{H}_0$ the chambers have the same strict sign, hence $D_H(C)=D_H(C^{\mathrm{\prime }})$ by step 3.1. In the formula for $X_{C,C^{\mathrm{\prime }}}$, the requirement $x\in D_{H_0}(C)$ implies $x\mathrm{\notin }{H}_0$. Thus whenever its implication has true antecedent $x\in H$, the two half-spaces in that implication agree. Interchanging $C,C^{\mathrm{\prime }}$ gives the reverse ordered set. [F6, step 1.1, step 2.1, step 3.1]

5.1 For a fixed $H$, failure of the implication defining $X_C$ is the condition $(x,y)\in H\times (E\setminus D_H(C))$. Both factors are closed by steps 2.1 and 3.1. Their product is closed in $E\times E$: its complement is the union of the open sets $(E\setminus H)\times E$ and $E\times D_H(C)$. A finite union of these failure sets is closed, so by [F3] and step 1.1 its complement in ${\mathcal{X}}_J^{\mathbb{C}}$ is open. This is precisely $X_C$. For $X_{C,C^{\mathrm{\prime }}}$, replace $D_H(C)$ in this argument by the open intersection $D_H(C)\cap D_H(C^{\mathrm{\prime }})$, and additionally intersect with the open product $D_{H_0}(C)\times E$. The same reasoning proves openness of $X_{C,C^{\mathrm{\prime }}}$ and, with the chambers interchanged, of $X_{C^{\mathrm{\prime }},C}$. If the arrangement is empty, $X_C={\mathcal{X}}_J^{\mathbb{C}}=E\times E$, and there are no adjacent distinct chambers, so the same conclusions cover this case. [F3, step 1.1, step 2.1, step 3.1, step 4.1] ∎

The article reference attributes the chamber-set formulas only. The root-orbit formula and restricted root evaluations are supplied by the existing mathematical definitions [F1] and [F2]; the finiteness, existence and independence of the defining functionals, and openness are proved locally above. No claim or proof from the article is used to establish its own prerequisite.

The article reference attributes the chamber-set formulas only. The root-orbit formula and restricted root evaluations are supplied by the existing mathematical definitions [F1] and [F2]; the finiteness, existence and independence of the defining functionals, and openness are proved locally above. No claim or proof from the article is used to establish its own prerequisite.