Separation by a hyperplane

PDF p. 8: paragraph before Lemma 3.5

Use the finite-rank Coxeter realization, dual space $\mathrm{\Theta }$ with its Euclidean topology, and full Tits cone $\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{e}\subseteq \mathrm{\Theta }$ of Reflecting hyperplanes, fundamental chamber and Tits cone. Let $H\subseteq \mathrm{\Theta }$ be a linear hyperplane, and choose a nonzero linear functional $a:\mathrm{\Theta }\to \mathbb{R}$ with $H=\ker a$. Its two open half-spaces are $\{x\in \mathrm{\Theta }:a(x)>0\}$ and $\{x\in \mathrm{\Theta }:a(x)<0\}$.

For $f,g\in \mathrm{C}\mathrm{o}\mathrm{n}\mathrm{e}$, say that $H$ separates $f$ and $g$ if $a(f)$ and $a(g)$ have opposite strict signs. Equivalently, $f$ and $g$ belong to different connected components of $\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{e}\setminus H$, equipped with the subspace topology. In particular, a point of $H$ is not separated from any point by $H$.

Reflecting hyperplanes, fundamental chamber and Tits cone

Locally finite arrangements and relative face terminology

Linear functionals and the algebraic dual V^*=L(V,F)

Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets

Connected components, quasicomponents, and totally disconnected spaces

Given: The finite-rank realization, full Tits cone and linear hyperplane $H$ in the Definition, with the coordinate map and metric in [F1]. The existence of $H$ implies $n\ge 1$.

1.1 Choose $u\mathrm{\notin }H$. Since $H$ has codimension one, every $x\in \mathrm{\Theta }$ has a unique expression $x=h+tu$ with $h\in H$ and $t\in \mathbb{R}$. Set $a(h+tu)=t$; this is a nonzero linear functional with kernel $H$. If $b$ is another nonzero linear functional with kernel $H$, then $b(u)\mathrm{\ne }0$ and $b(h+tu)=tb(u)=b(u)a(h+tu)$. Thus $b=b(u)a$, so the opposite-strict-sign condition is independent of the chosen defining functional. [F1, F3, algebra, choose]

2.1 Apply [F6] to the linear map $a\circ c^{-1}:{\mathbb{R}}^n\to \mathbb{R}$ and choose $K=K_0+1>0$. By linearity, $\mathrm{\mid }a(y)-a(x)\mathrm{\mid }\le K{d}_{\mathrm{\Theta }}(x,y)$. If $a(x)\mathrm{\ne }0$ and $d_{\mathrm{\Theta }}(x,y)<\mathrm{\mid }a(x)\mathrm{\mid }\mathrm{/}(2K)$, then $\mathrm{\mid }a(y)-a(x)\mathrm{\mid }<\mathrm{\mid }a(x)\mathrm{\mid }\mathrm{/}2$, so $a(y)$ has the same strict sign as $a(x)$. Hence each strict-sign half-space contains an open ball about each of its points and is open by [F1]. [F1, F3, F6, step 1.1, algebra]

2.2 Put $P=\{x\in \mathrm{C}\mathrm{o}\mathrm{n}\mathrm{e}:a(x)>0\}$ and $N=\{x\in \mathrm{C}\mathrm{o}\mathrm{n}\mathrm{e}:a(x)<0\}$. Convexity in [F2] and the identity $a((1-t)x+ty)=(1-t)a(x)+ta(y)$ show that both $P$ and $N$ are convex. The linear map $c$ therefore sends each to a convex subset of ${\mathbb{R}}^n$, connected by [F5] in its Euclidean subspace topology. The coordinate map transports the subspace topology by [F1]: a partition of $P$ or $N$ into two relatively open nonempty pieces would give such a partition of its coordinate image, contradicting connectedness. Thus every nonempty one of $P,N$ is connected. [F1, F2, F3, F4, F5, step 1.1]

3.1 The sets $P,N$ are disjoint, cover $\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{e}\setminus H$, and are relatively open there by step 2.1. A connected subset of $\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{e}\setminus H$ cannot meet both, since their traces would be a separation in the sense of [F4]. For a point in $P$, every connected subset containing it is therefore contained in $P$, while $P$ itself is connected by step 2.2. The component formula in [F4] consequently gives exactly $P$ as its component; the same argument applies to $N$. The nonempty sign intersections are precisely the connected components. This proves the stated equivalence, including at boundary points of the full Tits cone. Points of $H$ are excluded from the complement and never have opposite strict signs. [F1, F4, step 2.1, step 2.2] ∎