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Sign cells and local finite-arrangement reductions

Statement

Let $E$ be a finite-dimensional real vector space with its Euclidean topology, and let $T \subseteq E$ be open and convex. Let $A$ be a set of distinct linear hyperplanes, each meeting $T$ , locally finite in $T$ : each point of $T$ has a neighbourhood meeting only finitely many members. For each $H \in A$ , choose a nonzero linear functional $a_{H} : E \to R$ with kernel $H$ . A chamber is a connected component of $T ∖ ⋃_{H \in A} H$ .

For any assignment $ϵ_{H} \in {1, - 1}$ , the corresponding strict sign cell is $C_{ϵ} = {x \in T : ϵ_{H} a_{H} (x) > 0 for every H \in A}$ . Every nonempty sign cell is a convex chamber, and every chamber is a sign cell.

For a chamber $C = C_{ϵ}$ and a point $p \in \overline{C} \cap T$ , there is an open convex neighbourhood $U$ of $p$ contained in $T$ that meets $C$ and only finitely many hyperplanes. For every such $U$ , writing $A_{U} = {H \in A : H \cap U \neq \emptyset}$ , one has $\overline{C} \cap U = {x \in U : ϵ_{H} a_{H} (x) \geq 0 for all H \in A_{U}}$ . Here $\overline{C}$ denotes closure in $E$ . In particular, if a wall is an intersection $\overline{C} \cap H_{0}$ of affine dimension $\dim E - 1$ for $H_{0} \in A$ , its intersection with $U$ is described by these finitely many weak inequalities together with $a_{H_{0}} = 0$ ; it is empty if $H_{0}$ does not meet $U$ . No neighbourhood assertion at a boundary point outside $T$ is made.

Statement references

Locally finite arrangements and relative face terminology

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

A convex subset of ℝᵐ contains every line segment between two of its points

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

Connected components, quasicomponents, and totally disconnected spaces

Facts & Assumptions

[F1]

The arrangement terminology, chamber as a connected component, closure in $E$ , and wall as the specified hyperplane intersection are defined in Locally finite arrangements and relative face terminology.

[F2]

A defining functional is linear, so $a_{H} ((1 - t) x + t y) = (1 - t) a_{H} (x) + t a_{H} (y)$ ; see Linear functionals and the algebraic dual V^*=L(V,F). A set is convex when it contains every segment between its points, as in A convex subset of ℝᵐ contains every line segment between two of its points.

[F3]

Connectedness means absence of a separation into two nonempty disjoint relatively open sets; singletons are connected, as in Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets. The connected component of a point is the largest connected subset containing that point, as established in Connected components, quasicomponents, and totally disconnected spaces.

[F4]

Every convex subset of $R^{n}$ for $n \geq 1$ , with its usual topology, is connected, by claim 1 of Every convex subset of ℝⁿ, in particular every ball and ℝⁿ itself, is path-connected and hence connected.

[F5]

For a linear map $L : R^{m} \to R^{k}$ , there is a constant $K \geq 0$ such that $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for every $h$ , by Every Euclidean linear map has a unique matrix and satisfies ‖Lh‖₂≤ K‖h‖₂ for some K≥0.

Proof

Given: The space, open convex domain, locally finite hyperplanes, defining functionals and signs in the statement.

Goal: Identify the chambers with the nonempty sign cells and establish the stated finite local description.

1.1F1F2F3F4F5given

Write $n = \dim E$ and let $ϕ : E \to R^{n}$ be the chosen linear coordinates transporting the Euclidean topology, as in [F1]. If $n \geq 1$ , apply [F5] to the linear map $a_{H} \circ ϕ^{- 1} : R^{n} \to R$ to obtain $∣ a_{H} (x) - a_{H} (y) ∣ \leq K_{H} ∥ ϕ (x) - ϕ (y) ∥_{2}$ . For each $η > 0$ , taking $δ = η / (K_{H} + 1)$ shows that coordinate distance less than $δ$ implies $∣ a_{H} (x) - a_{H} (y) ∣ < η$ ; thus each defining functional is continuous. Also, a separation of a convex subset of $E$ would transport under $ϕ$ to a separation of its convex coordinate image, contrary to [F4]; hence every convex subset of $E$ is connected. If $n = 0$ , every subset of $E = {0}$ is empty or a singleton and is connected by [F3]; there are no nonzero linear functionals and no linear hyperplanes, so the functional assertions are vacuous. These arguments establish the needed continuity and connectedness in all dimensions.

2.1F1F2F3step 1.1given

For two points in a sign cell, the linearity in [F2] makes every prescribed signed value strictly positive along their connecting segment; convexity of $T$ keeps that segment in $T$ . Thus the cell is convex and is connected by step 1.1. Conversely, on any connected subset of the arrangement complement, each $a_{H}$ has a constant strict sign: by step 1.1 its positive and negative inverse images are disjoint relatively open sets whose union is the subset, and if both were nonempty they would give a separation forbidden by [F3]. Each nonempty sign cell is therefore a connected subset containing every connected subset of the complement that meets it. For a point in the cell, its component contains the cell and is contained in it by [F3] and this constant-sign property. Thus the cell is a chamber, and every chamber is obtained in this way.

3.1F1F2step 1.1step 2.1given∎

Let $p \in \overline{C} \cap T$ . Local finiteness and openness of $T$ allow an open coordinate ball $U$ around $p$ contained in $T$ and meeting only finitely many hyperplanes. It is convex, and it meets $C$ because $p \in \overline{C}$ . Now let $U$ be any open convex neighbourhood with these properties and choose $q \in C \cap U$ . If $H$ misses $U$ , the continuous function $a_{H}$ has constant nonzero sign on the connected set $U$ , so its sign there is $ϵ_{H}$ . If $x \in \overline{C} \cap U$ , continuity gives $ϵ_{H} a_{H} (x) \geq 0$ for each $H \in A_{U}$ . Conversely, suppose $x \in U$ satisfies those finitely many weak inequalities. The preceding observation gives strict signed positivity at $x$ for every other hyperplane. For $0 < t \leq 1$ , the point $x_{t} = (1 - t) x + t q$ lies in $U \subseteq T$ , and linearity gives $ϵ_{H} a_{H} (x_{t}) > 0$ for every $H \in A$ , including those whose signed value at $x$ is zero. Hence $x_{t} \in C$ by step 2.1, and $x_{t} \to x$ , proving the closure formula. Intersecting that formula with $H_{0} = \ker a_{H_{0}}$ gives the wall description; if $H_{0}$ misses $U$ the intersection is empty. This proves the claims within $T$ .

Depends on

Used by · 3

Dependency tree · 2 levels

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