pith. sign in

arxiv: math/0608591 · v2 · submitted 2006-08-23 · 🧮 math.CO

Chambers of Arrangements of Hyperplanes and Arrow's Impossibility Theorem

classification 🧮 math.CO
keywords epsilonmathcalmapstexttimesadmissiblearrangementevery
0
0 comments X
read the original abstract

Let ${\mathcal A}$ be a nonempty real central arrangement of hyperplanes and ${\rm \bf Ch}$ be the set of chambers of ${\mathcal A}$. Each hyperplane $H$ defines a half-space $H^{+} $ and the other half-space $H^{-}$. Let $B = \{+, -\}$. For $H\in {\mathcal A}$, define a map $\epsilon_{H}^{+} : {\rm \bf Ch} \to B$ by $\epsilon_{H}^{+} (C)=+ \text{(if} C\subseteq H^{+}) \text{and} \epsilon_{H}^{+} (C)= - \text{(if} C\subseteq H^{-}).$ Define $\epsilon_{H}^{-}=-\epsilon_{H}^{+}.$ Let ${\rm \bf Ch}^{m} = {\rm \bf Ch}\times{\rm \bf Ch}\times...\times{\rm \bf Ch} (m\text{times}).$ Then the maps $\epsilon_{H}^{\pm}$ induce the maps $\epsilon_{H}^{\pm} : {\rm \bf Ch}^{m} \to B^{m} $. We will study the admissible maps $\Phi : {\rm \bf Ch}^{m} \to {\rm \bf Ch}$ which are compatible with every $\epsilon_{H}^{\pm}$. Suppose $|{\mathcal A}|\geq 3$ and $m\geq 2$. Then we will show that ${\mathcal A}$ is indecomposable if and only if every admissible map is a projection to a omponent. When ${\mathcal A}$ is a braid arrangement, which is indecomposable, this result is equivalent to Arrow's impossibility theorem in economics. We also determine the set of admissible maps explicitly for every nonempty real central arrangement.

This paper has not been read by Pith yet.

discussion (0)

Sign in with ORCID, Apple, or X to comment. Anyone can read and Pith papers without signing in.