English

Chambers of Arrangements of Hyperplanes and Arrow's Impossibility Theorem

Combinatorics 2007-07-05 v2

Abstract

Let A{\mathcal A} be a nonempty real central arrangement of hyperplanes and Ch{\rm \bf Ch} be the set of chambers of A{\mathcal A}. Each hyperplane HH defines a half-space H+H^{+} and the other half-space HH^{-}. Let B={+,}B = \{+, -\}. For HAH\in {\mathcal A}, define a map ϵH+:ChB\epsilon_{H}^{+} : {\rm \bf Ch} \to B by ϵH+(C)=+(ifCH+)andϵH+(C)=(ifCH).\epsilon_{H}^{+} (C)=+ \text{(if} C\subseteq H^{+}) \text{and} \epsilon_{H}^{+} (C)= - \text{(if} C\subseteq H^{-}). Define ϵH=ϵH+.\epsilon_{H}^{-}=-\epsilon_{H}^{+}. Let Chm=Ch×Ch×...×Ch(mtimes).{\rm \bf Ch}^{m} = {\rm \bf Ch}\times{\rm \bf Ch}\times...\times{\rm \bf Ch} (m\text{times}). Then the maps ϵH±\epsilon_{H}^{\pm} induce the maps ϵH±:ChmBm\epsilon_{H}^{\pm} : {\rm \bf Ch}^{m} \to B^{m} . We will study the admissible maps Φ:ChmCh\Phi : {\rm \bf Ch}^{m} \to {\rm \bf Ch} which are compatible with every ϵH±\epsilon_{H}^{\pm}. Suppose A3|{\mathcal A}|\geq 3 and m2m\geq 2. Then we will show that A{\mathcal A} is indecomposable if and only if every admissible map is a projection to a omponent. When A{\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.

Keywords

Cite

@article{arxiv.math/0608591,
  title  = {Chambers of Arrangements of Hyperplanes and Arrow's Impossibility Theorem},
  author = {Hiroaki Terao},
  journal= {arXiv preprint arXiv:math/0608591},
  year   = {2007}
}