Equidecomposability of Jordan domains under groups of isometries

M. Laczkovich¹

¹ Department of Analysis Eötvös Loránd University Pázmány Péter sétány 1/C 1117 Budapest, Hungary and Department of Mathematics University College London Gower Street London, WC1E 6BT, England

Fundamenta Mathematicae, Tome 177 (2003) no. 2, pp. 151-173

Résumé

Let $G_d$ denote the isometry group of ${\mathbb R}^d.$ We prove that if $G$ is a paradoxical subgroup of $G_d$ then there exist $G$-equidecomposable Jordan domains with piecewise smooth boundaries and having different volumes. On the other hand, we construct a system ${\cal F}_d$ of Jordan domains with differentiable boundaries and of the same volume such that ${\cal F}_d$ has the cardinality of the continuum, and for every amenable subgroup $G$ of $G_d,$ the elements of ${\cal F}_d$ are not $G$-equidecomposable; moreover, their interiors are not $G$-equidecomposable as geometric bodies. As a corollary, we obtain Jordan domains $A,B\subset {\mathbb R}^2$ with differentiable boundaries and of the same area such that $A$ and $B$ are not equidecomposable, and $\mathop {\rm int} A$ and $\mathop {\rm int} B$ are not equidecomposable as geometric bodies. This gives a partial solution to a problem of Jan Mycielski.

Détail

DOI : 10.4064/fm177-2-4

Keywords: denote isometry group mathbb prove paradoxical subgroup there exist g equidecomposable jordan domains piecewise smooth boundaries having different volumes other construct system cal jordan domains differentiable boundaries volume cal has cardinality continuum every amenable subgroup elements cal g equidecomposable moreover their interiors g equidecomposable geometric bodies corollary obtain jordan domains subset mathbb differentiable boundaries area equidecomposable mathop int mathop int equidecomposable geometric bodies gives partial solution problem jan mycielski