Tiling space with notched cubes
A notched cube of dimension \(n\) is obtained from a unit \(n\)-cube by deleting a rectangular box from one of its corners. To be precise, let \(\mathbb{R}^ n\) be the \(n\)-dimensional Euclidean space, let \[ I^ n= \bigl\{(x_ 1,x_ 2,\dots, x_ n)\in \mathbb{R}^ n: 0\leq x_ i\leq 1\quad\text{for each } i=1,2,\dots, n\bigr\} \] be the unit \(n\)-cube, and let \({\mathbf a}= (a_ 1,a_ 2,\dots, a_ n)\) be a point in the interior of \(I^ n\). Then the \({\mathbf a}\)-notched \(n\)-cube is the polytope \[ K= \bigl\{(x_ 1,x_ 2,\dots, x_ n)\in I^ n: 0\leq x_ i\leq a_ i\quad\text{for some } i\bigr\}. \] The notation used here is a variant of the notation in \textit{S. Stein} [\((*)\) Discrete Math. 80, No. 3, 335-337 (1990; Zbl 0747.05028)]. Here \(a_ i\) is Stein's \(1- a_ i\). \textit{S. Stein} \((*)\) showed that \(\mathbb{R}^ n\) can be tiled by translates of the \({\mathbf a}\)-notched \(n\)-cube. Moreover \textit{S. Stein} \((*)\) explicitely exhibited \((n-1)\)! lattice tilings of \(\mathbb{R}^ n\) by translates of the notched \(n\)-cube which are equivalent under translation. But he remarked that, for \(n= 1,2,3\) there are precisely \((n-1)\)! inequivalent tilings, and left open whether, for \(n\geq 4\), there are additional tilings. In the paper under review it is shown that in all dimensions there are no other inequivalent tilings of \(\mathbb{R}^ n\) by translates of the notched cube that were discovered by \textit{S. Stein} \((*)\).
scientific article; zbMATH DE number 681701