Continuously many bounded displacement non-equivalences in substitution tiling spaces (Q2657685)
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| Language | Label | Description | Also known as |
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| English | Continuously many bounded displacement non-equivalences in substitution tiling spaces |
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Continuously many bounded displacement non-equivalences in substitution tiling spaces (English)
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14 March 2021
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Given an inflation factor \(\xi > 1\) and a set \(F = \{T_1, \dots, T_n\}\) of tiles in \(\mathbb R^d\), a substitution rule \(\varrho\) on \(F\) is a fixed way to tile each element of \(\xi F\) by tiles of \(F\). A patch is called legal if it is a subpatch of \(\varrho^m (T)\). A substitution tiling space \(\mathbb X_{\varrho}\) is the collection of tilings of \(\mathbb R^d\) with the property that every patch of them is legal. Let \(X\) and \(Y\) be two discrete sets in \(\mathbb R^d\). \(X\) is said to be bounded displacement (BD) equivalent to \(Y\) if there exists a bijection \(\phi\) that satisfies \(\sup_{x\in X}\Vert x - \phi(x) \Vert < \infty\). For tilings by tiles of bounded diameter and inradius bounded away from zero, the BD-equivalence is defined with the help of point sets obtained by placing a point in each tile. In the paper under review the author considers substitution tilings in \(\mathbb R^d\) that give rise to point sets which are not BD-equivalent to a lattice. He studies the cardinality of the set of distinct BD class representatives in the corresponding tiling space. The author proves a sufficient condition under which the tiling space contains continuously many distinct BD classes and presents such an example in the plane. In particular, in the paper for the first time it is shown that this cardinality can be greater than one.
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substitution tilings
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mathematical quasicrystals
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bounded displacement
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uniformly spread
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