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Let $\mathcal{P}$ a Penrose tiling built by a substitution $\omega$ with two triangles.

It is claimed, for instance, in the article of Anderson and Putnam "Topological invariants for substitution tilings and their $C^{\star}$-algebras" that $\omega$ is bijective.

Why is it true?

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This was proved in much greater generality in the paper by B. Solomyak "Nonperiodicity implies unique composition for self-similar translationally finite Tilings", Disc. Comp. Geom. 20 (1998), 265–279.

The main theorem says that every non-periodic translationally-fi nite self-similar tiling (which includes Penrose tilings) has the unique composition property (which implies that your $\omega$ is a bijection).

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  • $\begingroup$ Thank you Gjergji. In fact my aim is to prove properly that any Penrose tiling is non-peridodic but the Anderson-Putnam proof, which seems standard, is based on the fact that $\omega$ is an homeomorphism. $\endgroup$ May 28, 2015 at 15:11
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    $\begingroup$ @BjörnMonson a far simpler way to show the Penrose tiling is aperiodic is to show that the ratio between different tiles is irrational. This ratio is given by the entries in the Perron-Frobenius eigenvector for the associated incidence matrix for the substitution. The half-kite/half dart substitution has matrix (modulo rotated copies) $\left( \begin{smallmatrix}2&1\\1&1\end{smallmatrix}\right)$. This method is only a sufficient test for aperiodicity, as some aperiodic substitutions have rational ratios between tile frequencies (For example the Thue-Morse substitution). $\endgroup$
    – Dan Rust
    Dec 23, 2015 at 17:25

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