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What is known about the existence of a single tile, that tiles R^n only aperiodically?
Has such a tile been found/proven to exist/not exist for any R^n?

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2 Answers 2

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Seems to depend a bit on how aperiodic you want your tiling to be. Quoting Wikipedia:

In 1988, Peter Schmitt discovered a single aperiodic prototile in 3-dimensional Euclidean space. While no tiling by this prototile admits a translation as a symmetry, it has tilings with a screw symmetry, the combination of a translation and a rotation through an irrational multiple of $\pi$. This was subsequently extended by John Horton Conway and Ludwig Danzer to a convex aperiodic prototile, the Schmitt–Conway–Danzer tile. Because of the screw axis symmetry, this resulted in a reevaluation of the requirements for periodicity. Chaim Goodman-Strauss suggested that a protoset be considered strongly aperiodic if it admits no tiling with an infinite cyclic group of symmetries, and that other aperiodic protosets (such as the SCD tile) be called weakly aperiodic.

The URL is http://en.wikipedia.org/wiki/Aperiodic_tiling

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I want completely aperiodic, ie every two points are distinguishable –  user1708 Jul 5 '11 at 7:01
I don't know what you mean by "every two points are distinguishable." –  Gerry Myerson Jul 5 '11 at 7:30
i just mean that the tiling cant be made to coincide by a translation and any kind of rotations, is this what is meant by strongly aperiodic? what is "an infinite cyclic group of symmetries " they refer to? –  user1708 Jul 5 '11 at 7:41
OK then, you can't accept the SCD tile. "Infinite cyclic group of symmetries" means there is some rigid motion you can do to the tiling that makes it coincide with itself, and no matter how many times you do that rigid motion the individual tiles won't come back to where they started. But now you've got me curious - if you haven't done enough group theory to know what an infinite cyclic group is, what is your interest in aperiodic tilings? What are you hoping to do with them? –  Gerry Myerson Jul 5 '11 at 10:54
ok, is rigid motion=(translation+rotation)? I am interested because all proofs of tiling use some symmetry, but with no symmetry, how can one find a finite proof? . i want to learn this math if it exist. i dont think the proof would use group theory. –  user1708 Jul 5 '11 at 16:31
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The answer is known to be true in 3 dimensions or higher, the SCD tiling is one solution in three dimensions, and can easily be generalized to higher dimensions. The SCD tiling is basically a roof over an upside-down roof, which at every level is periodic, but the two roofs are skewed at an irrational angle. Thus, when on the "next" level, the upside-down roof fills the hole between two roofs, the next level is rotated by an irrational angle with respect to $\pi$, thus killing any possible translation symmetry.

Also, it was known for a very long time that in dimension 1 such tile cannot exists.

The hardest case $n=2$ was open until couple years ago, when a partial answer was found. The Socolor-Taylor tile tiles the plane only aperiodically, but it is not connected (or if it is made connected, it has some parts which are only boundary). If by a tile you understand a single piece which is closure of it's interior, the question is still open in dimension 2.

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