8
$\begingroup$

The question is in the title, but let me elaborate a little.

Background

Penrose Tilings are really pretty and satisfy some remarkable properties. For instance, I believe the following is true: even though a Penrose tiling is aperiodic by definition, given a point and an arbitrarily large neighborhood, there exists another point with precisely the same neighborhood (up to rigid motion). While many such fascinating properties are well-documented (see the linked Wikipedia article and the references therein), it is not so clear to me how the tilings themselves should be defined in the first place.

There are three specific tilings attributed to Penrose: P1 (generated by $5$ tiles), P2 and P3 (generated by two quadrilaterals each). My first question is purely one of terminology,

does the term "Penrose tiling" mean precisely one of these three specific constructions, or is it used as a blanket term for any aperiodic tiling that satisfies some axioms (such as the self similarity property mentioned above)?

Relationship with Quasicrystals

I am slightly tentative about mentioning Quasicrystals here since they may not lie within the strict purview of research-level math. However, it seems that MO has been kind to such questions before: see here and here. I'm not an expert on QCs beyond reading the odd survey article or two, so I am not too attached to any particular "definition". Here's the question:

What, if anything, is the precise mathematical connection between Penrose tilings and Quasicrystals?

In particular, it is just a question of using a mathematical object to model the (growth of the) physical object because of shared properties, or are there some concrete insights into the nature of Quasicrystals that can be attained by the study of Penrose tilings?

$\endgroup$
2
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ Sep 13, 2012 at 19:23
  • 1
    $\begingroup$ For this question Wikipedia is not only useless but also annoying since it presents a non-mathematical treatment of a mathematical subject. $\endgroup$
    – Misha
    Sep 13, 2012 at 20:12

3 Answers 3

3
$\begingroup$

As explained here, there is an infinite number of distinct tilings that can be constructed using the three sets of tiles introduced by Roger Penrose (rhomb, kite-dart, boat-star). The distinction between tiles and tilings is often not made, and one informally speaks of three types of Penrose tilings.

Quasicrystals are real physical objects, without any mathematical connection to Penrose tilings. I don't know of any naturally ocurring quasicrystal that is accurately approximated by a Penrose tiling. Physicists would call a Penrose tiling a "toy model" of a quasicrystal. If you are interested in how quasicrystals can be modeled by a Penrose tiling, you might want to read Mackay's paper on What has the Penrose tiling to do with the icosahedral phases?

$\endgroup$
3
  • $\begingroup$ There are an infinite number of sets that can be constructed using ZF axioms, that hardly makes set theory devoid of mathematical content. Maybe I have misunderstood what you are saying. $\endgroup$ Sep 13, 2012 at 20:11
  • $\begingroup$ I was not clear, I meant to distinguish tiles and tilings. $\endgroup$ Sep 13, 2012 at 20:36
  • 1
    $\begingroup$ "without any mathematical connection to Penrose tilings" sounds like a strong statement... with which I tend to disagree. $\endgroup$ Nov 16, 2013 at 21:36
4
$\begingroup$

Usually in the mathematics of long range aperiodic order, the Penrose Tiling refers to any tiling obtained from $P_2$ or $P_3$.

We don't differentiate between those, because they have a property called "mutually locally derivable", which says basically that there is a simple way of deriving one from the other. Basically they are the same tiling [the exact meaning of this is that dynamical systems of those tilings are topologically conjugate, which shows that most of their properties are the same].

The tiling $P_1$ is as far as I know not mentioned too often, but this is also "the same" (mutually locally derivable) as $P_2$ and/or $P_3$. This means that $P_1$ is harder to study physically, but it has the same long range properties of $P_2/ P_3$, so it can be studied by studying $P_2$ or $P_3$ instead.

Relation to quasicrystals

If you mark a point in each of the two (or 5) Penrose tiles, you get a point set which you can think as the positions of atoms in an idealized (infinite) solid.

If you calculate the diffraction measure of this model, you get a diffraction diagram which is discrete (i.e. consisting of sharp Bragg peaks) and with 10-fold symmetry. This is almost identical to the Diffraction of the first discovered quasicrystal, but as it was mentioned in another answer we don't know if there is any quasicrystal which is produced by the Penrose rules.

The mathematical models we have for quasicrystals (we don't know if this is the right model) are the model sets (or the larger class of Meyer sets if you also allow for a "small" diffuse background in the diffraction, as nothing in nature is perfect), produced by the cut and project schemes. Those models have diffraction properties identical to the discovered quasicrystals.

If by a "mathematical quasicrystal" you understand a model set, than it was shown by Ammann/de Bruijn that the Penrose tiling actually fits in this class. So in this sense there is a connection between the two concepts.

$\endgroup$
1
  • 2
    $\begingroup$ Great answer Nick. I do have a small correction. You say MLD is equivalent to the tiling space dynamical systems being topologically conjugate, but this isn't quite correct (MLD is a stronger condition). Clark, Sadun, When Shape Matters. $\endgroup$
    – Dan Rust
    Jun 5, 2014 at 14:45
3
$\begingroup$

at a purely mathematical level: take a look here

$\endgroup$
1
  • $\begingroup$ Just in case the link doesn't work at some point, the reference is APERIODIC ORDER AND QUASICRYSTALS: SPECTRAL PROPERTIES by DANIEL LENZ AND PETER STOLLMANN. $\endgroup$
    – Watson
    Apr 11, 2020 at 20:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.