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I came across an aperiodic tiling yesterday, and I was wondering, being just a stupid engineer and not a mathematician: Is it possible to map this into a periodic tiling using a continuous function? My guess is "No", but I thought I'd still ask, mathematics has the ability to surprise.

NNN
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  • This tile is homeomorphic to a disk. A square is also homeomorphic to a disk. – Trebor Mar 25 '23 at 06:30
  • So, you're saying such a mapping exists? – NNN Mar 25 '23 at 06:44
  • That was about the mapping of one tile (which is probably not what you want). As for the mapping of the entire tiling... what does that even mean? – Ivan Neretin Mar 25 '23 at 08:55
  • @IvanNeretin Just wondering if an aperiodic arrangement could be converted to a periodic one through a mapping. Probably just plain wrong thinking. – NNN Mar 25 '23 at 10:18
  • @IvanNeretin What it means is "Does there exist a homeomorphism of the plane such that each tile in one tiling is mapped to some tile in the other one?" Of course, there are simple invariants such that the number of neighbors of neighbors, etc. but if you cannot get away with something like that (and since the periodic tiling is not specified, even this is not obvious), the question becomes rather non-trivial. – fedja Mar 25 '23 at 18:22
  • @IvanNeretin For instance, are the tile (side) adjacency graphs isomorphic? – fedja Mar 25 '23 at 18:27
  • "being just a stupid engineer and not a mathematician:" This should read "being a stupid engineer and not not a stupid mathematician" :lol: At least I am more stupid than you. Proof: I know of aperiodic tilings for a long time and the natural question you asked never crossed my mind, while intelligent curiosity is certainly quite a noticeable portion of overall intelligence. I leave it to the others to decide whether they are qualified to join the local chapter of the "confederacy of dunces"... – fedja Mar 25 '23 at 19:22
  • Indeed, a continuous mapping of a plane should maintain the incidence relations - that is, transfer neighboring tiles to neighboring tiles, and non-neighbors to non-neighbors, and moreover, vertex-neighbors to vertex-neighbors. So the question is basically about the isomorphism of adjacency graphs, and as such, is surely meaningful and non-trivial. ("No" seems to be the answer, but I don't have a proof.) – Ivan Neretin Mar 25 '23 at 19:46
  • @IvanNeretin The only approach that comes to my mind is to take large regions in the graph sense (unfortunately, the metric sense can be distorted as much as one wants) and count the density of some constant size rooted subgraphs. If it is irrational for the aperiodic tiling, then we are done. But is it? – fedja Mar 25 '23 at 19:58
  • In the link I posted, some of the white tiles appear to have 6 edge-neighbors and some of the dark blue tiles appear to have 4 edge-neighbors. Is this enough to prove that at-least that tiling cannot be mapped to a periodic tiling? – NNN Mar 27 '23 at 10:44
  • I think the relevant notions you want to look into are "bounded distance to a lattice" and "bi-lipschitz equivalence with a lattice". These capture the more combinatorial aspects of the tilings without worrying so much about specific tile geometry via the standard method for associating a tiling with a Delone set and vice versa. – Dan Rust Mar 27 '23 at 11:37
  • @DanRust Ah, but for periodic tilings the graph distance is equivalent to some metric norm, so "Lipschitz" is unambiguous. But how do I know that this property holds for the aperiodic case (and does it hold?) – fedja Apr 01 '23 at 02:58
  • @fedja Did you look up the definitions? – Dan Rust Apr 01 '23 at 12:14
  • @DanRust I tried to google the words as you put them, but found only something like Spectra of suitably chosen Pisot-Vijayaraghavan numbers represent non-trivial examples of self-similar Delone point sets of finite local complexity. For the case of quadratic Pisot units we characterize, dependingly on digits in the corresponding numeration systems, the spectra which are bounded distance to an average lattice which is more or less Chinese for me. So, I decided to understand these English constructs literally and reacted to that :-) So, what is "bi-lipschitz equivalence with a lattice"? – fedja Apr 01 '23 at 14:25
  • A point-set is bi-Lipschitz equivalent with a lattice if there exists a lattice and a bijection from the point set to the lattice which is a bi-Lipschitz map. That is $A$ is bi-Lipschitz equivalent to the lattice $L$ if there exists a bijection $f \colon A \to L$ and a constant $K>0$ such that for all $a \neq a' \in A$, we have $\frac{1}{K}d(a,a') < d(f(a),f(a')) < K d(a,a')$. The point-set $A$ is bounded distance to $L$ if there exists a bijection $f \colon A \to L$ and a distance $R>0$ such that for all $a \in A$, $d(a,f(a))<R$. – Dan Rust Apr 27 '23 at 14:37

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