4
$\begingroup$

I want to represent an order-5 square tiling (image from Wikipedia; more text below image):

Order-5 square tiling

Obviously for a simple grid I can uniquely refer to a given square by its (x,y) position relative to the origin, in integer coordinates. To get to adjacent cells, one may simply increment or decrement x or y. However, on the hyperbolic plane, things get messy, and it's unclear how to refer to a given cell.

Is there a better visual representation of the order-5 square tiling that would allow intuitive understanding of how to refer to individual cells, and failing that, is there a simple method to give each cell an index (such as (x,y) for the Euclidean plane) that makes it easy to find its neighbors?

$\endgroup$
  • $\begingroup$ I think the problem is that it has order 5 (an odd number) maybe it becomes easier if you have an even order (preferably 4 or 8) $\endgroup$ – Willemien Jul 7 '14 at 21:42
  • $\begingroup$ Things are easier if you assume that each tile has extra symmetries, forming dihedral group of order 8 - this is the maximal symmetry your tile can allow. (This is not really needed, but simplifies life a lot.) If you are interested in the most symmetric case, then I will write down an answer. (Otherwise it is a bit too painful but the "coding" in the end is exactly the same.) $\endgroup$ – Moishe Kohan Jul 8 '14 at 0:55
1
$\begingroup$

A partial answer: For any square there are four “translations” of the hyperbolic plane, each sliding the given square onto one of its four neighbours. These translations generate a group which is finitely generated, namely, by the two translations sending one given square into either one of two adjacent neighbours. (Their inverses will map the given square to the neighbours on the opposite side.) Unfortunately, the given group is non-abelian, but it might be possible (I think) to describe it as the free group on two generators modulo a finite number of relations. I don't know how to determine those relations, though.

This group, call it $G$, acts transitively on the set of squares $S$ in the tiling. So you might expect to be able to label the squares by fixing one ($s_0\in S$), and then labeling $gs_0$ by $g\in G$. But that won't quite work, for the mapping $gs_0\mapsto g$ is not well defined: Translate a square five times, around each of the five edges meeting at one point, and it comes back rotated $90^\circ$. So let $H$ be the stabilizer of $s_0$: It is cyclic of order $4$, and most likely a non-normal subgroup of $G$. Then the quotient $G/H$ labels the squares in the tiling.

Not the answer you hoped for, I am sure, but I don't think it can be done a lot more simply. (And I don't know how to complete the answer.)

$\endgroup$
  • 1
    $\begingroup$ Am i missing something? " Rotate a square five times, around each of the five edges meeting at one point, and it comes back rotated $90^o$" the angle of the hyperbolic square is not $ 90^o $ but only $72^o$ degrees so they come back rotated to $0^o$ $\endgroup$ – Willemien Jul 7 '14 at 23:16
  • $\begingroup$ This argument works under some extra conditions, for instance, the opposite sides of the tile have the same length. Then indeed the group that does the job is $<a,b: [a,b]^5=1>$, where $a$ and $b$ are hyperbolic translations matching opposite sides of the tile. $\endgroup$ – Moishe Kohan Jul 8 '14 at 1:23
  • $\begingroup$ @Willemien I had meant to write translate, not rotate. I fixed that; thanks. $\endgroup$ – Harald Hanche-Olsen Jul 8 '14 at 9:01
  • $\begingroup$ @studiosus Isn't that the definition of a square in hyperbolic geometry? A quadrilateral whose sides all have the same length and whose corners all have the same angle? Sorry if I got that wrong; I have never seriously studied this stuff. (My first inclination was to leave a comment rather than an answer, but it got way too long for a comment.) Your identification of the group matches my guess. Do you have a reference for it? If so, you should post it as an answer, perhaps. $\endgroup$ – Harald Hanche-Olsen Jul 8 '14 at 9:05
  • $\begingroup$ Oh, you are right: I missed the assumption that the tiles where "square". I will then write an answer when I have time. $\endgroup$ – Moishe Kohan Jul 8 '14 at 9:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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