# Representation of cell location in hyperbolic plane

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

• 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) – Willemien Jul 7 '14 at 21:42
• 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.) – Moishe Kohan Jul 8 '14 at 0:55

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.
• 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 $0^o$ – Willemien Jul 7 '14 at 23:16
• 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. – Moishe Kohan Jul 8 '14 at 1:23