Let $S$ be a Riemann Surface of genus 2. Is there a picture in the literature for a regular tiling of $S$ by 12 heptagons (where three heptagons meet at each vertex)? Also, apart from the obvious restriction given by the Euler characteristic $2-2g=f-nf/2+nf/v$ (where $g$ is the genus, $f$ is the number of faces, $v$ is the number of faces meeting at each vertex) are there any obstructions for the existence of a tiling of a surface of genus $g$ by $n$-gons (where the same number $v$ of $n$-gons meet at each vertex).

I know that such a tiling exists (for a surface of genus 2 by heptagons), but I am unable to make a drawing. The construction goes like this (thanks to Mladen Bestvina): Start with a sphere. View this as a octahedron so that you have 8 triangles. Pick a hexagon (start from the top go down, left, down come back up from the opposite side) We fatten up these 6 sides to look like circles with one ray coming from the middle. You should think of these as the quotient of 2 triangles attached along one edge <|> via rotation by 180 degrees (so you get two edges meeting in a point A and 2 edges plus a ray meeting in the other point B). We call the center of the circle C so that the ray goes from C to B.
You have to view the point $B_i$ (of the $i$-th side, $i=1,2,3,4,5,6$) as being attached to $A_{i+1}$ for $i=1,2,3,4,5,6$ (of course mod 6) so each of these vertices has valence $5+2=7$. Now you take the double cover with fixed points $C_1,...,C_6$ and you get a surface of genus 2 with 28 triangles (28=2x8+2x6 where 8 is the number of triangles and 6 are the circles with a ray that unwrap in to two triangles). The vertices still have valence 7 (because the cover is étale here). Take the dual tiling and you are done!

This is just a curiosity. I tried asking the first part of this on stackexchange with no luck. I am also aware of the beautiful pictures of the famous tiling of a surface of genus 3 by 24 heptagons but you can not use that to obtain the tiling of a genus two surface in an obvious way.


2 Answers 2


Addressing the second question positively, this is the Main Theorem of this paper:

Edmonds, Allan L.; Ewing, John H.; Kulkarni, Ravi S., Regular tessellations of surfaces and (p,q,2)-triangle groups, Ann. Math. (2) 116, 113-132 (1982). ZBL0497.57001. MR0662119

In fact, they show this also works for non-orientable surfaces, except for the non-existence of a degree 3 triangulation of the projective plane.


A picture is in the comments to John Baez' Blog Post. EDIT For the second part of the question, no there is no obstruction, as long as the implied angle ($2\pi/v$) is smaller than the angle of a Euclidean regular $n$-gon. This is true by an obvious argument I had originally heard from Bill Thurston: a very small hyperbolic regular $n$-gon has angles close to Euclidean, a very large regular hyperbolic $n$-gon has angles $0$ so, by continuity, somewhere in between you will have the angle you want.

  • $\begingroup$ I could see lot's of beautiful pictures of tilings of a genus 3 surface, but I couldn't find the genus 2, but there are many comments so maybe I missed it; could you tell me the date of the comment so I can find the picture? Also, do you know the answer to the second part of the question? $\endgroup$
    – Hacon
    Feb 20, 2015 at 16:02
  • $\begingroup$ @Hacon Sure, it's 29 May 2013. $\endgroup$
    – Igor Rivin
    Feb 20, 2015 at 16:28
  • $\begingroup$ @Hacon And I edited to answer the second part of the question. $\endgroup$
    – Igor Rivin
    Feb 20, 2015 at 16:31
  • $\begingroup$ I see it now. Very nice! thank you. $\endgroup$
    – Hacon
    Feb 20, 2015 at 16:41
  • $\begingroup$ Still that picture is somewhat incomplete in that the surface as shown is in fact one of infinitely many fundamental domains for a group action on an infinite (noncompact) covering of the needed surface. I wonder if there is a satisfactory picture of the surface itself somewhere. $\endgroup$ Jun 19, 2015 at 6:19

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