I am trying to construct a compact hyperbolic surface tessellated with hyperbolic right-angled polygons with $n \ge 5 $ edges. I found quite easily a way to do it for $n$ even, but the odd case seems more tricky. For example, if n=5, my idea was to take two right-angled pentagons, with edges say $a$, $b$, $c$, $d$, $e$, and glue the couples of edges $(a,a)$, $(c,c)$ and $(d,e)$ preserving the orientation, and $(e,d)$ reversing it. My guess is that I should obtain a torus without a disk, that is, with one boundary component. However, I cannot see the tessellation on the torus, which makes me wonder if my construction actually works.

  • $\begingroup$ Idle thought: can you construct a 'symmetric' tiling of a figure-eight (genus-2 surface) and then cut it down the middle to get the one you're after? $\endgroup$ Apr 20, 2020 at 21:53
  • $\begingroup$ The problem is that I cannot see the tiling I would like to have on the torus, so I cannot see it on the figure-eight either $\endgroup$ Apr 20, 2020 at 22:15
  • $\begingroup$ I am not sure if I understand your construction. Are there two corner points of each pentagon on the boundary after gluing? Then there are six right angles left to construct the interior vertices. That would be too much for one interior vertex, but not enough for two. $\endgroup$ Apr 22, 2020 at 11:34

1 Answer 1


If you glue four regular right-angled pentagons (side-lengths $\ell$) at one corner, you get a right-angled hyperbolic octogon with alternating side-lengths ($\ell$ and $2\ell$). You can glue either along pairs of opposite long edges, or along pairs of opposite short edges to get the figure you want.


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