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I would like to construct a genus $g$ surface regularly tiled by triangles (for example by 238 triangles). Edmunds-Ewing-Kulkarni prove that the only obstruction to doing this is Euler characteristic considerations.

However, I'd like to ensure that the fundamental domain for my tiling contains a ball of some radius. Ideally, I'd like a result of the form: given a radius $R>0$, there is some genus $g_N$ such that for every genus $g \geq g_N$ a surface of genus $g$ can be regularly tiled by triangles using a fundamental domain that contains a ball of radius $R$.

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    $\begingroup$ What do you mean by large here? Every triangle in $H^2$ has area bounded above by $\pi$, the area of an ideal triangle. However, you should be able to get surfaces with large injectivity radius by combining the facts that surface groups are residually, finite and with appropriate facts about the length spectrum. I think congruence covers of your surface should do the trick. $\endgroup$
    – Neil Hoffman
    Commented Dec 6, 2018 at 18:36
  • $\begingroup$ I think the following works for even Euler characteristics ($\chi$=2n): take a regular 12n-gon where each edge equals the radius of the inscribed circle. It can be split into 12n triangles. Glue 4n more of such triangles, each to edges k, 4n+k, 8n+k. (This is a generalization of Schmutz's M(2) surface.) $\endgroup$
    – Zeno Rogue
    Commented Dec 6, 2018 at 18:49
  • $\begingroup$ Sorry, the Euler characteristics is 4-8n, so not every odd genus works; and this is M(3) not M(2). For Euler characteristics 6-12n, a similar generalization of Schmutz's M(4). We also take a 12n-gon, split into 12n triangles, glue another triangle to each edge, and then glue the right edge of triangle number i to the left edge of the triangle number i+5 (numbered clockwise). $\endgroup$
    – Zeno Rogue
    Commented Dec 6, 2018 at 19:38
  • $\begingroup$ @Neil By large ball, I mean a ball of radius R triangles (measure out R triangles from the center). Congruence subgroups do give a large injectivity radius, which is one way of constructing the type of fundamental domains that I want. However, in general congruence subgroups are rather sparse. I want to show I can do this in every genus larger than some given genus, congruence subgroups will not give every genus $\endgroup$
    – Arielle Leitner
    Commented Dec 7, 2018 at 8:12
  • $\begingroup$ @Zeno I want the tiling to be preserved (say 238 triangulation) I don't think your construction does this $\endgroup$
    – Arielle Leitner
    Commented Dec 7, 2018 at 8:13

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One can answer your question positively, although I haven't tried to compute $g_R$ as a function of $R$.

One can tessellate a certain right-angled pentagon by $238$ triangles. Moreover, a genus 2 surface is an index 8 orbifold cover of the right-angle pentagon orbifold (i.e. an index 8 torsion free subgroup of the reflection group in a right-angled pentagon is a genus 2 surface group).

enter image description here

The area of a right-angled pentagon is $\pi/2$, and by Gauss-Bonnet the area of a genus $g$ surface is $4\pi(g-1)$, so one wants to find tessellations of a right-angled polygon by $g-1$ right-angled pentagons. The reflection group in this polygon will be an index $g-1$ subgroup of the right-angled pentagon reflection group. Then the induced index 8 subgroup will be a genus $g$ surface group.

Thus, we'd like to know for a given $g$, what is the right-angled polygon tessellated by right-angled pentagons which contains a disk of radius $R$? Eight copies of this polygon will give a fundamental domain for the surface group, and hence have the properties you desire.

As in Peter Scott's proof of the residual finiteness of surface groups, given a disk of radius $R$, we may take its convex hull with respect to the lines in the pentagonal tiling.

This will take some number of pentagons which is roughly exponential in $R$. As $R$ increases, we will get a sequence of linear growth $R_n$ with tilings containing $s_n$ pentagon tiles and containing a disk of radius $R_n$. In this diagram, the polygons are made of a convex union of pentagons containing each circle (note, the pentagon made of 238 triangles has a different shape since it's not regular, this is just for demonstration, but would still yield $R_n$ up to quasi-isometry):

enter image description here

To interpolate between the $s_n$, a simple thing one can do is add a string of right-pentagons onto a free boundary component. Doing this, we can get any number of pentagons between $s_n$ and $s_{n+1}$, and hence for any number.

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    $\begingroup$ What are "238 triangles"? My guess would be "triangles with angles $\pi/2,\pi/3,\pi/8$", which I would rather call "(2,3,8)-trangles", in particular to avoid confusion with cardinal 238. $\endgroup$
    – Victor Protsak
    Commented May 13, 2019 at 19:29
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    $\begingroup$ Yes, I was just keeping with the asker's terminology. $\endgroup$
    – Ian Agol
    Commented May 13, 2019 at 19:41

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