Timeline for fundamental domains in H^2 containing large balls
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 20, 2019 at 12:42 | vote | accept | Arielle Leitner | ||
| May 13, 2019 at 5:16 | answer | added | Ian Agol | timeline score: 3 | |
| Dec 7, 2018 at 14:47 | comment | added | Zeno Rogue | Indeed -- I did not understand what you meant by 238 (I thought it was the number of triangles) so I ignored this. | |
| Dec 7, 2018 at 8:13 | comment | added | Arielle Leitner | @Zeno I want the tiling to be preserved (say 238 triangulation) I don't think your construction does this | |
| Dec 7, 2018 at 8:12 | comment | added | Arielle Leitner | @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 | |
| Dec 6, 2018 at 19:38 | comment | added | Zeno Rogue | 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). | |
| Dec 6, 2018 at 18:49 | comment | added | Zeno Rogue | 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.) | |
| Dec 6, 2018 at 18:36 | comment | added | Neil Hoffman | 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. | |
| Dec 6, 2018 at 17:52 | history | asked | Arielle Leitner | CC BY-SA 4.0 |