Timeline for Regarding fundamental domain of 2 genus surface
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 4, 2022 at 11:17 | comment | added | მამუკა ჯიბლაძე | I don't know. Since all vertices of the octagon must come together into a single vertex, this may be only that where 14 triangles of the tiling meet. These 14 sectors must be grouped into 8 bunches, presumably in a symmetric way. There are not so many possibilities to try - e. g., going around the vertex, 1,2,2,2,1,2,2,2 sectors. Or, 1,2,1,3,1,2,1,3 or also 1,1,2,3,1,1,2,3. Or, 1,1,1,4,1,1,1,4. I think that's it. But I don't really understand how these remaining four hexagonal holes come together, so I cannot check whether those cuttings may be made compatible. | |
| Sep 2, 2022 at 6:13 | comment | added | KAK | @მამუკაჯიბლაძე will the sides of the octagon necessarily pass along the edges of the tiling? | |
| Aug 30, 2022 at 10:20 | comment | added | მამუკა ჯიბლაძე | No, it does not. Actually irregular such tilings are known. One of the answers to the question I linked contains a link to a picture - unfortunately with four holes left unglued | |
| Aug 30, 2022 at 9:46 | comment | added | KAK | @მამუკაჯიბლაძე Does this imply that there does not exist (2,3,7) tiling on genus two compact orientable surface? | |
| Aug 30, 2022 at 6:57 | comment | added | მამუკა ჯიბლაძე | In fact I believe the (2,3,7)-triangle group does not have any torsion free genus 2 subgroups. In the paper Geometric uniformization in genus 2 (T. Kuusalo and M. Näätänen, Ann. Acad. Sci. Fenn. Ser. A. I. 20, 1995, 401-418) it is proved that the only triangle groups having such subgroups are (2, 3, 8), (2, 4, 6), (2, 4, 8), (2, 5, 10), (2, 6, 6), (2, 8, 8), (3, 3, 4), (3, 4, 4), (3, 6, 6), (4, 4, 4) and (5, 5, 5). | |
| Aug 30, 2022 at 6:46 | comment | added | მამუკა ჯიბლაძე | Let me add that answers to the question regular tiling of a surface of genus 2 by heptagons might help | |
| Aug 30, 2022 at 6:40 | comment | added | KAK | Yes. The octagon need not be regular. | |
| Aug 30, 2022 at 6:39 | comment | added | მამუკა ჯიბლაძე | A trivial remark - if the fundamental domain is a regular (hyperbolic) octagon, it cannot be assembled from the whole (2,3,7)-triangles since angles of the octagon are $\pi/4$ and you cannot combine $\pi/2$, $\pi/3$ and $\pi/7$ angles into a $\pi/4$ | |
| Aug 30, 2022 at 3:36 | history | asked | KAK | CC BY-SA 4.0 |
