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Timeline for Quadrilateral fundamental domain

Current License: CC BY-SA 4.0

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Jul 14, 2019 at 13:51 comment added Arun Ok, I edited my question. However I got my answer from another version of Poncare polygon theorem. Thx anyway!
Jul 14, 2019 at 10:10 history edited Arun CC BY-SA 4.0
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Jul 14, 2019 at 9:34 comment added Misha Your last statement is unclear. Of course, I am not saying that "most" quadrilaterals are fundamental domains. What I said is that the space of such quadrilaterals is real 2-dimensional; from this one concludes that there exists a 1-parameter family of fundamental quadrilaterals whose angles are non-constant. Regardless, you should revise your question to make it answerable, in the present form, it does not make much sense.
Jul 14, 2019 at 9:15 comment added Arun Thx! Most of the polygons does not satisfy the angle sum condition, since side paring are not the only isometries that can generate a Fuchsian group, so a polygon may still be a fundamental domain,
Jul 14, 2019 at 8:48 comment added Misha Yes, this would be wrong. You are not taking into account all possible side-pairing patters. You can take (orientation-preserving) side-pairing isometries pairing opposite sides of the quadrilateral; the condition then is that the angle sum of $P$ is of the form $2\pi/n$ where $n$ is a natural number. This does not imply much about individual angles of $P$. All in all, you will have 2-dimensional space of congruence classes of $P$ which yield fundamental domains.
Jul 14, 2019 at 8:31 comment added Arun correct me if I am wrong, a polygon with an angle irrational multiple of $\pi$ can not be a fundamental polygon, but how about quadrilateral with angles $2\pi/3, 2\pi/6, 2 \pi/6, 2\pi/6$.
Jul 14, 2019 at 7:00 comment added Misha What angles did you find? Poincare's theorem used properly yields necessary and sufficient conditions.
Jul 14, 2019 at 6:10 history edited Arun CC BY-SA 4.0
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Jul 13, 2019 at 19:39 history edited YCor
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Jul 13, 2019 at 19:08 history asked Arun CC BY-SA 4.0