Let P be a hyperbolic quadrilateral. Poincare polygon theorem provides sufficient condition for P to be a fundamental domain of some Fuchsian group in term of its inner angles. I find the angle sum condition in the theorem very restrictive. Is there less restrictive condition on angle sum for P to be a fundamental domain? For example, $P_1$ with angle sum $\pi/2+ \pi/2 + \pi/2+ \pi/3=11 \pi/6$ is not of the form $2\pi/n$ for any natural number $n$, But $P_1$ is a fundamental domain for Fuchsian group generated by reflections on edges of $P_1$.
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$\begingroup$ What angles did you find? Poincare's theorem used properly yields necessary and sufficient conditions. $\endgroup$– MishaCommented Jul 14, 2019 at 7:00
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$\begingroup$ 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 $2\pi/3, 2\pi/6, 2 \pi/6, 2\pi/6$. $\endgroup$– ArunCommented Jul 14, 2019 at 8:31
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$\begingroup$ 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. $\endgroup$– MishaCommented Jul 14, 2019 at 8:48
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$\begingroup$ 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, $\endgroup$– ArunCommented Jul 14, 2019 at 9:15
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$\begingroup$ 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. $\endgroup$– MishaCommented Jul 14, 2019 at 9:34
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