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Here's a partial answer, which is all I have time for at the moment: if it is a subgroup, its index must equal 3. Probably that can be used to either narrow down the possibilities for $\alpha$, and perhaps to rule it out altogether. Here's the proof that the index equals $3$. By the Gauss-Bonnet theorem, the area of a hyperbolic triangle with angles $a,b,c$ ...

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If you wish to study flat structures on the torus $T^2 = \mathbb{R}^2 / \mathbb{Z}^2$, i.e. Euclidean metrics on $T^2$, it comes down to studying lattices in $\mathbb{R}^2$, i.e. discrete subgroups isomorphic to $\mathbb{Z}^2$. Lattices can be normalized in some fashion. One useful normalization is to rotate and scale the lattice so that $(1,0)$ is a ...

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Assume that $M$ is the midpoint of $XY$. Then $d(M,X)=d(M,Y)$ has to hold, so: $$\frac{XQ\cdot MP}{XP\cdot MQ} = \frac{MQ\cdot YP}{MP\cdot YQ}$$ has to hold, and we must have: $$\left(\frac{MP}{MQ}\right)^2 = \frac{XP\cdot YP}{XQ\cdot YQ}$$ so it is quite easy to construct the midpoint of $XY$ with straightedge and compass, but in general it is not the ...

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If by "characteristic lines" you mean "geodesics", then this does not happen. All triangles have angle sum strictly less than $180^\circ$, by the Gauss-Bonnet theorem. Added: Your comment seems to indicate that you are asking not about "geodesics" but simply about arbitrary smooth curves. If so, then there are almost no restrictions on the angle sum ...

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Yes, this is correct. To extend an isometry of $\mathbb{H}^2$ to one of $\mathbb{L}^3$, one approach is to note that the isometries of Minkowski space are in fact linear transformations of the underlying vector space, so that once you know how they act on three linearly independent vectors, you know how to extend to all of $\mathbb{L}^3$. (You can also ...

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I am sorry, my comments suggesting that there was no such subgroup were wrong, and I have deleted them. If $G = \langle a,b \mid a^3, b^3, (ab)^\alpha \rangle$, then the subgroup $H = \langle x,y \rangle$ with $x=ab$ and $y=ba$ has index $3$ in $G$, and has the presentation $\langle x,y \mid x^\alpha,y^\alpha,(xy)^\alpha \rangle$. This is is easy to prove ...

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Assume that the angles of your triangles are all of the form $\pi/n$, $n\in {\mathbb N} - \{1\}$, of course is allowed to be different for different angles and triangles. Given a triangle $T$ with angles of this form, consider first the reflection group $\Gamma=\Gamma_T< Isom(H^2)$ generated by reflections in the edges of $T$. Then $T$ is a fudnamental ...

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The expression $ds=\frac{|dz|}{(1+|z|^2)}$ is an instruction for how to measure the length of a smooth (actually rectifiable) curve $\gamma: [a,b]\rightarrow \mathbb{C}$. When $\gamma$ is smooth, this length is just $$L(\gamma)=\int_\gamma \ ds =\int_a^b \gamma^*ds=\int_a^b\frac{|\gamma'(t)|}{1+|\gamma(t)|^2}dt.$$ A conformal metric on a domain $U\subset ... 1 The previous answer points to the paper http://arxiv.org/abs/1005.4442, by Gemmer and Venkataramani [GV], which tells us that the radius requested is$1.2654\$. But the previous answer makes some claims about the GV paper that might mislead. The GV paper shows that hyperbolic disks of arbitrarily large radius can be isometrically and real-analytically ...

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