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Ivo, the point is that with the Lorentz metric the structure equations change. I'm going to give you the appropriate moving frames computation without using an explicit parametrization. Consider the isometric group of the metric, $O(2,1)$. Let $e_0,e_1,e_2$ be a moving frame on your surface $M$, so that $e_0$ projects to $x\in M$ and $e_1,e_2$ project to ...


2

First, this is not true universally. For this to be true one must require that $\Delta$ is not only discrete, but that it acts freely, meaning that each nontrivial element of $\Delta$ has no fixed points in $\mathbb{H}^2$. But second, as long as $\Delta$ is discrete and free, then this follows from the theory of covering spaces: the quotient map $f : ...


1

There is another common model where the points of the projective plane consist of the points of the Euclidean plane and equivalence classes of lines for the equivalence relation "is parallel to". The mental image is that the latter types of points are to be thought of as the "point at infinity" that the class of parallel lines intersects at. (and the lines ...


1

If you know the three angles, you can get the lengths of the sides, using the Hyperbolic analog of the formulas for spherical triangles. You do have to be careful, though, I think that in the Law of Cosines, there may be a minus sign slipped in. Sorry to be so sketchy in this answer, I have a ton of other things hanging over me.


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The group $\Gamma = \langle \gamma_1,\gamma_2\rangle$ is not discrete. To prove this, consider the "height map" $\mathbb{H}^2 \to \mathbb{R}$ defined by $f(x,y) = \ln(y)$. The inverse images of points under this map are the horocycles $y = \text{(constant)}$ all based at $\infty$, each of which is preserved by both $\gamma_1$ and $\gamma_2$, and therefore ...


1

Ideal quadrilaterals have vertices on the boundary of hyperbolic space, so in the upper half-plane model the vertices are real (or at $\infty$). Three of those you can always normalize by a hyperbolic isometry (i.e., a linear fractional transformation), the fourth one is then uniquely determined.


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I've just realized by skipping to the text ahead that the cross product with respect to the bilinear form b is not the same as the standard cross product on R3. How would, in general, one find a cross product with respect to the given bilinear form? It's not too hard to answer this, provided one doesn't mind a little abstract linear algebra. Given a ...


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The total perimeter will be infinite. You can see that like this: start with one polygon in the center of the Poincaré disk. Now add to that every adjacent polygon, i.e. every polygon which has at least one vertex in common with the central one. The outer boundary will be a sequence of hyperbolic line segments which goes once around your central polygon. So ...


1

Poincaré model I'm not sure what $\mathcal P$ is in your notation. The Poincaré disk, as a subset of $\mathbb C$? If you restrict yourself to the Poincaré disk, then the midpoint $M$ of the geodesic segment $AB$ would be a point on the line $AB$ which has equal and finite hyperbolic distance from both $A$ and $B$. In your specific case, the line $AB$ ...



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