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Yes, this is indeed a form of duality. The following outline follows the approach from Perspectives on Projective Geometry by J. Richter-Gebert. Cayley-Klein metric Probably the best way to understand this is probably using Cayley-Klein metrics. That's a very general way to measure distances and lengths in projective geometry. You start by fixing one ...

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The usefulness of the PoincarĂ© model (or of Klein's) is that it shows that if Euclidean geometry is consistent, then also hyperbolic geometry is. This is because the hyperbolic axioms are true in the model, so a contradiction in hyperbolic geometry would yield a contradiction in Euclidean geometry. It would be wrong to prove theorems in hyperbolic geometry ...

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No. There is an infinite-dimensional space of homeomorphisms of the circle; the space of Mobius transformations is the 3-dimensional $PSL_2(\Bbb R)$. But more pointedly, the restriction of a Mobius transformation to the ideal boundary can have at most two fixed points if it's not the identity; but any closed subset of $S^1$ is the set of fixed points of some ...

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As proposed in the comments, I posted this question on MathOverflow and it got an answer there. Here is the link: The question on MathOverflow (answered) Thanks to everyone who read the question and thought about it.

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That is not true. A simple example is the following one: consider the hyperbolic isometry \begin{equation*} \phi:= \left[ \begin{array}{l l} 4 &0\\ 0 &\frac{1}{4}\\ \end{array} \right] \in\text{PSL}(2,\mathbb{R})=\text{Isom}^+\left(\mathbb{H}^2\right) \end{equation*} and the finite order elliptic element \begin{equation*} \rho:= \left[ \begin{...

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Any closed surface $\Sigma$ of genus $g \ge 2$ admits a hyperbolic metric, and hence is a quotient $\mathbb{H}/\Gamma$ of the upper half plane $\mathbb{H}$ by a discrete group $\Gamma$ of isometries acting on $\mathbb{H}$ which can be identified, as an abstract group, with the fundamental group $\pi_1(\Sigma)$. The isometry group of $\mathbb{H}$ is $PSL_2(\... 2 There is an absolutely fascinating little booklet called "Hyperbolic Functions" by V. G. Shervatov in which the author develops circular and hyperbolic functions in parallel from a purely geometric viewpoint. It is from the "Russian Series In Mathematics" and was written decades ago (1950s, I think) and is out of print, but is still out there if you search ... 2 All in an earlier post Definition of hyperbolic lenght. I was wondering about the translation. What is an equidistand line? I guess you mean a hypercycle. https://en.wikipedia.org/wiki/Hypercycle_(geometry) But an hypercycle does not have a lot in common with a line. (see the wikipedia article) Calling an horocycle an "a circle with infinite radius ... 2 One way to calculate this is by considering suitable triangles inside the dodecahedron and using hyperbolic trigonometry. This is almost the same as you could do in euclidean geometry, only you start with some triangle with fixed angles which fixes its lengths in hyperbolic geometry. It is a valid approach, however this probably leads to very ugly ... 2 Some comments in the form of an answer The PoincarĂ© disk model is only a model of the hyperbolic geometry. There are other models as well. For example the Klein disk model uses euclidean straits falling within an open disk to model the hyperbolic straights. (if this answers your question.) We do not necessarily need euclidean objects if we want to study ... 2 First, note that$\{\frac{\partial}{\partial x_i}\}_{i=1}^n$is the basis of the tangent space at any point of$\mathbb{H}$. Hence, it follows immediately from the definition of the metric that $$g_{ij}=g(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j})=\frac{\delta_{ij}}{x_n^2}.$$ From this, we can calculate$\frac{\partial}{\partial x_k}g_{ij}$... 2 Because$\mathbb{H}$has many symmetries doing it for one geodesic is enough. You can use an appropriate Mobius transformation,$g$, to realize any geodesic$\gamma$of$\mathbb{H}$as$\gamma(t)=g\cdot ie^{-t}$. Edit: I saw in your question you didn't want to use explicit geodesics, so you could also do something like this (to see why the denominator ... 1 A bit silly, since as mentioned, you can prove it by hand, but: covers of complete manifolds are complete, and compact manifolds are automatically complete. There are a great deal of hyperbolic structures on$\Sigma_g$,$g>1$. 1$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$tl; dr: There's a hyperboloid model of the projective plane that is easily shown (using elementary linear algebra, see sketch below) to be homogeneous and isotropic under its isometry group. That's enough to guarantee its Gaussian curvature is constant. (In fact, every ... 1 All correct (I think I wrote that on wikipedia) $$d(O, Q) = \frac{1}{2} \ln \frac{1+r}{1-r} = \operatorname{artanh} r$$ see also the formulas for tanh and artanh at en.wikipedia.org/wiki/Hyperbolic_function I have added it to the wikipedia page ps related: For the Poincare disk model he hyperbolic distance between O the center of unit disk and a point ... 1 It is$\mathbb{H}^2 \times \mathbb{R}$geometry. In general, given two metric spaces$X$and$Y$with metrics denoted$d_X$and$d_Y$, the product metric space is defined as a set to be the Cartesian product$X \times Y$with the metric$d_{X \times Y}$which, for each$p=(x_0,x_1)$and$q=(y_0,y_1) \in X \times Y$is defined by the formula$\$d_{X \times Y}...

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