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In the Euclidean space, sides of polygons are lines. In the hyperbolic space, sides of polygons are hyperbolic lines; in the ball model these are the diameters of the ball and spherical arcs that meet the boundary orthogonally. The sides always bend in in the ball model, as compared to the Euclidean triangle with same corners. Isometries conserve hyperbolic ...

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$$\left( \begin{array}{rr} \sqrt k & 0 \\ 0 & \frac{1}{\sqrt k} \end{array} \right)$$

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The scaling map $z \mapsto kz$ for $k > 0$ can be identified with $\begin{pmatrix} \sqrt{k} & 0 \\ 0 & 1/\sqrt{k} \end{pmatrix} \in PSL(2,\mathbb{R})$ because it acts by fractional linear transformation as $$\begin{pmatrix} \sqrt{k} & 0 \\ 0 & 1/\sqrt{k} \end{pmatrix} z = \frac{\sqrt{k} z + 0}{0 + \frac{1}{\sqrt{k}}} = k z.$$

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In algebraic books, especially those dealing with quadratic forms, there is a usage in English for "hyperbolic plane" that has nothing to do with non-Euclidean Geometry. From Rational Quadratic Forms by Cassels, it means a 2-dimensional quadratic space $U,\phi$ with a basis $u_1,u_2$ such that $$\phi(u_1) = \phi(u_2) =0, \; \; \; \phi(u_1, u_2) = 1.$$ ...

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For infinitesimal lengths, hyperbolic and Euclidean metrics are the same. (This follows from the fact that the ratio between squared lengths and Gaussian curvature becomes zero.) So perhaps the author meant the following argument: for infinitesimal lengths, the hyperbolic metric and the Euclidean one agree, and the latter is proportional to first order, so ...

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The upper bound, though I have not seen it written in precisely this form, is a quick consequence of the thinness of triangles. Here is something from A Course in Metric Geometry by Burago, Burago, Ivanov: Lemma 8.4.2 A shortest path $[ab]$ belongs to the $(d(b,c)+\delta)$-neighborhood of a shortest path $[ac]$. The authors omit the proof, which ...

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