By E. Askwith

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Euclidean and Hyperbolic Geometry Lemma 10. Let a = 0 be an element of X and τ > 1 be a real number. 17) 1 + a2 . τ 2α = (τ − 1) Proof. Since {x√∈ X | x −√ce− + x − ce get with c := a τ , := ln τ , obviously, a = ce− , τ a = ce , 2α = (e − e− ) = 2 sinh · √ 1 + c2 } is B (c, ), we 1 + c2 = (τ − 1) 1 + a2 . τ Proposition 11. Suppose that B (c, ), B (c , ) are hyperbolic balls satisfying and B (c, ) ⊆ B (c , ). 18) . Proof. 18), c = c and > 0. a, a motion µ such that µ (c) = 0, µ (c ) = λj, λ > 0.

5. Balls, hyperplanes, subspaces 49 a euclidean hyperplane of X. If e ∈ X satisﬁes e2 = 1, if t ∈ R and ω1 , ω2 ∈ O (X), then ω1 Tt ω2 (e⊥ ) = {ω1 Tt ω2 (x) | x ∈ e⊥ } will be called a hyperbolic √ hyperplane, where {Tt | t ∈ R} is based on the axis e and the kernel sinh · 1 + h2 . Of course, mutatis mutandis, also the euclidean hyperplanes can be described this way. In Proposition 17 parametric representations of hyperbolic hyperplanes will be given. Proposition 12. If H (a, α) and H (b, β) are euclidean hyperplanes with H (a, α) ⊆ H (b, β), then H (a, α) = H (b, β) and there exists a real λ = 0 with b = λa and β = λα.

E. 42). 42). e. 41), α (1 + δβ) − β (1 + δα) 2 = ϕ2 (ξ − η). e. α (1 + δβ) > β (1 + δα). 45) 30 Chapter 1. Translation Groups holds true for all ξ > η ≥ 0. 45) also holds true for ξ = η ≥ 0. a), we will distinguish two cases, namely δ = 0 and δ > 0. 45) yields ϕ (ξ − η) = ϕ (ξ) − ϕ (η) for all ξ ≥ η ≥ 0. Given arbitrarily t, s ∈ R≥0 , put ξ := t + s and η := s. Hence ξ ≥ η ≥ 0 and thus ϕ (t + s) = ϕ (t) + ϕ (s). 46) for all t ∈ R≥0 with a constant l > 0, in view of ϕ (1) > ϕ (0) = 0. a we get ψ (h) = 1 for all h ∈ H.