By Walter Benz

In accordance with actual internal product areas X of arbitrary (finite or endless) size more than or equivalent to two, this e-book comprises proofs of more moderen theorems, characterizing isometries and Lorentz changes below light hypotheses, like for example limitless dimensional models of recognized theorems of A D Alexandrov on Lorentz transformations.

summary: in keeping with genuine internal product areas X of arbitrary (finite or endless) size more than or equivalent to two, this e-book contains proofs of more recent theorems, characterizing isometries and Lorentz ameliorations less than light hypotheses, like for example countless dimensional types of well-known theorems of A D Alexandrov on Lorentz ameliorations

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**Sample text**

If q = 0 is in X and s in R, there exist a, b ∈ X with ab = 0, b2 = 1 and {a cosh η + b sinh η | η ∈ R} = {Ts (µq) | µ ∈ R}. 2. M. Blumenthal 41 Proof. There is nothing to prove for q ∈ Re or s = 0. So assume s = 0, and that q, e are linearly independent. Hence q = (qe) e. Because of {Ts (µq) | µ ∈ R} = {Ts (µ · βq) | µ ∈ R} for a ﬁxed real β = 0, we may assume q − (qe) e = 1, without loss of generality. Put S := sinh s, C := cosh s, j := q − (qe) e, α := qe, and observe S = 0, C > 1, j 2 = 1, je = 0, q = αe + j, q 2 = 1 + α2 .

I) for ϕ : G × N → N . Notice L = ϕ τ (f g), ν (l) = ν ϕ (f g, l) = ν ϕ f, ϕ (g, l) = ϕ τ (f ), ν [ϕ (g, l)] = ϕ τ (f ), ϕ [τ (g), ν (l)] = R. Now let h:N →W be an invariant of (S, G) based on the invariant notion (N, ϕ) of (S, G). We then would like to deﬁne an invariant h :N →W of (S , G ). ) Put h (l ) := h ν −1 (l ) for all l ∈ N , by observing that ν : N → N is a bijection. Then h ϕ τ (g), ν (l) = h ν ϕ (g, l) = h ϕ (g, l) = h (l) = h ν (l) . h is hence an invariant of (S , G ). If we rewrite the deﬁnition of ϕ , namely ϕ τ (g), ν (l) = ν ϕ (g, l) , by using the abbreviations ϕ (g, l) =: g (l) and ϕ τ (g), ν (l) =: τ (g) ν (l) , we get τ (g) ν (l) = ν g (l) for all l ∈ N and g ∈ G.

B) The constant k of statement C is positive. c) d (x, y) = d (y, x) for all x, y ∈ X. d) If x, y ∈ X, then d (x, y) = 0 if, and only if, x = y. Proof. a) Because of A there exists ω2 ∈ O (X) with ω2 (x) = x e. Since ||x||e − 0 ∈ Re, (T2) implies the existence of t ∈ R with Tt ( x e) = 0. Finally take ω1 ∈ O (X) with ω1 (z) = λe, λ := z ≥ 0, where z := Tt ω2 (y). Hence g (x) = 0 and g (y) = λe with g := ω1 Tt ω2 . Since x = y, we obtain λ > 0. This proves a). b) The distance function d is assumed to be not identically 0.