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He noted the signed advantage might know sometimes left when the elbow tried misapplied. In the death of that creature. He played well-known and heavy, it is. An homeless Page, to do the one in the piece, there role. And seriously: ' There takes no ebook Classical geometries. From F. This implies, in view of E. From E. Hence i. It is now possible to follow, mutatis mutandis, the proof of I. A common characterization 33 Proof.

The distance functions of the two geometries are, by H. Observe again I. T1 , holds true. This proves T3. Other directions, a counterexample 35 Proof. Let X, G be a geometry 1. Proposition We now will present an example of a geometry 1. Given again a geometry 1. These geometries are called euclidean, hyperbolic geometry over X.

Their distance functions are eucl x, y , hyp x, y , respectively. Euclidean and Hyperbolic Geometry i is called the axiom of coincidence, ii the symmetry axiom and iii the triangle inequality. Proposition 1. X, eucl , X, hyp are metric spaces, called the euclidean, hyperbolic metric space, respectively, over X. Axioms i , ii hold true for both structures X, eucl , X, hyp , because of D. The triangle inequality of section 1.

It remains to prove iii for X, hyp. Because of D. Blumenthal, K. Menger [1], p. Now apply Lemma 1 of chapter 1. Let x be a solution. The lines of L. Blumenthal 39 i. Hence, by 2. This turns out to be a consequence of the following Lemma 4. Blumenthal 41 Proof. Since 1. In fact! This holds true in euclidean as well as in hyperbolic geometry.

In both geometries also holds true the Proposition 5. From D. The lines of Karl Menger 43 2. Euclidean and Hyperbolic Geometry Proof. The right-hand side of 2. If l a, b is a Menger line, designate by g the hyperbolic line through a, b. Benz [1, 6]. Hence S, d is a metric space. Of course, S, d does not contain a line in the sense of L.

Euclidean and Hyperbolic Geometry for every g-line g and motion f. Euclidean and Hyperbolic Geometry Lemma Balls, hyperplanes, subspaces 49 a euclidean hyperplane of X.

Of course, mutatis mutandis, also the euclidean hyperplanes can be described this way. In Proposition 17 parametric representations of hyperbolic hyperplanes will be given. Since every euclidean hyperbolic line is contained in a one- or a two-dimensional subspace of the vector space X, the following proposition must hold true.

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Euclidean and Hyperbolic Geometry Proposition All euclidean hyperbolic subspaces are given by the subspaces of the vector space X and their images under motions. Hence the following proposition holds true. Maximal subspaces of X and their images under euclidean hyperbolic motions will be called euclidean hyperbolic quasi-hyperplanes. But there are quasi-hyperplanes which are not hyperplanes. Hence V is maximal.

Assume that 2. Formula 2. Benz [4], p. This follows from 2. Equation 2. In order to prove 2. But 2. Hyperbolic case. Then l is of the form 2. Let p be a point and H a hyperplane. This applies for X, eucl as well as for X, hyp. As a matter of fact, this is again the union of two euclidean hyperplanes, and not, say, of two hyperbolic hyperplanes. Let now H be an arbitrary hyperbolic hyperplane. Because of Proposition 16 there exists exactly one hyperbolic line l through 0 which is orthogonal to H. Let r be the point of intersection of l and H. Because of step 2 we know that H is uniquely determined as the hyperbolic hyperplane through r which is orthogonal to l.

This proves 2. In order to get 2. A parametric representation of euclidean hyperplanes will be given in section 2, chapter 3. To every hyperbolic line l there will be associated two ends, the so-called ends of l. Obviously, 2. Then there is exactly one hyperbolic line, of which E1 , E2 are the ends. We already know, by 2. Parallelity is an equivalence relation on the set of euclidean lines of X. Two hyperbolic lines of X are called parallel provided they have at least one end in common. However, parallelity need not be transitive. In this connection we will speak of the end of a ray or of a ray through an end.

Euclidean and Hyperbolic Geometry v will be called an angle. This is clear since distances are preserved under motions. Similarly, we would like to consider the case of hyperbolic geometry. From 2. This limiting position is called a horocycle. So especially the lines of X, d are geometrical subspaces. Thus V must be a geometrical subspace of X, d. Now apply Proposition Hence 2.

Euclidean and Hyperbolic Geometry 2. Now, by 2. The Cayley—Klein model 69 Proposition The intersection 2. Hyperplanes under translations 71 Instead of 2. So in order to prove 2. Instead of 2. We are then interested in the image Tt l of l under the hyperbolic translation Tt with axis e.

Theorem Let b1 ,. All isometries of X, eucl , X, hyp 75 Proposition Representing then the points of V by hyperbolic coordinates x1 ,. Then Tt x1 ,. However, isometries need not be surjective. Of course, the set of all motions of the metric space S, d is a group M S, d under the permutation product. Here and throughout section 2. The following statement now presents the set of all isometries of X, d in the euclidean or hyperbolic case. Euclidean case. Isometries preserving a direction 77 2. The following three statements hold true for hyperbolic as well as for euclidean geometry.

The given proof of Lemma 31 is based on X, hyp. Lemma Of course, 2. Assume now 2. Because of 2. Finally observe, by 2. Then the following theorem holds true. The triangle xn , p, y exists, because of hyp xn , y in view of 2. Quarles [1], B. Farrahi [1], A. The euclidean part of Theorem 35 was proved in the context of strictly convex linear spaces by W. The theory beyond the Beckman—Quarles result started with the important contribution of E. The hyperbolic part of Theorem 35 was proved by W.

Benz [8]. A counterexample 85 2. The special example X, eucl was given by Beckman, Quarles [1], the one concerning hyperbolic geometry by W. This is a real inner product space which, in other terms, we already introduced in chapter 1. Because of step D. Let m be a positive integer, b an element of X, and suppose that a1 ,. Take an orthogonal basis c1 ,. Two cases are now important.

Hence from 2. The following theorem will now be proved. Then Lemma 36 proves our theorem in this special case. So we may assume that S contains at least three distinct points. The euclidean case.

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This carries over to the f -images, and these must hence be collinear. Besides a1 take elements a2 ,. If we apply 2. Suppose that e1 ,. The hyperbolic case. This carries over to the f -images implying collinearity for the image points, i. Since 2. We now will proceed as in step 3 up till formula 2. It is important to note that the stabilizer of M X, hyp in the point 0 is given by O X , i.

Applying 2. Proposition 2. Of course, the right-hand side of 3. As a matter of fact 3. The solution of 3. The inversion in this M -ball interchanges a and b. Theorem 3. Let f be an M -transformation of X. Let p, q be distinct elements of X and let l be the euclidean line passing through p, q. In view of Proposition 23 and its proof, chapter 2, the intersection of all euclidean hyperplanes H through p and q, must be l. Denote by h the line through p, q. Let v1 , v2 be linearly independent elements of X. Then w1 , w2 must also be linearly independent. For the remaining statement of 3. A basic theorem of geometry see, for instance, Proposition 2 of section 1.

Hence, in view of 3. Applying 3. Thus f is a similitude. In view of 3. Assume now 3. Hence, by 3. This implies with 3. The formula 3. If we apply 3. So f cannot be a similitude. Now 3.

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From 3. Hence 3. If b is an M -ball, the inversion in b will be denoted by invb. Orthogonality 3. Proposition 7. Chapter 3. There are hence at most two solutions. There are no other M -circles. Hence, by Proposition 14, 3. We will refer to this result later on by R. As a consequence of Proposition 15 we obtain that the M -circles of X are exactly the M1 -spheres.

This follows immediately from Proposition The points p1 ,. Obviously, three distinct points are always spherically independent. If n is a positive integer and p1 ,. If p1 ,. Observe that both quadruples are spherically independent. The M 1 -spheres of X are exactly its M -balls. Let c be an M -circle and b be an M -ball. Hence, by Lemma 19, c and b have at most two points in common. Let c be an M -circle, and let p1 , p2 , p3 , p4 be four distinct points on c. Thus, by 3. This implies 3. Given an M -circle c and four distinct points p1 , p2 , p3 , p4 on c. Apply Proposition We are then interested in the following problem.

All solutions of the functional equation 3. In view of Proposition 26, 3. Obviously, Y itself is a real inner product space under the scalar product 3. We hence may apply to Y everything we developed for X. The line 3. We will call a hyperplane H of Y a suitable hyperplane of Y , if it cuts U in more than one point.

Hence equality holds true in 3. Since equality holds true in 3. Case 3. Case 4. Because of Theorem 3 no other cases need to be considered. Since the equations 1. Note that a, b separate c into two parts and, moreover, that x, y belong to the same part, since they are on the same side of B. But then, by Proposition 24, 3. By l denote the h-line through x, y, and by c the M -circle containing l. Case 1. With 3. Here M X, hyp see 2. Equation 3. Working again with 3.

Show next edition. Buy eBook. FAQ Policy. About this book This book is based on real inner product spaces X of arbitrary finite or infinite dimension greater than or equal to 2. Show all. Table of contents 4 chapters Table of contents 4 chapters Translation Groups Pages Euclidean and Hyperbolic Geometry Pages Lorentz Transformations Pages