Hyperbolic Geometry For Non-Differential Topologists

The n-dimensional (real) hyperbolic space is a metric space

, where Hⁿ(ℝ) = ℝⁿ and d_h is a metric (called hyperbolic) given by

(see (1.1)).

The above formula has its origin in differential/Riemannian geometry (most often it is defined as the length of a geodesic arc – so, to get it one needs to find geodesics and compute integrals related to them), see, e.g., [14] (consult also [1] and [2: Subsection 1.10], where (2.1) appears without derivation). We underline here – at the very beginning of our presentation – that establishing the triangle inequality for d_h using elementary methods is undoubtedly the most difficult part in the whole of this approach.

Our definition of Hⁿ(ℝ) is actually a somewhat modified Minkowski model of hyperbolic spaces. Indeed, in the last mentioned model one introduces a bilinear form

and deals with an n-dimensional hyperboloid

equipped with a metric d(x,y) = cosh⁻¹ (B(x,y)). It is straightforward to check that the function (H_n(ℝ), d_h) ∋ x ↦ ([x], x) ∈ (H, d) is a well defined bijective isometry.

As we will see in Corollary 5.2, the metric space (H¹(ℝ), d_h) is isometric to (ℝ, d_e). This contrasts with all other cases, as for n > 1 the metric space (Hⁿ(ℝ), d_h) admits no dilations other than isometries (see Theorem 3.5 below). In particular, for any integer n > 1 and positive real r ≠ 1 the metric spaces (Hⁿ(ℝ), d_h) and (Hⁿ(ℝ), rd_h) are non-isometric but homothetic. Each of the metrics rd_h (with fixed r > 0) may serve as a ‘standard’ hyperbolic metric. Actually, everything that will be proved in this paper about the metric spaces (Hⁿ(ℝ), d_h) remains true when the metric is replaced by rd_h.

Note also that, similarly to Euclidean geometry, for j < k the space H^j(ℝ) can naturally be considered as the subspace ℝ^j × {0}^k–j of H^k(ℝ). Under this identification, the hyperbolic metric of H^j(ℝ) coincides with the metric induced from the hyperbolic one of H^k(ℝ). This is the main reason why we ‘forget’ the dimension n in the notations ‘d_h’ and ‘d_e’.

The aim of this section is to show that d_h is a metric on ℝⁿ equivalent to d_e.

For anyx,y ∈ Hⁿ(ℝ), d_h(x,y) is well defined, non-negative andd_h(x,y) = d_h(y,x). Moreover, d_h(x,y) = 0 iff x = y.

Proof. It follows from the Schwarz inequality that

Equivalently, ∥x∥²∥y∥² + ∥x∥² −2 ⟨x,y⟩ + ∥y∥² ≥ 〈x, y〉², which gives 1 + ||x||² + ||y||² + ||x||²||y||² ≥ 1 + 2 〈x, y〉 + 〈x, y〉² and hence (1 + ∥x∥²)(1 + ∥y∥²) ≥ (1 + 〈x, y〉)². Taking square roots from both sides shows that [x][y] – 〈x, y〉 ≥ 1 and thus d_h(x, y) is well defined (and, of course, non-negative). Further, we see from (2.2) that [x][y] – ⟨x, y⟩ = 1 iff x = y, which gives the last claim of the lemma. Symmetry is trivial. □

To establish the triangle inequality, first we show its special case, which will be used later in the proof of a general case.

For anyx, y ∈ Hⁿ(ℝ), d_h(x, y) ≤ d_h(x, 0) + d_h (0, y), and equality appears iff eitherx = ty or y = tx for some t ≤ 0.

Proof. Observe that

On the other hand,

and equality holds in the above iff ⟨x, y⟩ = −∥x∥ ∥y∥, or, equivalently, if either x = ty or y = tx for some t ≤ 0. So, to complete the whole proof, we only need to check that

which is left to the reader as an elementary exercise. □

To get the triangle inequality for general triples of elements of the hyperbolic space, we will use certain one-dimensional perturbations of the identity map, which turn out to be isometries (with respect to the hyperbolic distance). They are introduced in the following.

For any y ∈ Hⁿ(ℝ) let a map T_y : Hⁿ(ℝ) → Hⁿ(ℝ) be defined by

For anyy ∈ Hⁿ(ℝ) the mapT_yis bijective and fulfills the equation:

Moreover,T_−yis the inverse ofT_y.

Proof. We start from computing [T_y(u)] for u ∈ Hⁿ(ℝ):

and thus

For simplicity, putα(x)=[x]+〈x,y〉[y]+1 (then T_y(x) = x + α(x)y). Observe that (2.3) is equivalent to

which can easily be transformed to an equivalent form:

The right-hand side expression of the above equation can be transformed as follows:

which equals [T_y(a)] [T_y(b)], by (2.4). So, (2.3) is proved and, combined with Lemma 2.4, implies that T_y is one-to-one. To finish the whole proof, it suffices to check that T_−y ○ T_y coincides with the identity map on Hⁿ(ℝ) (because then, by symmetry, also T_y ○ T_−y will coincide with the identity map). To this end, we fix x ∈ Hⁿ(ℝ), and may and do assume that y ≠ 0. Then there exist a unique vector z and a unique real number β such that ⟨z, y⟩ = 0 and x = z + βy. Note that T_y (x) = z + ([x] + β[y])y and, consequently,

Finally, an application of (2.4) enables us continuing the above calculations to obtain:

□

The functiond_his a metric inducing the Euclidean topology.

Proof. To show the triangle inequality, consider arbitrary three points x, y and z of Hⁿ(ℝ). Since T_y(0) = y, it follows from Lemmas 2.7 and 2.5 that

Further, to establish the equivalence of the metrics, observe that

and that both T_y and T_−y are continuous with respect to the Euclidean metric (and thus they are homeomorphisms in this metric). Thus, for an arbitrary sequence xnn=1∞ of elements of ℝⁿ and any x ∈ ℝⁿ, we have (note that T_−x(x) = 0):

□

The following is an immediate consequence of Theorem 2.8 (and the fact that the collections of all closed balls around 0 with respect to d_h and d_e, respectively, coincide – only radii change when switching between d_h and d_e). We skip its simple proof.

For eachn, the hyperbolic spaceHⁿ(ℝ) is locally compact and connected and the metric d_h is proper (that is, all closed balls are compact) and, in particular, complete.

A metric space (X, d) is said to be absolutely (metrically) homogeneous if any isometric map ƒ₀ : (X₀, d) → (X, d) defined on a non-empty subset X₀ of X extends to an isometry ƒ: (X, d) → (X, d).

In this section, we will show that hyperbolic spaces are absolutely homogeneous. According to Theorem 7.1 (see Section 7 below), this property makes them highly exceptional among all connected locally compact metric spaces.

The following is a reformulation of Lemma 2.7.

For anyy ∈ Hⁿ(ℝ), the mapT_y : (Hⁿ(ℝ), d_h) → (Hⁿ(ℝ), d_h) is an isometry.

For a mapu : A → Hⁿ(ℝ), where A is a subset ofHⁿ(ℝ) containing the zero vector, the following conditions are equivalent:

1. uis isometric (with respect tod_h) and u(0) = 0;

2. ⟨u(x), u(y)⟩ = ⟨x, y⟩ for allx, y ∈ A.

Proof. Assume (i) holds and fix x, y ∈ A. Then [u(x)] = cosh(d_h(u(x), u(0))) = cosh(d_h(x, 0)) = [x]. Similarly, [u(y)] = [y] and thus 〈u(x), u(y)〉 = [u(x)] [u(y)] – cosh(d_h(u(x), u(y))) = [x] [y] – cosh(d_h(x, y)) = 〈x, y〉. Conversely, if (ii) holds and x, y ∈ A, then 〈u(x), u(x)〉 = 〈x, x〉 and hence [u(x)] = [x] (and, similarly, [u(y)] = [y]) and u(0) = 0. But then easily d_h(u(x), u(y)) = d_h(x, y) and we are done. □

For anynthe hyperbolic space (Hⁿ(ℝ), d_h) is absolutely homogeneous.

Proof. Fix an isometric map v: A → Hⁿ(ℝ) where A is a non-empty subset of Hⁿ (ℝ). Take any a ∈ A, put B = T_−a(A) and u = T_−v(a) ○ v ○ T_a : B → Hⁿ(ℝ), and observe that 0 ∈ B, u(0) = 0 and u is isometric (with respect to d_h). So, the map u satisfies condition (ii) from Lemma 3.3. It is well known (and easy to show) that each such a map extends to a linear map U: ℝⁿ → ℝⁿ that corresponds (in the canonical basis of ℝⁿ) to an orthogonal matrix. The last property means precisely that also U fulfills the equation from condition (ii) of Lemma 3.3. Hence, U is isometric with respect to d_h. Then T_v(a) ○ U ○ T_−a is an isometry that extends v. □

The above proof enables us to describe all isometries of the hyperbolic space Hⁿ(ℝ): all of them are of the form u = T_a ○ U where a is a vector and U: ℝⁿ → ℝⁿ is an orthogonal linear map (both a and U are uniquely determined by u). In particular, the isotropy groups stab(x) = {u ∈ Iso(Hⁿ(ℝ), d_h): u(x) = x} of elements x ∈ Hⁿ(ℝ) are pairwise isomorphic and stab(0) is precisely the group O_n of all orthogonal linear transformations of ℝⁿ. The group O_n also coincides with the isotropy group of the zero vector with respect to the isometry group of the n-dimensional Euclidean space. So, hyperbolic spaces are quite similar to Euclidean. However, the last spaces have many (bijective) dilations, which contrasts with hyperbolic geometry, as shown by

Forn > 1, every dilation on (Hⁿ(ℝ), d_h) is an isometry. More generally, ifu : Hⁿ(ℝ) → H^m(ℝ) is a dilation (and n > 1), thenuis isometric.

Proof. Assume u: Hⁿ(ℝ) → H^m(ℝ) satisfies d_h (u(x), u(y)) = cd_h(x, y) for all x, y ∈ Hⁿ(ℝ) and some constant c > 0. Our aim is to show that then c = 1. (That is all we need to prove, since any global isometric map on Hⁿ(ℝ) is onto – its inverse map is extendable to an isometry which means that actually it is an isometry.)

Replacing u by T_−u(0) ○ u, we may and do assume that u(0) = 0. Then it follows from (2.5) that for any x, y ∈ Hⁿ(ℝ):

Further, d_h(x, 0) = d_h(−x, 0) and d_h(x, –x) = d_h(x, 0) + d_h(0, –x). Consequently, d_h(u(x), 0) = d_h(u(–x), 0) and d_h(u(x), u(–x)) = d_h(u(x), 0) + d_h(0, u(–x)). So, we infer from Lemma 2.5 (and from (3.1)) that u(–x) = –u(x) for all x ∈ Hⁿ(ℝ).

Fix arbitrary two vectors x, y ∈ Hⁿ(ℝ) such that 〈x, y〉 = 0. Then

thus d_h (u(x), u(y)) = d_h(u(x), u(–y)) = d_h(u(x), –u(y)). This implies that

Now for arbitrary t ≥ 1 choose two vectors x, y ∈ Hⁿ(ℝ) such that [x] = [y] = t and 〈x, y〉 = 0. Then also 〈u(x), u(y)〉 = 0 and (by (3.1)) [u(y)] = [u(x)] = cosh(d_h(u(x),0)) = cosh(cd_h(x, 0)) = cosh(c cosh⁻¹(t)). It follows from the former property that c cosh⁻¹([x] [y]) = cd_h(x, y) = d_h(u(x), u(y)) = cosh⁻¹ ([u(x)] [u(y)]), which combined with the latter yields

The above equation is valid for every t ≥ 1 only if c = 1. Although this is a well-known fact, for the reader’s convenience we give its brief proof. Observe that

and

As a consequence, limt→∞coshccosh−1t2t2c=2c, whereas

So, if (3.2) holds for all t ≥ 1, then 2^c = 2^2c−1 and c = 1. □

We underline that the claim of Theorem 3.5 is false for n = 1 (see Corollary 5.2 below).

As an immediate consequence of the above result, we obtain:

For anyn > 1, the metric spaces (Hⁿ(ℝ), d_h) and (ℝⁿ, d_e) are non-isometric.

The iso-measurem_hon the hyperbolic spaceHⁿ(ℝ) and the n-dimensional Lebesgue measure are mutually absolutely continuous. More precisely, for any Borel setA ⊂ ℝⁿ:

Proof. We only need to check that the measure m_h defined by (4.1) is invariant under any isometry u of Hⁿ(ℝ). To this end, write u = T_a ○ U where U ∈ O_n and a ∈ Hⁿ(ℝ). We will show separately the invariance under U and T_a, applying the change-of-variables theorem for the Lebesgue integral.

For the orthogonal matrix U this is quite easy: substituting y = U(x), we get:

since [U⁻¹(y)] = [y]. For T_a, we set y = T_a (x) and use the following two properties:

• Ta−1=T−a;

• the Jacobian of T_−a at y equals 1−y[y]−a1+[a],a (and is positive).

The latter property easily follows from the fact that the derivative of T_−a at y is a linear map of the form

So, by (2.4):

□

Denoting by B¯ϱ(a,r) the closed ball around a of radius r with respect to a metric ϱ, we have (of course) mB¯de(a,r)=α∫0rtn−1dt where m stands for the Lebesgue measure on ℝⁿ and α is a constant that depends only on the dimension n of the space ℝⁿ. A counterpart of this formula for hyperbolic spaces is formulated in the next result and reveals another connection with hyperbolic functions (first such a connection is exposed in the formula for d_h).

For anyn > 0, a ∈ Hⁿ(ℝ) andR > 0:

Proof. For simplicity, set F(R)=mhB¯dh(a,R). By metric homogeneity of Hⁿ(ℝ) and invariance of m_h under any isometry, we have F(R)=mhB¯dh(0,R). Observe that B¯dh(0,R)=B¯de(0,sinh(R)). So, F(R)=∫B¯de(0,sinh(R))dx[x]. When n > 1, a usage of the hyperspherical variables transforms this integral to

and hence F(R)=γ∫0sinh(R)rn−11+r2dr (this formula is correct also for n = 1, with γ = 2). Now to finish the proof it remains to substitute r = sinh(t) where 0 ≤ t ≤ R and notice that then dr = cosh(t) dt and cosh(t)=1+sinh2(t). □

Letaandbbe two distinct points ofHⁿ(ℝ).

1. The metric segmentI (a, b) = {x ∈ Hⁿ (ℝ): d_h(a, b) = d_h(a, x) + d_h(x, b)} is a unique straight line segment inHⁿ(ℝ) that joinsaandb.

2. The set L(a, b) of allx ∈ Hⁿ(ℝ) such thatx, a, bare metrically collinear is a unique hyperbolic line passing througha and b.

3. Every isometric mapγ: ℝ → Hⁿ(ℝ) that sends 0 to a is of the formγ(t) = T_a (sinh(t)z) wherez ∈ Hⁿ(ℝ) is such that ∥z∥ = 1.

Proof. First assume a = 0. All the assertions of the theorem in this case will be shown in a few steps.

For any z ∈ Hⁿ(ℝ) with ∥z∥ = 1 denote by γ_z : ℝ → Hⁿ(ℝ) a map given by γ_z(t) = sinh(t)z. This map is isometric, which can be shown by a straightforward calculation:

Observe that the image of γ_z coincides with the linear span of the vector z. Thus, every Euclidean line passing through the zero vector is also a hyperbolic line.

Now let x, y, z ∈ Hⁿ(ℝ) satisfy d_h(x, z) = d_h(x, y) + d_h(y, z) and let 0 ∈ {x, y, z}. We claim that x, y, z lie on a Euclidean line. Indeed, if y = 0, it suffices to apply Lemma 2.5 and otherwise, we may assume, without loss of generality, that x = 0. In that case, we proceed as follows. The linear span L of y is a hyperbolic line (by the previous paragraph). So, there exists an isometry q: L → L that sends y to 0. By absolute homogeneity established in Theorem 3.4, there exists an isometry Q: Hⁿ(ℝ) → Hⁿ(ℝ) that extends q. So, Q(L) = L, Q(y) = 0 and d_h(Q(x), Q(z)) = d_h(Q(x), Q(y)) + d_h(Q(y), Q(z)). Again, Lemma 2.5 implies that Q(z) = tQ(x) for some t ≤ 0 (as Q(x) ≠ 0 = Q(y)). But Q(x) = Q(0) ∈ Q(L) = L and therefore also Q(z) ∈ L. So, x, y, z ∈ L and we are done.

Now we prove items (A)–(C) in the case a = 0. Let z ∈ Hⁿ(ℝ) be a vector such that ∥z∥ = 1 and b ∈ γ_z (ℝ). It follows from the last paragraph that each element x ∈ Hⁿ(ℝ) such that x, 0, b are metrically collinear belongs to γ_z (ℝ). We also know that γ_z (ℝ) is a hyperbolic line. This shows that L(0, b) = γ_z(ℝ) and proves (B), from which (A) easily follows. Finally, if γ: ℝ → Hⁿ(ℝ) is an isometric map such that γ(0) = 0, then for any t ∈ ℝ, the points 0, γ(1) and γ(t) are metrically collinear, so γ(t) ∈ L(0, γ(1)). It follows from the above argument that L(0, γ(1)) = γ_z(ℝ) for some unit vector z ∈ Hⁿ(ℝ). Then v=γz−1∘γ:ℝ→ℝ is a Euclidean isometry sending 0 to 0. Thus v(t) = t or v(t) = – t. In the former case, we get γ = γ_z, whereas in the latter we have γ = γ_−z, which finishes the proof of (C).

Now we consider a general case. When a is arbitrary, b ≠ a and γ: ℝ → Hⁿ(ℝ) is isometric and sends 0 to a, it is easy to verify that T_−a(I(a, b)) = I(0, T_−a(b)), T_−a (L(a, b)) = L(0, T_−a(b)) and T_−a ○ γ is an isometric map from ℝ into Hⁿ(ℝ) that sends 0 to 0. So, the first part of the proof implies that I (a, b) and L(a, b) are a unique straight line segment joining a and b, and – respectively – a unique hyperbolic line passing through a and b. Similarly, T_−a ○ γ = γ_z for some unit vector z. Then γ = T_a ○ γ_z and we are done.

The map (ℝ, d_e) ∋ t ↦ sinh(t) ∈ (H¹(ℝ), d_h) is an isometry.

The above result implies, in particular, that H¹(ℝ) admits many non-isometric dilations. Its assertion is the reason for considering (in almost whole existing literature) only hyperbolic spaces of dimension greater than one.

Another immediate consequence of Theorem 5.1 reads as follows.

Hyperbolic spaces are geodesic.

The next result may be named Failure of the Classical Euclid’s Fifth Postulate.

Let n > 1, Lbe a hyperbolic line inHⁿ(ℝ) and a ∉ L. There are infinitely many hyperbolic lines that pass through a and are disjoint fromL.

Proof. Thanks to the metric homogeneity of Hⁿ(ℝ), we may and do assume that a = 0. So, L is a hyperbolic line that does not pass through 0. We claim that then there exist two linearly independent vectors a and b such that

Indeed, we know that L = γ(ℝ) where γ = T_y ○ γ_z, y ≠ 0, ∥z∥ = 1 and γ_z is given by γ_z(t) = sinh(t)z. Then γ(t)=sinh(t)z+[sinh(t)z]+sinh(t)〈z,y〉[y]+1y=sinh(t)z+〈z,y〉[y]+1y+cosh(t)y.. So, substituting a=z+〈z,y〉[y]+1y and b = y, it remains to check that a and b are linearly independent to get (5.1). If a and b were not such, then L would be a subset of the linear span Y of y. But Y is a hyperbolic line and therefore we would obtain that L = Y and hence 0 ∈ L. This proves (5.1).

Now let μ ∈ ℝ be such that |μ| > 1. Put c = μa + b and let L_μ be the linear span of c. We claim that L_μ is disjoint from L. Indeed, let s and t be real and assume, on the contrary, that sc = sinh(t)a + cosh(t)b (cf. (5.1)). It follows from the linear independence of a and b that sinh(t) = sμ and cosh(t) = s. Consequently, |sinh(t)| = |cosh(t)μ| > |cosh(t)|, which is impossible.

So, for any real μ with |μ| > 1 the set L_μ is a hyperbolic line disjoint from L and passing through 0. Since a and b are linearly independent, these sets L_μ are all different. □

For any two hyperbolic lines inH²(ℝ) there exists a hyperbolic line disjoint from both of them.

Proof. Observe that for two orthogonal vectors a and b, the hyperbolic lines T_a(ℝb) and T_−a(ℝb) are contained in different components of ℝ² \ ℝb. This, by metric homogeneity, proves that for an arbitrary hyperbolic line, each component of its complement contains a hyperbolic line.

If the given two lines K and L are disjoint (or equal), we can pick a line from the component of ℝ \ K different than the one containing L.

Now proceed to the remaining case of the two intersecting lines. We may and do assume that they intersect at 0. Let them be the linear spans of independent unit vectors, a and b, respectively. We claim that there exist β > 0 such that T_−β(a+b)(ℝ(a – b)) is a hyperbolic line disjoint from ℝa and ℝb.

For convenience, let us transform the whole setting by T_β(a+b), obtaining hyperbolic lines {T_β(a+b)(ta): t ∈ ℝ}, {T_β(a+b)(tb): t ∈ ℝ} and {s(a – b): s ∈ ℝ}. We want to prove that for sufficiently large β neither T_β(a+b)(ta) = s(a – b) nor T_β(a+b)(tb) = s(a – b), regardless of the values of real parameters s and t. Here we focus only on the former equation (the latter can be treated analogously). Using the explicit formula for the isometry, we obtain (in the former equation):

Using linear independence of a and b, we obtain a system of two equations:

Adding the equations, we obtain

Since a and b are unit vectors, 2 〈a, a + b〉 = ∥a + b∥². Expanding [β(a + b)] and simplifying, we arrive at

Observe that the above equation is fully symmetric with respect to a and b.

Consider the asymptotic behaviour of the [ · ] function:

Since ∥a + b∥ < 2, for sufficiently large parameter β, 2β > [β(a + b)]. Then |−t[β(a+b)]|<2β|t|<2βt2+1,, hence the equation (5.3) holds for no t.

In the hyperbolic space (Hⁿ(ℝ), d_h), we define relations of equidistance and betweenness in the following way (a, b, c, d ∈ Hⁿ(ℝ)):

Observe that all the relations are strictly preserved by isometries, that means for any formula ϕ(p₁,…, p_k) of k free variables and fixed points x₁,…, x_k, the following are equivalent

1. ϕ(x₁ …, x_k),

2. ∃_{Φ∈Iso(Hⁿ (ℝ), de}ϕ(Φ(x₁),…,Φ(x_k),

3. ∀_{Φ∈Iso(Hⁿ(ℝ), de)}ϕ(Φ(x₁),…, Φ(x_k)).

We will show that with elementary relations introduced as above, our model of hyperbolic geometry satisfies all of the Tarski’s axioms except for the Euclid’s Axiom, what shall be expected. We focus on the two dimensional case (we have listed Tarski’s axioms particularly for dimension 2), while most of the proofs apply to the n-dimensional case.

Axioms 1–3 are obvious consequences of the general properties of a distance function.

Ad 4. As we mentioned before, we can reduce to the case when a = 0 (using the isometry T_−a). Then it suffices to take x = –q/∥q∥ sinh(d_h(b, c)), for q ≠ 0 (cf. item (C) in Theorem 5.1). If q = 0 = a, an x suitable for arbitrary q ≠ 0 would be appropriate.

Ad 5. The Fixe-Segment Axiom follows from the metric 3-homogeneity. Under its assumptions, we can find a global isometry Φ such that Φ(a) = a′, Φ(b) = b′ and Φ(c) = c′. If then Φ(d) = d′, the assertion of the axiom is satisfied. But since a ≠ b, the isometry Φ is uniquely determined on the line L(a, b), hence indeed Φ(d) = d′.

Ad 6. By the definition of betweenness, 2d_h(a, b) = d_h(a, a) = 0 and then a = b.

Ad 7. Since any line on a hyperbolic plane (H²(ℝ), d_h) can be mapped onto a one-dimensional vector subspace by an isometry of the plane, its complement has two connected components.

Let L be a line in (H²(ℝ), d_h) and letC₁andC₂be the two connected components of ℝ² \ L. Fix pointsx ∈ L, y ∈ C₁andz ∈ ℝ². Then

1. zbelongs toC₁ ∪ Lprovided thatB(xyz),

2. zdoes not belong toC₁provided thatB(yxz).

Proof. By metric homogeneity, we may reduce to the case when x = (0,0) and L = {(t, 0): t ∈ ℝ}. Since hyperbolic lines passing through 0 are simply vector subspaces, both cases are obvious. □

To prove that the Pasch Axiom is satisfied in (H²(ℝ), d_h), we observe that b is not in the same component of ℝ² \ L(p, c) as a and q. (One can easily verify the degenerated cases when [some of] those points lie on L(p, c).) Therefore I(b, q) has to intersect L(p, c) in a (unique) point, say x.

The same reasoning applied to c, L(q, b), a and p leads to the conclusion that I(c, p) intersects L(b, q):

Observe that the setting of the Pasch Axiom is contained within two intersecting lines, and therefore can be assumed to be contained in ℝ² × {0}ⁿ⁻² ⊂ Hⁿ(ℝ), which is naturally isometric with (H²(ℝ), d_h). Hence the axiom is satisfied in higher dimensions as well.

Ad 8. Axiom 8 is witnessed e.g. by the triple (0, 0), (0, 1) and (1, 0).

Ad 9. By an isometric transformation, we can reduce the issue to the case when b = –a. Observe that 2 〈x, a〉 = cosh(d_h(x, –a)) – cosh(d_h(x, a)), for any x ∈ ℝⁿ. Hence all the points equidistant from a and –a are vectors perpendicular to a. In H²(ℝ) the set of all such points is a straight line and the axiom 9 follows.

Ad 10. That the Euclid’s Axiom is violated in the hyperbolic plane, can be easily deduced from Theorem 5.4 or 5.5.

We pick two lines, K and L, intersecting in a point a. Then – thanks to Theorem 5.5 – we find a third line, M, disjoint from K and L. Then we pick an arbitrary point t on M and points b, c, d (with b ∈ K and c ∈ L close to a, and d ∈ I (a, t) ⋂ I (b, c)) to complete the setting of the axiom. Points x and y from the assertion cannot be found, since one of them has to belong to the other component of ℝ² \ M than K ⋃ L (by Lemma 6.2).

Ad 11. The continuity axiom is trivial if one of the sets consists of less than two points. In the other case both sets have to be contained in a hyperbolic line which is isometric to ℝ. On the real line the assertion is satisfied.

Each connected locally compact 3-point homogeneous metric space having more than one point is isometric to exactly one metric space (X, d) among all listed below (everywhere belowndenotes a positive integer).

• X = ℝⁿ, d = ω ○ d_ewhereω ∈ Ω andω(1)=min(1,12ω(∞)).

• X=Sn, d = ω ○ d_swhereω ∈ Ω₁.

• X = Hⁿ(ℝ) = d = ω ○ d_hwheren > 1 and ω ∈ Ω.

In particular:

• all connected locally compact 3-point homogeneous metric spaces are absolutely homogeneous, and each of them is homeomorphic either to a Euclidean space or to a Euclidean sphere (unless it has at most one point);

• each locally compact geodesic 3-point homogeneous metric space having more than one point is isometric to exactly one of the spaces: (ℝⁿ, d_e), (Sn, rds)(where r > 0), (Hⁿ(ℝ), rd_h) (where n > 1 and r > 0).

Busemann’s [5] and Birkhoff’s [3, 4] results have a similar spirit. However, both of them assume, besides metric convexity, a sort of uniqueness of straight line segments joining sufficiently close two points, and their statements are less general.

It seems that the assertion of Theorem 7.1 (in this form) has never appeared in the literature. Nevertheless, we consider this result as Freudenthal’s theorem – not ours. Such a thinking is justified by the fact that Theorem 7.1 easily follows from Freudenthal’s theorem, stated below, proved in [8].

Denoting by Pⁿ(ℝ) the n-dimensional real projective space (realised as the quotient space of Sn obtained by gluing antipodal points) equipped with a geodesic metric dp(u¯,v¯)=2πarccos(|〈u,v〉|) (where u,v∈Sn and u¯={u,−u} and v¯={v,−v} denote their equivalence classes belonging to Pⁿ(ℝ)), we can formulate Freudenthal’s Hauptsatz IV from [8] as follows.

Let (Z, λ) be a connected locally compact metric space that satisfies the following condition:

Then there is a unique space (M, ϱ) among (ℝⁿ, d_e) (n > 0), (Sn, ds) (n > 0), (Hⁿ(ℝ), d_h) (n > 1), (Pⁿ(ℝ), d_p) (n > 1) and a homeomorphismh: Z → Msuch that:

We prove Theorem 7.1 in few steps, each formulated as a separate lemma. All of them are already known but – for the reader’s convenience – we give their short proofs.

Forn > 1 the space (Pⁿ(ℝ), d_p) is 2-point, but not 3-point, homogeneous.

Proof. We use here the notation introduced in the paragraph preceding the formulation of Theorem 7.3.

Let (u, v) and (x, y) be two pairs of points of Sn such that dp(u¯,v¯)=dp(x¯,y¯). This means that there is ε ∈ {1, –1} such that d_s(u, v) = d_s(x, εy). Consequently, there exists A ∈ O_n+1 (which is automatically an isometry with respect to d_s) such that A(u) = x and A(v) = εy. Every member B of O_n+1 naturally induces an isometry B¯:Pn(ℝ)→Pn(ℝ) that satisfies B¯(z¯)=B(z)¯ for any z∈Sn. Moreover, the assignment

is surjective (this is classical, but non-trivial). So, A¯∈Iso(Pn(ℝ),dp) satisfies A¯(u¯)=x¯ and A¯(v¯)=y¯, which shows that Pⁿ(ℝ) is 2-point homogeneous.

To convince oneself that Pⁿ(ℝ) is not 3-point homogeneous for n > 1, it is enough to consider two triples (x, y, z₁) and (x, y, z₂) of points of Sn with x=(1,0,0,0→), y=22,22,0,0→, z1= 14,14,144,0→  and z2=14,−34,64,0→ where 0→ denotes the zero vector in ℝⁿ⁻² (which has to be omitted when n = 2). Observe that ⟨x, z₁⟩ = ⟨x, z₂⟩ and ⟨y, z₁) = – ⟨y, z₂⟩. Consequently, dpw¯,z¯1=dpw¯,z¯2 for any w ∈ {x, y}. If Pⁿ(ℝ) was 3-point homogeneous, there would exist an isometry of Pⁿ(ℝ) that fixes x and y and sends z₁ onto z₂, which is impossible. To see this, assume – on the contrary – there exists such an isometry. It then follows from the surjectivity of (7.2) that there exists B ∈ O_n+1 such that B(x) = ±x, B(y) = ±y and B(z₁) = ±z₂. Then, replacing if needed B by –B, we may and do assume B(x) = x and thus, since ⟨B(x), B(u)⟩ = ⟨x, u⟩ for any u ∈ ℝⁿ⁺¹, also B(y) = y and B(z₁) = z₂. But then ⟨y, z₂⟩ = ⟨B(y), B(z₁)⟩ = ⟨y, z₁⟩ which is false. □

Let (M, ϱ) be a 2-point homogeneous geodesic metric space having more than one point and I denoteϱ(M × M), and letƒ: I → [0, ∞) be a one-to-one function such thatϱ_ƒ = ƒ ○ ϱis a metric onMequivalent toϱ. Then:

1. Iis an interval andƒ ∈ Ω(I);

2. (M, ϱ_ƒ) is 2-point homogeneous;

3. Iso(M, ϱ_ƒ) = Iso(M, ϱ);

4. (M, ϱ_ƒ) is absolutely homogeneous iff so is (M, ϱ);

5. (M, ϱ_ƒ) is geodesic iffƒ (t) = ctfor some positive constantc (and allt ∈ I).

Proof. Since ƒ is one-to-one, any map u: A → M (where A ⊂ M) is isometric with respect to ϱ_ƒ if and only if it is isometric with respect to ϱ. This implies (v1), (v2) and (v3). It follows from 1-point homogeneity of (M, ϱ) that I = {ϱ(a, x): x ∈ M} where a is arbitrarily fixed element of M. But this, combined with continuity of ϱ_ƒ with respect to ϱ, yields that ƒ is continuous at 0. Further, since M is geodesic, I is an interval and for any x, y ∈ I with x + y ∈ I there are points a, b, c ∈ M such that ϱ(a, c) = x + y, ϱ(a, b) = x and ϱ(b, c) = y. Then ƒ (x + y) = ϱ_ƒ (a, c) ≤ ϱ_ƒ (a, b) + ϱ_ƒ (b, c) = ƒ (x) + ƒ (y). So, (ω2) holds. This implies that |ω(x) – ω(y)| ≤ ω(|x – y|) for any x, y ∈ I. Thus f, being continuous at 0, is continuous (at each point of I). Finally, a continuous one-to-one function vanishing at 0 satisfies (ω1) and therefore f ∈ Ω(I).

It remains to show that if ϱ_f is geodesic, then f is linear. Since f ∈ Ω(I), it is a bijection between I and J = f (I), and J is an interval. Moreover, ϱ = f⁻¹ ○ ϱ_f. So, assuming that ϱ_f is geodesic, it follows from the first part of the proof that also f⁻¹ ∈ Ω(J). But if f ∈ Ω(I) and f⁻¹ ∈ Ω(J), then f(x + y) = f(x) + f(y) for all x, y ∈ I with x + y ∈ I. The last equation implies that f is linear (for f is continuous). □

Assume (Z, λ) and (M, ϱ) are two 2-point homogeneous metric spaces having more than one point and satisfying the following three conditions:

• (M, ϱ) is geodesic;

• there existsR ∈ {1, ∞} such thatϱ(M × M) = {r ∈ ℝ: 0 ≤ r ≤ R}

• there exists a homeomorphismh: Z → Msuch that (7.1) holds.

Then there exists a unique ω ∈ Ω_Rsuch that λ = ω ○ ϱ ○ (h × h). Moreover, (Z, λ) is 3-point or absolutely homogeneous iff so is (M, ϱ).

Proof. Recall that h × h: Z × Z → M × M is given by (h × h)(x, y) = (h(x), h(y)). Further, for simplicity, denote I = ϱ(M × M). It follows from our assumptions that I = [0, 1] (if R = 1) or I = [0, ∞) (if R = ∞).

For any four points a, b, x and y in an arbitrary 2-point homogeneous metric space (Y, p) the following equivalence holds:

The above condition, applied for both (Z, λ) and (M, ϱ), combined with (7.1) yields that

We infer that there exists a one-to-one function ω: I → [0, ∞) such that λ = ω ○ ϱ ○ (h × h). Since ω ○ ϱ = λ ○ (h⁻¹ × h⁻¹) is a metric equivalent to ϱ, we conclude from Lemma 7.5 that ω ∈ Ω_R. The uniqueness of ω is trivial.

Finally, the remainder of the lemma (about 3-point or absolute homogeneity) follows from (7.3). Indeed, this condition implies that a map u: (A, λ) → (Z, λ) (where A ⊂ Z) is isometric iff the map h ○ u ○ h⁻¹|_h(A): (h(A), ϱ) → (M, ϱ) is isometric. □

Proof of Theorem 7.1.

First of all, since each of the spaces

is absolutely homogeneous, it follows from Lemma 7.5 that any space (X, d) listed in the statement of the theorem is also absolutely homogeneous (and, of course, connected, locally compact and has more than one point). Lemma 7.5 enables us recognizing those among them that are geodesic.

Further, if (Z, λ) is a connected locally compact 3-point homogeneous metric space having more than one point, we infer from Theorem 7.3 that there are a metric space (M, ϱ) and a homeomorphism h: Z → M such that (7.1) holds and either (M, ϱ) is listed in (7.4) or it is (Pⁿ(ℝ), d_p) with n > 1. However, since (M, ϱ) is 2-point homogeneous (cf. Lemma 7.4), it follows from Lemma 7.6 that it is also 3-point homogeneous (because (Z, λ) is so). Thus, Lemma 7.4 implies that (M, ϱ) is listed in (7.4), and we conclude from Lemma 7.6 that there is ω˜∈Ω(I) (where I = ϱ(M × M)) for which

Now if (M, ϱ) is not a Euclidean space, put ω=ω˜ and observe that (X, d) = (M, ω ○ ϱ) is listed in the statement of the theorem and h: (Z, λ) → (X, d) is an isometry, thanks to (7.5).

In the remaining case, when (M, ϱ) = (ℝⁿ, d_e), we proceed as follows. There is α > 0 such that ω˜(α)=min1,12ω˜(∞). We define ω ∊ Ω by ω(t)=ω˜(αt). Observe that then ω(∞)=ω˜(∞) and hence ω(1)=min1,12ω(∞). So, (X, d) = (ℝⁿ, ω ○ d_e) is listed in the statement of the theorem. What is more, the map

is an isometry, again by (7.5) (and the definition of ω).