# The Beckman–Quarles Theorem in Hyperbolic Geometry

Authors: Oğuzhan Demirel; Damla Topal; Leyla Aslan
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we present the counterpart of the Beckman–Quarles theorem in the Poincaré disc model of hyperbolic geometry to characterize the gyroisometries (hyperbolic isometries) with a single nonzero distancea∈0,1 satisfying a2∈ℚ.


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## Body

## 1. Introduction

LetX,d be a metric space and f be a function defined from X,d to itself. If dfP,fQ=dP,Q holds for all P,Q∈X, then f is called an isometry of X. An isometry is injective, but it need not be surjective. If an isometry is invertible, its inverse is also an isometry. An isometry of Euclidean space ℝn is bijective. The set Iso (ℝn) of all isometries of ℝn is a group under composition. It is well known that every isometry of ℝn is a composition of at most n+1 reflections. An isometry that fixes at least one point is a composition of at most n reflections.Let us consider the Euclidean spaceℝnn≥2 and f be a function defined from ℝn to itself. The Beckman–Quarles [1] theorem states that if f is a function defined from ℝnn≥2 to itself which preserves a distance k∈ℝ+, then it is an isometry of ℝn. In the literature, there are several proofs of this famous theorem, see [2, 3], and some authors [4–7] tried to find the counterpart of this theorem in various spaces.Hyperbolic geometry is a non-Euclidean geometry that rejects the validity of Euclid’s fifth postulate. The principles of hyperbolic geometry, however, admit the other four Euclidean postulates. Although many of the theorems of hyperbolic geometry are identical to those of Euclidean, others differ. For example, in Euclidean geometry, two parallel lines are taken to be everywhere equidistant. In hyperbolic geometry, two parallel lines are taken to converge in one direction and diverge in the other. In Euclidean geometry, the sum of the angles in a triangle is equal to two right angles; in hyperbolic geometry, the sum is less than two right angles. In Euclidean geometry, polygons of differing areas can be similar, and in hyperbolic geometry, similar polygons of differing areas do not exist. There are many principal hyperbolic geometry models, for instance, Weierstrass model, Beltrami–Klein model, Poincaré disc model, and Poincaré upper-half plane model. In this paper, we deal with the Poincaré disc model of hyperbolic geometry to get the desired results. The points of this model are the points of the complex unit disc:(1)D=z∈ℂ:z<1,and the hyperbolic lines are circular arcs orthogonal to the boundary circle of the disc including the diameters of D. Two arcs which do not meet correspond to parallel rays and the arcs which meet orthogonally correspond to perpendicular lines (see Figure 1). The hyperbolic angle between two hyperbolic lines is the usual Euclidean angle between Euclidean tangents to the circular arcs. The advantage of the Poincaré disc model is that it is conformal, namely, circles and angles are not distorted.Figure 1
A gyroline (hyperbolic line) passing through the pointsMand K is a circular arc that meets the boundary of the disc D orthogonally. The hyperbolic lines passing through the center of disc are also corresponding to chords of the disc.The classical hyperbolic distancedH in D is defined by(2)dHz1,z2=12ln1+dz1,z21−dz1,z2,where(3)dz1,z2=z1−z21−z1¯z2,for allz1,z2∈D.A Möbius transformationf of the extended complex plane ℂ∪∞ is a rational function of the form(4)fz=az+bcz+d,of one complex variable z; here, the coefficients a,b,c,d are complex numbers satisfying ad−bc≠0. Möbius transformations are named in honor of August Ferdinand Möbius; they are also variously named bilinear transformations, homographic transformations, or fractional linear transformations. The set of all Möbius transformations forms a group under composition. Möbius transformations preserve the measures of the angles with orientation. Euclidean transformations, Euclidean rotations, inversions, and similarities are well-known Möbius transformations.The most general Möbius transformation of the complex unit discD in the complex plane to itself(5)z↦eiθz0+z1+z0¯z=eiθz0⊕z,defines the Möbius addition in the disc, which allows the Möbius transformation of the disc to be viewed as a Möbius left translation [8, 9]:(6)z↦z0⊕z=z0+z1+z0¯z,where θ∈ℝ, z0∈D, and z0¯ is the complex conjugate of z0. A left Möbius translation is also called a left gyrotranslation [8]. It is known that the Möbius addition “⊕” is analogous to the common vector addition “+” in Euclidean plane geometry. Möbius addition ⊕ is neither commutative nor associative. By defining the gyrator(7)gyr:D×D⟶AutD,⊕,gyra,b=a⊕bb⊕a=1+ab¯1+ba¯,where AutD,⊕ is the automorphism group of the Möbius groupoid D,⊕, the following group-like properties of D can be verified by straightforward algebra for all a,b,c∈D:G1.a⊕b=gyra,bb⊕a (gyrocommutative law)G2.a⊕b⊕c=a⊕bgyra,bc (left gyroassociative law)G3.a⊕b⊕c=a⊕bgyrb,ac (right gyroassociative law)G4.gyra,b=gyra⊕b,b (left loop property)G5.gyra,b=gyra,b⊕a (right loop property)G6.a⊕b=gyra,bb⊕a (gyrocommutative law)The Möbius gyrodistance functiondM in D is(8)dMz,w=z⊖w=z−w1−z¯w,which is closely related to classical hyperbolic distance dH as follows:(9)tanhdHz,w=z−w1−z¯w=dMz,w.The operations “⊕” and “⊖” are called Möbius addition and Möbius subtraction, respectively. dM is not a metric on D since the triangle inequality is not provided. The gyrodistances are invariant under left gyrotranslations and rotations about origin. Clearly, dHz1,z2∈0,∞ and dMz1,z2∈0,1for all z1,z2∈D. The prefix “gyro” stems from Thomas gyration which is, in turn, the mathematical abstraction of a special relativistic effect known as Thomas precession [8, 9]. In gyrolanguage, we prefix a gyro to any term describes a concept in Euclidean geometry and in associative algebra to mean the analogous concept in hyperbolic geometry and nonassociative algebra. In full analogy with Euclidean geometry, a gyrotriangle consists of three line segments called sides or edges and three points called vertices (see Figure 2). The following theorems about gyrotriangles play major roles in our results.Figure 2
A gyrotriangle (hyperbolic triangle) in the complex unit discD.Theorem 1.
LetABC be a gyrotriangle in D with vertices A,B,C corresponding gyroangles α,β,γ and side gyrolengthsa,b,c. The sides of the gyrotriangle ABC are determined by its gyroangles by(10)a2=cosα+cosβ+γcosα+cosβ−γ,b2=cosβ+cosα+γcosβ+cosα−γ,c2=cosγ+cosα+βcosγ+cosα−β.
Theorem1 presents a most important disanalogy with Euclidean triangle similarity. For the proof of Theorems 1 and 2, we refer the reader to [9].Theorem 2.
LetABC be an equilateral gyrotriangle in D with vertices A,B,C satisfying ∠ABC=∠BCA=∠CAB=α, the side gyrolengths of which are a, and then a=2cosα−1.


## 2. The Beckman–Quarles Theorem in Hyperbolic Geometry

Throughout the paper, we denote byX′ the image of X under f, by P,Q the gyrosegment that links the points P and Q, by PQthe gyroline through the points P and Q, by PQR the gyrotriangle with three ordered verticesP,Q,R, by dP,Q the Möbius gyrodistance between P and Q, by PQRS the gyroquadrilateral with four ordered vertices P,Q,R,S, and by ∠PQR the gyroangle between P,Q and R,Q. We say that f preserves s if any points A,B∈D such that dA,B=salso satisfy dfA,fB=s.Lemma 1.
Iff:D⟶D preserves some gyrodistance a∈0,1, then it is single valued.Proof.
Assume thatf be a nonsingle valued and A be a point in D such thatfA=U,fA=V,U≠V. Now, construct an equilateral gyrotriangle ABC satisfying dA,B=dB,C=dA,C=a. Then, hypothesis A′B′C′ is also an equilateral hyperbolic triangle with dA′,B′=dA′,C′=dB′,C′=a. Assume ∠ABC=α. Then, we get a2=2cosα−1 which implies cosα=a2+1/2 by Theorem 2. Now, construct the gyrorhombus ABDC with the help of the gyrotriangle ABC. Hence, by Theorem 1, we get(11)dA,D=cos2α+cosαcos2α+1=1+a2−1a2+12.
Since there are no points other thanU and V at distance a from B′ to C′, one can easily get D′=U or D′=V. Let gθ be the hyperbolic rotation with respect to A for an appropriate θ∈ℝ satisfying dD,gθD=a and ∠gθDAD=θ. Clearly, gθx=A⊕eiθ⊖A⊕X, ⊖A=−A for X∈D. Let us denotegθD=E. Clearly, dA,D=dA,E holds and the gyroquadrilateral AgθBEgθC is a gyrorhombus congruent to ABCD. Since dgθB,A=a,dgθC,A=a, we get dfgθB,U=dfgθB,V=a anddfgθC,U=dfgθC,V=a. This yields fgθB,fgθC=B′,C′, and this implies E′=U or E′=V. Therefore, we get dD′,E′∈0,1+a2−1/a2+12, but since dD,E=a, this is a contradiction.Corollary 1.
Iff:D⟶D is a map which preserves some gyrodistance a∈0,1, then it preserves the gyrodistance 1+a2−1/a2+12.Proof.
Let us consider the gyrorhombusesABDC and AgθBEgθC in Lemma 1. Clearly, dA,D=dA,E=1+a2−1/a2+12 and this implies dA′,D′=0 or A′,D′=1+a2−1/a2+12. Assume dA′,D′=0. Hence, we get A′=D′ and this implies dA′,E′=dD′,E′=0 or dA′,E′=dD′,E′=1+a2−1/a2+12. Since dD,E=a holds, this is a contradiction. Thus, we get A′,D′=1+a2−1/a2+12.
Let us defineψa=1+a2−1/a2+12 and ψna=ψψ…,ψa,…n−times for all n∈ℕ. Clearly, a<ψa and ψia<ψja hold if i<j.Corollary 2.
Iff:D⟶D is a map which preserves some gyrodistance a∈0,1, then it preserves the gyrodistances ψna for all n∈ℕ.Theorem 3.
Iff:D⟶D is a map which preserves some gyrodistance a∈0,1 satisfying a2∈ℚ, then f preserves all gyrodistances.Proof.
Step 1. We claim that f preserves the gyrodistances less than a. Let P and Q be two distinct points in D satisfying dP,Q<a. Clearly, there exist two points in D, say C and E, such that dP,C=a=dE,C and Q∈C,E. Let SC,a be the gyrocircle centered at C with radius a. Then, there exist two points on SC,a, say R1 and R2 such that dP,R1=dP,R2=a. Thus, we construct two equilateral gyrotriangles PR1C,PR2C, the side gyrolengths of which are a. Here, the point Q can be thought to be on the gyrosegment P,R1, otherwise a new configuration is needed to provide this feature, which is not difficult to construct. Assume ∠PR1C=∠PR2C=α. Now, construct the sequence AiCAi+1i=1,2,… with dC,Ai=dC,Ai+1=dAi,Ai+1=a, A1=P,A2=R1,∠AiCAi+1=α. It is clear that there is another sequence BiCBi+1 (i = 1, 2, …) with d(C, Bi) = d(C, Bi+1) = d(Bi, Bi+1) = a, B1 = P, B2 = R2, ∠BiCBi+1 = α. If it can be established that α∉πℚ, then there is no nonzero integer k such that A1=Ak. If there is such a k satisfying A1=Ak, then one can easily see that kα=2π, i.e., α=2π/k∈πℚ. In [3], it was shown that if ρ=m/nπ, where m,n∈ℤ satisfying cosρ∈ℚ, then cosρ∈0,±1/2,±1. Since cosα=a2+1/2∉0,∓1/2,∓1 for all a∈0,1 satisfying a2∈ℚ, we immediately get that α∉πℚ. Hence, the gyrotriangles which define the sequence AiCAi+1touch the gyroarc arc(PR1) at an infinite number of points. Thus, [P, R1] has been divided into gyrosegments as desired, and the images of the points that provide this disintegration are gyrocollinear. Moreover, the gyrodistances between the points that provide this disintegration are also preserved. Thus, f preserves the gyrodistances less than a. Hence, we obtain that dP,Q=dP′,Q′ holds.Step 2. We claim that f preserves the gyrodistance 2⊗a=a⊕a. Let K and L be two distinct points in the gyrocircle SC,a satisfying dK,L=2⊗a. Since f preserves the measure of the equilateral gyrotriangles with their gyroangle α, constructing a similar sequence in Step 1 above, we immediately get dK′,L′=2⊗a. Similarly, it is clear that f also preserves the gyrodistances n⊗a=a⊕⋯⊕a (n−times) for all n∈ℕ. Notice that since a∈0,1 and a2∈ℚ, this implies n⊗a∈0,1 and n⊗a2∈ℚ. Moreover, for each equilateral gyrotriangle ABC with dA,B=dA,C=dB,C=n⊗aa2∈ℚ, if ∠ABC:=β, then cosβ∈ℚ and β∉πℚ.Step 3. f preserves the gyrodistances greater than a. Let P and Q be two distinct points in D satisfying dP,Q>a. Let k be a positive integer such that dP,Q<k⊗a. As in Step 1, one can easily construct a gyrocircle with radius k⊗a passing through P. Denote this gyrocircle by D,k⊗a where D is its center. Then, there exists a point W on SD,k⊗a such that DPW is an equilateral gyrotriangle and denote ∠DPW=ϕ. Since cosϕ∈ℚ by Step 2, f preserves all the gyrodistances less than k⊗a. Hence, we see that f preserves all gyrodistances.


## 3. Conclusions

We have proved that iff:D⟶D is a mapping which preserves some gyrodistance a∈0,1 satisfying a2∈ℚ, then f is a gyroisometry; that is, it preserves all gyrodistances. This implies that f is also a hyperbolic isometry. Naturally, one may wonder whether Theorem 3 is valid for an arbitrary gyrodistance a satisfying a2∈0,1∖ℚ. We leave the solution of this problem to the reader’s attention.



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