The Beckman–Quarles Theorem in Hyperbolic Geometry

Let (X, d) be a metric space and f be a function defined from (X, d) to itself. If d(f(P), f(Q)) � d(P, Q) holds for all P, Q ∈ X, thenf 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 R is bijective. +e set Iso (R) of all isometries of R is a group under composition. It is well known that every isometry ofR 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 spaceR(n≥ 2) and f be a function defined fromR to itself.+e Beckman–Quarles [1] theorem states that if f is a function defined from R(n≥ 2) to itself which preserves a distance k ∈ R, then it is an isometry of R. 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. +e 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. +ere 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. +e points of this model are the points of the complex unit disc:


Introduction
Let (X, d) be a metric space and f be a function defined from (X, d) to itself. If d(f(P), f(Q)) � d(P, 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 R n is bijective. e set Iso (R n ) of all isometries of R n is a group under composition. It is well known that every isometry of R n 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 space R n (n ≥ 2) and f be a function defined from R n to itself. e Beckman-Quarles [1] theorem states that if f is a function defined from R n (n ≥ 2) to itself which preserves a distance k ∈ R + , then it is an isometry of R n . In the literature, there are several proofs of this famous theorem, see [2,3], and some authors [4][5][6][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. e 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.
ere 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. e points of this model are the points of the complex unit disc: 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). e hyperbolic angle between two hyperbolic lines is the usual Euclidean angle between Euclidean tangents to the circular arcs. e advantage of the Poincaré disc model is that it is conformal, namely, circles and angles are not distorted. e classical hyperbolic distance d H in D is defined by for all z 1 , z 2 ∈ D.
A Möbius transformation f of the extended complex plane C ∪ ∞ { } is a rational function of the form 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]: where θ ∈ R, z 0 ∈ D, and z 0 is the complex conjugate of z 0 . 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 where Aut(D, ⊕) 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: which is closely related to classical hyperbolic distance d H as follows:  Journal of Mathematics e prefix "gyro" stems from omas gyration which is, in turn, the mathematical abstraction of a special relativistic effect known as omas 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). e following theorems about gyrotriangles play major roles in our results.
eorem 1 presents a most important disanalogy with Euclidean triangle similarity. For the proof of eorems 1 and 2, we refer the reader to [9].

The Beckman-Quarles Theorem in Hyperbolic Geometry
roughout the paper, we denote by X ′ 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 d(P, 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 en, we get a 2 � 2 cos α − 1 which implies cos α � (a 2 + 1)/2 by eorem 2. Now, construct the gyrorhombus AB DC with the help of the gyrotriangle ABC. Hence, by eorem 1, we get Since there are no points other than U and V at distance a from B ′ to C ′ , one can easily get D ′ � U or D ′ � V. Let g θ be the hyperbolic rotation with respect to A for an appropriate θ ∈ R satisfying d(D, g θ (D)) � a and ∠g θ (D)A D � θ. Clearly, g θ (x) � A⊕(e iθ (⊖A⊕X)), (⊖A � −A) for X ∈ D. Let us denoteg θ (D) � E. Clearly, d(A, D) � d(A, E) holds and the gyroquadrilateral Ag θ (B)Eg θ (C) is a gyrorhombus congruent to ABC D.  this is a  contradiction. us, we get (A ′ ,D ′ ) � ������������������ 1 + ((a 2 − 1)/(a 2 + 1) 2 ).
Theorem 3. If f: D ⟶ D is a map which preserves some gyrodistance a ∈ (0, 1) satisfying a 2 ∈ 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 d(P, Q) < a. Clearly, there exist two points in D, say C and E, such that d(P, C) � a � d(E, C) and Q ∈ [C, E]. Let S(C, a) be the gyrocircle centered at C with radius a.
en, there exist two points on S(C, a), say R 1 and R 2 such that d(P, R 1 ) � d(P, R 2 ) � a. us, we construct two equilateral gyrotriangles PR 1 C, PR 2 C, the side gyrolengths of which are a. Here, the point Q can be thought to be on the gyrosegment [P, R 1 ], otherwise a new configuration is needed to provide this feature, which is not difficult to construct. Assume If it can be established that α ∉ πQ, then there is no nonzero integer k such that A 1 � A k . If there is such a k satisfying A 1 � A k , then one can easily see that kα � 2π, i.e., α � (2π/k) ∈ π Q. In [3], it was shown that if ρ � (m/n)π, where m, n ∈ Z satisfying cos ρ ∈ Q, then cos ρ ∈ 0, { } for all a ∈ (0, 1) satisfying a 2 ∈ Q, we immediately get that α ∉ πQ. Hence, the gyrotriangles which define the sequence A i CA i+1 touch the gyroarc arc(PR 1 ) at an infinite number of points. us, [P, R 1 ] 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.
us, f preserves the gyrodistances less than a. Hence, we obtain that d(P, Q) � d(P ′ , 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 S(C, a) satisfying d(K, 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 d(K ′ , 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 a 2 ∈ Q, this implies n ⊗ a ∈ (0, 1) and (n ⊗ a) 2 ∈ Q. Moreover, for each equilateral gyrotriangle ABC with d(A, B) � d(A, C) � d(B, C) � n ⊗ a (a 2 ∈ Q), if ∠ABC: � β, then cos β ∈ Q and β ∉ πQ.
Step 3. f preserves the gyrodistances greater than a. Let P and Q be two distinct points in D satisfying d(P, Q) > a. Let k be a positive integer such that d(P, 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. en, there exists a point W on S(D, k ⊗ a) such that DPW is an equilateral gyrotriangle and denote ∠DPW � ϕ. Since cos ϕ ∈ Q by Step 2, f preserves all the gyrodistances less than k ⊗ a. Hence, we see that f preserves all gyrodistances.

Conclusions
We have proved that if f: D ⟶ D is a mapping which preserves some gyrodistance a ∈ (0, 1) satisfying a 2 ∈ Q, then f is a gyroisometry; that is, it preserves all gyrodistances. is implies that f is also a hyperbolic isometry. Naturally, one may wonder whether eorem 3 is valid for an arbitrary gyrodistance a satisfying a 2 ∈ (0, 1)∖Q. We leave the solution of this problem to the reader's attention.

Data Availability
No data were used to support this study.

Conflicts of Interest
e authors declare that they have no conflicts of interest.