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Bull. Math. Soc. Sci. Math. Roumanie Tome 61 (109), No. 4, 2018, 409–415

Degenerate Lambert and M¨obiustransformations by Oguzhan˘ Demirel

Abstract We present a new characterization of M¨obiustransformations by using a new geo- metric concept that we will call “degenerate Lambert ”. Our proof is based on a geometric approach.

Key Words: Lambert quadrilateral, M¨obiustransformation. 2010 Mathematics Subject Classification: Primary 51B10, Secondary 30F45, 51M09, 51M10, 51M25.

1 Introduction

M¨obiustransformations, also known as linear fractional transformations, are rational func- az+b tions of the form f(z) = cz+d satisfying ad − bc 6= 0, where a, b, c, d ∈ C. M¨obiustrans- formations are the automorphisms of C = C ∪ {∞}, namely the meromorphic bijections f : C → C. M¨obiustransformations are also directly conformal homeomorphisms of C onto itself. There are well-known elementary proofs that if f is a continuous injective map of the extended complex plane C into itself that preserves cross-ratios, or that maps circles into circles, then f is a M¨obius transformation. The connection between M¨obius transformations, cross-ratios and the preservation of circles is well known. az+b The transformations f(z) = cz+d with ad − bc 6= 0, where a, b, c, d ∈ C are known as conjugate M¨obius transformations of C. Each conjugate M¨obiustransformation f is the composition of complex conjugation with a M¨obiustransformation, since both of these are homeomorphisms of C onto itself (complex conjugation being given by reflection in the plane through R ∪ {∞}), so is f. Clearly, the composition of a conjugate M¨obiustransformation with a M¨obiustransformation is a conjugate M¨obiustransformation and the composition of two conjugate M¨obiustransformations is a M¨obiustransformation. There is a topological distinction between M¨obiustransformations and conjugate M¨obiustransformations in that M¨obiustransformations preserve the orientation of C while conjugate M¨obiustransforma- tions reverse it. To see more details about conjugate M¨obiustransformations, we refer [9]. C. Carath´eodory [3] proved that every arbitrary one to one correspondence between the points of a circular disc C and a bounded point set C0 by which circles lying completely in C are transformed into circles lying in C0 must always be either a M¨obiustransformation f(z) or f(z). Carath´eodory’s theorem is generalized by R. H¨ofer[8] to arbitrary dimensions. R. H¨oferproved that for a domain D of Rn, if any injective mapping f : D → Rn which takes hyperspheres whose interior is contained in D to hyperspheres in Rn, then f is the 410 Degenerate Lambert quadrilaterals and M¨obiustransformations restriction of a M¨obiustransformation. For more details about sphere preserving maps, see [2]. M¨obiustransformations are well known and fundamental in hyperbolic since they act as isometries on hyperbolic space. There are many characterizations of M¨obius transformations by using Lambert and Saccheri quadrilaterals [14] and [15], hyperbolic regular [4], hyperbolic regular star polygons [5], Apollonius quadrilaterals [7], tri- angular domains [10], polygons having type A [11] and others. Moreover, in [6], O. Demirel presented a characterization of M¨obiustransformations by use of mappings preserving hy- perbolic with their hyperbolic areas.

π Definition 1 ([1]). The Lambert quadrilateral is a hyperbolic quadrilateral with 2 , π π 2 , 2 , θ. Clearly a Lambert quadrilateral is a hyperbolic convex quadrilateral with three right angles and one acute . Distorting of the non-adjacent right angles of a Lambert quadrilateral, degenerate Lam- bert quadrilateral concept is defined as follows:

Definition 2. A degenerate Lambert quadrilateral is a hyperbolic convex quadrilateral with π π π π π θ ordered angles 2 + , 2 , 2 − , θ where 0 < θ < 2 and 0 <  < 2 − 2 . Notice that, for a degenerate Lambert quadrilateral, the sum of the measures of mutual π π distorted angles which are 2 + and 2 − is π and the sum of the measures of other mutual angles is less than π. π In this paper we call the degenerate Lambert quadrilaterals having ordered angles 2 +, π π 2 2 , 2 − , θ briefly as −Lambert quadrilaterals. We consider the hyperbolic plane B = 2 2 4|dz| {z : |z| < 1} with length differential ds = (1−|z|2)2 . Throughout of the paper we denote by X0 the image of X under f , by [P,Q] the geodesic segment between points P and Q, by PQ the geodesic through points P and Q, by P QR the hyperbolic with three ordered vertices P,Q and R, by P QRS the hyperbolic quadrilateral with four ordered vertices P, Q, R and S, and by ∠P QR the angle between [Q, P ] and [Q, R].

2 A Characterization of M¨obiusTransformations by use of Degenerate Lambert Quadrilaterals

In [14], S. Yang and A. Fang proved the following result:

Theorem 1. [14] Let f : B2 → B2 be a continuous bijection. Then f is M¨obiusif and only if f preserves Lambert quadrilaterals in B2.

In the expression of this result, by the Lambert quadrilateral preserving property of π π π functions, this means that if ABCD is a Lambert quadrilateral having angles 2 , 2 , 2 and 0 0 0 0 π π π 0 θ, then A B C D is a Lambert quadrilateral having angles 2 , 2 , 2 and θ . Obviously, if f preserves Lambert quadrilaterals in B2, this implies that it preserves three right angles. The π 0 0 0 authors proved that if ∠DAB = ∠ABC = ∠BCD = 2 and ∠CDA = θ, then ∠D A B = 0 0 0 0 0 0 π 0 0 0 ∠A B C = ∠B C D = 2 and ∠C D A = θ. For more details, we refer [14] and [15]. O˘guzhanDemirel 411

Similarly to [14], we use often the −Lambert quadrilaterals preserving property of functions. This means that if ABCD is a −Lambert quadrilateral having ordered angles π π π 0 0 0 0 π 2 +, 2 , 2 −, θ, then A B C D is a −Lambert quadrilateral having ordered angles 2 +, π π 0 2 , 2 − , θ . Lemma 1. Let f : B2 → B2 be a mapping which preserves all −Lambert quadrilaterals π where 0 <  < 2 . Then f preserves right angles of −Lambert quadrilaterals. Proof. Step 1: Firstly, we claim that f is injective. Let us take two different points P and Q in B2. Then by constructing a −Lambert quadrilateral P QRS, one can easily get that P 0Q0R0S0 is also a −Lambert quadrilateral by the property of f. Therefore, the points P 0 and Q0 must be different which implies that f is injective. Step 2: We claim that f preserves the collinearity and betweenness properties of the points. Let P and Q be two different points in B2 and assume that S is an interior point of [P,Q]. Let S be the set of all −Lambert quadrilaterals whose vertices are P and Q. Notice that, if XYPQ ∈ S, then there are four possible cases for the measure of ∠YPQ π π π such as ∠YPQ = 2 + , ∠YPQ = 2 − , ∠YPQ = 2 and ∠YPQ = β. By the definition of S, the −Lambert quadrilaterals in S have a common side [P,Q] which contains S. Hence S0 must be on all elements of S0. By the property of f, the set S0 must be consists of −Lambert quadrilaterals with common vertices P 0 and Q0. Notice that by the injectivity of f, P 0 6= S0 6= Q0 since P 6= S 6= Q. Therefore, S0 must be an interior point of [P 0,Q0] which implies that f preserves the collinearity and betweenness of the points. Step 3: f preserves right angles of all −Lambert quadrilaterals. Let ABCD be a π π π −Lambert quadrilateral with ∠ABC = 2 + , ∠BCD = 2 , ∠CDA = 2 − , ∠DAB = θ and E be a point on [B,C]. Now construct a new −Lambert quadrilateral with the help of the points E,C,D. Firstly draw a hyperbolic geodesic passing through E, say q, such π that the measure of the angle between the hyperbolic geodesics q and BC is 2 + . We suppose that q must intersect [A, D] but not at A and D. Indeed, if q intersects [A, B] at a point, say W , then sum of the measures of interior angles of the WBE π π is ∠WBE + ∠BEW + ∠EWB = ( 2 + ) + ( 2 − ) + ∠EW B > π which is impossible for a hyperbolic triangle. If q intersects [C,D] at a point, say V , then the sum of the interior π π angles of the hyperbolic triangle VEC is ∠VEC +∠ECV +∠CVE = ( 2 +)+ 2 +∠CV E > π which is impossible for a hyperbolic triangle. Therefore, q must intersect [A, D] but not at A and D. Let us denote the common point of the hyperbolic geodesics q and AD by F . Hence we obtain a new −Lambert quadrilateral CDFE. By Step 2, the points E0 and F 0 must be interior points of [B0,C0] and [A0,D0], respectively. Now we consider four cases: 0 0 0 π Case 1. Assume ∠A B C = 2 . Hence, by the definition of −Lambert quadrilateral, 0 0 0 π 0 0 0 π 0 0 0 π we have ∠B C D = 2 +  or ∠B C D = 2 − . If ∠B C D = 2 + , then we get 0 0 0 0 0 0 π 0 0 0 0 0 0 0 π ∠D A B = ∠D F E = 2 − , ∠C D F := θ and ∠F E C = 2 since the quadrilaterals A0B0C0D0 and C0D0F 0E0 are −Lambert quadrilaterals. Therefore, the sum of the measures 0 0 0 0 0 0 0 π 0 0 0 0 0 0 π of angles of A B E F is 2π. If ∠B C D = 2 − , then ∠D A B = ∠D F E = 2 + , 0 0 0 0 0 0 0 π ∠C D F := θ and ∠F E C = 2 which implies that the sum of the measures of angles of A0B0E0F 0 is 2π. In both cases, this is not possible in since the sum of the measures of the angles of a hyperbolic quadrilateral must be less than 2π. Therefore 0 0 0 π ∠A B C 6= 2 . 0 0 0 π 0 0 0 π 0 0 0 π Case 2. Assume ∠D A B = 2 . Thus we get ∠C D A = 2 +  or ∠C D A = 2 − . In both cases, similarly to first case above, the sum of the measures of angles of A0B0E0F 0 412 Degenerate Lambert quadrilaterals and M¨obiustransformations is 2π which is a contradiction. 0 0 0 π Case 3. Assume ∠C D A = 2 . Let us take a point G on the geodesic segment [A, B] π and construct the hyperbolic triangle GCD. Because of the fact that ∠GCD := α1 < 2 π and ∠CDG := α2 < 2 hold true, there exists a point, say H, on the geodesic [C,D] such π π that ∠GHD = 2 . Obviously, there exists a point K on [H,D] such that ∠GKC = 2 −  π π π since ∠GHD = 2 , ∠CDA = 2 −  and ∠CDG = α2 < 2 . Hence we get a new −Lambert π π π quadrilateral GBCK satisfying ∠GKC = 2 − , ∠GBC = 2 + , ∠BCK = 2 , ∠KGB := 0 0 0 0 α3. By Step 2, the points G and K must be lie on the geodesic segments [A ,B ] and [C0,D0], respectively and by the property of f, the quadrilateral G0B0C0K0 is a −Lambert 0 0 0 π 0 0 0 0 quadrilateral. By the assumption of ∠C D A = 2 , one can easily get ∠C B A = θ and 0 0 0 π 0 0 0 π 0 0 0 π ∠C K G = 2 . Thus we have ∠B C K = 2 +  or ∠B C K = 2 − . In both cases, the sum of the measures of angles of A0G0K0D0 is 2π which is a contradiction. 0 0 0 π Consequently, the fourth case must be hold, that is ∠B C D = 2 .

Lemma 2. Let f : B2 → B2 be a mapping which preserves all −Lambert quadrilaterals π π π where 0 <  < 2 . Then f preserves the measures of degenerate angles 2 +  and 2 − . π π Proof. Let ABCD be a −Lambert quadrilateral with ∠ABC = 2 + , ∠BCD = 2 , π 0 0 0 π ∠CDA = 2 − , and ∠DAB := θ. Then from Lemma 1, we get ∠B C D = 2 and 0 0 0 0 0 0 0 π 0 0 0 π 0 0 0 ∠D A B := θ which implies ∠A B C = 2 − or ∠A B C = 2 +. Now assume ∠A B C = π 0 0 0 π 2 − . Therefore, we get ∠C D A = 2 + . Since all the angles of the triangle ACD are π acute angles, there exist a point on the geodesic [C,D], say H, satisfying ∠AHD = 2 . Let Q be the intersection of the unit disc B2 with the geodesic CD such that D lies on [Q, C]. Now draw a hyperbolic geodesic p passing through A such that the measure π of the angle between AH and p is 2 + . The existence of p is clear since ∠HAD < π π θ < 2 . Pick a point M on p satisfying that ∠MAH = 2 + . Let X be a point on [H,Q] moving from H to Q, then the angle ∠MXH goes from ∠MHQ which is an obtuse π angle to 0 continuously. Since ∠MDH > 2 − , there exists a unique point N on [Q, D] π such that ∠MNH = 2 − . Hence we get a new −Lambert quadrilateral MNHA with π π π ∠MAH = 2 + , ∠AHN = 2 , ∠HNM = 2 − , and ∠NMA := θ1. Since f preserves the −Lambert quadrilaterals and the right angles of −Lambert quadrilaterals, then the 0 0 0 0 0 0 0 0 0 0 0 π quadrilaterals A B C D and M N H A are −Lambert quadrilaterals with ∠B C D = 2 0 0 0 π 0 0 0 and ∠A H N = 2 . Therefore, the sum of the interior angles of hyperbolic triangle A H D 0 0 0 π 0 0 0 π is greater than π since ∠A H D = 2 and ∠H D A = 2 +. This is a contradiction. Hence 0 0 0 π we get ∠A B C = 2 + .

Lemma 3. Let f : B2 → B2 be a mapping which preserves all −Lambert quadrilaterals π where 0 <  < 2 . Then f preserves the angles of all −Lambert quadrilaterals. π π Proof. Let ABCD be a −Lambert quadrilateral with ∠ABC = 2 + , ∠BCD = 2 , π ∠CDA = 2 − , and ∠DAB := θ. Now draw a hyperbolic P QRS with center is θ O, where O is the origin, satisfying ∠P QR = ∠QRS = ∠RSP = ∠SPQ = 2 . Without loss of generality, we may assume that the points P and R lie on the y−axis and the points Q and S lie on the x−axis. Let the points M1, M2, M3 and M4 be the midpoints of the geodesic segments of [P,Q], [Q, R], [R,S] and [S, P ], respectively. By the property of a hyperbolic square, M1M3 is perpendicular to PQ and RS, and M2M4 is perpendicular O˘guzhanDemirel 413

to QR and SP . Moreover, M1M3 is perpendicular to M2M4 at O. Clearly, PM1OM4 is a Lambert quadrilateral. Pick a point on [P,M4] and denote it by Xi. Obviously, π π ∠OXiP > 2 . Thus we represent the measure of this angle by ∠OXiP = 2 + i. Now, pick a point on [M1,Q], say Yi, such that such that dH (Xi,P ) = dH (Yi,Q) where dH is the hyperbolic distance function. Therefore, we get a hyperbolic quadrilateral OXiPYi. We claim that OXiPYi is a i−Lambert quadrilateral. To see this, it is enough to consider the hyperbolic triangles OXiP and OYiQ. Because of the fact that dH (Xi,P ) = dH (Yi,Q), θ dH (O,P ) = dH (O,Q) and ∠XiPO = 4 = ∠YiQO hold true, by side-angle-side theorem for hyperbolic triangles, see [12], the hyperbolic triangles OXiP and OYiQ are congruent. π π Hence we get that ∠OYiQ = 2 +i, which implies ∠PYiO = 2 −i and we get that 0XiPYi is a i−Lambert quadrilateral. It is possible that for each point Xs on [P,M4] satisfying π ∠OXsP := 2 + s, following the construction of i−Lambert quadrilateral 0XiPYi, one can easily construct a s−Lambert quadrilateral OXsPYs where Ys is an interior point of π [Q, M1] satisfying ∠OYsP = 2 − s. It is easy to see that there exist a unique point on θ π [P,M4], say Xj, such that ∠OXjM4 = θ, since ∠OPXj = 4 and ∠OM4P = 2 . Since π π ∠OXjP = π − θ = 2 + ( 2 − θ), we get a new degenerate Lambert quadrilateral OXjPYj π π which is a ( 2 − θ)−Lambert quadrilateral. Notice that 2 − θ = j. The angle ∠BAD of the −Lambert quadrilateral ABCD can be moved to the point Xj by an appropriate hyperbolic isometry g, such that the points g(B) and g(D) lie on the geodesics XjO and XjM4, π 0 0 0 0 respectively. Since f preserves −Lambert quadrilaterals for all 0 <  < 2 , then O XjP Yj π 0 0 0 π π is a ( 2 −θ)−Lambert quadrilateral. By Lemma 2, we get ∠OXjP = ∠O XjP = 2 +( 2 −θ) 0 0 0 0 0 0 which implies ∠BAD = ∠OXjM4 = ∠O XjM4 = ∠B A D = θ.

Lemma 4. Let f : B2 → B2 be a mapping which preserves all −Lambert quadrilaterals π where 0 <  < 2 . Then f preserves hyperbolic distance. Proof. Let A and B be two different points in B2. It is not hard to see that these points define two hyperbolic . Let us consider one of them and denote it by ABCD. Assume ∠ABC = ∠BCD = ∠CDA = ∠DAB := θ and denote it center by M. Similar to the proof π of Lemma 3, for all 0 <  < 2 , ABCD contains four −Lambert quadrilaterals with common π vertex M. Since f preserves the angles of −Lambert quadrilaterals for all 0 <  < 2 by 0 0 0 0 0 0 0 0 0 0 0 0 Lemma 1, Lemma 2 and Lemma 3, then ∠A B C = ∠B C D = ∠C D A = ∠D A B = θ which implies that A0B0C0D0 is a hyperbolic square. Because of the fact that the angles 0 0 at the vertices of a hyperbolic square define its lengths, we get dH (A, B) = dH (A ,B ), see [13].

Theorem 2. Let f : B2 → B2 be a surjective transformation. Then f is a M¨obiustrans- formation or a conjugate M¨obiustransformation if and only if f preserves all −Lambert π quadrilaterals where 0 <  < 2 . Proof. The “only if” part is obvious because f is an isometry. Conversely, we may assume π 2 that f preserves −Lambert quadrilaterals for all 0 <  < 2 in B and f(O) = O by composing an hyperbolic isometry if necessary. Let x, y be two arbitrary points in B2. By 0 0 0 Lemma 4, we have dH (O, x) = dH (O, x ) and dH (O, y) = dH (O, y ), namely |x| = |x | and 414 Degenerate Lambert quadrilaterals and M¨obiustransformations

|y| = |y0|, where | · | denotes the Euclidean norm. Thus we get |x − y| = |x0 − y0| by since f preserves angular sizes by Lemma 3. As

2hx, yi = |x|2 + |y|2 − |x − y|2 = |x0|2 + |y0|2 − |x0 − y0|2 = 2hx0, y0i, f preserves inner-products and then is the restriction on B2 of an orthogonal transfor- mation, that is, f is a M¨obiustransformation or a conjugate M¨obiustransformation by Carath´eodory’s theorem.

Corollary 1. Let f : B2 → B2 be a conformal (angle preserving with sign) surjective transformation. Then f is a M¨obiustransformation if and only if f preserves all −Lambert π quadrilaterals where 0 <  < 2 . Acknowledgement. The author would like to thank to the referee for carefully reading of the paper and for the significant suggestions.

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Received: 04.02.2017 Revised: 27.04.2017 Accepted: 08.05.2017

Department of Mathematics, Faculty of Science and Literature, Afyon Kocatepe University, 03200, Afyonkarahisar, Turkey E-mail: [email protected]