Bull. Math. Soc. Sci. Math. Roumanie Tome 61 (109), No. 4, 2018, 409{415 Degenerate Lambert quadrilaterals 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 quadrilateral". 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 2 C. M¨obiustrans- formations are the automorphisms of C = C [ f1g, 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 2 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 [ f1g), 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 geometry 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 polygons [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 triangles with their hyperbolic areas. π Definition 1 ([1]). The Lambert quadrilateral is a hyperbolic quadrilateral with angles 2 , π π 2 , 2 , θ. Clearly a Lambert quadrilateral is a hyperbolic convex quadrilateral with three right angles and one acute angle. 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 4jdzj fz : jzj < 1g with length differential ds = (1−|zj2)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 triangle 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 2 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 hyperbolic triangle W BE π π is \W BE + \BEW + \EW B = ( 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 V EC is \V EC +\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 CDF E. 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.