The hyperbolic polygons of type (ϵ, n) and Möbius transformations

Theorem 1

[1] If f : ℂ → ℂ is a circle preserving map, then f is a Möbius transformation if and only if f is a bijection.

The transformations f(z) = az¯+bcz¯+d, where a, b, c, d ∈ ℂ with ad − bc ≠ 0 are known as conjugate Möbius transformations of ℂ. It is easy to see that each conjugate Möbius transformation f is the composition of complex conjugation with a Möbius transformation. Since the complex conjugate transformation and Möbius transformations are homeomorphisms of ℂ onto itself (complex conjugation is given by reflection in the plane through ℝ ∪ {∞}), conjugate Möbius transformations are homeomorphisms of ℂ onto itself. Notice that the composition of a conjugate Möbius transformation with a Möbius transformation is a conjugate Möbius transformation and composition of two conjugate Möbius transformations is a Möbius transformation. There is topological distinction between Möbius transformations and conjugate Möbius transformations: Möbius transformations preserve the orientation of ℂ, whereas conjugate Möbius transformations reverse it. To see more details about conjugate Möbius transformations, we refer the reader to [2]. The following definitions are well known and fundamental in hyperbolic geometry.

Definition 2

[3] A Lambert quadrilateral is a hyperbolic quadrilateral with ordered interior angles π2,π2,π2 and θ, where 0 < θ < π2 .

Definition 3

[3] A Saccheri quadrilateral is a hyperbolic quadrilateral with ordered interior angles π2,π2, θ, θ, where 0 < θ < π2 .

A Möbius invariant property is naturally related to hyperbolic geometry. To see the characteristics of Möbius transformations involving Lambert quadrilaterals and Saccheri quadrilaterals, we refer the reader to [4]. Moreover, there are many characterizations of Möbius transformations by using various hyperbolic polygons; see, for instance, [5, 6, 7].

In [8, 9], O. Demirel presented some characterizations of Möbius transformations by using new classes of geometric hyperbolic objects called “degenerate Lambert quadrilaterals” and “degenerate Saccheri quadrilaterals”, respectively.

Definition 4

[9] A degenerate Lambert quadrilateral is a hyperbolic convex quadrilateral with ordered interior angles π2+ϵ,π2,π2−ϵ, θ, where 0 < θ < π2 and 0 < ϵ < π2−θ2.

Theorem 5

[9] Let f : B² → B² be a surjective transformation. Then f is a Möbius transformation or a conjugate Möbius transformation if and only if f preserves all ϵ-Lambert quadrilaterals.

Definition 6

[8] A degenerate Saccheri quadrilateral is a hyperbolic convex quadrilateral with ordered angles π2−ϵ,π2+ϵ, θ, θ, where 0 < θ < π2 and 0 < ϵ < π2−θ2.

Theorem 7

[8] Let f : B² → B² be a surjective transformation. Then f is a Möbius transformation or a conjugate Möbius transformation if and only if f preserves all ϵ-Saccheri quadrilaterals.

In the theorems above, B² is the open unit disc in the complex plane. Naturally, one may wonder whether the counterpart of Theorem 7 exists for hyperbolic polygons instead of using degenerate Saccheri quadrilaterals. Before giving the affirmative answer of this question let us state the following definition:

Definition 8

Let n be a positive integer satisfying n ≥ 5. An n-sided hyperbolic polygon of type (ϵ, n) is a convex hyperbolic polygon with ordered interior angles π2 + ϵ, θ₁, θ₂, …, θ_n−2, π2 − ϵ or π2 − ϵ, θ₁, θ₂, …, θ_n−2, π2 + ϵ, where 0 < ϵ < π2 and 0 < θ_i < π satisfying

and θ_i + θ_i+1 ≠ π (1 ≤ i ≤ n − 3), θ₁ + ( π2 + ϵ) ≠ π, θ_n−2 + ( π2 − ϵ) ≠ π. Notice that the sides of the n-sided hyperbolic polygon of type (ϵ, n) we mentioned here are hyperbolic line segments.

The existence of n-sided hyperbolic polygons of type (ϵ, n) is clear by the following result:

Lemma 9

[3] Let (θ₁, θ₂, ⋯, θ_n) be any ordered n-tuple with 0 ≤ θ_j < π, j = 1, 2, …, n. Then there exists a polygon P with interior angles θ₁, θ₂, …, θ_n, occurring in this order around ∂ P, if and only if ∑i=1n θ_i < (n − 2)π.

This paper presents a new characterization of Möbius transformations by use of mappings which preserve n-sided hyperbolic polygons of type (ϵ, n). To do so, we need Carathéodory’s theorem which plays a major role in our results. C. Carathéodory [10] proved that every arbitrary one-to-one correspondence between the points of a circular disc C and a bounded point set C′ such that which circles lying completely in C are transformed into circles lying in C′ must always be either a Möbius transformation or a conjugate Möbius transformation.

Throughout the paper we denote by X′ 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 PQR the hyperbolic triangle with vertices P, Q and R, by ∠PQR the angle between [P, Q] and [P, R] and by d_H(P, Q) the hyperbolic distance between points P and Q. We consider the hyperbolic plane B² = {z ∈ ℂ: ∣z∣ < 1} with length differential ds2=4|dz|2(1−|z|2)2. The Poincaré disc model of hyperbolic geometry is built on B²; more precisely the points of this model are points of B² and the hyperbolic lines of this model are Euclidean semicircular arcs that intersect the boundary of B² orthogonally including diameters of B². Given two distinct hyperbolic lines that intersect at a point, the measure of the angle between these hyperbolic lines is defined by the Euclidean tangents at the common point.

Lemma 10

Let f : B² → B² be a mapping which preserves n-sided hyperbolic polygons of type (ϵ, n) for all 0 < ϵ < π2 . Then f is injective.

Proof

Let A₁ and A₂ be two distinct points in B². It is clear that there exists an (2n − 4)-sided hyperbolic regular polygon (n > 4, n ∈ ℕ), say A₁A₂ ⋯ A_2n−4. By β denote the interior angles of A₁A₂ ⋯ A_2n−4. Let M and N be the midpoints of [A_2n−4, A₁] and [A_n−2, A_n−1], respectively. Then the hyperbolic polygons MA₁A₂ ⋯ A_n−2N and NA_n−1A_n ⋯ A_2n−4M are n-sided hyperbolic polygons satisfying MN ⊥ A₁A_2n−4 and MN ⊥ A_n−2A_n−1. Let P be a point on [M, A₁] and let Q be a point on [N, A_n−1] satisfying d_H(P, A₁) = d_H(Q, A_n−1). By ψ denote the angle ∠ QPA₁. Since A₁A₂A₃ ⋯ A_2n−4 is an (2n − 4)-sided hyperbolic regular polygon, we immediately get ∠ PQA_n−1 = ψ. If ψ > π2 let’s denote ψ = π2 + α and if ψ < π2 let’s denote ψ = π2 − α. Hence we see that PA₁A₂ ⋯ A_n−2Q is an n-sided hyperbolic polygon of type (α, n). By assumption, we obtain that P′A1′A2′⋯An−2′Q′ is also an n-sided hyperbolic polygon of type (α, n), which implies A1′≠A2′. Thus f is injective.□

Lemma 11

Let f : B² → B² be a mapping which preserves n-sided hyperbolic polygons of type (ϵ, n) for all 0 < ϵ < π2 . Then f preserves the collinearity and betweenness property of the points.

Proof

Let P and Q be two distinct points in B² and assume that S is an interior point of [P, Q]. Let Δ be the set of all n-sided hyperbolic polygons of type (ϵ, n) such that the points P and Q are two adjacent vertices of these hyperbolic polygons. Then S belongs to all elements of Δ. By the property of f, the images of the elements of Δ are n-sided hyperbolic polygons of type (ϵ, n) whose vertices contain P′ and Q′. Moreover, the images of the elements of Δ must contain S′. Since f is injective by Lemma 10, we get P′ ≠ S′ ≠ Q′. Therefore, S′ must be an interior point of [P′, Q′], which implies that f preserves the collinearity and betweenness of the points.□

Lemma 12

Let f : B² → B² be a mapping which preserves n-sided hyperbolic polygons of type (ϵ, n) for all 0 < ϵ < π2 . Then f preserves the angles π2 + ϵ and π2 − ϵ.

Proof

Let A₁A₂ ⋯ A_n be an n-sided hyperbolic polygon of type (ϵ, n) such that ∠A_nA₁A₂ = π2 − ϵ, ∠ A_i−1A_iA_i+1 = θ_i (2 ≤ i ≤ n − 1) and ∠ A_n−1A_nA₁ = π2 + ϵ. Let us denote the midpoint of A₁ and A_n by M. Clearly, the hyperbolic line A₁A₂ and the complex unit disc B² meet at two points, say P and Q. Assume A₁ ∈ [P, A₂] and ∠ MA₂Q = ρ. Let X be a point on [P, A₂] moving from P to A₂. It is easy to see that if X moves from P to A₂, the angle ∠MXQ increases continuously from 0 to ρ. Let H be a point on [A₁, A₂] such that ∠ MHA₂ < π2 . Now take a point on the hyperbolic line A_nA_n−1, say S, satisfying A_n ∈ [S, A_n−1] and d_H(S, A_n) = d_H(H, A₁). It is easy to see that d_H(M, A₁) = d_H(M, A_n), ∠ SA_nM = ∠ HA₁M, and d_H(A₁, H) = d_H(A_n, S) hold, which implies that the hyperbolic triangles HA₁M and SA_nM are congruent triangles by the hyperbolic side-angle-side theorem. Hence we get ∠A₁MH = ∠A_nMS and this yields that the points H, M and S must be collinear and ∠ MHA₁ = ∠ MSA_n. Since ∠ MHA₂ < π2 , we may assume the representation ∠ MHA₂ = π2 − α, where 0 < α < π2 , which implies ∠ MSA_n = π2 + α. Notice that α must be less than ϵ since π2 − ϵ < π2 − α. Therefore, one can easily see that HA₂ ⋯ A_n−1S is an n-sided hyperbolic polygon of type (α, n). By the property of f, the images of the hyperbolic polygons A₁A₂ ⋯ A_n and HA₂ ⋯ A_n−1S are n-sided hyperbolic polygons of type (ϵ, n) and type (α, n), respectively. The hyperbolic polygons A1′A2′⋯An′ and H′A2′⋯An−1′S′ have n − 2 common angles, which implies ∠An′A1′A2′=π2±ϵ and ∠S′H′A2′=π2±α. Now assume ∠An′A1′A2′=π2+ϵ. By Lemma 11, we get H′∈[A1′,A2′],M′∈[A1′,An′] and M′ ∈ [H′, S′]. Since ∠S′H′ A2′=π2 ± α and α < ϵ hold, we get that the sum of the measures of interior angles of the hyperbolic triangle M′A1′H′ is ∠A1′M′H′+(π2+ϵ)+(π2±α), which is greater then π. This is not possible in hyperbolic geometry and so we get ∠An′A1′A2′=π2−ϵ, which yields ∠An−1′An′A1′=π2+ϵ. □

Lemma 13

Let f : B² → B² be a mapping which preserves n-sided hyperbolic polygons of type (ϵ, n) for all 0 < ϵ < π2 . Then f preserves the hyperbolic distance.

Proof

Let P and Q be two distinct points in B². Take a point S such that PQS forms a hyperbolic equilateral triangle. By β denote its angles ∠ PQS = ∠ QSP = ∠ SPQ := β. Since β < π2 let’s use the representation β = π2 − α with 0 < α < π2 . By Lemma 9, there exists an n-sided hyperbolic polygon of type (α, n), say A₁A₂ ⋯ A_n, such that ∠A₁A₂A₃ = γ₂, ∠A₂A₃A₄ = γ₃, …, ∠ A_n−2A_n−1A_n = γ_n−1, ∠ A_n−1A_nA₁ = π2 + α and ∠A_nA₁A₂ = π2 − α. Then the angle ∠A_nA₁A₂ of the hyperbolic polygon A₁A₂ ⋯ A_n can be moved to the point P by using an appropriate hyperbolic isometry g such that g(A₂) ∈ [P, S] (or S ∈ [P, g(A₂)]) and g(A_n) ∈ [P, Q] (or Q ∈ [P, g(A_n)]). Since f preserves the angles π2 + ϵ and π2 − ϵ of n-sided hyperbolic polygons of type (ϵ, n) for all 0 < ϵ < π2 by Lemma 12, we get

which implies ∠ PQS = ∠P′Q′S′ and ∠ QSP = ∠Q′S′P′. Because of the fact that the angles at the vertices of a hyperbolic triangle determine its lengths, we get d_H(P, Q) = d_H(P′, Q′); see [11, 12].□

Corollary 14

Let f : B² → B² be a mapping which preserves n-sided hyperbolic polygons of type (ϵ, n) for all 0 < ϵ < π2 . Then f preserves the measures of all angles.

Theorem 15

Let f : B² → B² be a surjective transformation. Then f is a Möbius transformation or a conjugate Möbius transformation if and only if f preserves n-sided hyperbolic polygons of type (ϵ, n) for all 0 < ϵ < π2 .

Proof

The “only if” part is clear since f is an isometry. Conversely, we may assume that f preserves all n-sided hyperbolic polygons of type (ϵ, n) in B² and f (O) = O by composing an hyperbolic isometry if necessary. Let us take two distinct points in B² and denote them by x, y. By Lemma 13, we immediately get d_H(O, x) = d_H(O, x′) and d_H(O, y) = d_H(O, y′), namely ∣x∣ = ∣x′∣ and ∣y∣ = ∣y′∣, where ∣⋅∣ denotes the Euclidean norm. Therefore, we get ∣x − y∣ = ∣x′ − y′∣ since f preserves angular sizes by Corollary 14. As

f preserves the Euclidean inner-product this implies that f is a restriction of an orthogonal transformation on B², that is, f is a Möbius transformation or a conjugate Möbius transformation by Carathéodory’s theorem.□

Corollary 16

Let f : B² → B² be a conformal (angle preserving with sign) surjective transformation. Then f is a Möbius transformation if and only if f preserves n-sided hyperbolic polygons of type (ϵ, n) for all 0 < ϵ < π2 .

Corollary 17

Let f : B² → B² be an angle reversing surjective transformation. Then f is a conjugate Möbius transformation if and only if f preserves n-sided hyperbolic polygons of type (ϵ, n) for all 0 < ϵ < π2 .