New York Journal of Mathematics

New York J. Math. 15 (2009) 423–429.

A new characterization for isometries by triangles

Baokui Li and Yuefei Wang

Abstract. Let ℝ^{n} be an n-dimensional Euclidean space and 𝔻^{n} be an n-dimensional hyperbolic space with the Poincaré metric for n > 1. In this paper, we shall prove the following results. (i) A bijection f : 𝔻^{n} → 𝔻^{n} is an isometry (Möbius transformation) if and only if f is triangle preserving. (ii) A bijection f : ℝ^{n} → ℝ^{n} is an affine transformation if and only if f is triangle preserving.

Contents

1. Introduction 423 2. Hyperbolic space 425

2.1. Triangle domain preserving maps for n = 2 425 2.2. Triangle domain preserving maps for n > 2 425 2.3. Triangle preserving maps for n = 2 426 2.4. Triangle preserving maps for n > 2 427

3. Euclidean space 428 References 428

1. Introduction

Let ℝ^{n} be an n-dimension Euclidean space and

𝔻^{n} = {(x_{1}, x_{2}, ⋯, x_{n}) ∈ ℝ^{n} | x_{1}^{2} + x_{2}^{2} + ⋯ + x_{n}^{2} < 1}

be an n-dimension hyperbolic space with the Poincaré metric for n > 1.

A Möbius transformation f : 𝔻^{n} → 𝔻^{n}, which is equivalently an isometry under the Poincaré metric, has many beautiful properties. For example, f

Received July 1, 2009; revised September 9, 2009. Mathematics Subject Classification. 30C35, 51F99. Key words and phrases. Isometry, geodesic, affine transformation, triangle preserving,

triangle domain preserving.

The research was supported by NSFC and MSBRD Program of China.

423^{ISSN 1076-9803/09}

is bijective, holomorphic, circle-preserving, sphere-preserving, and geodesicpreserving . Moreover, f preserves angles and polygons.

An affine transformation f : ℝ^{n} → ℝ^{n} is line-preserving, super-plane-preserving, and keeps the parallel relation of two lines, and the rate of two segments in the same (parallel) line. The following results are known.

Theorem A ([3]). Suppose that f : ℝ^{n} → ℝ^{n} (n > 1) is surjective and line into line. Then f is an affine transformation.

Theorem B ([8]). Suppose that f : 𝔻^{n} → 𝔻^{n} (n > 1) is a bijection that preserves geodesics. Then f is an isometry.

Theorem C ([8]). Suppose that f : ℝ^{n} → ℝ^{n} (n > 1) is a bijection that preserves lines. Then f is an affine transformation.

Here we say that f preserves lines if for any line l, f(l) is a line.

Theorem D ([9]). Suppose that f : 𝔻^{n} → 𝔻^{n} is geodesic-preserving. Then f is an isometry if and only if f is nondegenerate.

Theorem E ([9]). Suppose that f : ℝ^{n} → ℝ^{n} is line-preserving. Then f is an affine transformation if and only if f is nondegenerate.

Here, f is nondegenerate if the image of the whole space under f is more than a line or geodesic.

In the hyperbolic plane 𝔻, Haruki and Rassias [6] gave a characterization of isometries by using Apollonius quadrilaterals and proved that if f is meromorphic and sends Apollonius quadrilaterals to Apollonius quadrilaterals, then f is Möbius. Here an Apollonius quadrilateral ABCD satisfies |AB| · |CD| = |BC| · |DA|, where |AB| denotes the length of the segment joining A and B. See [4] [5] [7] for other related results.

Yang and Fang ([13]) gave a characterization of isometries on 𝔻 by using Lambert quadrilaterals.

Theorem F ([13]). Suppose that f : 𝔻 → 𝔻 is a continuous bijection. Then f is Möbius if and only if f preserves Lambert quadrilaterals in 𝔻.

Recently, Yang ([12]) proved the following result on triangles in 𝔻.

Theorem G ([12]). Suppose that f : 𝔻 → 𝔻 is an injection. Then f is an isometry if and only if for some 0 < θ < π, f preserves triangles with an interior angle equal to θ.

In this paper, we shall prove the following theorems.

Theorem 1. Suppose that f : 𝔻^{n} → 𝔻^{n} is a bijection. Then f is an isometry if and only if f is triangle domain preserving.

Here, a triangle domain is a closed domain bounded by a triangle.

Theorem 2. Suppose that f : ℝ^{n} → ℝ^{n} is a bijection. Then f is an affine transformation if and only if f is triangle domain preserving.

As an application of Theorems 1 and 2, we obtain our main results.

Theorem 3. Suppose that f : 𝔻^{n} → 𝔻^{n} is a bijection. Then f is an isometry if and only if f is triangle preserving.

Theorem 4. Suppose that f : ℝ^{n} → ℝ^{n} is a bijection. Then f is an affine transformation, if and only if f is triangle preserving.

2. Hyperbolic space

In this section, we shall first prove Theorem 1.

2.1. Triangle domain preserving maps for n = 2. We shall prove Theorem 1 for n = 2.

Denote points by A, B, C, … in 𝔻, the images under f by A′, B′, C′, …, the geodesic passing through A, B by L_{AB}, the segment ended by A, B by AB, a triangle domain by △, the boundary of a triangle domain △ by ∂△, and the image triangle domain of a triangle domain △ by △′.

We have the following lemmas.

Lemma 2.1. Suppose that f : 𝔻 → 𝔻 is a triangle domain preserving injection. Then for any triangle domain △, f(∂△) ⊂ ∂△′.

In fact, for any P ∈ ∂△, we can choose another triangle domain △_{1}, such that △ ∩ △_{1} = {P }. So △′ ∩ △′_{1} = {P′}, and P′ ∈ ∂△′.

Lemma 2.2. Suppose that f : 𝔻 → 𝔻 is a triangle domain preserving injection. Then the image of a segment under f is a segment.

Proof. Given a segment L, one can find two triangle domains △_{1} and △_{2}, such that L = △_{1} ∩ △_{2}. So the image △′_{1} ∩ △′_{2} = L′ is a convex set by the convexity of triangle domains.

On the other hand, as L ⊂ ∂△_{1} ∩ ∂△_{2}, L′ ⊂ ∂△′_{1} ∩ ∂△′_{2} by Lemma 2.1. Therefore L′ is a segment, and the proof is complete. □

Proof of Theorem 1 for n = 2. It follows from Lemma 2.2 that f : 𝔻 → 𝔻 is segment to segment, and geodesic to geodesic. Moreover, for any three collinear points P′_{1}, P′_{2}, P′_{3}, with P′_{2} ∈ P′_{1}P′_{3}, we denote the inverse image points P_{1}, P_{2}, P_{3}. The image of the segment P_{1}P_{3} is a segment containing P′_{1} and P′_{3}, and so P′_{2} ∈ f(P_{1}P_{3}). That is, P_{2} ∈ P_{1}P_{3}, and f^{−1} : 𝔻 → 𝔻 is geodesic to geodesic. Then f is geodesic-preserving. Therefore f is an isometry by Theorem B or D. □

2.2. Triangle domain preserving maps for n > 2. In this part, we shall prove Theorem 1 for n > 2. First we have the following lemmas.

Lemma 2.3. Suppose that f : 𝔻^{n} → 𝔻^{n} is a triangle domain preserving bijection. Then the image of any 2-dimensional plane under f is in a 2dimensional plane.

Proof. For any 2-dimensional plane 𝔻, we choose a triangle domain △ ⊂ 𝔻, and denote the image triangle domain by △′ in some 2-dimensional plane 𝔻′.

For any point P ∈ 𝔻\△, we can choose A, B, A_{1}, B_{1} ∈ △, such that A_{1} ∈ P A, B_{1} ∈ P B, and L_{P A} ∩ L_{P B} = P.

Case 1. A′, B′, A′_{1}, B′_{1} are noncollinear. Then the image of triangle domain △_{P AB} is a triangle domain passing through A′, B′, A′_{1}, B′_{1}, which is in 𝔻′.

Case 2. A′, B′, A′_{1}, B′_{1} is collinear. Choose a point C′ ∈ △′ noncollinear with them. Suppose that C is noncollinear with P, A. Then the image of triangle △_{P AC} is a triangle domain passing through A′, A′_{1}, C′, which is in 𝔻′.

So P′ ∈ 𝔻′ and the proof is complete. □

Proof of Theorem 1 for n > 2. By Lemma 2.3, the image of any 2-di-mensional plane D is in a 2-dimensional plane. By composing some isometry, we may suppose that f : 𝔻 → 𝔻.

Therefore f : 𝔻 → 𝔻 is a triangle domain preserving injection. By Lemma 2.2, f : 𝔻 → 𝔻 is segment onto segment, and f : 𝔻^{n} → 𝔻^{n} is segment onto segment. As in the proof in the case of n = 2, we conclude that f and f^{−1} are geodesic to geodesic. So f is geodesic preserving. By Theorem B or D, f is an isometry, and the proof is complete. □

In the following we shall prove Theorem 3.

2.3. Triangle preserving maps for n = 2. In this part, we shall prove Theorem 3 for n = 2.

Denote the boundary of a triangle domain △ by ∂△, the image triangle of ∂△ by ∂′△. It is easy to see that ∂△ divides 𝔻(ℝ^{2}) into two portions, the inner ∂△^{∘} and the outer ∂△^{c}.

It is obvious that ∂△^{∘} is connected in 𝔻\∂△, which means that P, Q can be joined by some segments (triangle) in 𝔻\∂△ for any P, Q ∈ ∂△^{∘}. The image of any connected component in 𝔻\∂△ is connected in 𝔻\∂′△.

Let NC(∂△) be the number of connected components of 𝔻\∂△. Obviously,

2 ≤ NC(∂′△) ≤ NC(∂△) ≤ 4.

Then we have the following lemma.

Lemma 2.4. Suppose that f : 𝔻 → 𝔻 is a triangle preserving bijection. Then f is triangle domain preserving.

Proof. For any triangle domain △, choose three noncollinear points A′, B′, C′ in ∂′△, such that A, B, C ∈ ∂△ are noncollinear (except the vertices).

In fact, we can choose any three points A′, B′, C′ from three different sides of ∂′△ (not the vertices of ∂′△ and the image of the vertices of ∂△).

Suppose that A, B, C are in the same side of ∂△. Choose a point A_{1} in one of the other sides of ∂△ (not the vertices or the inverse image of the vertices of ∂′△), and A′_{1}, B′, C′ are noncollinear.

It is obvious that ∂△_{ABC} ⊂ △, and ∂△_{ABC} ∩ ∂△^{∘} ≠ ∅.

Note that the image triangle ∂′△_{ABC} passing through A′, B′, C′ must have crossing points with ∂′△^{∘}(As in Figure 1). That is, f(∂△^{∘}) ∩ ∂′△^{∘} ≠ ∅. So f(∂△^{∘}) ⊂ ∂′△^{∘}.

[figure: ∂△^{∘}; ∂△^{c}; A; B; C; ∂′△^{∘}; ∂′△^{c}; A′; B′; C′] Figure 1

Case 1. NC(∂△) = 2. Then f(∂△^{∘}) = ∂′△^{∘}, and f(△) is a triangle domain with boundary ∂′△, denoted by △′.

Case 2. NC(∂△) = 3 (or 4). The triangle domain △ can be separated into two triangle domains by some segment S, say △_{1}, △_{2}, such that NC(∂△_{1}) = NC(∂△_{2}) = 2. Denote the image triangle domains △′_{1} and △′_{2}.

Since the segment S = ∂△_{1} ∩ ∂△_{2} and S = △_{1} ∩ △_{2}, the image S′ = ∂△′_{1} ∩ ∂△′_{2} and S′ = △′_{1} ∩ △′_{2} is a segment. So the image triangle ∂′△ = ∂(△′_{1} ∪ △′_{2}). Note that f(△) = △′_{1} ∪ △′_{2}. Therefore f(△) is a triangle domain.

Therefore, for any triangle domain △, f(△) is a triangle domain. This completes the proof. □

Proof of Theorem 3 for n = 2. In this case, Theorem 3 follows from Lemma 2.4 and Theorem 1. □

2.4. Triangle preserving maps for n > 2. In this part, we shall prove Theorem 3 for n > 2.

Lemma 2.5. Suppose that f : 𝔻^{n} → 𝔻^{n} is a triangle preserving bijection. Then the image of any 2-dimensional plane is in a 2-dimensional plane.

Proof. For any 2-dimensional plane 𝔻, we choose a triangle ∂△ ⊂ 𝔻, and denote the image triangle by ∂′△ in some 2-dimensional plane 𝔻′.

For any point P ∈ 𝔻, we can choose Q ∈ 𝔻, such that L_{QP} ∩ ∂△ = {A, A_{1}}, and P, A_{1} ∈ QA. Choose B ∈ ∂△, such that L_{QB} ∩∂△ = {B, B_{1}}, and B_{1} ∈ QB.

Case 1. A′, B′, A′_{1}, B′_{1} are noncollinear. Then the image of triangle ∂△_{QAB} is a triangle passing through A′, B′, A′_{1}, B′_{1}, which is in 𝔻′.

Case 2. A′, B′, A′_{1}, B′_{1} is collinear. Choose a point C′ ∈ ∂△′ noncollinear with them. Then the image of triangle ∂△_{QAC} is a triangle passing through A′, A′_{1}, C′, which is in 𝔻′.

We have shown that P′ ∈ 𝔻′ for all cases. The proof is complete. □

Lemma 2.6. Suppose that f : 𝔻^{n} → 𝔻^{n} is a triangle preserving bijection. Then for any 2-dimensional plane 𝔻, f(𝔻) is a 2-dimensional plane.

Proof. Choose a triangle ∂△ ⊂ 𝔻 and the image triangle ∂′△ ⊂ 𝔻′. By Lemma 2.3, f(𝔻) ⊂ 𝔻′. Therefore we only need to prove that f : 𝔻 → 𝔻′ is surjective.

Suppose that there exists P′ ∈ 𝔻′\f(𝔻). Then we can have its inverse image point P ∈ 𝔻^{n}\𝔻 since f is surjection. Without loss of generality, we may suppose that P′ is in the outside of ∂′△.

Denote 3-dimensional space containing 𝔻 and P by 𝔻^{3}.

Choose A ∈ ∂△, such that L_{A′P′} ∩ ∂△ = {A′, A′_{1}}, and A′_{1} ∈ P′A′. Given any C ∈ ∂△\{A, A_{1}}, the image triangle of ∂△_{P AC}, passing through three noncollinear points P′, A′, C′, is in 𝔻′. So f(P A) ⊂ 𝔻′.

Choose E′ ∈ ∂′△\{A′}, such that any triangle in 𝔻′, passing through P′, E′, has more than one crossing points with ∂′△ (in fact, E′ is in the interior of the farther side of P′). Choose Q ∈ P A\{P, A} and then

∂△_{P QE} ∩ ∂△ = {E},

and their image triangles have more than one crossing point. This is a contradiction, and the proof is complete. □

Proof of Theorem 3 for n > 2. For any 2-dimensional plane 𝔻 ∈ 𝔻^{n}, f(𝔻) is a 2-dimensional plane by Lemma 2.6. By composing some suitable isometry, we may suppose that f : 𝔻 → 𝔻 is a triangle preserving bijection. By the result of the case n = 2, f : 𝔻 → 𝔻 is an isometry and f : 𝔻 → 𝔻 is geodesic preserving. So f : 𝔻^{n} → 𝔻^{n} is geodesic preserving. By Theorem B or D, f is an isometry. This completes the proof. □

3. Euclidean space

By the same methods as in the case of hyperbolic space, we can prove Theorems 2 and 4 similarly. We omit the details.

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Department of Mathematics, Beijing Institute of Technology, 100081, China henan lbk@bit.edu.cn

Institute of Mathematics, AMSS, Chinese Academy of Sciences, 100190, China wangyf@math.ac.cn

This paper is available via http://nyjm.albany.edu/j/2009/15-23.html.