Reflection-Like Maps in High-Dimensional Euclidean Space

Abstract

In this paper, we introduce reflection-like maps in n-dimensional Euclidean spaces, which are affinely conjugated to $\theta :(x_1,x_2,\dots ,x_n)\to \left({\displaystyle \frac{1}{x_1}},{\displaystyle \frac{x_2}{x_1}},\dots ,{\displaystyle \frac{x_n}{x_1}}\right).$ We shall prove that reflection-like maps are line-to-line, cross ratios preserving on lines and quadrics preserving. The goal of this article was to consider the rigidity of line-to-line maps on the local domain of ${\mathbb{R}}^n$ by using reflection-like maps. We mainly prove that a line-to-line map $\eta$ on any convex domain satisfying ${\eta }^{\circ 2}=id$ and fixing any points in a super-plane is a reflection or a reflection-like map. By considering the hyperbolic isometry in the Klein Model, we also prove that any line-to-line bijection $f:{\mathbb{D}}^n\mapsto {\mathbb{D}}^n$ is either an orthogonal transformation, or a composition of an orthogonal transformation and a reflection-like map, from which we can find that reflection-like maps are important elements and instruments to consider the rigidity of line-to-line maps.

1. Introduction

The research of rigidity of line-to-line maps has a long history (see Reference [1,2,3,4,5], etc.) from different perspectives. We say that a map $f:{\mathbb{R}}^n\mapsto {\mathbb{R}}^n$ is line-to-line, if $f\left(l\right)$ is contained in some line for any line l in ${\mathbb{R}}^n$. Similarly, we say that a circle in Möbious space ${\widehat{\mathbb{R}}}^n={\mathbb{R}}^n\cup \{\infty \}$ (or a geodesic in hyperbolic space ${\mathbb{H}}^n=\{(x_1,\cdots ,x_{n-1},x_n)\in {\mathbb{R}}^n|x_n>0\}$) is a line. For example, in Reference [4], J. Jeffers proves that a circle-to-circle bijection $f:{\widehat{\mathbb{R}}}^n\mapsto {\widehat{\mathbb{R}}}^n$ is a Möbious transformation, a geodesic-to-geodesic bijection $f:{\mathbb{H}}^n\mapsto {\mathbb{H}}^n$ is a hyperbolic isometry and a line-to-line bijection $f:{\mathbb{R}}^n\mapsto {\mathbb{R}}^n$ is an affine transformation. Various geometries are considered in mathematical researches of different transformations, such as complex curves, were studied using Laguerre planes and Grünwald planes in Reference [6].

It is well known that any Möbious transformation is a composition of finite inversions in $n-$dimensional spherical space ${\widehat{\mathbb{R}}}^n$ (see Reference [7] for details). We can say that inversions are basic elements of Möbious transformations. Let

$${\mathbb{H}}^{n+1}=\{(x_1,\dots ,x_n,x_{n+1})\in {\mathbb{R}}^{n+1}|x_{n+1}>0\}$$

be $(n+1)-$dimensional hyperbolic space with hyperbolic metric ${\rho }_H={\displaystyle \frac{1}{x_{n+1}}}$. A reflection on ${\mathbb{H}}^{n+1}$ is an isometry which fixes an $n-$hyperplane in ${\mathbb{H}}^{n+1}$ and any hyperbolic isometry is a composition of finite reflections in ${\mathbb{H}}^{n+1}$. We can say that reflections are basic elements of hyperbolic isometries. Similarly,

$${\mathbb{S}}_{+}^n=\{(x_1,\dots ,x_n,x_{n+1})\in {\mathbb{R}}^{n+1}|x_1^2+\cdots +x_n^2+x_{n+1}^2=1,x_{n+1}>0\}$$

(1)

can be seen as an $n-$dimensional hyperbolic subspace of ${\mathbb{H}}^{n+1}$. Let

$${\mathbb{D}}^n=\{(x_1,x_2,\dots ,x_n)\in {\mathbb{R}}^n|x_1^2+x_2^2+\cdots +x_n^2<1\}$$

(2)

be the Klein Model of hyperbolic space defined by the natural projection

$$\begin{array}{cc}\hfill \tau :& {\mathbb{S}}_{+}^n\mapsto {\mathbb{D}}^n\hfill \\ \hfill & (x_1,\dots ,x_n,x_{n+1})\to (x_1,\dots ,x_n).\hfill \end{array}$$

(3)

Obviously, a map $F:{\mathbb{S}}_{+}^n\mapsto {\mathbb{S}}_{+}^n$ is a hyperbolic isometry, if and only if the transformation $f=\tau \circ F\circ {\tau }^{-1}:{\mathbb{D}}^n\mapsto D^n$ is a hyperbolic isometry in Klein Model ${\mathbb{D}}^n$ in the following commutative diagram

$$\begin{array}{ccc}{\mathbb{S}}_{+}^n& \stackrel{ F }{\to }& {\mathbb{S}}_{+}^n\\ \tau \downarrow & & \downarrow \tau \\ {\mathbb{D}}^n& \stackrel{f}{\to }& {\mathbb{D}}^n\end{array}.$$

(4)

A geodesic in Klein Model ${\mathbb{D}}^n$ is a segment which is the projection of a geodesic in ${\mathbb{S}}_{+}^n$ under $\tau$, since any geodesic in ${\mathbb{S}}_{+}^n$ is an arc perpendicular to $\partial {\mathbb{S}}_{+}^n\subset \partial {\mathbb{H}}^{n+1}$.

For any subset $\mathsf{\Omega }\subset {\mathbb{R}}^n$, we call L a line in $\mathsf{\Omega }$, if there exists a line l in ${\mathbb{R}}^n$, such that $L=l\cap \mathsf{\Omega }$. We say that two lines $L_1,L_2$ in $\mathsf{\Omega }$ are parallel, if $l_1,l_2$ are parallel. We say that three lines $L_1,L_2,L_3$ in $\mathsf{\Omega }$ are concurrent, if $l_1,l_2,l_3$ have a common point in ${\mathbb{R}}^n$. We say that a map $f:\mathsf{\Omega }\mapsto {\mathbb{R}}^n$ is line-to-line, if the image points of any collinear points are collinear and $f:\mathsf{\Omega }\mapsto {\mathsf{\Omega }}^{\prime }$ is line-onto-line, if $f\left(L\right)$ is a line in ${\mathsf{\Omega }}^{\prime }\subset {\mathbb{R}}^n$ for any line L in $\mathsf{\Omega }$.

One can find that f is a line-to-line bijection in ${\mathbb{D}}^n$ because the isometry F is a geodesic-to-geodesic bijection in ${\mathbb{S}}_{+}^n$ in diagram (4). Especially, if the isometry $F:{\mathbb{S}}_{+}^n\mapsto {\mathbb{S}}_{+}^n$ is a reflection, then the line-to-line map $f:{\mathbb{D}}^n\mapsto {\mathbb{D}}^n$ satisfies $f^{\circ 2}=id$ and its fixed-points set is an $(n-1)-$dimensional superplane in ${\mathbb{D}}^n$. Obviously, f may not be an affine transformation. This is the origin of reflection-like maps considered in this paper. We shall show that reflection-like maps are basic elements and instruments to consider the rigidity of line-to-line maps.

In Reference [8], B. Li et al., introduce $g-$reflection maps in ${\mathbb{R}}^2$, which are affinely conjugated to the map

$$(x,y)\to \left(-\frac{x}{1+x},\frac{y}{1+x}\right)$$

(5)

for any point in $\{(x,y)\in {\mathbb{R}}^2|x\ne -1\}$ and give the following result.

Theorem 1

([8]). Suppose that $\mathcal{D}\subset {\mathbb{R}}^2$ is a convex domain and a map $f:\mathcal{D}\mapsto \mathcal{D}$ is line-to-line and satisfies $f^{\circ 2}=id$. If f is not the restriction to $\mathcal{D}$ of an affine transformation of ${\mathbb{R}}^2$, then f is a restriction of $g-$reflection map to $\mathcal{D}$.

In Reference [9], B. Li et al., use $g-$reflection maps on the rigidity of line-to-line maps in the upper plane $\mathbb{H}\subset {\mathbb{R}}^2$ and prove that

Theorem 2

([9]). Suppose that $f:\mathbb{H}\mapsto \mathbb{H}$ is a line-to-line surjection. Then, f is either an affine transformation, or a composition of an affine transformation and a g-reflection map.

In Reference [10], B. Li et al., prove that any $g-$refection map preserves the cross ratios

$$[z_1,z_2,z_3,z_4]={\displaystyle \frac{(z_1-z_3)(z_2-z_4)}{(z_1-z_4)(z_2-z_3)}}$$

of any four collinear points $z_1,z_2,z_3,z_4\in \mathbb{C}$ and the following result.

Theorem 3

([10], Theorem 3.6). Suppose that $\mathcal{D}\subset {\mathbb{R}}^2$ is a domain and a line-to-line map $f:\mathcal{D}\mapsto {\mathbb{R}}^2$ is injective and non-degenerate. Then, f is either an affine transformation, or a composition of a g-reflection map and an affine transformation.

Here, a line-to-line map $f:\mathcal{D}\mapsto {\mathbb{R}}^2$ is degenerate (see Reference [11]), if the image space $f\left(\mathcal{D}\right)$ is contained in some line (otherwise, it is non-degenerate).

The goal of this article is to consider the rigidity of line-to-line maps on local domains in ${\mathbb{R}}^n$. We shall introduce the case in $n-$dimensional space ${\mathbb{R}}^n$ of $g-$reflection maps, named reflection-like maps in this paper, and prove the following main results.

Theorem 4.

Suppose that Ω is any convex domain in ${\mathbb{R}}^n$ and ${\mathcal{A}}^{\eta }$ is a super-plane such that $\mathsf{\Omega }\cap {\mathcal{A}}^{\eta }\ne \varnothing$. A line-to-line map $\eta :\mathsf{\Omega }\mapsto \mathsf{\Omega }$ satisfies ${\eta }^{\circ 2}=id$ and $\eta \left(P\right)=P$ for any $P\in \mathsf{\Omega }\cap {\mathcal{A}}^{\eta }$. Then, η is a reflection or a reflection-like map.

Theorem 5.

Suppose that ${\mathbb{D}}^n\subset {\mathbb{R}}^n$ is a Klein Model of $n-$dimensional hyperbolic space and a map $f:{\mathbb{D}}^n\mapsto {\mathbb{D}}^n$ is a hyperbolic isometry. Then, f is either an orthogonal transformation, or a composition of an orthogonal transformation and a reflection-like map.

In the next sections, we shall prove that reflection-like maps are line-to-line and linearly conjugated to each other. Moreover, the image of three parallel lines under reflection-like maps are parallel or concurrent. The absolute cross ratios may not be preserved by reflection-like maps. But, we shall prove that refection-like maps preserve the absolute cross ratios of any four distinct collinear points, something like projective maps preserve the cross ratios of four points in a projective line in projective geometries. We shall also prove that refection-like maps transfer spheres to quadrics, from which we can obtain that they map quadrics to quadrics. Especially, if the image of a sphere is a sphere, then it is invariant.

2. Reflection-Like Maps in High Dimension Space ${\mathbb{R}}^n$

In this section, we shall give the definition of reflection-like maps firstly and prove invariant properties under affine conjugation. We mainly prove Theorem 4, the rigidity of reflection-like maps in local domain of $n-$dimensional space.

Denote points in ${\mathbb{R}}^n$ by $X(x_1,x_2,\cdots ,x_n)$, $Y(y_1,y_2,\cdots ,y_n)$ and the line passing through $X,Y$ by $L_{XY}$, the Euclidean distance between $X,Y$ by $|X-Y|$. Denote the vector from X to Y by $\overrightarrow{XY}$.

Let $\mathcal{A},\mathcal{B}$ be two $(n-1)-$dimensional planes (superplanes) in ${\mathbb{R}}^n$ and $\mathcal{P}$ be a point

$$\begin{array}{cc}\hfill \mathcal{A}& =\{(x_1,x_2,\dots ,x_n)\in {\mathbb{R}}^n|x_1=1\},\hfill \\ \hfill \mathcal{B}& =\{(x_1,x_2,\dots ,x_n)\in {\mathbb{R}}^n|x_1=0\},\hfill \\ \hfill \mathcal{P}& =(-1,0,\dots ,0).\hfill \end{array}$$

(6)

Obviously, $\mathcal{P}$ and $\mathcal{A}$ have equal Euclidean distances to $\mathcal{B}$. The map

$$\begin{array}{cc}\hfill \theta :& {\mathbb{R}}^n\backslash \mathcal{B}\mapsto {\mathbb{R}}^n\backslash \mathcal{B}\hfill \\ \hfill & (x_1,x_2,\dots ,x_n)\to \left(\frac{1}{x_1},\frac{x_2}{x_1},\dots ,\frac{x_n}{x_1}\right)\hfill \end{array}$$

(7)

satisfies ${\theta }^{\circ 2}=id$. Moreover, $\left\{\mathcal{P}\right\}\cup \mathcal{A}$ is the fixed-point set of $\theta$ and the two components of ${\mathbb{R}}^n$ divided by $\mathcal{B}$ are invariant under $\theta$.

Definition 1.

We say that a map η is a reflection-like map in ${\mathbb{R}}^n$, if it is affinely conjugated to θ. That is, one can find an affine transformation $g:{\mathbb{R}}^n\mapsto {\mathbb{R}}^n$, such that $\eta =g\circ \theta \circ g^{-1}:{\mathbb{R}}^n\backslash g\left(\mathcal{B}\right)\mapsto {\mathbb{R}}^n\backslash g\left(\mathcal{B}\right)$.

Obviously, θ defined in (7) is a refection-like map in ${\mathbb{R}}^n$.

Moreover, we say that $\mathcal{A}$ defined in (6) is Axis, $\mathcal{B}$ is Boundary, and $\mathcal{P}$ is Base point of the refection-like map θ.

Generally, given any affine transformation g, the reflection-like map $\eta =g\circ \theta \circ g^{-1}:{\mathbb{R}}^n\backslash g\left(\mathcal{B}\right)\mapsto {\mathbb{R}}^n\backslash g\left(\mathcal{B}\right)$ has Boundary ${\mathcal{B}}^{\eta }=g\left(\mathcal{B}\right)$, Axis ${\mathcal{A}}^{\eta }=g\left(\mathcal{A}\right)$, and Base point ${\mathcal{P}}^{\eta }=g\left(\mathcal{P}\right)$. Obviously, ${\eta }^{\circ 2}=id$, $\left\{{\mathcal{P}}^{\eta }\right\}\cup {\mathcal{A}}^{\eta }$ is the fixed-point set of $\eta$, ${\mathcal{B}}^{\eta }$ is parallel to ${\mathcal{A}}^{\eta }$, and ${\mathcal{P}}^{\eta }$ and ${\mathcal{A}}^{\eta }$ have equal Euclidean distances to ${\mathcal{B}}^{\eta }$. Moreover, the two components of ${\mathbb{R}}^n$ divided by ${\mathcal{B}}^{\eta }$ are invariant under $\eta$.

Definition 2.

We call L a line in ${\mathbb{R}}^n\backslash \mathcal{B}$, if there exists a line l in ${\mathbb{R}}^n$, such that $L=l\cap {\mathbb{R}}^n\backslash \mathcal{B}$. If $l\cap \mathcal{B}=\left\{\tilde{P}\right\}$, then we say that L has boundary point $\tilde{P}$.

Proposition 1.

The reflection-like map $\theta :{\mathbb{R}}^n\backslash \mathcal{B}\mapsto {\mathbb{R}}^n\backslash \mathcal{B}$ is a line-onto-line bijection.

Proof. 

Let us prove that f is line-to-line in ${\mathbb{R}}^n\backslash \mathcal{B}$ firstly. That is, for any three collinear points $X(x_1,x_2,\cdots ,x_n)$, $Y(y_1,y_2,\cdots ,y_n)$, $Z(z_1,z_2,\cdots ,z_n)$, their image points $X^{\prime }(x_1^{\prime },x_2^{\prime },\cdots ,x_n^{\prime })$, $Y^{\prime }(y_1^{\prime },y_2^{\prime },\cdots ,y_n^{\prime })$, $Z^{\prime }(z_1^{\prime },z_2^{\prime },\cdots ,z_n^{\prime })$ are collinear. There exists some $\lambda \in \mathbb{R}\backslash \{0,1\}$, such that $Z=\lambda X+(1-\lambda )Y$. That is, $z_i=\lambda x_i+(1-\lambda )y_i$, for any $i=1,2,\cdots ,n$. We have $x_1^{\prime }={\displaystyle \frac{1}{x_1}}$, $y_1^{\prime }={\displaystyle \frac{1}{y_1}}$, and

$$\begin{array}{cc}\hfill z_1^{\prime }={\displaystyle \frac{1}{z_1}}=& {\displaystyle \frac{1}{\lambda x_1+(1-\lambda )y_1}}\hfill \\ \hfill =& {\displaystyle \frac{x_1^{\prime }{y}_1^{\prime }}{\lambda y_1^{\prime }+(1-\lambda )x_1^{\prime }}}\hfill \\ \hfill =& {\displaystyle \frac{\lambda y_1^{\prime }}{\lambda y_1^{\prime }+(1-\lambda )x_1^{\prime }}}{x}_1^{\prime }+{\displaystyle \frac{(1-\lambda )x_1^{\prime }}{\lambda y_1^{\prime }+(1-\lambda )x_1^{\prime }}}{y}_1^{\prime }.\hfill \end{array}$$

Let ${\lambda }^{\prime }={\displaystyle \frac{\lambda y_1^{\prime }}{\lambda y_1^{\prime }+(1-\lambda )x_1^{\prime }}}$, and then $z_1^{\prime }={\lambda }^{\prime }{x}_1^{\prime }+(1-{\lambda }^{\prime })y_1^{\prime }$. Meanwhile, $x_i^{\prime }={\displaystyle \frac{x_i}{x_1}}$, $y_i^{\prime }={\displaystyle \frac{y_i}{y_1}}$ for any $i=2,3,\dots ,n$ and

$$\begin{array}{cc}\hfill z_i^{\prime }={\displaystyle \frac{z_i}{z_1}}=& {\displaystyle \frac{\lambda x_i+(1-\lambda )y_i}{\lambda x_1+(1-\lambda )y_1}}\hfill \\ \hfill =& {\displaystyle \frac{\lambda y_1^{\prime }}{\lambda y_1^{\prime }+(1-\lambda )x_1^{\prime }}}{x}_i^{\prime }+{\displaystyle \frac{(1-\lambda )x_1^{\prime }}{\lambda y_1^{\prime }+(1-\lambda )x_1^{\prime }}}{y}_i^{\prime }\hfill \\ \hfill =& {\lambda }^{\prime }{x}_i^{\prime }+(1-{\lambda }^{\prime })y_i^{\prime }.\hfill \end{array}$$

Thus, $Z^{\prime }={\lambda }^{\prime }{X}^{\prime }+(1-{\lambda }^{\prime })Y^{\prime }$, which follows that $X^{\prime },Y^{\prime },Z^{\prime }$ are collinear. Hence, $\theta$ is line-to-line. Moreover, one can find that $\theta$ is bijective and $\theta \left(L\right)$ is a line in ${\mathbb{R}}^n\backslash \mathcal{B}$ for any line L in ${\mathbb{R}}^n\backslash \mathcal{B}$, since ${\theta }^{\circ 2}=id$. That is, $\theta$ is a line-onto-line bijection and the proof is completed. □

Proposition 2.

For any line L in ${\mathbb{R}}^n\backslash \mathcal{B}$, the reflection-like map $\theta :{\mathbb{R}}^n\backslash \mathcal{B}\mapsto {\mathbb{R}}^n\backslash \mathcal{B}$ satisfies the following.

(i)

If $L\not\subset \mathcal{A}$, then $\theta \left(L\right)=L$, if and only if Base point $\mathcal{P}\in L$;

(ii)

If $\theta \left(L\right)\ne L$, then $\theta \left(L\right)$ is parallel to L, if and only if L is parallel to Axis $\mathcal{A}$.

Proof. 

$\left(\mathrm{i}\right).$ We only need to prove that $\mathcal{P}\in L_{X{X}^{\prime }}$ for any point $X\in {\mathbb{R}}^n\backslash \mathcal{B}$ and $X^{\prime }=\theta \left(X\right)\ne X$, since $\theta$ is line-to-line and satisfies ${\theta }^{\circ 2}=id$.

Let $X(x_1,x_2,\dots ,x_n)$, $X^{\prime }\left({\displaystyle \frac{1}{x_1}},{\displaystyle \frac{x_2}{x_1}},\dots ,{\displaystyle \frac{x_n}{x_1}}\right)$ and $\lambda ={\displaystyle \frac{1}{1-x_1}}$, then one can find that $\mathcal{P}=\lambda X+(1-\lambda )X^{\prime },$ which means $\mathcal{P}\in L_{X{X}^{\prime }}$.

$\left(\mathrm{ii}\right).$ For two distinct points $X(x_1,x_2,\cdots ,x_n)$, $Y(y_1,y_2,\cdots ,y_n)$ in L, denote the image points under $\theta$ by $X^{\prime }\left({\displaystyle \frac{1}{x_1}},{\displaystyle \frac{x_2}{x_1}},\cdots ,{\displaystyle \frac{x_n}{x_1}}\right)$, $Y^{\prime }\left({\displaystyle \frac{1}{y_1}},{\displaystyle \frac{y_2}{y_1}},\cdots ,{\displaystyle \frac{y_n}{y_1}}\right)$. Obviously, L is parallel to Axis $\mathcal{A}$, if and only if $x_1=y_1(\ne 0)$. Then, $\theta \left(L\right)$ is parallel to L by

$$\overrightarrow{X^{\prime }{Y}^{\prime }}=\left(0,{\displaystyle \frac{y_2-x_2}{y_1}},\cdots ,{\displaystyle \frac{y_n-x_n}{y_1}}\right)={\displaystyle \frac{1}{y_1}}\overrightarrow{XY}.$$

On the other side, suppose that $\theta \left(L\right)$ is parallel to L and $x_1\ne y_1$. We can obtain that $y_i={\displaystyle \frac{1+y_1}{1+x_1}}{x}_i$, for any $i=2,\cdots ,n$ by $\overrightarrow{X^{\prime }{Y}^{\prime }}//\overrightarrow{XY}$. Let $\lambda ={\displaystyle \frac{1+y_1}{y_1-x_1}}$, then $\mathcal{P}=\lambda X+(1-\lambda )Y$, which means that $\mathcal{P}\in L$, and $\theta \left(L\right)=L$ by the result of $\left(\mathrm{i}\right)$. This is a contradiction, and the proof is completed. □

Corollary 1.

The image of a parallelogram under a reflection-like map is a parallelogram, if and only if the parallelogram is parallel to Axis of the reflection-like map. Moreover, the image of a square is a square, if the square is parallel to Axis.

Proposition 3.

For any two lines $L_1,L_2$ in ${\mathbb{R}}^n\backslash \mathcal{B}$, not parallel to $\mathcal{A}$, the reflection-like map $\theta :{\mathbb{R}}^n\backslash \mathcal{B}\mapsto {\mathbb{R}}^n\backslash \mathcal{B}$ satisfies the followings.

(i)

$\theta \left(L_1\right)$ and $\theta \left(L_2\right)$ share a common boundary point if $L_1$ is parallel to $L_2$;

(ii)

$\theta \left(L_1\right)$ is parallel to $\theta \left(L_2\right)$ if $L_1$ and $L_2$ share a common boundary point.

Proof. 

We only need to prove that $L_1$ is a line passing through $\mathcal{P}$. From Proposition 2, we have $\theta \left(L_1\right)=L_1$. Denote the boundary point of $L_1$ by $\tilde{X}(0,x_2,\cdots ,x_n)$, then the vector $\overrightarrow{\mathcal{P}\tilde{X}}=(1,x_2,\cdots ,x_n)\subset L_1$.

(i)

Suppose that $L_2$ is any line parallel to $L_1$. For any point $Y(y_1,y_2,\cdots ,y_n)$ in $L_2$, one can obtain $L_2=\{(y_1+t,y_2+t{x}_2,\cdots ,y_n+t{x}_n)|t\in \mathbb{R}\backslash \{-y_1\}\}$ and

$$\theta \left(L_2\right)=\left\{\left({\displaystyle \frac{1}{y_1+t}},{\displaystyle \frac{y_2+t{x}_2}{y_1+t}},\cdots ,{\displaystyle \frac{y_n+t{x}_n}{y_1+t}}\right)|t\in \mathbb{R}\backslash \{-y_1\}\right\}.$$

It follows that $\tilde{X}$ is the limit point of $\theta \left(L_2\right)$ as t tends to ∞. That is, $\theta \left(L_2\right)$ and $\theta \left(L_1\right)$ share common boundary point if $L_2$ is parallel to $L_1$.

(ii)

Suppose that $L_2$ is any line sharing common boundary point $\tilde{X}(0,x_2,\cdots ,x_n)$ with $L_1$. For any point $Y(y_1,y_2,\cdots ,y_n)\in L_2$, we can find that the vector

$$\overrightarrow{\tilde{X}Y}=(y_1,y_2-x_2,\cdots ,y_n-x_n)\subset L_2.$$

So, we have

$$L_2=\{((1+t)y_1,y_2+t(y_2-x_2),\cdots ,y_n+t(y_n-x_n))|t\in \mathbb{R}\backslash \{-1\}\},$$

and

$$\theta \left(L_2\right)=\left\{\left({\displaystyle \frac{1}{(1+t)y_1}},{\displaystyle \frac{y_2+t(y_2-x_2)}{(1+t)y_1}},\cdots ,{\displaystyle \frac{y_n+t(y_n-x_n)}{(1+t)y_1}}\right)|t\in \mathbb{R}\backslash \{-1\}\right\}.$$

As t tends to ∞, we obtain its boundary point $\tilde{Y}\left(0,{\displaystyle \frac{y_2-x_2}{y_1}},\cdots ,{\displaystyle \frac{y_n-x_n}{y_1}}\right)$.

Denote $\theta \left(Y\right)=Y^{\prime }\left({\displaystyle \frac{1}{y_1}},{\displaystyle \frac{y_2}{y_1}},\cdots ,{\displaystyle \frac{y_n}{y_1}}\right)\in \theta \left(L_2\right)$, then the vector

$$\overrightarrow{\tilde{Y}{Y}^{\prime }}=\left({\displaystyle \frac{1}{y_1}},{\displaystyle \frac{x_2}{y_1}},\cdots ,{\displaystyle \frac{x_n}{y_1}}\right)\subset \theta \left(L_2\right),$$

which follows that $\theta \left(L_2\right)$ is parallel to $\theta \left(L_1\right)$ for $\overrightarrow{\mathcal{P}\tilde{X}}=y_1\overrightarrow{\tilde{Y}{Y}^{\prime }}$. □

Moreover, we can have the following.

Lemma 1.

Suppose that a reflection-like map η has the same Base point and Axis as θ. Then, $\eta =\theta$.

Proof. 

We need only prove that the reflection-like map is uniquely determined by Base point $\mathcal{P}$ and Axis $\mathcal{A}$. One can know that Boundary $\mathcal{B}$ is parallel to $\mathcal{A}$ and lies between $\mathcal{P}$ and $\mathcal{A}$ with equal distances. For any point $X\in {\mathbb{R}}^n\backslash \mathcal{B}$, let $L_1$ be the line in ${\mathbb{R}}^n\backslash \mathcal{B}$ passing through X and $\mathcal{P}$, then $\eta \left(L_1\right)=L_1$ by Proposition 2. Choose any point $Y\in \mathcal{A}\backslash L_1$ and let $L_2$ be the line in ${\mathbb{R}}^n\backslash \mathcal{B}$ passing through X and Y, then it is easy to find $Y\in \eta \left(L_2\right)$. Let $L_3$ be the line passing through $\mathcal{P}$ and parallel to $L_2$, then $\eta \left(L_3\right)=L_3$. So $\eta \left(L_2\right),\eta \left(L_3\right)$ share common boundary point, denoted by $\tilde{Y}$ by Proposition 3. It follows that $\eta \left(L_2\right)$ is the line passing through Y and having boundary point $\tilde{Y}$. Then, $\eta \left(X\right)=L_1\cap \eta \left(L_2\right)$ is determined uniquely. That is, the reflection-like map $\eta$ is determined by $\mathcal{A}$ and $\mathcal{P}$. Therefore, we have $\eta =\theta$. □

A transformation $g:{\mathbb{R}}^n\mapsto {\mathbb{R}}^n$ is linear, if it is a composition of translations, scaling and orthogonal transformations on ${\mathbb{R}}^n$. We say that a reflection-like map $\eta$ is linearly conjugated to $\theta$, if one can find a linear map g, such that $\eta =g\circ \theta \circ g^{-1}$.

For any super-plane $\mathsf{\Pi }$ and a point $P\notin \mathsf{\Pi }$, one can find a linear transformation g such that $g\left(\mathcal{A}\right)=\mathsf{\Pi }$ and $g\left(\mathcal{P}\right)=P$. Then, $\eta =g\circ \theta \circ g^{-1}$ is a reflection-like map with Base point P and Axis $\mathsf{\Pi }$. So, we can obtain the following by Lemma 1.

Theorem 6.

Any reflection-like map is linearly conjugated to θ.

By conjugating affine transformation $g:(x_1,x_2,\cdots ,x_n)\to (x_1-1,x_2,\cdots ,x_n)$,

$${\theta }^{\prime }=g\circ \theta \circ g^{-1}:(x_1,x_2,\dots ,x_n)\to \left(-\frac{x_1}{1+x_1},\frac{x_2}{1+x_1},\cdots ,\frac{x_n}{1+x_1}\right)$$

is the general form of the $g-$reflection map defined in (5) on $n-$dimensional space.

Proof of Theorem 4.

Let $\mathcal{U}$ and ${\mathcal{U}}^{\prime }$ be the two components of $\mathsf{\Omega }$ divided by ${\mathcal{A}}^{\eta }$. We claim that there exist $P\in \mathcal{U}$ and $P^{\prime }\in {\mathcal{U}}^{\prime }$ such that $\eta \left(P\right)=P^{\prime }$. Otherwise, suppose that we have $X,X^{\prime }\in \mathcal{U}$, such that $\eta \left(X\right)=X^{\prime }$ (as in the Figure 1a). For any $P^{\prime }\in {\mathcal{U}}^{\prime }\backslash L_{X{X}^{\prime }}$, denote $Y_1\in L_{X{P}^{\prime }}\cap {\mathcal{A}}^{\eta }$, $Y_2\in L_{X^{\prime }{P}^{\prime }}\cap {\mathcal{A}}^{\eta }$ and $P\in L_{X{Y}_2}\cap L_{X^{\prime }{Y}_1}$, then $P\in \mathcal{U}$ and $P^{\prime }=\eta \left(P\right)$.

We shall prove that the line-to-line map is uniquely determined by $P,P^{\prime }$ and ${\mathcal{A}}^{\eta }\cap \mathsf{\Omega }$. Let $\mathcal{V}$ denote the smallest convex domain containing $P,P^{\prime }$ and $\mathsf{\Omega }\cap {\mathcal{A}}^{\eta }$. For any point $X\in \mathcal{V}\backslash L_{P{P}^{\prime }}$ (as in the Figure 1b), let $Y_1\in L_{XP}\cap {\mathcal{A}}^{\eta }$, $Y_2\in L_{X{P}^{\prime }}\cap {\mathcal{A}}^{\eta }$, we can find that $X^{\prime }=\eta \left(X\right)\in L_{P{Y}_2}\cap L_{P^{\prime }{Y}_1}$ is unique. Moreover, the line-to-line map on $\mathsf{\Omega }$ will be uniquely determined by the mapping on its sub-domain $\mathcal{V}\backslash L_{P{P}^{\prime }}$.

Next, we shall prove the existence of $\eta$. By conjugating some suitable affine transformation, we can suppose that ${\mathcal{A}}^{\eta }=\left\{(x_1,x_2,\cdots ,x_n)\right|x_1=0\}$, $P=(-1,0,\cdots ,0)$ and $P^{\prime }=(k,0,\cdots ,0)$ $(k>0)$. If $k=1$, then $\eta$ is a reflection about ${\mathcal{A}}^{\eta }$

$$\eta :(x_1,x_2,\cdots ,x_n)\to (-x_1,x_2,\cdots ,x_n).$$

Otherwise, let ${\mathcal{P}}^{\eta }=(-{\displaystyle \frac{2k}{k-1}},0,\cdots ,0)$ and $K={\displaystyle \frac{k-1}{k}}$, then

$$\eta :(x_1,x_2,\cdots ,x_n)\to \left(-{\displaystyle \frac{x_1}{1+K{x}_1}},{\displaystyle \frac{x_2}{1+K{x}_1}},\cdots ,{\displaystyle \frac{x_n}{1+K{x}_1}}\right)$$

is the reflection-like map with Axis ${\mathcal{A}}^{\eta }$ and Base point ${\mathcal{P}}^{\eta }$ such that $\eta \left(P\right)=P^{\prime }$. □

Corollary 2.

Suppose that θ is the reflection-like map defined in (7). Given any positive integer $1<r<n$, let Π be any $r-$dimensional plane in ${\mathbb{R}}^n\backslash \mathcal{B}$ passing through $\mathcal{P}$, then $\theta \left(\mathsf{\Pi }\right)=\mathsf{\Pi }$. Moreover, if $\mathsf{\Pi }\cap \mathcal{A}\ne \varnothing$, then ${\theta |}_{\mathsf{\Pi }}:\mathsf{\Pi }\mapsto \mathsf{\Pi }$ is a reflection-like map with Axis $\mathsf{\Pi }\cap \mathcal{A}$ and Base point $\mathcal{P}$.

Remark 1.

We give an example ($n=3$) to show that Theorem 1Adoes not hold in the case of reflection-like maps in ${\mathbb{R}}^n$ ($n>2$). That is, a line-to-line map $f:\mathsf{\Omega }\mapsto \mathsf{\Omega }$ on a convex domain $\mathsf{\Omega }\subset {\mathbb{R}}^n$ satisfying $f^{\circ 2}=id$ may not be an affine transformation or a reflection-like map.

Example 1.

Let $\mathcal{B}=\{(x_1,x_2,x_3)\in {\mathbb{R}}^3|x_1=0\}$ and $f:{\mathbb{R}}^3\backslash \mathcal{B}\mapsto {\mathbb{R}}^3\backslash \mathcal{B}$ be defined as

$$f:(x_1,x_2,x_3)\to \left({\displaystyle \frac{1}{x_1}},{\displaystyle \frac{x_2}{x_1}},-{\displaystyle \frac{x_3}{x_1}}\right).$$

Obviously, $f^{\circ 2}=id$ and f is line-to-line, since f is a composition of an orthogonal transformation and a reflection-like map, while f cannot be a reflection-like map since its fixed-point set is $L_1\cup L_2$, where $L_1=\{(x_1,x_2,x_3)\in {\mathbb{R}}^3|x_1=1,x_3=0\}$ and $L_2=\{(x_1,x_2,x_3)\in {\mathbb{R}}^3|x_1=-1,x_2=0\}$.

3. The Absolute Cross ratios in High Dimension Space ${\mathbb{R}}^n$

For any four distinct points $X(x_1,x_2,\cdots ,x_n)$, $Y(y_1,y_2,\cdots ,y_n)$, $Z(z_1,z_2,\cdots ,z_n)$, $W(w_1,w_2,\cdots ,w_n)$ in ${\mathbb{R}}^n$, the absolute cross ratio is defined as

$$|X,Y,Z,W|=\frac{|X-Z|·|Y-W|}{|X-W|·|Y-Z|}.$$

It is very important in high dimensional space. Especially, if $Z=\infty$, we can define it by the limit as Z tends to ∞

$$|X,Y,\infty ,W|=\frac{|Y-W|}{|X-W|}.$$

It is well known that, for any subdomain $\mathsf{\Omega }\subset {\mathbb{R}}^n$, a map $f:\mathsf{\Omega }\mapsto {\mathbb{R}}^n$ is a Möbious transformation, if and only if f preserves the absolute cross ratios. The cross ratio is defined on four collinear points in projective geometry, and a projective transformation preserves cross ratios (see Reference [2,12] for details). While a reflection-like map considers one more dimension than a projectivity, it does not preserve absolute cross ratios.

For example (as in Figure 2), let $X(1,0)$, $Y(1,1)$, $Z(2,1)$ and $W(2,0)\in {\mathbb{R}}^2$, then $\theta \left(X\right)=X^{\prime }(1,0)$, $\theta \left(Y\right)=Y^{\prime }(1,1)$, $\theta \left(Z\right)=Z^{\prime }\left({\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}}\right)$ and $\theta \left(W\right)=W^{\prime }\left({\displaystyle \frac{1}{2}},0\right)$. We have that $\theta$ maps the square $XYZW$ to the quadrilateral $X^{\prime }{Y}^{\prime }{Z}^{\prime }{W}^{\prime }$ since $\theta$ is line-to-line. It is easy to calculate that $|X,Y,Z,W|=2$ and $|X^{\prime },Y^{\prime },Z^{\prime },W^{\prime }|=\sqrt{5}$.

In this section, we shall prove that reflection-like maps preserve the absolute cross ratios of any four collinear points. In fact, for any collinear points $X,Y,Z,W$, if $x_i\ne y_i$ for some $i=1,\cdots ,n$, then we can have

$$|X,Y,Z,W|=\frac{|x_i-z_i|·|y_i-w_i|}{|x_i-w_i|·|y_i-z_i|}.$$

Theorem 7.

Suppose that η is a reflection-like map with boundary ${\mathcal{B}}^{\eta }$. Then, for any four distinct collinear points $X,Y,Z,W$ in ${\mathbb{R}}^n\backslash {\mathcal{B}}^{\eta }$, the absolute cross ratio $|X,Y,Z,W|$ is invariant under η. That is,

$$|X,Y,Z,W|=\left|\eta \right(X),\eta (Y),\eta (Z),\eta (W\left)\right|.$$

Proof. 

By conjugating some suitable linear transformation, we can suppose that the reflection-like map is $\theta$ defined in (7). Then, we have $\theta \left(X\right)=X^{\prime }\left({\displaystyle \frac{1}{x_1}},{\displaystyle \frac{x_2}{x_1}},\cdots ,{\displaystyle \frac{x_n}{x_1}}\right)$, $\theta \left(Y\right)=Y^{\prime }\left({\displaystyle \frac{1}{y_1}},{\displaystyle \frac{y_2}{y_1}},\cdots ,{\displaystyle \frac{y_n}{y_1}}\right)$, $\theta \left(Z\right)=Z^{\prime }\left({\displaystyle \frac{1}{z_1}},{\displaystyle \frac{z_2}{z_1}},\cdots ,{\displaystyle \frac{z_n}{z_1}}\right)$ and $\theta \left(W\right)=W^{\prime }\left({\displaystyle \frac{1}{w_1}},{\displaystyle \frac{w_2}{w_1}},\cdots ,{\displaystyle \frac{w_n}{w_1}}\right)$ are collinear. If $x_1\ne y_1$, we have ${\displaystyle \frac{1}{x_1}}\ne {\displaystyle \frac{1}{y_1}}$ and

$$\begin{array}{cc}\hfill |X^{\prime },Y^{\prime },Z^{\prime },W^{\prime }|=& \frac{|{\displaystyle \frac{1}{x_1}}-{\displaystyle \frac{1}{z_1}}|·|{\displaystyle \frac{1}{y_1}}-{\displaystyle \frac{1}{w_1}}|}{|{\displaystyle \frac{1}{x_1}}-{\displaystyle \frac{1}{w_1}}|·|{\displaystyle \frac{1}{y_1}}-{\displaystyle \frac{1}{z_1}}|}\hfill \\ \hfill =& \frac{|x_1-z_1|·|y_1-w_1|}{|x_1-w_1|·|y_1-z_1|}\hfill \\ \hfill =& |X,Y,Z,W|.\hfill \end{array}$$

If $x_1=y_1$, then there exists some i, such that $x_i\ne y_i$. Thus, ${\displaystyle \frac{x_i}{x_1}}\ne {\displaystyle \frac{y_i}{y_1}}$ and

$$\begin{array}{cc}\hfill |X^{\prime },Y^{\prime },Z^{\prime },W^{\prime }|=& \frac{|{\displaystyle \frac{x_i}{x_1}}-{\displaystyle \frac{z_i}{z_1}}|·|{\displaystyle \frac{y_i}{y_1}}-{\displaystyle \frac{w_i}{w_1}}|}{|{\displaystyle \frac{x_i}{x_1}}-{\displaystyle \frac{w_i}{w_1}}|·|{\displaystyle \frac{y_i}{y_1}}-{\displaystyle \frac{z_i}{z_1}}|}\hfill \\ \hfill =& \frac{|x_i-z_i|·|y_i-w_i|}{|x_i-w_i|·|y_i-z_i|}\hfill \\ \hfill =& |X,Y,Z,W|.\hfill \end{array}$$

We complete the proof. □

4. Reflection-Like Maps and Quadrics

In this section, we shall prove that $\theta$ maps spheres to quadrics, from which we can obtain that reflection-like maps transfer quadrics to quadrics. Especially, if the image of a sphere is a sphere, then it is invariant.

Definition 3.

Given any reflection-like map, we say that the line passing its Base point and perpendicular to its Axis is its Equator.

For example, the Equator of $\theta$ is

$$\mathcal{L}=\left\{(x_1,0,\dots ,0)\right|x_1\in \mathbb{R},x_1\ne 0\}.$$

One can find that, given any affine transformation, the Equator of $\eta =g·\theta ·g^{-1}$ may not be $g\left(\mathcal{L}\right)$, while, if g is linear, the Equator of $\eta$ is $g\left(\mathcal{L}\right)$.

Theorem 8.

The reflection-like map θ maps any sphere to a quadric.

If both $\mathbb{S}$ and $\theta \left(\mathbb{S}\right)$ are $(n-1)-$dimensional spheres, then $\theta \left(\mathbb{S}\right)=\mathbb{S}$.

Moreover, if $\theta \left(\mathbb{S}\right)=\mathbb{S}$, then the center of $\mathbb{S}$ lies in the equator $\mathcal{L}$ of θ.

For any $P\in \mathcal{L}$, such that $P^{\prime }=\theta \left(P\right)\ne P$, let $\mathbb{S}$ be the $(n-1)-$dimensional sphere with diameter $P{P}^{\prime }$, then $\theta \left(\mathbb{S}\right)=\mathbb{S}$.

Proof. 

Suppose that $\mathbb{S}$ is a sphere with radius r and center $C(c_1,c_2,\dots ,c_n)$. Then, any point $X(x_1,x_2,\dots ,x_n)\in \mathbb{S}$ satisfies

$$\mathbb{S}:{(x_1-c_1)}^2+{(x_2-c_2)}^2+\cdots +{(x_n-c_n)}^2=r^2.$$

Denote the image point $\theta \left(X\right)=X^{\prime }(x_1^{\prime },x_2^{\prime },\cdots ,x_n^{\prime })\in \theta \left(\mathbb{S}\right)$, then $\theta \left(X^{\prime }\right)=X$ since ${\theta }^{\circ 2}=id$. It follows $\theta \left(X^{\prime }\right)=\left({\displaystyle \frac{1}{x_1^{\prime }}},{\displaystyle \frac{x_2^{\prime }}{x_1^{\prime }}},\cdots ,{\displaystyle \frac{x_n^{\prime }}{x_1^{\prime }}}\right)\in \mathbb{S}$, that is

$$\theta \left(\mathbb{S}\right)=\left\{(x_1^{\prime },x_2^{\prime },\cdots ,x_n^{\prime })|{\left(\frac{1}{x_1^{\prime }}-c_1\right)}^2+{\left(\frac{x_2^{\prime }}{x_1^{\prime }}-c_2\right)}^2+\cdots +{\left(\frac{x_n^{\prime }}{x_1^{\prime }}-c_n\right)}^2=r^2\right\}.$$

Obviously, it is a quadric

$$\theta \left(\mathbb{S}\right):{(1-c_1{x}_1^{\prime })}^2+{(x_2^{\prime }-c_2{x}_1^{\prime })}^2+\cdots +{(x_n^{\prime }-c_n{x}_1^{\prime })}^2=r^2{x}_1^2.$$

Then, $\theta \left(\mathbb{S}\right)$ is a sphere, if and only if $c_2=c_3=\cdots =c_n=0$ and $c_1^2-r^2=1$, since $-2c_i$ is the coefficient of the term $x_1^{\prime }{x}_i^{\prime }$$(i=2,\cdots ,n)$ and $c_1^2-r^2$ is the coefficient of the term $x_1^{\prime 2}$. It follows that, if $\theta \left(\mathbb{S}\right)$ is also a sphere, then

$$\theta \left(\mathbb{S}\right):{(x_1^{\prime }-c_1)}^2+x_2^{\prime 2}+\cdots +x_n^{\prime 2}=r^2.$$

Thus, $\theta \left(\mathbb{S}\right)=\mathbb{S}$ and the center $C(c_1,0,\cdots ,0)\in \mathcal{L}$ (as in Figure 3).

For any $P\in \mathcal{L}$ satisfying $P^{\prime }=\theta \left(P\right)\ne P$, let $\mathbb{S}$ be the $(n-1)-$dimensional sphere with diameter $P{P}^{\prime }$ (as in Figure 3). Denote $P(p_1,\cdots ,0)$, $P^{\prime }\left({\displaystyle \frac{1}{p_1}},\cdots ,0\right)$, $c_1={\displaystyle \frac{1}{2}}\left(p_1+{\displaystyle \frac{1}{p_1}}\right)$ and $r={\displaystyle \frac{1}{2}}|p_1-{\displaystyle \frac{1}{p_1}}|$, then $\mathbb{S}$ has radius r and center $C(c_1,0,\cdots ,0)\in \mathcal{L}$. One can find that $\theta \left(\mathbb{S}\right)=\mathbb{S}$ since $c_1^2-r^2=1$. □

Obviously, the invariant sphere $\mathbb{S}$ lies in one component of ${\mathbb{R}}^n\backslash \mathcal{B}$ and the interior $\mathsf{\Omega }$ of $\mathbb{S}$ is invariant under $\theta$ by the continuity of reflection-like maps, which shows that $\theta :\mathsf{\Omega }\mapsto \mathsf{\Omega }$ is a line-to-line bijection. Moveover, if $\mathsf{\Omega }$ is a Klein Model of hyperbolic space, then $\theta :\mathsf{\Omega }\mapsto \mathsf{\Omega }$ is an isometry.

5. Reflection-Like Maps and Hyperbolic Isometries in Klein Model

In this section, we shall prove Theorem 5, the rigidity of line-to-line maps in a local domain of ${\mathbb{R}}^n$ by hyperbolic isometry on Klein Model defined by projection $\tau :{\mathbb{S}}_{+}^n\mapsto {\mathbb{D}}^n$ as in Equations (1)–(3).

Lemma 2.

Suppose that $F:{\mathbb{S}}_{+}^n\mapsto {\mathbb{S}}_{+}^n$ is a reflection. Then, $f=\tau \circ F\circ {\tau }^{-1}:{\mathbb{D}}^n\mapsto {\mathbb{D}}^n$ is a refection-like map or a reflection.

Proof. 

Suppose that $F:{\mathbb{S}}_{+}^n\mapsto {\mathbb{S}}_{+}^n$ is a reflection relative to $(n-1)-$hyperbolic plane $S\subset {\mathbb{S}}_{+}^n$. Then $F^{\circ 2}=id$ and $F\left(P\right)=P$ for any $P\in S$. It follows that $\tau \left(S\right)$ is an $(n-1)-$dimensional plane in ${\mathbb{D}}^n$ and $f=\tau \circ F\circ {\tau }^{-1}:{\mathbb{D}}^n\mapsto {\mathbb{D}}^n$ is a line-onto-line bijection, satisfying $f^{\circ 2}=id$ and $f\left(X\right)=X$ for any $X\in \tau \left(S\right)$. Then, f is the restriction of a refection-like map or a reflection by Theorem 4. Specifically, f is a reflection if the origin point $O\in \tau \left(S\right)$; otherwise, f is a reflection-like map. □

For any two distinct points $P,Q\in {\mathbb{S}}_{+}^n$, one can always get a unique reflection $F:{\mathbb{S}}_{+}^n\mapsto {\mathbb{S}}_{+}^n$, satisfying that $F\left(P\right)=Q$. We can obtain the following Corollary.

Corollary 3.

For any point $X\in {\mathbb{D}}^n\backslash \left\{O\right\}$, there is a reflection-like map η satisfying that $\eta \left({\mathbb{D}}^n\right)={\mathbb{D}}^n$ and $\eta \left(O\right)=X$. Moveover, denote Axis of η by ${\mathcal{A}}^{\eta }$, then ${\mathcal{A}}^{\eta }\cap {\mathbb{D}}^n\ne \varnothing$.

Proof of Theorem 5.

If $f\left(O\right)=O$, then $f:{\mathbb{D}}^n\mapsto {\mathbb{D}}^n$ is the restriction to ${\mathbb{D}}^n$ of an orthogonal transformation on ${\mathbb{R}}^n$.

If $f\left(O\right)\ne O$, then there exists a reflection-like map $\eta$ such that $\eta \left({\mathbb{D}}^n\right)={\mathbb{D}}^n$ and $\eta \left(O\right)=f^{-1}\left(O\right)$ by Corollary 3, which follows $g=f\circ \eta :{\mathbb{D}}^n\mapsto {\mathbb{D}}^n$ is a hyperbolic isometry satisfying $g\left(O\right)=O$. Thus, $g:{\mathbb{D}}^n\mapsto {\mathbb{D}}^n$ is the restriction to ${\mathbb{D}}^n$ of an orthogonal transformation on ${\mathbb{R}}^n$. It implies that $f=g\circ \eta$.

Above all, any hyperbolic isometry in Klein Model is either an orthogonal transformation, or a composition of an orthogonal transformation and a reflection-like map. □

From Theorem 5, one can deduce that any line-to-line bijection on ${\mathbb{D}}^n$ can be extended line-to-line to ${\mathbb{R}}^n$ (or except a superplane).