Elliptic and Hyperbolic Rotational Motions on General Hyperboloids

Abstract

This study proposes a new way to represent elliptic and hyperbolic motions on any general hyperboloids of one or two sheets using the famous Rodrigues, Cayley, and Householder transformations. These transformations are used within the generalized Minkowski 3-space which extends the usual Lorentzian geometry by introducing a generalized scalar product. The study is carried out by considering the unit sphere defined in this generalized space along with the use of three-dimensional generalized Lorentzian skew-symmetric matrices that naturally generate continuous rotational motions. The obtained results provide rotational motions on the sphere in Minkowski 3-space as well as elliptic and hyperbolic motions on general hyperboloids in Euclidean 3-space. A numerical example is provided for each of the explored rotation methods.

1. Introduction

Rotation transformations are fundamental in various scientific and practical fields where precise control and prediction of rotational behavior play a critical role, including robotics [1], aerospace [2], and computer graphic [3]. This concept is crucial in physics and engineering, as it explains the phenomena behind the dynamics of spinning objects ranging from the Earth’s rotation to gyroscopes [4,5]. This widespread applicability underscores the essential role of rotation transformations in advancing both theoretical and applied sciences. In mathematics, rotations are represented by special orthogonal matrices that preserve distances and orientations, forming the SO(3) group in three-dimensional space.

Rotations in Lorentz geometry differ fundamentally from those in Euclidean space [6], as the Lorentzian scalar product distinguishes between the timelike, spacelike, and lightlike directions. Rotation preserves the causal structure in addition to the norm and orientation. In Lorentz geometry, refs. [7,8] studied the rotation matrices. Rotation matrices with a lightlike axis have also been studied in [9]. The Lorentzian scalar product is essential for understanding three-dimensional Lorentzian geometry, which is the geometry of space–time, as well as for studying the structure of space–time in special relativity. It is used for determining rotational motions on special hyperboloids with the following equation:

$$-x^2+y^2+z^2=\pm r^2$$

which are the spheres of three-dimensional Lorentzian geometry. Generalizations of the Lorentzian scalar product in three and two dimensions were studied by [10,11,12], respectively. In this study, we investigate the rotations in the generalized Lorentzian scalar product space defined in [13], where the spheres are general hyperboloids of one or two sheets with the following equation:

$$A{x}^2+B{y}^2+C{z}^2+2Dxy+2Exz+2Fyz+Gx+Hy+Iz+J=0$$

where the rotations are elliptic and hyperbolic. In fact, the rotations are closely related to number systems; these motions were studied in [14] and determined in [13] using a number system known as generalized split quaternions. However, the formulas derived to date are complicated and require the use of a number system. The aim of our study is to obtain useful and simple formulas for elliptic and hyperbolic rotation matrices on general hyperboloids without using a number system. For this purpose, we use the well-known Rodrigues formula, Cayley map, and Householder maps in the generalized Lorentzian scalar product space.

The rest of this paper is organized as follows: first, we introduce the generalized Lorentzian scalar product space where the sphere corresponds to a hyperboloid of one or two sheets; subsequently, we derive formulas for non-parabolic conical rotations occurring on any given hyperboloid of one or two sheets; finally, we present several numerical examples to demonstrate the practical application of these formulas.

2. Preliminaries

The generalized Minkowski 3-space (or three-dimensional generalized Lorentzian space) is the Euclidean space endowed with the following generalized Lorentzian scalar product for vectors $\mathbf{u}=(u_1,u_2,u_3)$, $\mathbf{w}=(w_1,w_2,w_3)\in {\mathbb{R}}^3$:

$${\mathcal{B}}_{\Omega }(\mathbf{u},\mathbf{w})={\mathbf{u}}^T\Omega \mathbf{w}=A{u}_1{w}_1+B{u}_2{w}_2+C{u}_3{w}_3+D(u_1{w}_2+u_2{w}_1)+E(u_1{w}_3+u_3{w}_1)+F(u_2{w}_3+u_3{w}_2),$$

where

$$\Omega =\left[\begin{array}{ccc}A& D& E\\ D& B& F\\ E& F& C\end{array}\right]$$

is a real symmetric matrix having a negative determinant and for which all eigenvalues are not of the same sign [13]. It is denoted by ${\mathbb{R}}_{{\mathcal{B}}_{\Omega }}^3$, and has properties similar to the classical Minkowski 3-space [15,16]. The constant of the matrix associated with the ${\mathcal{B}}_{\Omega }$-scalar product and the ${\mathcal{B}}_{\Omega }$-norm are respectively defined by

$$\Delta =\sqrt{\left|det\Omega \right|}=\sqrt{A{F}^2+B{E}^2+C{D}^2-ABC-2FDE},$$

$${∥\mathbf{u}∥}_{{\mathcal{B}}_{\Omega }}=\sqrt{\left|{\mathcal{B}}_{\Omega }(\mathbf{u},\mathbf{u})\right|}=\sqrt{\left|A{u}_1^2+B{u}_2^2+C{u}_3^2+2D{u}_1{u}_2+2E{u}_1{u}_3+2F{u}_2{u}_3\right|}.$$

As usual, a vector $\mathbf{u}$ is called unit if its ${\mathcal{B}}_{\Omega }$-norm is 1. The vectors of ${\mathbb{R}}_{{\mathcal{B}}_{\Omega }}^3$ are categorized as follows:

(i)

  If ${\mathcal{B}}_{\Omega }(\mathbf{u}$$,\mathbf{u}$$)>0$ or $\mathbf{u}=\mathbf{0}$, then $\mathbf{u}$ is called a ${\mathcal{B}}_{\Omega }$-spacelike vector.

(ii)

 If ${\mathcal{B}}_{\Omega }(\mathbf{u}$$,\mathbf{u}$$)<0$, then $\mathbf{u}$ is called a ${\mathcal{B}}_{\Omega }$-timelike vector.

(iii)

If ${\mathcal{B}}_{\Omega }(\mathbf{u}$$,\mathbf{u}$$)=0$ and $\mathbf{u}\ne \mathbf{0}$, then $\mathbf{u}$ is called a ${\mathcal{B}}_{\Omega }$-lightlike or ${\mathcal{B}}_{\Omega }$-null vector.

For a positive real number r, the set

$$S_{{\mathcal{B}}_{\Omega }}^2\left(r\right)=\left\{\mathbf{u}\in {\mathbb{R}}_{{\mathcal{B}}_{\Omega }}^3:{\mathcal{B}}_{\Omega }(\mathbf{u},\mathbf{u})=r^2\right\}$$

is called the ${\mathcal{B}}_{\Omega }$-pseudosphere with radius r, which is a general hyperboloid of one sheet. The set

$$H_{{\mathcal{B}}_{\Omega }}^2\left(r\right)=\left\{\mathbf{u}\in {\mathbb{R}}_{{\mathcal{B}}_{\Omega }}^3:{\mathcal{B}}_{\Omega }(\mathbf{u},\mathbf{u})=-r^2\right\}$$

is called the ${\mathcal{B}}_{\Omega }$-hyperbolic sphere with radius r, which is a general hyperboloid of two sheets. These spheres are both called the generalized Lorentzian (or ${\mathcal{B}}_{\Omega }$-) sphere.

Moreover, the generalized Lorentzian vector product is defined as follows:

$$\begin{array}{ccc}\hfill \mathbf{u}{\times }_{\Omega }\mathbf{w}& =& -\left(u_2{w}_3-u_3{w}_2\right)\widehat{\mathbf{i}}-\left(u_3{w}_1-u_1{w}_3\right)\widehat{\mathbf{j}}-\left(u_1{w}_2-u_2{w}_1\right)\widehat{\mathbf{k}}\hfill \\ & =& \left|\begin{array}{ccc}-\widehat{\mathbf{i}}& -\widehat{\mathbf{j}}& -\widehat{\mathbf{k}}\\ u_1& u_2& u_3\\ w_1& w_2& w_3\end{array}\right|\hfill \end{array}$$

where $\widehat{\mathbf{i}}={\Delta }_3\mathbf{i}+{\Delta }_6\mathbf{j}+{\Delta }_5\mathbf{k}$, $\widehat{\mathbf{j}}={\Delta }_6\mathbf{i}+{\Delta }_2\mathbf{j}+{\Delta }_4\mathbf{k}$, $\widehat{\mathbf{k}}={\Delta }_5\mathbf{i}+{\Delta }_4\mathbf{j}+{\Delta }_1\mathbf{k}$ and ${\Delta }_1=(AB-D^2)/\Delta$, ${\Delta }_2=(AC-E^2)/\Delta$, ${\Delta }_3=(BC-F^2)/\Delta$, ${\Delta }_4=(DE-AF)/\Delta$, ${\Delta }_5=(DF-BE)/\Delta ,{\Delta }_6=(EF-CD)/\Delta$.

Similar to the classical Minkowski 3-space [17], the ${\mathcal{B}}_{\Omega }$-measure of the angle between linearly independent ${\mathcal{B}}_{\Omega }$-spacelike vectors $\mathbf{u}$ and $\mathbf{w}$ is determined as follows:

(i)

If $\mathbf{u}{\times }_{\Omega }\mathbf{w}$ is ${\mathcal{B}}_{\Omega }$-spacelike and ${\mathcal{B}}_{\Omega }(\mathbf{u},\mathbf{w})>0$, which are equivalent to ${\mathcal{B}}_{\Omega }(\mathbf{u},\mathbf{w})>{∥\mathbf{u}∥}_{\Omega }{∥\mathbf{w}∥}_{\Omega }$, then

$${\mathcal{B}}_{\Omega }\left(\mathbf{u},\mathbf{w}\right)={∥\mathbf{u}∥}_{\Omega }{∥\mathbf{w}∥}_{\Omega }cosh\theta .$$

(ii)

If $\mathbf{u}{\times }_{\Omega }\mathbf{w}$ is ${\mathcal{B}}_{\Omega }$-timelike, which is equivalent to $\left|{\mathcal{B}}_{\Omega }(\mathbf{u},\mathbf{w})\right|<{∥\mathbf{u}∥}_{\Omega }{∥\mathbf{w}∥}_{\Omega }$, then

$${\mathcal{B}}_{\Omega }\left(\mathbf{u},\mathbf{w}\right)={∥\mathbf{u}∥}_{\Omega }{∥\mathbf{w}∥}_{\Omega }cos\theta$$

and the ${\mathcal{B}}_{\Omega }$-angle between $\mathbf{u}$ and $\mathbf{w}$ is $\theta$.

The ${\mathcal{B}}_{\Omega }$-measure of the angle between linearly independent ${\mathcal{B}}_{\Omega }$-timelike vectors $\mathbf{u}$ and $\mathbf{w}$ satisfying ${\mathcal{B}}_{\Omega }(\mathbf{u},\mathbf{w})<0$ is determined by the formula

$${\mathcal{B}}_{\Omega }\left(\mathbf{u},\mathbf{w}\right)=-{∥\mathbf{u}∥}_{\Omega }{∥\mathbf{w}∥}_{\Omega }cosh\theta ,$$

and the ${\mathcal{B}}_{\Omega }$-angle between $\mathbf{u}$ and $\mathbf{w}$ is $\theta$. In addition, $\mathbf{u}$ and $\mathbf{w}$ are called ${\mathcal{B}}_{\Omega }$-orthogonal if ${\mathcal{B}}_{\Omega }(\mathbf{u},\mathbf{w})=0$.

In the generalized Minkowski 3-space, matrices satisfying the condition

$${\mathcal{B}}_{\Omega }(S\mathbf{u},\mathbf{w})=-{\mathcal{B}}_{\Omega }(\mathbf{u},S\mathbf{w})$$

are called ${\mathcal{B}}_{\Omega }$-skew symmetric and have the following form:

$$S_{\mathbf{u}}=\left[\begin{array}{ccc}{\Delta }_5{u}_2-{\Delta }_6{u}_3& {\Delta }_3{u}_3-{\Delta }_5{u}_1& {\Delta }_6{u}_1-{\Delta }_3{u}_2\\ {\Delta }_4{u}_2-{\Delta }_2{u}_3& {\Delta }_6{u}_3-{\Delta }_4{u}_1& {\Delta }_2{u}_1-{\Delta }_6{u}_2\\ {\Delta }_1{u}_2-{\Delta }_4{u}_3& {\Delta }_5{u}_3-{\Delta }_1{u}_1& {\Delta }_4{u}_1-{\Delta }_5{u}_2\end{array}\right]$$

(1)

where $\mathbf{u}=(u_1,u_2,u_3)\in {\mathbb{R}}^3$ [13].

3. Rotations on a General Hyperboloid

It is known that the rotations can be represented by orthogonal matrices with determinants of 1, since these are the only linear transformations which preserve the norm and direction of the vector product. The ${\mathcal{B}}_{\Omega }$-orthogonal matrices and ${\mathcal{B}}_{\Omega }$-rotation matrices were defined in [13]. A matrix R is defined as ${\mathcal{B}}_{\Omega }$-orthogonal if and only if it satisfies the relation $R^t\Omega R=\Omega$. Moreover, when a ${\mathcal{B}}_{\Omega }$-orthogonal matrix R has determinant equal to 1, it is referred to as a ${\mathcal{B}}_{\Omega }$-rotation matrix. The ${\mathcal{B}}_{\Omega }$-rotation about a given vector u, corresponding to the ${\mathcal{B}}_{\Omega }$-angle $\theta$, is denoted by $R_{\theta }^{\mathbf{u}}$.

Note that ${\mathcal{B}}_{\Omega }$-rotations occur on Lorentzian spheres which are general hyperboloids of one or two sheets. However, throughout this study we only consider the central hyperboloids of one or two sheets. The ${\mathcal{B}}_{\Omega }$-rotation matrices can be easily generalized to general hyperboloids. If $R_{\theta }^{\mathbf{u}}$ is the matrix of a ${\mathcal{B}}_{\Omega }$-rotation on the hyperboloid with the equation

$$A{x}^2+B{y}^2+C{z}^2+2Dxy+2Exz+2Fyz=\pm r^2,$$

then

$$R_{\theta }^{\mathbf{u}}\left[\begin{array}{c}x-m\\ y-n\\ z-l\end{array}\right]+\left[\begin{array}{c}m\\ n\\ l\end{array}\right]$$

is ${\mathcal{B}}_{\Omega }$-rotation matrix which occurs on a general hyperboloid with the equation

$$g(x,y,z)=A{x}^2+B{y}^2+C{z}^2+2Dxy+2Exz+2Fyz+Gx+Hy+Iz+J=0,$$

where

$${\displaystyle \frac{\partial g}{\partial x}}(m,n,l)=0,{\displaystyle \frac{\partial g}{\partial y}}(m,n,l)=0,{\displaystyle \frac{\partial g}{\partial z}}(m,n,l)=0.$$

If we translate the general hyperboloid by $(-m,-n,-l)$, then we obtain the central hyperboloid

$$A{x}^2+B{y}^2+C{z}^2+2Dxy+2Exz+2Fyz+g(m,n,l)=0,$$

which is the central ${\mathcal{B}}_{\Omega }$-sphere with $r=\sqrt{\left|g(m,n,l)\right|}.$

3.1. The Rodrigues Rotation Formula

In Lorentzian geometry, the skew symmetric matrices are used to generate rotation matrices in [18,19] by the Rodrigues rotation formula. Similarly, we use ${\mathcal{B}}_{\Omega }$-skew symmetric matrices in those formulas. Note that

$$P\left(x\right)=x^3-{\mathcal{B}}_{\Omega }(\mathbf{u},\mathbf{u})x$$

is the characteristic polynomial of $S_{\mathbf{u}}$; in addition, if $\mathbf{u}$ is unit ${\mathcal{B}}_{\Omega }$-spacelike, then $S^3=S$, while if $\mathbf{u}$ is unit ${\mathcal{B}}_{\Omega }$-timelike, then $S^3=-S$. Thus, the Rodrigues formula can be considered in ${\mathbb{R}}_{{\mathcal{B}}_{\Omega }}^3$.

Theorem 1.

Let $S_{\mathbf{u}}$ be the ${\mathcal{B}}_{\Omega }$-skew symmetric matrix with respect to the unit vector $\mathbf{u}=(u_1,u_2,u_3)\in {\mathbb{R}}^3$.

(i)

If $\mathbf{u}$ is unit ${\mathcal{B}}_{\Omega }$-spacelike, then

$$e^{S\theta }=I+(sinh\theta )S_{\mathbf{u}}+(cosh\theta -1)S_{\mathbf{u}}^2,$$

(2)

which provides the ${\mathcal{B}}_{\Omega }$-rotation around $\mathbf{u}$ by the ${\mathcal{B}}_{\Omega }$-angle θ, for which the trajectory is a hyperbola on the hyperboloid of one or two sheets $A{x}^2+B{y}^2+C{z}^2+2Dxy+2Exz+2Fyz=\mp r^2$.

(ii)

If $\mathbf{u}$ is unit ${\mathcal{B}}_{\Omega }$-timelike, then

$$e^{S\theta }=I+(sin\theta )S_{\mathbf{u}}+(1-cos\theta )S_{\mathbf{u}}^2,$$

(3)

which provides the ${\mathcal{B}}_{\Omega }$-rotation around $\mathbf{u}$ by the ${\mathcal{B}}_{\Omega }$-angle θ, the trajectory of which is an ellipse on the hyperboloid of one or two sheets $A{x}^2+B{y}^2+C{z}^2+2Dxy+2Exz+2Fyz=\mp r^2$.

Proof. 

Defining ${\Delta }_{i,j}^{\prime }=\left({\Delta }_i{u}_2-{\Delta }_j{u}_3\right)$, ${\Delta }_{i,j}^{\prime \prime }=\left({\Delta }_i{u}_3-{\Delta }_j{u}_1\right)$, ${\Delta }_{i,j}^{\prime \prime \prime }=\left({\Delta }_i{u}_1-{\Delta }_j{u}_2\right)$, the matrix (1) is then

$$S_{\mathbf{u}}=\left[\begin{array}{ccc}{\Delta }_{5,6}^{\prime }& {\Delta }_{3,5}^{\prime \prime }& {\Delta }_{6,3}^{\prime \prime \prime }\\ {\Delta }_{4,2}^{\prime }& {\Delta }_{6,4}^{\prime \prime }& {\Delta }_{2,6}^{\prime \prime \prime }\\ {\Delta }_{1,4}^{\prime }& {\Delta }_{5,1}^{\prime \prime }& {\Delta }_{4,5}^{\prime \prime \prime }\end{array}\right].$$

Then, we can determine the matrix $S_{\mathbf{u}}^2$ as follows:

$$S_{\mathbf{u}}^2=\left[\begin{array}{ccc}{\mathcal{B}}_{\Omega }(\mathbf{u},\mathbf{u})-{\delta }_{u_1}^{\prime }& -{\delta }_{u_1}^{\prime \prime }& -{\delta }_{u_1}^{\prime \prime \prime }\\ -{\delta }_{u_2}^{\prime }& {\mathcal{B}}_{\Omega }(\mathbf{u},\mathbf{u})-{\delta }_{u_2}^{\prime \prime }& -{\delta }_{u_2}^{\prime \prime \prime }\\ -{\delta }_{u_3}^{\prime }& -{\delta }_{u_3}^{\prime \prime }& {\mathcal{B}}_{\Omega }(\mathbf{u},\mathbf{u})-{\delta }_{u_3}^{\prime \prime \prime }\end{array}\right]$$

with ${\delta }_{u_i}^{\prime }=u_i\left(A{u}_1+D{u}_2+E{u}_3\right)$, ${\delta }_{u_i}^{\prime \prime }=u_i\left(D{u}_1+B{u}_2+F{u}_3\right)$, ${\delta }_{u_i}^{\prime \prime \prime }=u_i\left(E{u}_1+F{u}_2+C{u}_3\right)$.

(i)

If $\mathbf{u}$ is unit ${\mathcal{B}}_{\Omega }$-spacelike, then ${\mathcal{B}}_{\Omega }(\mathbf{u},\mathbf{u})=1$ and $S_{\mathbf{u}}^3=S_{\mathbf{u}}$. Thus, we can obtain the matrix $R_{\theta }^{\mathbf{u}}$ of the ${\mathcal{B}}_{\Omega }$-rotation by the ${\mathcal{B}}_{\Omega }$-angle $\theta$ with the following computations:

$$\begin{gathered} e^{S\theta }=I+(sinh\theta )S_{\mathbf{u}}+(cosh\theta -1)S_{\mathbf{u}}^2 \\ =\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]+(sinh\theta )\left[\begin{array}{ccc}{\Delta }_{5,6}^{\prime }& {\Delta }_{3,5}^{\prime \prime }& {\Delta }_{6,3}^{\prime \prime \prime }\\ {\Delta }_{4,2}^{\prime }& {\Delta }_{6,4}^{\prime \prime }& {\Delta }_{2,6}^{\prime \prime \prime }\\ {\Delta }_{1,4}^{\prime }& {\Delta }_{5,1}^{\prime \prime }& {\Delta }_{4,5}^{\prime \prime \prime }\end{array}\right]+(cosh\theta -1)\left[\begin{array}{ccc}1-{\delta }_{u_1}^{\prime }& -{\delta }_{u_1}^{\prime \prime }& -{\delta }_{u_1}^{\prime \prime \prime }\\ -{\delta }_{u_2}^{\prime }& 1-{\delta }_{u_2}^{\prime \prime }& -{\delta }_{u_2}^{\prime \prime \prime }\\ -{\delta }_{u_3}^{\prime }& -{\delta }_{u_3}^{\prime \prime }& 1-{\delta }_{u_3}^{\prime \prime \prime }\end{array}\right] \\ =\left[\begin{array}{ccc}1-{\mu }_s+{\delta }_{u_1}^{\prime }{\mu }_s+{\Delta }_{5,6}^{\prime }{\rho }_s& {\delta }_{u_1}^{\prime \prime }{\mu }_s+{\Delta }_{3,5}^{\prime \prime }{\rho }_s& {\delta }_{u_1}^{\prime \prime \prime }{\mu }_s+{\Delta }_{6,3}^{\prime \prime \prime }{\rho }_s\\ {\delta }_{u_2}^{\prime }{\mu }_s+{\Delta }_{4,2}^{\prime }{\rho }_s& 1-{\mu }_s+{\delta }_{u_2}^{\prime \prime }{\mu }_s+{\Delta }_{6,4}^{\prime \prime }{\rho }_s& {\delta }_{u_2}^{\prime \prime \prime }{\mu }_s+{\Delta }_{2,6}^{\prime \prime \prime }{\rho }_s\\ {\delta }_{u_3}^{\prime }{\mu }_s+{\Delta }_{1,4}^{\prime }{\rho }_s& {\delta }_{u_3}^{\prime \prime }{\mu }_s+{\Delta }_{5,1}^{\prime \prime }{\rho }_s& 1-{\mu }_s+{\delta }_{u_3}^{\prime \prime \prime }{\mu }_s+{\Delta }_{4,5}^{\prime \prime \prime }{\rho }_s\end{array}\right] \end{gathered}$$

where ${\mu }_s=1-cosh\theta$ and ${\rho }_s=sinh\theta$. Substituting all values in the matrix and with the aid of a computer calculation, it can be seen that the matrix $R_{\theta }^{\mathbf{u}}$ is ${\mathcal{B}}_{\Omega }$-orthogonal and $det\left(R_{\theta }^{\mathbf{u}}\right)=1$. On the other hand, the eigenvalues of $R_{\theta }^{\mathbf{u}}$ are $e^{\theta },$$e^{-\theta }$, 1 and the eigenvector corresponding to 1 is $\mathbf{u}$. Hence, the axis of the ${\mathcal{B}}_{\Omega }$-rotation is $\mathbf{u}$.

(ii)

If $\mathbf{u}$ is unit ${\mathcal{B}}_{\Omega }$-timelike, then ${\mathcal{B}}_{\Omega }(\mathbf{u},\mathbf{u})=-1$ and $S_{\mathbf{u}}^3=-S_{\mathbf{u}}$. Thus, we can obtain the matrix $R_{\theta }^{\mathbf{u}}$ of the ${\mathcal{B}}_{\Omega }$-rotation by the ${\mathcal{B}}_{\Omega }$-angle $\theta$ with the following computations:

$$\begin{gathered} e^{S\theta }=I+(sin\theta )S_{\mathbf{u}}+(1-cos\theta )S_{\mathbf{u}}^2 \\ =\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]+(sin\theta )\left[\begin{array}{ccc}{\Delta }_{5,6}^{\prime }& {\Delta }_{3,5}^{\prime \prime }& {\Delta }_{6,3}^{\prime \prime \prime }\\ {\Delta }_{4,2}^{\prime }& {\Delta }_{6,4}^{\prime \prime }& {\Delta }_{2,6}^{\prime \prime \prime }\\ {\Delta }_{1,4}^{\prime }& {\Delta }_{5,1}^{\prime \prime }& {\Delta }_{4,5}^{\prime \prime \prime }\end{array}\right]+(1-cos\theta )\left[\begin{array}{ccc}-1-{\delta }_{u_1}^{\prime }& -{\delta }_{u_1}^{\prime \prime }& -{\delta }_{u_1}^{\prime \prime \prime }\\ -{\delta }_{u_2}^{\prime }& -1-{\delta }_{u_2}^{\prime \prime }& -{\delta }_{u_2}^{\prime \prime \prime }\\ -{\delta }_{u_3}^{\prime }& -{\delta }_{u_3}^{\prime \prime }& -1-{\delta }_{u_3}^{\prime \prime \prime }\end{array}\right] \\ =\left[\begin{array}{ccc}1+{\mu }_t+{\delta }_{u_1}^{\prime }{\mu }_t+{\Delta }_{5,6}^{\prime }{\rho }_t& {\delta }_{u_1}^{\prime \prime }{\mu }_t+{\Delta }_{3,5}^{\prime \prime }{\rho }_t& {\delta }_{u_1}^{\prime \prime \prime }{\mu }_t+{\Delta }_{6,3}^{\prime \prime \prime }{\rho }_t\\ {\delta }_{u_2}^{\prime }{\mu }_t+{\Delta }_{4,2}^{\prime }{\rho }_t& 1+{\mu }_t+{\delta }_{u_2}^{\prime \prime }{\mu }_t+{\Delta }_{6,4}^{\prime \prime }{\rho }_t& {\delta }_{u_2}^{\prime \prime \prime }{\mu }_t+{\Delta }_{2,6}^{\prime \prime \prime }{\rho }_t\\ {\delta }_{u_3}^{\prime }{\mu }_t+{\Delta }_{1,4}^{\prime }{\rho }_t& {\delta }_{u_3}^{\prime \prime }{\mu }_t+{\Delta }_{5,1}^{\prime \prime }{\rho }_t& 1+{\mu }_t+{\delta }_{u_3}^{\prime \prime \prime }{\mu }_t+{\Delta }_{4,5}^{\prime \prime \prime }{\rho }_t\end{array}\right] \end{gathered}$$

where ${\mu }_t=cos\theta -1$ and ${\rho }_t=sin\theta$. Substituting all values in the matrix and with the aid of a computer calculation, it can be seen that the matrix $R_{\theta }^{\mathbf{u}}$ is ${\mathcal{B}}_{\Omega }$-orthogonal and $det\left(R_{\theta }^{\mathbf{u}}\right)=1$. In addition, the eigenvalues of $R_{\theta }^{\mathbf{u}}$ are $e^{i\theta }$, $e^{-i\theta }$, 1 and the eigenvector corresponding to 1 is $\mathbf{u}$. Hence, the axis of the ${\mathcal{B}}_{\Omega }$-rotation is $\mathbf{u}$.

□

Example 1.

Given the hyperboloid of one sheet

$$x^2+y^2+4z^2-6xy-4yz=1$$

having the following parameterization:

$$\alpha \left(\theta ,\beta \right)=(\frac{1}{2}sinh\theta -\frac{1}{2}cosh\theta sin\beta ,\frac{1}{2}sinh\theta -\frac{1}{3}cosh\theta cos\beta +\frac{1}{6}cosh\theta sin\beta ,\frac{1}{3}\left(cosh\theta \right)\left(cos\beta +sin\beta \right))$$

where $\theta \in \mathbb{R}$ and $\beta \in [0,2\pi )$, this hyperboloid is the unit ${\mathcal{B}}_{\Omega }$-pseudosphere, where

$$\Omega =\left[\begin{array}{ccc}1& -3& 0\\ -3& 1& -2\\ 0& -2& 4\end{array}\right].$$

Considering the two unit ${\mathcal{B}}_{\Omega }$-spacelike vectors

$$\begin{array}{ccc}\hfill \alpha \left(ln2,\pi /3\right)& =& \left(\frac{6-5\sqrt{3}}{16},\frac{5\sqrt{3}+8}{48},\frac{5\sqrt{3}+5}{24}\right)=\mathbf{v},\hfill \\ \hfill \alpha \left(ln2+ln3,\pi /3\right)& =& \left(\frac{70-37\sqrt{3}}{48},\frac{37\sqrt{3}+136}{144},\frac{37\sqrt{3}+37}{72}\right)=\mathbf{w}\hfill \end{array}$$

on the hyperboloid, we can find the ${\mathcal{B}}_{\Omega }$-rotation matrix $R_{\theta }^{\mathbf{u}}$, where $\mathbf{u}=\frac{\mathbf{w}{\times }_{\Omega }\mathbf{v}}{{∥\mathbf{w}{\times }_{\Omega }\mathbf{v}∥}_{\Omega }}$ and $\theta =ln3$. Here, with some computations, we have $\Delta =6$, ${\Delta }_1=-{\displaystyle \frac{4}{3}}$, ${\Delta }_2={\displaystyle \frac{2}{3}}$, ${\Delta }_3=0$, ${\Delta }_4={\displaystyle \frac{1}{3}}$, ${\Delta }_5=1$, ${\Delta }_6=2$ and obtain

$$\mathbf{u}=\left(\frac{1}{4},-\frac{2\sqrt{3}+1}{12},\frac{\sqrt{3}-1}{6}\right),$$

which is unit ${\mathcal{B}}_{\Omega }$-spacelike; then, we obtain the ${\mathcal{B}}_{\Omega }$-rotation matrix as follows:

$$R_{ln3}^{\mathbf{u}}=\left[\begin{array}{ccc}{\displaystyle \frac{23-9\sqrt{3}}{12}}& {\displaystyle \frac{\sqrt{3}-3}{12}}& {\displaystyle \frac{9-2\sqrt{3}}{12}}\\ {\displaystyle \frac{11-5\sqrt{3}}{36}}& {\displaystyle \frac{13\sqrt{3}+33}{36}}& {\displaystyle \frac{16\sqrt{3}+27}{36}}\\ {\displaystyle \frac{2\sqrt{3}+1}{9}}& {\displaystyle \frac{2\sqrt{3}+3}{9}}& {\displaystyle \frac{7\sqrt{3}+27}{18}}\end{array}\right].$$

It can be easily seen that $R_{ln3}^{\mathbf{u}}\left(\mathbf{v}\right)=\mathbf{w}$. Note that the rotation plane is

$$(1+\sqrt{3})x-(1+\sqrt{3})y+(2\sqrt{3}-1)z=0,$$

which is ${\mathcal{B}}_{\Omega }$-orthogonal to $\mathbf{u}$, and that the trajectory is a hyperbola.

Example 2.

Given the hyperboloid of two sheets

$$x^2+y^2+4z^2-6xy-4yz=-1$$

for which one sheet has the following parametric equation:

$$\alpha \left(\theta ,\beta \right)=(\frac{1}{2}cosh\theta -\frac{1}{2}sinh\theta sin\beta ,\frac{1}{2}cosh\theta -\frac{1}{3}sinh\theta cos\beta +\frac{1}{6}sinh\theta sin\beta ,\frac{1}{3}sinh\theta \left(cos\beta +sin\beta \right))$$

where $\theta \in \mathbb{R}$ and $\beta \in [0,2\pi )$, this hyperboloid is the unit ${\mathcal{B}}_{\Omega }$-hyperbolic sphere, where Ω is the same as in the previous example. Considering the two unit ${\mathcal{B}}_{\Omega }$-timelike vectors

$$\begin{array}{ccc}\hfill \alpha \left(ln2,\pi /3\right)& =& \left(\frac{10-3\sqrt{3}}{16},\frac{\sqrt{3}+8}{16},\frac{\sqrt{3}+1}{8}\right)=\mathbf{v},\hfill \\ \hfill \alpha \left(ln2+ln3,\pi /3\right)& =& \left(\frac{74-35\sqrt{3}}{48},\frac{35\sqrt{3}+152}{144},\frac{35\sqrt{3}+35}{72}\right)=\mathbf{w}\hfill \end{array}$$

on the hyperboloid, we can find the rotation matrix $R_{\theta }^{\mathbf{u}}$, where $\mathbf{u}=\frac{\mathbf{v}{\times }_{\Omega }\mathbf{w}}{{∥\mathbf{v}{\times }_{\Omega }\mathbf{w}∥}_{{\mathcal{B}}_{\Omega }}}$ and $\theta =ln3$. Using the same parameters as in the previous example, we obtain

$$\mathbf{u}=\left(\frac{1}{4},-\frac{2\sqrt{3}+1}{12},\frac{\sqrt{3}-1}{6}\right),$$

which is unit ${\mathcal{B}}_{\Omega }$-spacelike, and obtain the ${\mathcal{B}}_{\Omega }$-rotation matrix as follows:

$$R_{ln3}^{\mathbf{u}}=\left[\begin{array}{ccc}{\displaystyle \frac{23-9\sqrt{3}}{12}}& {\displaystyle \frac{\sqrt{3}-3}{12}}& {\displaystyle \frac{9-2\sqrt{3}}{12}}\\ {\displaystyle \frac{11-5\sqrt{3}}{36}}& {\displaystyle \frac{13\sqrt{3}+33}{36}}& {\displaystyle \frac{16\sqrt{3}+27}{36}}\\ {\displaystyle \frac{2\sqrt{3}+1}{9}}& {\displaystyle \frac{2\sqrt{3}+3}{9}}& {\displaystyle \frac{7\sqrt{3}+27}{18}}\end{array}\right].$$

It can be seen that $R_{ln3}^{\mathbf{u}}\left(\mathbf{v}\right)=\mathbf{w}$. Note that the rotation plane is the same and that the trajectory is also a hyperbola.

Example 3.

Given the hyperboloid of one sheet

$$-3y^2+z^2-6xy-4xz-2yz=1$$

having the parameterization

$$\alpha \left(\theta ,\beta \right)=(\frac{1}{3}sinh\theta +\frac{1}{3}cosh\theta cos\beta +\frac{2}{3}cosh\theta sin\beta ,\frac{1}{3}sinh\theta -\frac{2}{3}cosh\theta cos\beta -\frac{1}{3}cosh\theta sin\beta ,cosh\theta cos\beta ),$$

where $\theta \in \mathbb{R}$ and $\beta \in [0,2\pi )$, this hyperboloid is the unit ${\mathcal{B}}_{\Omega }$-pseudosphere, where

$$\Omega =\left[\begin{array}{ccc}0& -3& -2\\ -3& -3& -1\\ -2& -1& 1\end{array}\right].$$

Considering the two unit ${\mathcal{B}}_{\Omega }$-spacelike vectors

$$\begin{array}{c}\hfill \alpha \left(0,\pi /3\right)=\left(\frac{2\sqrt{3}+1}{6},-\frac{\sqrt{3}+2}{6},\frac{1}{2}\right)=\mathbf{v},\\ \hfill \alpha \left(0,\pi /3+\pi /4\right)=\left(\frac{3\sqrt{2}+\sqrt{6}}{12},\frac{\sqrt{6}-3\sqrt{2}}{12},\frac{\sqrt{2}-\sqrt{6}}{4}\right)=\mathbf{w}\end{array}$$

on the hyperboloid, we can find the rotation matrix $R_{\theta }^{\mathbf{u}}$, where $\mathbf{u}=\frac{\mathbf{w}{\times }_{\Omega }\mathbf{v}}{{∥\mathbf{w}{\times }_{\Omega }\mathbf{v}∥}_{{\mathcal{B}}_{\Omega }}}$ and $\theta =\pi /4$. Here, with some calculations, we have $\Delta =3$, ${\Delta }_1=-3$, ${\Delta }_2=-{\displaystyle \frac{4}{3}}$, ${\Delta }_3=-{\displaystyle \frac{4}{3}}$, ${\Delta }_4=2$, ${\Delta }_5=-1$, ${\Delta }_6={\displaystyle \frac{5}{3}}$ and obtain

$$\mathbf{u}=\left(\frac{1}{3},\frac{1}{3},0\right),$$

which is unit ${\mathcal{B}}_{\Omega }$-timelike. Then, we obtain the ${\mathcal{B}}_{\Omega }$-rotation matrix as follows:

$$R_{\pi /4}^{\mathbf{u}}=\left[\begin{array}{ccc}{\displaystyle \frac{\sqrt{2}+2}{6}}& {\displaystyle \frac{4-\sqrt{2}}{6}}& {\displaystyle \frac{\sqrt{2}+1}{3}}\\ {\displaystyle \frac{\sqrt{2}+2}{6}}& {\displaystyle \frac{4-\sqrt{2}}{6}}& {\displaystyle \frac{1-2\sqrt{2}}{3}}\\ -{\displaystyle \frac{\sqrt{2}}{2}}& {\displaystyle \frac{\sqrt{2}}{2}}& \sqrt{2}\end{array}\right].$$

It can be seen that $R_{\pi /4}^{\mathbf{u}}\left(\mathbf{v}\right)=\mathbf{w}$. Note that the rotation plane is

$$x+2y+z=0,$$

which is ${\mathcal{B}}_{\Omega }$-orthogonal to $\mathbf{u}$, and that the trajectory is an ellipse.

Remark 1.

Note that for the ${\mathcal{B}}_{\Omega }$-spacelike vectors $\mathbf{v}$ and $\mathbf{w}$, the ${\mathcal{B}}_{\Omega }$-rotation which transforms $\mathbf{v}$ to $\mathbf{w}$ has a positive ${\mathcal{B}}_{\Omega }$-angle and has the axis $\mathbf{w}{\times }_{\Omega }\mathbf{v}$. For the ${\mathcal{B}}_{\Omega }$-timelike vectors $\mathbf{v}$ and $\mathbf{w}$,the ${\mathcal{B}}_{\Omega }$-rotation which transforms $\mathbf{v}$ to $\mathbf{w}$ has a positive ${\mathcal{B}}_{\Omega }$-angle and the axis $\mathbf{v}{\times }_{\Omega }\mathbf{w}$.

3.2. The Cayley Rotation Formula

The Cayley rotation formula does not contain trigonometric functions. Rotation matrices using the Cayley formula were studied in three-dimensional space in [20] and in four dimensional space in [21]. If S is a $3\times 3$ skew-symmetric matrix and I is the $3\times 3$ identity matrix, where $(I-S)$ is invertible, then the Cayley map

$$Cay\left(S\right)=(I+S){(I-S)}^{-1}={(I-S)}^{-1}(I+S)$$

transforms the matrix S into a rotation matrix. We use the ${\mathcal{B}}_{\Omega }$-skew symmetric matrix $S_{\mathbf{u}}$ in the Cayley map to generate ${\mathcal{B}}_{\Omega }$-rotation matrices.

Theorem 2.

Let $S_{\mathbf{u}}$ be the ${\mathcal{B}}_{\Omega }$-skew symmetric matrix with respect to the unit vector $\mathbf{u}\in {\mathbb{R}}_{{\mathcal{B}}_{\Omega }}^3$. Then, the matrix

$$R^{\mathbf{u}}=(I+S_{\mathbf{u}}){(I-S_{\mathbf{u}})}^{-1}$$

(4)

is the ${\mathcal{B}}_{\Omega }$-rotation around the vector $\mathbf{u}$ on the hyperboloids of $A{x}^2+B{y}^2+C{z}^2+2Dxy+2Exz+2Fyz=\pm r^2$.

Proof. 

Using the ${\mathcal{B}}_{\Omega }$-skew symmetric matrix condition $S_{\mathbf{u}}^t\Omega =-\Omega S_{\mathbf{u}}$, we obtain

$$\begin{array}{cc}\hfill {(I+S_{\mathbf{u}})}^t\Omega & =\Omega (I-S_{\mathbf{u}}),\hfill \\ \hfill {(I-S_{\mathbf{u}})}^t\Omega & =\Omega (I+S_{\mathbf{u}}).\hfill \end{array}$$

Then, we have

$${\left(R^{\mathbf{u}}\right)}^t\Omega \left(R^{\mathbf{u}}\right)={\left((I+S_{\mathbf{u}}){(I-S_{\mathbf{u}})}^{-1}\right)}^t\Omega (I+S_{\mathbf{u}}){(I-S_{\mathbf{u}})}^{-1}=\Omega .$$

Thus, $R^{\mathbf{u}}$ is a ${\mathcal{B}}_{\Omega }$-orthogonal matrix. In addition, we can also obtain

$$det\left(I+S_{\mathbf{u}}\right)=det\left(I-S_{\mathbf{u}}\right)=1-{\mathcal{B}}_{\Omega }(\mathbf{u},\mathbf{u}),$$

meaning that $det\left(R^{\mathbf{u}}\right)=1$. Hence, the matrix $R^{\mathbf{u}}$ is a ${\mathcal{B}}_{\Omega }$-rotation matrix. □

The next theorem provides the ${\mathcal{B}}_{\Omega }$-rotation matrix about a ${\mathcal{B}}_{\Omega }$-non-lightlike axis using the ${\mathcal{B}}_{\Omega }$-angle $\theta$.

Theorem 3.

Let $S_{\mathbf{u}}$ be the ${\mathcal{B}}_{\Omega }$-skew symmetric matrix with respect to the unit vector $\mathbf{u}\in {\mathbb{R}}_{{\mathcal{B}}_{\Omega }}^3$. Then:

(i)

If $\mathbf{u}$ is unit ${\mathcal{B}}_{\Omega }$-spacelike,

$$Cay\left(S_{\mathbf{u}}tanh\theta \right)=\left(I+S_{\mathbf{u}}tanh\theta \right){\left(I-S_{\mathbf{u}}tanh\theta \right)}^{-1}$$

(5)

is the ${\mathcal{B}}_{\Omega }$-rotation about the axis $\mathbf{u}$ using the ${\mathcal{B}}_{\Omega }$-angle θ, that is, $R_{\theta }^{\mathbf{u}}$.

(ii)

If $\mathbf{u}$ is unit ${\mathcal{B}}_{\Omega }$-timelike, then

$$Cay\left(S_{\mathbf{u}}tan{\displaystyle \frac{\theta }{2}}\right)=\left(I+S_{\mathbf{u}}tan{\displaystyle \frac{\theta }{2}}\right){\left(I-S_{\mathbf{u}}tan{\displaystyle \frac{\theta }{2}}\right)}^{-1}$$

(6)

is the ${\mathcal{B}}_{\Omega }$-rotation about the axis $\mathbf{u}$ using the ${\mathcal{B}}_{\Omega }$-angle θ, that is, $R_{\theta }^{\mathbf{u}}$.

Proof. 

(i) Let $\mathbf{u}$ be unit ${\mathcal{B}}_{\Omega }$-spacelike. The matrix $a{S}_{\mathbf{u}}$ is a ${\mathcal{B}}_{\Omega }$-skew symmetric matrix for all $a\in \mathbb{R}$. Then,

$$Cay\left(a{S}_{\mathbf{u}}\right)=\left(I+a{S}_{\mathbf{u}}\right){\left(I-a{S}_{\mathbf{u}}\right)}^{-1}$$

(7)

is a ${\mathcal{B}}_{\Omega }$-rotation matrix. To find the value of a, consider that $Cay\left(a{S}_{\mathbf{u}}\right)=R_{\theta }^{\mathbf{u}}$. Then, we obtain

$$R_{\theta }^{\mathbf{u}}\left(I-a{S}_{\mathbf{u}}\right)=I+a{S}_{\mathbf{u}},$$

and per Formula (2) we have

$$\left(I+(sinh\theta )S_{\mathbf{u}}+(cosh\theta -1)S_{\mathbf{u}}^2\right)\left(I-a{S}_{\mathbf{u}}\right)=I+a{S}_{\mathbf{u}}.$$

(8)

If $\mathbf{u}$ is unit ${\mathcal{B}}_{\Omega }$-spacelike, then $S_{\mathbf{u}}^3=S_{\mathbf{u}}$. From (8), we obtain

$$I+\left(sinh\theta -acosh\theta \right)S_{\mathbf{u}}+\left(cosh\theta -asinh\theta -1\right)S_{\mathbf{u}}^2=I+a{S}_{\mathbf{u}}.$$

This equation implies that

$$\begin{array}{ccc}\hfill sinh\theta -acosh\theta & =& a,\hfill \end{array}$$

(9)

$$\begin{array}{ccc}\hfill cosh\theta -asinh\theta & =& 1.\hfill \end{array}$$

(10)

Substituting (9) into (10), we obtain

$$a=tanh\theta$$

by calculation. Substituting this value into (7), we obtain

$$Cay\left(S_{\mathbf{u}}tanh\theta \right)=\left(I+S_{\mathbf{u}}tanh\theta \right){\left(I-S_{\mathbf{u}}tanh\theta \right)}^{-1}=R_{\theta }^{\mathbf{u}}.$$

(ii) Let $\mathbf{u}$ be unit ${\mathcal{B}}_{\Omega }$-timelike. The matrix $b{S}_{\mathbf{u}}$ is also ${\mathcal{B}}_{\Omega }$-skew symmetric for all $b\in \mathbb{R}$. Then,

$$Cay\left(b{S}_{\mathbf{u}}\right)=\left(I+b{S}_{\mathbf{u}}\right){\left(I-b{S}_{\mathbf{u}}\right)}^{-1}$$

(11)

is a ${\mathcal{B}}_{\Omega }$-rotation matrix. To find the value of b, we can consider that $Cay\left(b{S}_{\mathbf{u}}\right)=R_{\theta }^{\mathbf{u}}$. Then, we obtain

$$R_{\theta }^{\mathbf{u}}\left(I-b{S}_{\mathbf{u}}\right)=I+b{S}_{\mathbf{u}},$$

and per Formula (3) we have

$$\left(I+(sin\theta )S_{\mathbf{u}}+(1-cos\theta )S_{\mathbf{u}}^2\right)\left(I-b{S}_{\mathbf{u}}\right)=I+b{S}_{\mathbf{u}}.$$

If $\mathbf{u}$ is unit ${\mathcal{B}}_{\Omega }$-timelike, then $S^3=-S$, and we obtain

$$I+\left(sin\theta -bcos\theta \right)S_{\mathbf{u}}+\left(1-bsin\theta -cos\theta \right)S_{\mathbf{u}}^2=I+b{S}_{\mathbf{u}}.$$

This equation implies that

$$\begin{array}{cc}\hfill sin\theta -bcos\theta & =b,\hfill \end{array}$$

(12)

$$\begin{array}{cc}\hfill bsin\theta +cos\theta & =1.\hfill \end{array}$$

(13)

Substituting (12) into (13), we obtain

$$b=tan{\displaystyle \frac{\theta }{2}}$$

by calculation. Substituting this value into (11), we obtain

$$Cay\left(S_{\mathbf{u}}tan{\displaystyle \frac{\theta }{2}}\right)=\left(I+S_{\mathbf{u}}tan{\displaystyle \frac{\theta }{2}}\right){\left(I-S_{\mathbf{u}}tan{\displaystyle \frac{\theta }{2}}\right)}^{-1}=R_{\theta }^{\mathbf{u}}.$$

□

It can be seen that the inverse of the Cayley map can be expressed as

$$Ca{y}^{-1}\left(R_{\theta }^{\mathbf{u}}\right)={\left(R_{\theta }^{\mathbf{u}}+I\right)}^{-1}\left(R_{\theta }^{\mathbf{u}}-I\right)=\left(R_{\theta }^{\mathbf{u}}-I\right){\left(R_{\theta }^{\mathbf{u}}+I\right)}^{-1},$$

(14)

which is consistent with its standard form provided in [22]. Using this fact, we can prove the following theorem, which can then be used to determine the ${\mathcal{B}}_{\Omega }$-rotation angle of a ${\mathcal{B}}_{\Omega }$-rotation matrix $R_{\theta }^{\mathbf{u}}$.

Theorem 4.

Let $R_{\theta }^{\mathbf{u}}$ be a ${\mathcal{B}}_{\Omega }$-rotation matrix and let $S_{\mathbf{u}}$ be the ${\mathcal{B}}_{\Omega }$-skew symmetric matrix with respect to the unit vector $\mathbf{u}\in {\mathbb{R}}_{{\mathcal{B}}_{\Omega }}^3$. Then:

(i)

If $\mathbf{u}$ is unit ${\mathcal{B}}_{\Omega }$-spacelike,

$$S_{\mathbf{u}}tanh\theta ={\left(R_{\theta }^{\mathbf{u}}+I\right)}^{-1}\left(R_{\theta }^{\mathbf{u}}-I\right)$$

(15)

is satisfied.

(ii)

If $\mathbf{u}$ is unit ${\mathcal{B}}_{\Omega }$-timelike,

$$S_{\mathbf{u}}tan{\displaystyle \frac{\theta }{2}}={\left(R_{\theta }^{\mathbf{u}}+I\right)}^{-1}\left(R_{\theta }^{\mathbf{u}}-I\right)$$

(16)

is satisfied.

Proof. 

Let $R_{\theta }^{\mathbf{u}}$ be a rotation matrix in ${\mathbb{R}}_{\Omega }^3$.

(i)

Per Equation (5), we have

$$R_{\theta }^{\mathbf{u}}=Cay\left(S_{\mathbf{u}}tanh\theta \right)=\left(I+S_{\mathbf{u}}tanh\theta \right){\left(I-S_{\mathbf{u}}tanh\theta \right)}^{-1}.$$

Using the inverse Cayley map (14), we obtain

$$Ca{y}^{-1}\left(R_{\theta }^{\mathbf{u}}\right)={\left(R_{\theta }^{\mathbf{u}}+I\right)}^{-1}\left(R_{\theta }^{\mathbf{u}}-I\right)=S_{\mathbf{u}}tanh\theta .$$

(ii)

Per Equation (6), we have

$$R_{\theta }^{\mathbf{u}}=Cay\left(S_{\mathbf{u}}tan{\displaystyle \frac{\theta }{2}}\right)=\left(I+S_{\mathbf{u}}tan{\displaystyle \frac{\theta }{2}}\right){\left(I-S_{\mathbf{u}}tan{\displaystyle \frac{\theta }{2}}\right)}^{-1}.$$

Similarly, we obtain

$$Ca{y}^{-1}\left(R_{\theta }^{\mathbf{u}}\right)={\left(R_{\theta }^{\mathbf{u}}+I\right)}^{-1}\left(R_{\theta }^{\mathbf{u}}-I\right)=S_{\mathbf{u}}tan{\displaystyle \frac{\theta }{2}}.$$

□

Example 4.

Given the hyperboloid of one sheet

$$-3y^2+z^2-6xy-4xz-2yz=1$$

which is the unit ${\mathcal{B}}_{\Omega }$-pseudosphere with

$$\Omega =\left[\begin{array}{ccc}1& -3& 0\\ -3& 1& -2\\ 0& -2& 4\end{array}\right],$$

and given the unit ${\mathcal{B}}_{\Omega }$-timelike vector $\mathbf{u}=\left({\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{3}},0\right)$, the matrix $S_{\mathbf{u}}$ with respect to $\mathbf{u}=\left({\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{3}},0\right)$ can be computed as follows:

$$S_{\mathbf{u}}=\left[\begin{array}{ccc}-{\displaystyle \frac{1}{3}}& {\displaystyle \frac{1}{3}}& 1\\ {\displaystyle \frac{2}{3}}& -{\displaystyle \frac{2}{3}}& -1\\ -1& 1& 1\end{array}\right].$$

Using the Cayley formula from (4), we can obtain the ${\mathcal{B}}_{\Omega }$-rotation matrix around the vector $\mathbf{u}$ as follows:

$$R^{\mathbf{u}}=\left[\begin{array}{ccc}0& 1& {\displaystyle \frac{4}{3}}\\ 1& 0& -{\displaystyle \frac{2}{3}}\\ -1& 1& 1\end{array}\right].$$

We can then find the rotation angle using (16), as follows:

$$S_{\mathbf{u}}tan{\displaystyle \frac{\theta }{2}}=Ca{y}^{-1}\left(R_{\theta }^{\mathbf{u}}\right)={\left(R_{\theta }^{\mathbf{u}}+I\right)}^{-1}\left(R_{\theta }^{\mathbf{u}}-I\right)=\left[\begin{array}{ccc}-{\displaystyle \frac{1}{3}}& {\displaystyle \frac{1}{3}}& 1\\ {\displaystyle \frac{2}{3}}& -{\displaystyle \frac{2}{3}}& -1\\ -1& 1& 1\end{array}\right]$$

which provides the ${\mathcal{B}}_{\Omega }$-angle of the ${\mathcal{B}}_{\Omega }$-rotation as $\theta ={\displaystyle \frac{\pi }{2}}$. It is easy to check that the ${\mathcal{B}}_{\Omega }$-rotation matrix $R^{\mathbf{u}}$ can be derived using the axis $\mathbf{u}=\left({\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{3}},0\right)$ and the ${\mathcal{B}}_{\Omega }$-angle $\theta ={\displaystyle \frac{\pi }{2}}$ in the Rodrigues rotation formula from (3).

3.3. The Householder Transformation

In Euclidean space, $\mathbf{v}\in {\mathbb{R}}^n$ for a non-isotropic vector, which means that $〈\mathbf{v},\mathbf{v}〉\ne 0$. Then, the well-known Householder map is provided by

$${\mathcal{H}}_{\mathbf{v}}\left(\mathbf{u}\right)=\mathbf{u}-{\displaystyle \frac{2{\mathbf{vv}}^t}{{\mathbf{v}}^t\mathbf{v}}}\mathbf{u},$$

which determines the reflection about the hyperplane perpendicular to $\mathbf{v}$. Here, we define the ${\mathcal{B}}_{\Omega }$-Householder map that provides the ${\mathcal{B}}_{\Omega }$-reflection, which we then use to produce ${\mathcal{B}}_{\Omega }$-rotation matrices in 3D generalized Lorentzian space. For a non-${\mathcal{B}}_{\Omega }$-isotropic vector $\mathbf{v}$ in the space ${\mathbb{R}}_{{\mathcal{B}}_{\Omega }}^3$, we obtain the ${\mathcal{B}}_{\Omega }$-Householder map from [23,24], as follows:

$${\mathcal{H}}_{\mathbf{v}}^{{\mathcal{B}}_{\Omega }}\left(\mathbf{u}\right)=\mathbf{u}-{\displaystyle \frac{2{\mathbf{vv}}^t\Omega }{{\mathbf{v}}^t\Omega \mathbf{v}}}\mathbf{u}=\left(I-{\displaystyle \frac{2{\mathbf{vv}}^t\Omega }{{\mathbf{v}}^t\Omega \mathbf{v}}}\right)\mathbf{u}.$$

In addition, we obtains the matrix of the map as follows:

$${\mathcal{H}}_{\mathbf{v}}^{{\mathcal{B}}_{\Omega }}={\displaystyle \frac{1}{{\mathcal{B}}_{\Omega }(\mathbf{v},\mathbf{v})}}\left[\begin{array}{ccc}{\mathcal{B}}_{\Omega }(\mathbf{v},\mathbf{v})-2{\delta }_{u_1}^{\prime }& -2{\delta }_{u_1}^{\prime \prime }& -2{\delta }_{u_1}^{\prime \prime \prime }\\ -2{\delta }_{u_2}^{\prime }& {\mathcal{B}}_{\Omega }(\mathbf{v},\mathbf{v})-2{\delta }_{u_2}^{\prime \prime }& -2{\delta }_{u_2}^{\prime \prime \prime }\\ -2{\delta }_{u_3}^{\prime }& -2{\delta }_{u_3}^{\prime \prime }& {\mathcal{B}}_{\Omega }(\mathbf{v},\mathbf{v})-2{\delta }_{u_3}^{\prime \prime \prime }\end{array}\right]$$

where ${\delta }_{u_i}^{\prime }=u_i\left(A{u}_1+D{u}_2+E{u}_3\right)$, ${\delta }_{u_i}^{\prime \prime }=u_i\left(D{u}_1+B{u}_2+F{u}_3\right)$, ${\delta }_{u_i}^{\prime \prime \prime }=u_i\left(E{u}_1+F{u}_2+C{u}_3\right)$. It can be seen that ${\mathcal{H}}_{\mathbf{v}}^{{\mathcal{B}}_{\Omega }}$ is ${\mathcal{B}}_{\Omega }$-symmetric, ${\mathcal{B}}_{\Omega }$-orthogonal, and involutory, and that $det\left({\mathcal{H}}_{\mathbf{v}}^{{\mathcal{B}}_{\Omega }}\right)=-1$. Hence, ${\mathcal{H}}_{\mathbf{v}}^{{\mathcal{B}}_{\Omega }}$ describes ${\mathcal{B}}_{\Omega }$-reflection about the plane that is ${\mathcal{B}}_{\Omega }$-ortho-gonal to $\mathbf{v}$ and passing through the origin. It is known that the composition of an even number of reflections is a rotation by the Cartan–Dieudonné theorem (see [25,26,27]). In addition, it can be seen by the following theorem that the composition of two ${\mathcal{B}}_{\Omega }$-Householder maps determines a ${\mathcal{B}}_{\Omega }$-rotation.

Theorem 5.

The composition of two ${\mathcal{B}}_{\Omega }$-Householder maps determines a ${\mathcal{B}}_{\Omega }$-rotation.

Proof. 

Let ${\mathcal{H}}_{\mathbf{v}}^{{\mathcal{B}}_{\Omega }}$ and ${\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}$ be two ${\mathcal{B}}_{\Omega }$ -reflection matrices. These are both ${\mathcal{B}}_{\Omega }$-orthogonal matrices, and their determinants are equal to $-1$. It can be easily seen that

$${\left({\mathcal{H}}_{\mathbf{v}}^{{\mathcal{B}}_{\Omega }}{\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}\right)}^t\Omega \left({\mathcal{H}}_{\mathbf{v}}^{{\mathcal{B}}_{\Omega }}{\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}\right)={\left({\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}\right)}^t{\left({\mathcal{H}}_{\mathbf{v}}^{{\mathcal{B}}_{\Omega }}\right)}^t\Omega {\mathcal{H}}_{\mathbf{v}}^{{\mathcal{B}}_{\Omega }}{\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}={\left({\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}\right)}^t\Omega {\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}=\Omega$$

and $det\left({\mathcal{H}}_{\mathbf{v}}^{{\mathcal{B}}_{\Omega }}{\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}\right)=1.$ Hence, ${\mathcal{H}}_{\mathbf{v}}^{{\mathcal{B}}_{\Omega }}{\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}$ determines a ${\mathcal{B}}_{\Omega }$-rotation. □

We now present two theorems providing the ${\mathcal{B}}_{\Omega }$-reflection and ${\mathcal{B}}_{\Omega }$-rotation that transform a given vector into another given vector on the same hyperboloid.

Theorem 6.

Let $\mathbf{v}$ and $\mathbf{w}$ be two different non-${\mathcal{B}}_{\Omega }$-isotropic vectors on the same ${\mathcal{B}}_{\Omega }$-sphere which is a hyperboloid of one or two sheets. Then,

$$\mathcal{Y}\left(\mathbf{u}\right)=\left\{\begin{array}{cc}{\mathcal{H}}_{\mathbf{v}}^{{\mathcal{B}}_{\Omega }}\left(\mathbf{u}\right)\hfill & \mathrm{if}\mathbf{v}+\mathbf{w}=\mathbf{0}\hfill \\ {\mathcal{H}}_{\mathbf{v}-\mathbf{w}}^{{\mathcal{B}}_{\Omega }}\left(\mathbf{u}\right)\hfill & \mathrm{if}\mathbf{v}+\mathbf{w}\ne \mathbf{0}\mathrm{and}\mathbf{v}-\mathbf{w}\mathrm{is}\mathrm{not}{\mathcal{B}}_{\Omega }-\mathrm{isotropic}\hfill \\ {\mathcal{H}}_{\mathbf{v}+\mathbf{w}}^{{\mathcal{B}}_{\Omega }}\left(-\mathbf{u}\right)\hfill & \mathrm{if}\mathbf{v}+\mathbf{w}\ne \mathbf{0}\mathrm{and}\mathbf{v}-\mathbf{w}\mathrm{is}{\mathcal{B}}_{\Omega }-\mathrm{isotropic}\hfill \end{array}\right.$$

is the ${\mathcal{B}}_{\Omega }$-reflection such that $\mathcal{Y}\left(\mathbf{v}\right)=\mathbf{w}$.

Proof. 

Consider the given transformation:

(i)

Suppose that $\mathbf{v}+\mathbf{w}=\mathbf{0}$; then,

$${\mathcal{H}}_{\mathbf{v}}^{{\mathcal{B}}_{\Omega }}\left(\mathbf{v}\right)=\mathbf{v}-{\displaystyle \frac{2{\mathbf{vv}}^t\Omega }{{\mathbf{v}}^t\Omega \mathbf{v}}}\mathbf{v}=-\mathbf{v}=\mathbf{w}.$$

(ii)

Assume that $\mathbf{v}+\mathbf{w}\ne \mathbf{0}$. It is clear that $\mathbf{v}-\mathbf{w}\ne \mathbf{0}$. If $\mathbf{v}-\mathbf{w}$ is not ${\mathcal{B}}_{\Omega }$-null; thus,

$$\begin{array}{ccc}\hfill {\mathcal{H}}_{\mathbf{v}-\mathbf{w}}^{{\mathcal{B}}_{\Omega }}\left(\mathbf{v}\right)& =& \mathbf{v}-{\displaystyle \frac{2\left(\mathbf{v}-\mathbf{w}\right){\left(\mathbf{v}-\mathbf{w}\right)}^t\Omega }{{\left(\mathbf{v}-\mathbf{w}\right)}^t\Omega \left(\mathbf{v}-\mathbf{w}\right)}}\mathbf{v}\hfill \\ & =& \mathbf{v}-{\displaystyle \frac{2\left(\mathbf{v}-\mathbf{w}\right)\left({\mathcal{B}}_{\Omega }(\mathbf{v},\mathbf{v})-{\mathbf{w}}^t\Omega \mathbf{v}\right)}{2{\mathcal{B}}_{\Omega }(\mathbf{v},\mathbf{v})-2{\mathbf{w}}^t\Omega \mathbf{v}}}\hfill \\ & =& \mathbf{v}-\left(\mathbf{v}-\mathbf{w}\right)=\mathbf{w}.\hfill \end{array}$$

(iii)

Assume that $\mathbf{v}+\mathbf{w}\ne \mathbf{0}$ and that $\mathbf{v}-\mathbf{w}$ is ${\mathcal{B}}_{\Omega }$-null. Using ${\mathcal{B}}_{\Omega }(\mathbf{v}-\mathbf{w},\mathbf{v}-\mathbf{w})=0$, we can see that

$${\mathcal{B}}_{\Omega }(\mathbf{v}+\mathbf{w},\mathbf{v}+\mathbf{w})\ne 0.$$

Then, $\mathbf{v}+\mathbf{w}$ is not ${\mathcal{B}}_{\Omega }$-isotropic, and we have

$$\mathcal{H}\left(\mathbf{v}\right)={\mathcal{H}}_{\mathbf{v}+\mathbf{w}}^{{\mathcal{B}}_{\Omega }}\left(-\mathbf{v}\right)=\mathbf{w}$$

by similar calculations.

□

Theorem 7.

Let $\mathbf{v}$ and $\mathbf{w}$ be two different vectors on the same ${\mathcal{B}}_{\Omega }$-sphere which is a hyperboloid of one or two sheets. Then,

$$\mathcal{R}\left(\mathbf{u}\right)=\left\{\begin{array}{cc}{\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}{\mathcal{H}}_{\mathbf{v}}^{{\mathcal{B}}_{\Omega }}\left(-\mathbf{u}\right)\hfill & \mathrm{if}\mathbf{v}+\mathbf{w}=\mathbf{0}\hfill \\ {\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}{\mathcal{H}}_{\mathbf{v}-\mathbf{w}}^{{\mathcal{B}}_{\Omega }}\left(-\mathbf{u}\right)\hfill & \mathrm{if}\mathbf{v}+\mathbf{w}\ne \mathbf{0}\mathrm{and}\mathbf{v}-\mathbf{w}\mathrm{is}\mathrm{not}{\mathcal{B}}_{\Omega }-\mathrm{isotropic}\hfill \\ {\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}{\mathcal{H}}_{\mathbf{v}+\mathbf{w}}^{{\mathcal{B}}_{\Omega }}\left(\mathbf{u}\right)\hfill & \mathrm{if}\mathbf{v}+\mathbf{w}\ne \mathbf{0}\mathrm{and}\mathbf{v}-\mathbf{w}\mathrm{is}{\mathcal{B}}_{\Omega }-\mathrm{isotropic}\hfill \end{array}\right.$$

is the ${\mathcal{B}}_{\Omega }$-rotation such that $\mathcal{R}\left(\mathbf{v}\right)=\mathbf{w}$.

Proof. 

It is clear that $\mathcal{R}$ is a ${\mathcal{B}}_{\Omega }$-rotation. In addition, we can use Theorem 4 to obtain the following:

(i)

If $\mathbf{v}+\mathbf{w}=\mathbf{0}$, then

$${\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}{\mathcal{H}}_{\mathbf{v}}^{{\mathcal{B}}_{\Omega }}\left(-\mathbf{v}\right)={\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}\left(-\mathbf{w}\right)=\mathbf{w}.$$

(ii)

If $\mathbf{v}+\mathbf{w}\ne \mathbf{0}$ and $\mathbf{v}-\mathbf{w}$ is not ${\mathcal{B}}_{\Omega }$-null, then

$${\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}{\mathcal{H}}_{\mathbf{v}-\mathbf{w}}^{{\mathcal{B}}_{\Omega }}\left(-\mathbf{v}\right)={\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}(-\mathbf{w})=\mathbf{w}.$$

(iii)

If $\mathbf{v}+\mathbf{w}\ne \mathbf{0}$ and $\mathbf{v}-\mathbf{w}$ is ${\mathcal{B}}_{\Omega }$-null, then $\mathbf{v}+\mathbf{w}$ is not ${\mathcal{B}}_{\Omega }$-null and

$${\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}{\mathcal{H}}_{\mathbf{v}+\mathbf{w}}^{{\mathcal{B}}_{\Omega }}\left(\mathbf{v}\right)={\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}(-\mathbf{w})=\mathbf{w}.$$

□

Example 5.

Consider the hyperboloid of one sheet

$$-3y^2+z^2-6xy-4xz-2yz=2$$

which is the ${\mathcal{B}}_{\Omega }$-pseudosphere with $r=\sqrt{2}$, where

$$\Omega =\left[\begin{array}{ccc}0& -3& -2\\ -3& -3& -1\\ -2& -1& 1\end{array}\right].$$

Given two vectors on the hyperboloid $\mathbf{v}=(-1/4,0,1)$ and $\mathbf{w}=\left(-5/6,1,0\right)$, we can find the ${\mathcal{B}}_{\Omega }$-rotation $\mathcal{R}$ such that $\mathcal{R}\left(\mathbf{v}\right)=\mathbf{w}$. Because $\mathbf{v}+\mathbf{w}=\left(-13/12,1,1\right)\ne \mathbf{0}$ and $\mathbf{v}-\mathbf{w}$ is not ${\mathcal{B}}_{\Omega }$-isotropic, we can use the ${\mathcal{B}}_{\Omega }$-Householder transformations

$${\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}=\left[\begin{array}{ccc}-{\displaystyle \frac{3}{2}}& -{\displaystyle \frac{5}{12}}& {\displaystyle \frac{5}{9}}\\ 3& {\displaystyle \frac{3}{2}}& -{\displaystyle \frac{2}{3}}\\ 0& 0& 1\end{array}\right]\mathrm{and}{\mathcal{H}}_{\mathbf{v}+\mathbf{w}}^{{\mathcal{B}}_{\Omega }}=\left[\begin{array}{ccc}-{\displaystyle \frac{24}{41}}& -{\displaystyle \frac{39}{164}}& {\displaystyle \frac{169}{246}}\\ {\displaystyle \frac{60}{41}}& {\displaystyle \frac{50}{41}}& -{\displaystyle \frac{26}{41}}\\ {\displaystyle \frac{60}{41}}& {\displaystyle \frac{9}{41}}& {\displaystyle \frac{15}{41}}\end{array}\right]$$

to determine the matrix $\mathcal{R}={\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}{\mathcal{H}}_{\mathbf{v}+\mathbf{w}}^{{\mathcal{B}}_{\Omega }}$ as follows:

$$\mathcal{R}=\left[\begin{array}{ccc}{\displaystyle \frac{133}{123}}& -{\displaystyle \frac{29}{984}}& -{\displaystyle \frac{277}{492}}\\ -{\displaystyle \frac{22}{41}}& {\displaystyle \frac{159}{164}}& {\displaystyle \frac{71}{82}}\\ {\displaystyle \frac{60}{41}}& {\displaystyle \frac{9}{41}}& {\displaystyle \frac{15}{41}}\end{array}\right].$$

It is easy to ensure that ${\mathcal{R}}^t\Omega \mathcal{R}=\Omega$, $det\left(\mathcal{R}\right)=1$, and $\mathcal{R}\left(\mathbf{v}\right)=\mathbf{w}$. In addition, the axis of the ${\mathcal{B}}_{\Omega }$-rotation is

$$\mathbf{w}\times \mathbf{v}=\left(\frac{7}{36},-\frac{19}{18},\frac{1}{12}\right)$$

and the ${\mathcal{B}}_{\Omega }$-angle is ${cos}^{-1}{\displaystyle \frac{17}{24}}$. Note that the rotation plane is

$$-12x-10y-3z=0,$$

which is ${\mathcal{B}}_{\Omega }$-orthogonal to the axis, and that the rotation is elliptic, as its cross-section of the hyperboloid is an ellipse. It is easy to check that the ${\mathcal{B}}_{\Omega }$-rotation matrix $R_{\theta }^{\mathbf{u}}$ can be derived by using the axis $\mathbf{u}=\left(\frac{7}{36},-\frac{19}{18},\frac{1}{12}\right)$ and the ${\mathcal{B}}_{\Omega }$-angle $\theta ={cos}^{-1}{\displaystyle \frac{17}{24}}=1.1463$ in the Rodrigues rotation formula from (3).

We can continue the same example with different vectors such that the rotation occurs in a hyperbola on the same hyperboloid. Consider the ${\mathcal{B}}_{\Omega }$-rotation transforming $\mathbf{v}=(-1/4,0,1)$ to ${\mathbf{w}}^{\prime }=\left(-5/14,1,2\right)$ on the hyperboloid. Similarly, we can determine the matrix ${\mathcal{H}}_{{\mathbf{w}}^{\prime }}^{{\mathcal{B}}_{\Omega }}{\mathcal{H}}_{\mathbf{v}+{\mathbf{w}}^{\prime }}^{{\mathcal{B}}_{\Omega }}$ for vectors ${\mathbf{w}}^{\prime }$ and $\mathbf{v}+{\mathbf{w}}^{\prime }=\left(-17/28,1,3\right)$. Using ${\mathcal{B}}_{\Omega }$-Householder transformation matrix, we obtain

$${\mathcal{H}}_{{\mathbf{w}}^{\prime }}^{{\mathcal{B}}_{\Omega }}=\left[\begin{array}{ccc}-{\displaystyle \frac{3}{2}}& -{\displaystyle \frac{275}{196}}& {\displaystyle \frac{30}{49}}\\ 7& {\displaystyle \frac{69}{14}}& -{\displaystyle \frac{12}{7}}\\ 14& {\displaystyle \frac{55}{7}}& -{\displaystyle \frac{17}{7}}\end{array}\right]\mathrm{and}{\mathcal{H}}_{\mathbf{v}+{\mathbf{w}}^{\prime }}^{{\mathcal{B}}_{\Omega }}=\left[\begin{array}{ccc}0& -{\displaystyle \frac{13}{28}}& {\displaystyle \frac{5}{14}}\\ {\displaystyle \frac{28}{17}}& {\displaystyle \frac{30}{17}}& -{\displaystyle \frac{10}{17}}\\ {\displaystyle \frac{84}{17}}& {\displaystyle \frac{39}{17}}& -{\displaystyle \frac{13}{17}}\end{array}\right].$$

Then, we obtain

$${\mathcal{H}}_{{\mathbf{w}}^{\prime }}^{{\mathcal{B}}_{\Omega }}{\mathcal{H}}_{\mathbf{v}+{\mathbf{w}}^{\prime }}^{{\mathcal{B}}_{\Omega }}=\left[\begin{array}{ccc}{\displaystyle \frac{5}{7}}& -{\displaystyle \frac{3}{8}}& -{\displaystyle \frac{5}{28}}\\ -{\displaystyle \frac{6}{17}}& {\displaystyle \frac{103}{68}}& {\displaystyle \frac{31}{34}}\\ {\displaystyle \frac{16}{17}}& {\displaystyle \frac{61}{34}}& {\displaystyle \frac{38}{17}}\end{array}\right].$$

Again, it is easy to check that $\mathcal{R}\left(\mathbf{v}\right)={\mathbf{w}}^{\prime }$, $det\left(\mathcal{R}\right)=1$ and ${\mathcal{R}}^t\Omega \mathcal{R}=\Omega$. In addition, the axis of the ${\mathcal{B}}_{\Omega }$-rotation is

$${\mathbf{w}}^{\prime }{\times }_{\Omega }\mathbf{v}=\left(\frac{51}{28},-\frac{33}{14},\frac{57}{28}\right)$$

and the ${\mathcal{B}}_{\Omega }$-angle is ${cosh}^{-1}{\displaystyle \frac{97}{56}}$. Note that the rotation plane is

$$12y-84x-21z=0,$$

which is ${\mathcal{B}}_{\Omega }$-orthogonal to the axis, and that the rotation is hyperbolic, as its cross-section of the hyperboloid is a hyperbola. In fact, it is possible to find a ${\mathcal{B}}_{\Omega }$-rotation example that occurs on two parallel lines, since the intersection of the plane and a hyperboloid of one sheet can also be two parallel lines. It is easy to check that the ${\mathcal{B}}_{\Omega }$-rotation matrix $R_{\theta }^{\mathbf{u}}$ can be derived by using the axis $\mathbf{u}=(\frac{51}{28},-\frac{33}{14},\frac{57}{28})$ and the ${\mathcal{B}}_{\Omega }$-angle $\theta ={cosh}^{-1}{\displaystyle \frac{97}{56}}=1.1463$ in the Rodrigues rotation formula from (2).

Example 6.

Given the hyperboloid of two sheets

$$-3y^2+z^2-6xy-4xz-2yz=-2,$$

which is a central ${\mathcal{B}}_{\Omega }$-sphere with $r=\sqrt{2}$, we have the same Ω as in the example from (5). For the vectors $\mathbf{v}=(3/4,0,1)$ and $\mathbf{w}=\left(-1/6,1,0\right)$ on the hyperboloid, we need to obtain the matrix $\mathcal{R}={\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}{\mathcal{H}}_{\mathbf{v}+\mathbf{w}}^{{\mathcal{B}}_{\Omega }}$, where $\mathcal{R}\left(\mathbf{v}\right)=\mathbf{w}$. Using the ${\mathcal{B}}_{\Omega }$-Householder transformation matrix, we obtain

$${\mathcal{H}}_{\mathbf{w}}^{{\mathcal{B}}_{\Omega }}=\left[\begin{array}{ccc}{\displaystyle \frac{3}{2}}& {\displaystyle \frac{5}{12}}& {\displaystyle \frac{1}{9}}\\ -3& -{\displaystyle \frac{3}{2}}& -{\displaystyle \frac{2}{3}}\\ 0& 0& 1\end{array}\right]\mathrm{and}{\mathcal{H}}_{\mathbf{v}+\mathbf{w}}^{{\mathcal{B}}_{\Omega }}=\left[\begin{array}{ccc}{\displaystyle \frac{24}{59}}& -{\displaystyle \frac{161}{236}}& -{\displaystyle \frac{49}{354}}\\ -{\displaystyle \frac{60}{59}}& -{\displaystyle \frac{10}{59}}& -{\displaystyle \frac{14}{59}}\\ -{\displaystyle \frac{60}{59}}& -{\displaystyle \frac{69}{59}}& {\displaystyle \frac{45}{59}}\end{array}\right].$$

Then, we have

$$R_{\theta }^{\mathbf{u}}=\left[\begin{array}{ccc}{\displaystyle \frac{13}{177}}& -{\displaystyle \frac{1733}{1416}}& -{\displaystyle \frac{157}{708}}\\ {\displaystyle \frac{58}{59}}& {\displaystyle \frac{727}{236}}& {\displaystyle \frac{31}{118}}\\ -{\displaystyle \frac{60}{59}}& -{\displaystyle \frac{69}{59}}& {\displaystyle \frac{45}{59}}\end{array}\right].$$

It is easy to ensure that $\mathcal{R}\left(\mathbf{v}\right)=\mathbf{w}$, $det\left(\mathcal{R}\right)=1$ and ${\mathcal{R}}^t\Omega \mathcal{R}=\Omega$. In addition, the axis of the ${\mathcal{B}}_{\Omega }$-rotation is

$$\mathbf{v}{\times }_{\Omega }\mathbf{w}=\left(-\frac{11}{36},-\frac{1}{18},\frac{19}{12}\right)$$

and the ${\mathcal{B}}_{\Omega }$-angle is ${cosh}^{-1}\left(35/24\right)$. It is easy to check that the ${\mathcal{B}}_{\Omega }$-rotation matrix $R_{\theta }^{\mathbf{u}}$ can be derived using the axis $\mathbf{u}=\left(-\frac{11}{36},-\frac{1}{18},\frac{19}{12}\right)$ and the ${\mathcal{B}}_{\Omega }$-angle $\theta ={cosh}^{-1}\left(35/24\right)=0.92418$ in the Rodrigues rotation formula from (2).

4. Conclusions

In this study, we have investigated elliptic and hyperbolic rotational motions on arbitrary hyperboloids of one or two sheets as rotations in the generalized Minkowski 3-space using the Cayley, Householder, and Rodrigues transformations. Without relying on affine transformations or split quaternions, we obtain formulas that describe these rotational motions in three-dimensional space. These types of motions can have real-life applications in several areas: computer graphics, where simulating non-Euclidean transformations can enhance visual realism; robotics, where it is important to model the movements of robots; in aerospace engineering, where understanding rotational dynamics in relativistic settings is crucial for navigation systems; and in theoretical physics, particularly special relativity and general relativity, where space–time is modeled as Minkowski 3-space. For future work, generalized lightlike vectors should primarily be considered as axes of rotations. This can further contribute to our understanding of lightlike trajectories, as the axes of the rotations in this work are limited to generalized spacelike and timelike vectors. Future work could also consider interpolations, which are closely related to rotational transformations.