Homothetic Motions and Surfaces in E$^{4}$

Abstract

In this paper, we determine a surface $M$ by means of homothetic motion in $\mathbb {R}^{4}$ and reparametrize this surface $M$ with bicomplex numbers. Also, by using curves and surfaces which are obtained by homothetic motion, we give some special subgroups of the Lie group $P$.

1 Introduction

A one-parameter homothetic motion of a rigid body in Euclidean n-space is given analytically by

$$\begin{aligned} X^{\prime }=h(t)A(t)X+C(t) \end{aligned}$$

(1)

in which $X^{\prime }$ and $X$ are the position vectors of the same point with respect to the rectangular coordinate frames of the fixed space $R^{\prime }$ and the moving space $R,$ respectively. $A$ is an orthonormal $n\times n$ matrix, $C$ is a translation vector, and $h$ is the homothetic scale of the motion. Also $h,\,A$, and $C$ are continuously differentiable function of a real parameter $t.$ If we take an arbitrary position vector of a curve instead of the point $X$ at one-parameter homothetic motion equation which is given by (1), we obtain a surface.

In mathematics, a Lie group is a group which is also a differentiable manifold with the property that the group operations are differentiable. A manifold $M$ carrying n linearly independent non-vanishing vector fields is called parallelisable and a Lie group is parallelisable. The spheres that admit the structure of a Lie group are the 0-sphere $S^{0}$ (real numbers with absolute value $1$), the circle $S^{1}$ (complex numbers with absolute value $1$), the 3-sphere $S^{3}$ (the set of quaternions of unit form), and $S^{7}.$ For even $n>1\,S^{n}$ is not a Lie group because it can not be parallelisable as a differentiable manifold. Thus, $S^{n}$ is parallelisable if and only $n=0,1,3,7$.

Özkaldı and Yaylı [7] showed that a hyperquadric $P$ in $\mathbb {R}^{4}$ is a Lie group by using bicomplex number product. They determined some special subgroups of this Lie group $P,$ by using the tensor product surfaces of Euclidean planar curves.

In this paper, we determine a homothetic motion by using a rotation matrix which is given by Moore [5] and obtain a surface $M$ by means of this homothetic motion in $\mathbb {R}^{4}$. If we take as the homothetic scale $h(t)=1$ and the translation vector $C(t)=0$, we obtain a rotational surface [5, 9]. Even if, in special cases, we get some tensor product surfaces by means of this homothetic motion [3, 7]. We reparametrize this surface $M$ with bicomplex number product and addition. To establish group structure on the surface is quite difficult. How should we choose the position vector of the curve at homothetic motion given by (2) that the surface $M$ be a Lie subgroup of the hyperquadric $P$. In this study, we answer this question and by using surface $M$ which is obtained by homothetic motion, we determine some special Lie subgroups of this Lie group $P$. Furthermore, we mention $C^{\infty }$ -action of the Lie group $P$ onto the manifold $\mathbb {R}^{4}$ and define an action of $\mathbb {R}$ on $P$ by using orthonormal matrix at homothetic motion. Also, we determine a Lie subgroup of $P$ with this action and give some results.

2 Preliminaries

Bicomplex number is defined by the basis $\left\{ 1,i,j,ij\right\}$ where $i,j,ij$ satisfy $i^{2}=-1,\,j^{2}=-1,\,ij=ji.$ Thus, any bicomplex number $x$ can be expressed as $x=x_{1}1+x_{2}i+x_{3}j+x_{4}ij$, $\forall x_{1},x_{2},x_{3},x_{4}\in \mathbb {R}.$ We denote the set of bicomplex numbers by $C_{2}.$ For any $x=x_{1}1+x_{2}i+x_{3}j+x_{4}ij$ and $y=y_{1}1+y_{2}i+y_{3}j+y_{4}ij$ in $C_{2}$ the bicomplex number addition is defined as

$$\begin{aligned} x+y=\left( x_{1}+y_{1}\right) +\left( x_{2}+y_{2}\right) i+\left( x_{3}+y_{3}\right) j+\left( x_{4}+y_{4}\right) ij. \end{aligned}$$

The multiplication of a bicomplex number $x=x_{1}1+x_{2}i+x_{3}j+x_{4}ij$ by a real scalar $\lambda$ is defined as

$$\begin{aligned} \lambda x=\lambda x_{1}1+\lambda x_{2}i+\lambda x_{3}j+\lambda x_{4}ij. \end{aligned}$$

So the set of bicomplex numbers $C_{2}$ is a vector space over the real numbers with above the addition and scalar multiplication operations

Let us denote the bicomplex number product over $C_2$ by $\times$. The bicomplex number product is given by

$$\begin{aligned} x\times y&= \left( x_{1}y_{1}-x_{2}y_{2}-x_{3}y_{3}+x_{4}y_{4}\right) +\left( x_{1}y_{2}+x_{2}y_{1}-x_{3}y_{4}-x_{4}y_{3}\right) i \\&+\left( x_{1}y_{3}+x_{3}y_{1}-x_{2}y_{4}-x_{4}y_{2}\right) j+\left( x_{1}y_{4}+x_{4}y_{1}+x_{2}y_{3}+x_{3}y_{2}\right) ij. \end{aligned}$$

the set of bicomplex numbers $C_2$ is a real algebra with above the bicomplex number product operation $\times$.

We can consider any bicomplex number as $4\times 4$ real matrix as follows:

$$\begin{aligned} Q=\left\{ \left( \begin{array}{cccc} x_{1} &{} -x_{2} &{} -x_{3} &{} x_{4} \\ x_{2} &{} x_{1} &{} -x_{4} &{} -x_{3} \\ x_{3} &{} -x_{4} &{} x_{1} &{} -x_{2} \\ x_{4} &{} x_{3} &{} x_{2} &{} x_{1} \end{array} \right) ;\qquad \qquad x_{i}\in \mathbb {R}, \qquad 1\le i\le 4\right\} . \end{aligned}$$

The set $Q$ together with matrix addition and scalar matrix multiplication is a real vector space. Furthermore, the vector space together with matrix product is an algebra [7].

The transformation

$$\begin{aligned} g:C_{2}\rightarrow Q \end{aligned}$$

given by

$$\begin{aligned} g\left( x=x_{1}1+x_{2}i+x_{3}j+x_{4}ij\right) =\left( \begin{array}{cccc} x_{1} &{} -x_{2} &{} -x_{3} &{} x_{4} \\ x_{2} &{} x_{1} &{} -x_{4} &{} -x_{3} \\ x_{3} &{} -x_{4} &{} x_{1} &{} -x_{2} \\ x_{4} &{} x_{3} &{} x_{2} &{} x_{1} \end{array} \right) \end{aligned}$$

is one to one and onto. Morever $\forall x,y\in C_{2}$ and $\lambda \in \mathbb {R},$ we have

$$\begin{aligned} g\left( x+y\right)&= g\left( x\right) +g\left( y\right) \\ g\left( \lambda x\right)&= \lambda g\left( x\right) \\ g\left( xy\right)&= g\left( x\right) g\left( y\right) . \end{aligned}$$

Thus the algebras $C_{2}$ and $Q$ are isomorphic.

Let $x\in C_{2}.$ Then $x$ can be expressed as $x=\left( x_{1}+x_{2}i\right) +\left( x_{3}+x_{4}i\right) j.$ In that case, there is three different conjugations for bicomplex numbers as follows:

$$\begin{aligned} x^{t_{1}}&= \left[ \left( x_{1}+x_{2}.i\right) +\left( x_{3}+x_{4}.i\right) j\right] ^{t_{1}}=\left( x_{1}-x_{2}.i\right) +\left( x_{3}-x_{4}.i\right) j \\ x^{t_{2}}&= \left[ \left( x_{1}+x_{2}.i\right) +\left( x_{3}+x_{4}.i\right) j\right] ^{t_{2}}=\left( x_{1}+x_{2}.i\right) -\left( x_{3}+x_{4}.i\right) j \\ x^{t_{3}}&= \left[ \left( x_{1}+x_{2}.i\right) +\left( x_{3}+x_{4}.i\right) j\right] ^{t_{3}}=\left( x_{1}-x_{2}.i\right) -\left( x_{3}-x_{4}.i\right) j . \end{aligned}$$

And we can write

$$\begin{aligned} xx^{t_{1}}&= \left( x_{1}^{2}+x_{2}^{2}-x_{3}^{2}-x_{4}^{2}\right) +2\left( x_{1}x_{3}+x_{2}x_{4}\right) j \\ xx^{t_{2}}&= \left( x_{1}^{2}-x_{2}^{2}+x_{3}^{2}-x_{4}^{2}\right) +2\left( x_{1}x_{2}+x_{3}x_{4}\right) i \\ xx^{t_{3}}&= \left( x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}\right) +2\left( x_{1}x_{4}-x_{2}x_{3}\right) ij. \end{aligned}$$

3 Homothetic Motions and Surfaces in $E^{4}$

In this section, we define a surface by using the homothetic motion as follows:

$$\begin{aligned} \varphi \left( t,s\right) =h(t) \begin{pmatrix} \cos t &{} -\sin t &{} 0 &{} 0 \\ \sin t &{} \cos t &{} 0 &{} 0 \\ 0 &{} 0 &{} \cos t &{} -\sin t \\ 0 &{} 0 &{} \sin t &{} \cos t \end{pmatrix} \left( \begin{array}{c} \alpha _{1}(s) \\ \alpha _{2}(s) \\ \alpha _{3}(s) \\ \alpha _{4}(s) \end{array} \right) +\left( \begin{array}{c} C_{1}\left( t\right) \\ C_{2}\left( t\right) \\ C_{3}\left( t\right) \\ C_{4}\left( t\right) \end{array} \right) , \end{aligned}$$

(2)

where $h(t)$ is the homothetic scale of the motion, $C\left( t\right) =(C_{1}\left( t\right) ,C_{2}\left( t\right) ,C_{3}\left( t\right) ,C_{4}\left( t\right) )$ is the translation vector and $\alpha \left( s\right) =\left( \alpha _{1}\left( s\right) ,\alpha _{2}\left( s\right) ,\alpha _{3}\left( s\right) ,\alpha _{4}\left( s\right) \right)$ is a profile curve.

Now, we can reparametrize this surface by using bicomplex number product and addition.

Proposition 1

Let $\varphi :M\rightarrow E^{4}$ be an immersion of a surface $M$ in the Euclidean 4-space. If $M$ is a surface in $E^{4}$ given by the parametrization (2), then $M$ can be reparametrized by $\varphi \left( t,s\right) =\beta \left( t\right) \times \alpha \left( s\right) +C\left( t\right) ,$ where “$\times$” bicomplex product, “$+$” bicomplex addition, $\beta \left( t\right) =(h(t)\cos t,h(t)\sin t,0,0),\,\alpha \left( s\right) =\left( \alpha _{1}(s),\alpha _{2}(s),\alpha _{3}(s),\alpha _{4}(s)\right)$ are the curves and $C\left( t\right) =(C_{1}\left( t\right) ,C_{2}\left( t\right) ,C_{3}\left( t\right) ,C_{4}\left( t\right) )$ is the translation vector.

Proof

We can consider the curves $\beta ,\,\alpha$ and the translation vector $C$ as bicomplex numbers. Then we can rewrite them as follows:

$$\begin{aligned} \beta \left( t\right)&= h(t)\cos t+\left( h(t)\sin t\right) i \\ \alpha \left( s\right)&= \alpha _{1}(s)+\alpha _{2}(s)i+\alpha _{3}(s)j+\alpha _{4}(s)ij \\ C\left( t\right)&= C_{1}\left( t\right) +C_{2}\left( t\right) i+C_{3}\left( t\right) j+C_{4}\left( t\right) ij. \end{aligned}$$

By using the bicomplex product and addition, we obtain $\varphi \left( t,s\right) =\beta \left( t\right) \times \alpha \left( s\right) +C\left( t\right)$.$\square$

Corollary 1

Let $M_{\beta \left( t\right) }$ be the matrix representation of bicomplex $\beta \left( t\right) =h(t)\cos t+\left( h(t)\sin t\right) i.$ Then we get the surface $M$ given by the parametrization (2) as $\varphi \left( t,s\right) =M_{\beta \left( t\right) }\alpha \left( s\right) +C\left( t\right)$.

The surface $M$ given by the parametrization (2) is reparametrized as bicomplex product of two curves in four dimensional Euclidean space. Now, we can reparametrize this surface $M$ as bicomplex product of a curve and a surface

Corollary 2

Let $\varphi :M\rightarrow E^{4}$ be an immersion of a surface $M$ in the Euclidean 4-space and $M$ be a surface given by the parametrization (2). Then, the surface $M$ can be reparametrized by $\varphi \left( t,s\right) =\gamma \left( t\right) \times r\left( t,s\right) +C\left( t\right) ,$ where $\gamma \left( t\right) =\left( \cos t,\sin t,0,0\right)$ is a circle$,\,r\left( t,s\right) =h(t)\alpha (s)$ is a surface and $C\left( t\right) =(C_{1}\left( t\right) ,C_{2}\left( t\right) ,C_{3}\left( t\right) ,C_{4}\left( t\right) )$ is the translation vector$.$

Corollary 3

Let $M_{\gamma \left( t\right) }$ be the matrix representation of bicomplex $\gamma \left( t\right) =\cos t+\left( \sin t\right) i.$ Then the surface $M$ given by the parametrization (2) can be written as $\varphi \left( t,s\right) =M_{\gamma \left( t\right) }r\left( t,s\right) +C\left( t\right) .$

Remark 1

Let $M$ be a surface in $E^{4}$ given by the parametrization (2). In particular, if we take as the homothetic function $h(t)=1$ and the translation vector $C\left( t\right) =0,$ we obtain a rotation surface given by Moore [5]$.$

Remark 2

Let $M$ be a surface in $E^{4}$ given by the parametrization (2). In particular, if we take as the homothetic function $h(t)=1$ and the translation vector $C\left( t\right) =0$ and the profile curve $\alpha \left( s\right) =\left( r(s)\cos s,0,r(s)\sin s,0\right) ,$ we obtain a rotation surface which is called Vranceanu surface [9].

4 Lie Groups, $C^{\infty }$ Action of the Lie Groups and Some Special Lie Subgroups

4.1 Lie Groups

In this subsection, by Theorem 1, we mention that the hyperquadric $P$ is a Lie group with bicomplex number product.

$$\begin{aligned} P=\left\{ x=\left( x_{1},x_{2},x_{3},x_{4}\right) \ne 0;\qquad x_{1}x_{4}=x_{2}x_{3}\right\} . \end{aligned}$$

We can consider $P$ as the set of bicomplex numbers as follows:

$$\begin{aligned} P=\left\{ x=x_{1}1+x_{2}i+x_{3}j+x_{4}ij\;\,\, x_{1}x_{4}=x_{2}x_{3},\,x\ne 0\right\} . \end{aligned}$$

Also, let the matrix representation of $P$ be given by

$$\begin{aligned} \tilde{P}=\left\{ M_{x}=\left( \begin{array}{cccc} x_{1} &{} -x_{2} &{} -x_{3} &{} x_{4} \\ x_{2} &{} x_{1} &{} -x_{4} &{} -x_{3} \\ x_{3} &{} -x_{4} &{} x_{1} &{} -x_{2} \\ x_{4} &{} x_{3} &{} x_{2} &{} x_{1} \end{array} \right) ;\,x_{1}x_{4}=x_{2}x_{3},\,x\ne 0\right\} . \end{aligned}$$

Theorem 1

([7]) The set of $P$ together with the bicomplex number product is a Lie group

Proof

$\tilde{P}$ is a differentiable manifold and at the same time a group with group operation given by matrix multiplication. The group function

$$\begin{aligned} .:\tilde{P}\times \tilde{P}\rightarrow \tilde{P} \end{aligned}$$

defined by $\left( x,y\right) \rightarrow x.y$ is differentiable. So $($P$,.)$ can be made a Lie group so that $g$ is a isomorphism.$\square$

Let us denote the set of all unit bicomplex numbers on P by $P_1$. We can consider $P_1$ as follow:

$$\begin{aligned} P_{1}=\left\{ x\in P\,;\qquad \left\| x\right\| _{t_{3}}=1\right\} . \end{aligned}$$

And we denote $\tilde{P}_{1}$ the matrix form of the group $P_{1}$

$$\begin{aligned} \tilde{P}_{1}=\left\{ x\in \tilde{P}\,;\,\left\| x\right\| _{t_{3}}=1\right\} . \end{aligned}$$

$P_{1}$ is a subgroup of $P$ with the group operation of bicomplex multiplication

Lemma 1

([7]) $P_{1}$ is 2-dimensional Lie subgroup of $P$.

Remark 3

$S^{3}$ is a Lie group with the quaternion multiplication. We can write the set $P_{1}$ as $P_{1}=P\cap S^{3}$ and $P_{1}$ is a Lie group with bicomplex multiplication. Even though $P_{1}$ is a subset of the sphere $S^{3}$ and $P_{1}$ is a Lie group, $P_{1}$ is not a Lie subgroup of $S^{3}$.

4.2 $\mathbf {C}^{\infty }$ Action of the Lie Groups

In this subsection, we mention $C^{\infty }-$actions of the Lie groups $\tilde{P}$ and $\tilde{P}_{1}$ onto the manifold $\mathbb {R}^{4}$ and we define an action of $\mathbb {R}$ on $\tilde{P}$ and $\tilde{P}_{1}$ Lie groups. We give some special Lie subgroups of $P$ and $P_{1}$ by means of the action of $\mathbb {R}$ on $\tilde{P}$ and $\tilde{P}_{1}.$

Let us consider the mapping

$$\begin{aligned} \theta :\tilde{P}\times \mathbb {R}^{4}\rightarrow \mathbb {R}^{4} \end{aligned}$$

for any $A\in \tilde{P}$ and $X\in \mathbb {R}^{4}$ given by

$$\begin{aligned} (A,X)\rightarrow \theta (A,X)=AX. \end{aligned}$$

Theorem 2

([7]) The mapping $\theta ,$ defined above, is a $C^{\infty }-$action of the Lie group $\tilde{P}$ onto the manifold $\mathbb {R}^{4}.$ This action is transitive and effective.

Theorem 3

Let $f:\mathbb {R\rightarrow }\tilde{P}$ be the mapping which sends every $t\in \mathbb {R}$ to

$$\begin{aligned} t\rightarrow f(t)=e^{bt}\left( \begin{array}{cccc} \cos t &{} -\sin t &{} 0 &{} 0 \\ \sin t &{} \cos t &{} 0 &{} 0 \\ 0 &{} 0 &{} \cos t &{} -\sin t \\ 0 &{} 0 &{} \sin t &{} \cos t \end{array} \right) , \end{aligned}$$

and the mapping $\psi$ be given by

$$\begin{aligned}&\psi :\mathbb {R\times }\tilde{P}\rightarrow \tilde{P}\\&(t,A)\rightarrow \psi \left( t,A\right) =Af(t). \end{aligned}$$

Then the mapping $\psi$ is an action of $\mathbb {R}$ on $\tilde{P}.$

Proof

Let the mapping $\psi :\mathbb {R\times }\tilde{P}\rightarrow \tilde{P}$ be given by

$$\begin{aligned} \psi \left( t,A\right) =Af(t) \end{aligned}$$

Since $f$ is a homomorphism, it can be easily seen that $\psi$ satisfies

• (i) $\psi \left( 0,A\right) =A$

• (ii) $\psi \left( t_{1}+t_{2},A\right) =\psi \left( t_{1},\psi \left( t_{2},A\right) \right)$

Hence, the mapping $\psi$ is an action of $\mathbb {R}$ on $\tilde{P}$.$\square$

Corollary 4

The image of the mapping $f$ determines a one-parameter Lie-subgroup of $P$.

Proof

Since the mapping $f:\mathbb {R\rightarrow }\tilde{P}$ is a homomorphism, homomorphic image $H=f(\mathbb {R})$ is a subgroup of $\tilde{P}$ and since $g$ is a isomorphism, $H$ is a spiral curve as $\alpha (t)=e^{bt}\left( \cos t,\sin t,0,0\right)$ in $P.$ So, it is a one-parameter Lie-subgroup of $P$.$\square$

Corollary 5

Let $\psi :\mathbb {R\times }\tilde{P}\rightarrow \tilde{P}$ be an action of $\mathbb {R}$ on $\tilde{P}$. The infinitesimal generator associated with the mapping $\psi$ is $X_{x}=\left( bx_{1}-x_{2},bx_{2}+x_{1},bx_{3}-x_{4},bx_{4}+x_{3}\right)$ and $\alpha (t)=e^{bt}\left( \cos t,\sin t,0,0\right)$ is an integral curve of $X_{x}$.

Proof

Let $\psi :\mathbb {R\times }\tilde{P}\rightarrow \tilde{P}$ be an action of $\mathbb {R}$ on $\tilde{P}.$ The infinitesimal generator at $x\in \tilde{P}$ is given by

$$\begin{aligned} X_{x}=\dot{\psi }\left( 0,x\right) =\left( bx_{1}-x_{2},bx_{2}+x_{1},bx_{3}-x_{4},bx_{4}+x_{3}\right) , \end{aligned}$$

where $\dot{\psi }\left( 0,x\right) =\left( \frac{\partial \psi }{\partial t} \left( t,x\right) \right) _{t=0}.$ It can be easily seen that $\alpha (t)=e^{bt}\left( \cos t,\sin t,0,0\right)$ is an integral curve of $X_{x}$.$\square$

Corollary 6

$\theta _{\tilde{P}_{1}}:\tilde{P}_{1}\times \mathbb {R} ^{4}\rightarrow \mathbb {R}^{4}$ defines a $C^{\infty }$ action of $\tilde{P} _{1}$ on $\mathbb {R}^{4}$.

Proof

Since $\tilde{P}_{1}$ is a Lie-subgroup of $\tilde{P}$ and the inclusion map $i:\tilde{P}_{1}\rightarrow \tilde{P}$ is $C^{\infty }.$ The restriction $\theta _{\tilde{P}_{1}}=\theta \circ i:\tilde{P}_{1}\times \mathbb {R} ^{4}\rightarrow \mathbb {R}^{4}$ defines a $C^{\infty }$ action of $\tilde{P} _{1}$ on $\mathbb {R}^{4}$.$\square$

Theorem 4

Let $f_{1}:\mathbb {R\rightarrow }\tilde{P}_{1}$ be the mapping which sends every $t\in \mathbb {R}$ to

$$\begin{aligned} t\rightarrow f_{1}(t)=\left( \begin{array}{cccc} \cos t &{} -\sin t &{} 0 &{} 0 \\ \sin t &{} \cos t &{} 0 &{} 0 \\ 0 &{} 0 &{} \cos t &{} -\sin t \\ 0 &{} 0 &{} \sin t &{} \cos t \end{array} \right) , \end{aligned}$$

and the mapping $\psi _{1}$ be given by

$$\begin{aligned}&\psi _{1}:\mathbb {R\times }\tilde{P}_{1}\rightarrow \tilde{P}_{1}\\&(t,A)\rightarrow \psi _{1}\left( t,A\right) =Af_{1}(t). \end{aligned}$$

Then, the mapping $\psi _{1}$ is an action of $\mathbb {R}$ on $\tilde{P}_{1}$.

Proof

Since $f_{1}$ is a homomorphism, it can be easily seen that $\psi _{1}$ satisfies

• (i) $\psi _{1}\left( 0,A\right) =A$

• (ii) $\psi _{1}\left( t_{1}+t_{2},A\right) =\psi _{1}\left( t_{1},\psi _{1}\left( t_{2},A\right) \right)$

Hence, the mapping $\psi _{1}$ is an action of $\mathbb {R}$ on $\tilde{P} _{1}$.$\square$

Corollary 7

The image of the mapping $f_{1}$ determines a one-parameter Lie-subgroup of $P_{1}$.

Proof

Since the mapping $f_{1}:\mathbb {R\rightarrow }\tilde{P}_{1}$ is a homomorphism, homomorphic image $H_{1}=f_{1}(\mathbb {R})$ is a subgroup of $\tilde{P}_{1}$ and since $g$ is a isomorphism, $H_{1}$ is a circle as $\alpha (t)=\left( \cos t,\sin t,0,0\right)$ in $P_{1}.$ So, it is a one-parameter Lie-subgroup of $P_{1}$.$\square$

Corollary 8

Let $\psi _{1}:\mathbb {R\times }\tilde{P}_{1}\rightarrow \tilde{P}_{1}$ be an action of $\mathbb {R}$ on $\tilde{P}_{1}.$ The infinitesimal generator associated with the mapping $\psi _{1}$ is $X_{x}=\left( -x_{2},x_{1},-x_{4},x_{3}\right)$ and $\alpha (t)=\left( \cos t,\sin t,0,0\right)$ is an integral curve of $X_{x}$.

Proof

Let $\psi _{1}:\mathbb {R\times }\tilde{P}_{1}\rightarrow \tilde{P}_{1}$ be an action of $\mathbb {R}$ on $\tilde{P}_{1}.$ The infinitesimal generator at $x\in \tilde{P}_{1}$ is given by

$$\begin{aligned} X_{x}=\dot{\psi }_{1}\left( 0,x\right) =\left( -x_{2},x_{1},-x_{4},x_{3}\right) , \end{aligned}$$

where $\dot{\psi }_{1}\left( 0,x\right) =\left( \frac{\partial \psi _{1}}{ \partial t}\left( t,x\right) \right) _{t=0}.$ It can be easily seen that $\alpha (t)=\left( \cos t,\sin t,0,0\right)$ is an integral curve of $X_{x}$.$\square$

Corollary 9

$f_{1}$ induces an action on $S^{3}=\left\{ \left( x_{1},x_{2},x_{3},x_{4}\right) \in \mathbb {R}^{4};\,x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=1\right\}$.

Proof

$$\begin{aligned} \bar{\psi }\left( t,x_{1},x_{2},x_{3},x_{4}\right) =\left( \begin{array}{c} x_{1}\cos t-x_{2}\sin t,x_{1}\sin t+x_{2}\cos t, \\ x_{3}\cos t-x_{4}\sin t,x_{3}\sin t+x_{4}\cos t \end{array} \right) . \end{aligned}$$

It is obvious that $\bar{\psi }$ is an action on $S^{3}$.$\square$

4.3 Some Special Lie Subgroups

Özkaldı and Yaylı showed that a hyperquadric $P$ in $\mathbb {R} ^{4}$ is a Lie group using bicomplex number product. Also, they determined some special subgroups of Lie group $P,$ using the tensor product surfaces of Euclidean planar curves [7].

Our aim in this subsection is to determine some special subgroups of this Lie group $P$ using the surface $M$ which is obtained with homothetic motion. In this case, how should we choose the position vector of the curve at homothetic motion given by (2) that the surface M be a Lie subgroup of the hyperquadric $P$. We answer this question. If we take the profile curve $\alpha (s)=\left( \alpha _{1}(s),\alpha _{2}(s),\alpha _{3}(s),\alpha _{4}(s)\right)$ such that $\alpha _{1}(s)\alpha _{4}(s)=\alpha _{2}(s)\alpha _{3}(s)$ and the translation vector $C(t)=0,$ then the surface $M$ is given by the parametrization (2) is subset of $P$.

Theorem 5

Let $\gamma$ be a curve which is obtained using the homothetic motion with the homothetic function $h(t)=e^{at}$ and the profile curve $\alpha \left( t\right) =e^{bt}\left( \cos t,\sin t,0,0\right)$ where $a,b$ are real constants. Then curve $\gamma$ is a one-parameter subgroup in a Lie group $P$.

Proof

We can write the curve $\gamma$ as follows:

$$\begin{aligned} \gamma \left( t\right)&= e^{at}\left( \begin{array}{cccc} \cos t &{} -\sin t &{} 0 &{} 0 \\ \sin t &{} \cos t &{} 0 &{} 0 \\ 0 &{} 0 &{} \cos t &{} -\sin t \\ 0 &{} 0 &{} \sin t &{} \cos t \end{array} \right) \left( \begin{array}{c} e^{bt}\cos t \\ e^{bt}\sin t \\ 0 \\ 0 \end{array} \right) \\&= e^{(a+b)t}\left( \cos 2t,\sin 2t,0,0\right) . \end{aligned}$$

It can be easily seen that

$$\begin{aligned} \gamma \left( t_{1}\right) \times \gamma \left( t_{2}\right) =\gamma \left( t_{1}+t_{2}\right) \end{aligned}$$

for all $t_{1},t_{2}\in \mathbb {R}.$ Hence, $\left( \gamma \left( t\right) ,\times \right)$ is one-parameter Lie subgroup of $\left( P,\times \right)$ .$\square$

Remark 4

From Corollary (4), we know that $\alpha \left( t\right) =e^{bt}\left( \cos t,\sin t,0,0\right)$ is a one-parameter Lie subgroup of $P.$ In Theorem (5), we show that the trajector of the curve $\alpha$ under the homothetic motion is a one-parameter Lie subgroup of $P$ too.

Theorem 6

Let $\gamma$ be a curve which is obtained using the homothetic motion with the homothetic function $h(t)=e^{at}$ and the profile curve $\alpha \left( t\right) =e^{bt}\left( \cos t,0,\sin t,0\right)$ where $a,b$ are real constants. Then, the curve $\gamma$ is a one-parameter Lie-subgroup in Lie group $P$.

Proof

We can write the curve $\gamma$ as follows:

$$\begin{aligned} \gamma \left( t\right)&= e^{at}\left( \begin{array}{cccc} \cos t &{} -\sin t &{} 0 &{} 0 \\ \sin t &{} \cos t &{} 0 &{} 0 \\ 0 &{} 0 &{} \cos t &{} -\sin t \\ 0 &{} 0 &{} \sin t &{} \cos t \end{array} \right) \left( \begin{array}{c} e^{bt}\cos t \\ 0 \\ e^{bt}\sin t \\ 0 \end{array} \right) \\&= e^{(a+b)t}\left( \cos ^{2}t,\sin t\cos t,\sin t\cos t,\sin ^{2}t\right) . \end{aligned}$$

It can be easily seen that

$$\begin{aligned} \gamma \left( t_{1}\right) \times \gamma \left( t_{2}\right) =\gamma \left( t_{1}+t_{2}\right) \end{aligned}$$

for all $t_{1},t_{2}\in \mathbb {R}.$ Hence $\left( \gamma \left( t\right) ,\times \right)$ is one-parameter Lie subgroup of $\left( P,\times \right)$.$\square$

Remark 5

The above curve $\gamma$ can be expressed as tensor product of two spirals with the same parameter, that is, let $\beta :\mathbb {R} \rightarrow E^{2},\,\beta \left( t\right) =e^{at}\left( \cos t,\sin t\right)$ and $\delta \left( t\right) =e^{bt}\left( \cos t,\sin t\right)$ be two spirals. Then the curve $\gamma$ can be written as $\gamma \left( t\right) =\beta \left( t\right) \otimes \delta \left( t\right)$.

Corollary 10

Let $\gamma$ be a curve which is obtained using the homothetic motion with the homothetic function $h(t)=1$ and the profile curve $\alpha \left( t\right) =\left( \cos t,\sin t,0,0\right)$. Then, the curve $\gamma$ is a one-parameter Lie-subgroup in Lie group $P_{1}$.

Proof

For $h(t)=1$ and the profile curve $\alpha \left( t\right) =\left( \cos t,\sin t,0,0\right) ,$ we get

$$\begin{aligned} \gamma \left( t\right) =\left( \cos 2t,\sin 2t,0,0\right) \end{aligned}$$

Since $\left\| \gamma \left( t\right) \right\| _{t_{3}}=1,$ it follows that $\gamma \left( t\right) \subset P_{1}.$ So it is a one-parameter Lie subgroup in Lie group $P_{1}$. $\square$

Corollary 11

Let $\gamma$ be a curve which is obtained using the homothetic motion with the homothetic function $h(t)=1$ and the profile curve $\alpha \left( t\right) =\left( \cos t,0,\sin t,0\right)$. Then the curve $\gamma$ is a one-parameter Lie subgroup in Lie group $P_{1}$.

Proof

For $h(t)=1$ and the profile curve $\alpha \left( t\right) =\left( \cos t,0,\sin t,0\right) ,$ we get

$$\begin{aligned} \gamma \left( t\right) =\left( \cos ^{2}t,\sin t\cos t,\sin t\cos t,\sin ^{2}t\right) \end{aligned}$$

Since $\left\| \gamma \left( t\right) \right\| _{t_{3}}=1,$ it follows that $\gamma \left( t\right) \subset P_{1}.$ So, it is a one-parameter Lie-subgroup in Lie group $P_{1}$. $\square$

Remark 6

The above curve $\gamma$ can be expressed as tensor product of two circles with the same parameter, that is, let $\beta :\mathbb {R} \rightarrow E^{2},\,\beta \left( t\right) =\left( \cos t,\sin t\right)$ and $\delta \left( t\right) =\left( \cos t,\sin t\right)$ be circles. Then, the curve $\gamma$ can be written as $\gamma \left( t\right) =\beta \left( t\right) \otimes \delta \left( t\right)$.

Theorem 7

Let $M$ be a surface which is obtained using the homothetic motion with the homothetic function $h(t)=e^{at}$ and the profile curve $\alpha \left( s\right) =e^{bs}\left( \cos s,0,\sin s,0\right)$. Then, the surface $M$ is a 2-dimensional Lie-subgroup of $P$.

Proof

$$\begin{aligned} \varphi \left( t,s\right)&= e^{at}\left( \begin{array}{cccc} \cos t &{} -\sin t &{} 0 &{} 0 \\ \sin t &{} \cos t &{} 0 &{} 0 \\ 0 &{} 0 &{} \cos t &{} -\sin t \\ 0 &{} 0 &{} \sin t &{} \cos t \end{array} \right) \left( \begin{array}{c} e^{bs}\cos s \\ 0 \\ e^{bs}\sin s \\ 0 \end{array} \right) \\&= e^{at+bs}\left( \cos s\cos t,\cos s\sin t,\sin s\cos t,\sin s\sin t\right) . \end{aligned}$$

Since $\varphi \left( t,s\right)$ is both a subgroup and submanifold of Lie group $P$, we obtain that $\varphi \left( t,s\right)$ is a 2-dimensional Lie subgroup of $P$.$\square$

Remark 7

The above surface $M$ can be expressed as tensor product surface of two spirals, that is, let $\beta :\mathbb {R}\rightarrow E^{2},\, \beta \left( s\right) =e^{as}\left( \cos s,\sin s\right)$ and $\delta \left( t\right) =e^{bt}\left( \cos t,\sin t\right)$ be two spirals. Then the surface $M$ can be written as $\varphi \left( t,s\right) =\beta \left( s\right) \otimes \delta \left( t\right)$.

Corollary 12

Let $M$ be a Vranceanu surface with the profile curve $\alpha \left( s\right) =e^{bs}\left( \cos s,0,\sin s,0\right) .$ Then the surface $M$ is a 2-dimensional Lie subgroup of $P$.

Proof

If we take as $a=0$ in Theorem (7), we obtain a rotation surface which is called Vranceanu surface in $E^{4}.$ Then, Vranceanu surface is a 2-dimensional Lie subgroup of $P$.$\square$

Corollary 13

Clifford torus is a 2-dimensional Lie-subgroup of $P_{1}$.

Proof

By using the homothetic function $h(t)=1$ and the profile curve $\alpha \left( s\right) =\left( \cos s,0,\sin s,0\right)$, we obtain a rotation surface which is called Clifford Torus. This surface is product of two plane circles with the same radius, that is,

$$\begin{aligned} \varphi \left( t,s\right) =\left( \cos s\cos t,\cos s\sin t,\sin s\cos t,\sin s\sin t\right) . \end{aligned}$$

Since $\left\| \varphi \left( t,s\right) \right\| _{t_{3}}=1,\, \varphi \left( t,s\right)$ is subset of $P_{1}$. Hence, Clifford Torus is a 2-dimensional Lie subgroup of $P_{1}$.$\square$

Remark 8

The Clifford Torus is a subset of $S^{3}$ and it is a Lie group with bicomplex number product, but it is not a Lie subgroup of $S^{3}.$ Also, since the Clifford Torus is a Lie group, it is parallelisable.