On Translation Curves and Geodesics in ${\mathrm{Sol}}_1^4$

Abstract

A translation curve in a homogeneous space is a curve such that for a given unit vector at the origin, translation of this vector is tangent to the curve in its every point. Translation curves coincide with geodesics in most Thurston spaces, but not in twisted product Thurston spaces. Moreover, translation curves often seem more intuitive and simpler than geodesics. In this paper, we determine translation curves in ${\mathrm{Sol}}_1^4$ space. Their curvature properties are discussed and translation spheres are presented. Finally, characterization of geodesics in ${\mathrm{Sol}}_1^4$ space is given.

1. Introduction

A homogeneous geometry is a pair $(G,X)$ consisting of a smooth manifold X, equipped with the transitive action of a Lie group G. The manifold X defines the underlying homogeneous space, and the group G defines the set of allowable motions.

A Thurston geometry $(G,X)$ is a homogeneous space such that X is connected and simply connected, G acts transitively on X with compact point stabilizers, G is not contained in any larger group of diffeomorphisms of X, and there is at least one compact manifold modeled on $(G,X)$. More about eight three-dimensional Thurston spaces can be found in [1,2].

There are 19 homogeneous four-dimensional model spaces according to Filipkiewicz [3]. Our ambient space ${\mathrm{Sol}}_1^4,$ is among them. Furthermore, it is one of the 14 spaces admitting a complex structure compatible with the geometric structure (cf. [4]). ${\mathrm{Sol}}_1^4$ possesses a locally conformal Kahler (LCK) structure. This structure in [5] facilitated a study of minimal invariant, totally real, and CR-submanifolds of ${\mathrm{Sol}}_1^4$. On the other hand, J-trajectories, which represent a generalization of geodesics, are investigated in [6].

Some homogeneous spaces allow a specific translation different from the geodesic translation. This specific translation carries a unit vector given at the origin to any point in space by its tangent mapping. The corresponding curve is called the translation curve. Molnár and Szilágyi in [7] initiated the study of translation curves along with the investigation of translation spheres in three-dimensional product and twisted product Thurston geometries. In four-dimensional Thurston spaces, translation curves have first been considered in [8]. More about geodesics and their generalizations in some other four-dimensional Thurston spaces can be found in [9,10,11].

Here, we classify translation curves and investigate geodesics in ${\mathrm{Sol}}_1^4$ space.

This paper is organized as follows. First, we recall basic properties of ${\mathrm{Sol}}_1^4$ space; then, we determine its translation curves and discuss their curvature properties. Finally, we consider geodesics in ${\mathrm{Sol}}_1^4.$

2. The Model Space ${\mathrm{Sol}}_{\mathbf{1}}^{\mathbf{4}}$

2.1. Lie Group

The underlying manifold of the model space ${\mathrm{Sol}}_1^4$ is the simply connected solvable Lie group $G_8$ described in [3] (p. 101).

The Lie group multiplication is given by:

$$(x_1,y_1,z_1,t_1)\ast (x_2,y_2,z_2,t_2)=(x_1+e^{t_1}{x}_2,y_1+e^{-t_1}{y}_2,z_1+z_2+e^{-t_1}{x}_1{y}_2,t_1+t_2).$$

(1)

This operation can be derived from the following matrix multiplication:

$$\begin{array}{c}\left(\begin{array}{cccc}1& 0& e^{-t_1}{x}_1& z_1\\ 0& e^{t_1}& 0& x_1\\ 0& 0& e^{-t_1}& y_1\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& e^{-t_2}{x}_2& z_2\\ 0& e^{t_2}& 0& x_2\\ 0& 0& e^{-t_2}& y_2\\ 0& 0& 0& 1\end{array}\right)=\hfill \\ \hfill =\left(\begin{array}{cccc}1& 0& e^{-(t_1+t_2)}(x_1+e^{t_1}{x}_2)& z_1+z_2+e^{-t_1}{x}_1{y}_2\\ 0& e^{t_1+t_2}& 0& x_1+e^{t_1}{x}_2\\ 0& 0& e^{-(t_1+t_2)}& y_1+e^{-t_1}{y}_2\\ 0& 0& 0& 1\end{array}\right)\end{array}$$

by the identification:

$$(x,y,z,t):=\left(\begin{array}{cccc}1& 0& e^{-t}x& z\\ 0& e^t& 0& x\\ 0& 0& e^{-t}& y\\ 0& 0& 0& 1\end{array}\right).$$

(2)

Multiplication (1) can be interpreted as a translation T of $(x_2,y_2,z_2,t_2)$ to $(x_1,y_1,z_1,t_1).$ The neutral element of (1) is $(0,0,0,0)$. The inverse element of $(x,y,z,t)$ is given by:

$${(x,y,z,t)}^{-1}=(-e^{-t}x,-e^{t}y,-z+xy,-t).$$

(3)

2.2. Metric and Basis

Using the inverse translation (3), by pullback of coordinate differentials,

$$\left(\begin{array}{cccc}1& 0& -x& xy-z\\ 0& e^{-t}& 0& -e^{-t}x\\ 0& 0& e^t& -e^{t}y\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}0& 0& e^{-t}(dx-xdt)& dz\\ 0& e^{t}dt& 0& dx\\ 0& 0& -e^{-t}dt& dy\\ 0& 0& 0& 0\end{array}\right)=\left(\begin{array}{cccc}0& 0& e^{-t}dx& dz-xdy\\ 0& dt& 0& e^{-t}dx\\ 0& 0& -dt& e^{t}dy\\ 0& 0& 0& 0\end{array}\right)$$

we obtain the left invariant Riemannian metric g of ${\mathrm{Sol}}_1^4$:

$$g=e^{-2t}d{x}^2+e^{2t}d{y}^2+{(dz-xdy)}^2+d{t}^2.$$

Hence, the orthonormal coframe $\{{\vartheta }^1,{\vartheta }^2,{\vartheta }^3,{\vartheta }^4\}$ is given by:

$${\vartheta }^1=e^{-t}dx,{\vartheta }^2=e^{t}dy,{\vartheta }^3=dz-xdy,{\vartheta }^4=dt.$$

Thus, the metrically dual left invariant basis vector fields are:

$$e_1=e^t\frac{\partial }{\partial x},e_2=e^{-t}\left(\frac{\partial }{\partial y}+x\frac{\partial }{\partial z}\right),e_3=\frac{\partial }{\partial z},e_4=\frac{\partial }{\partial t}.$$

Remark 1.

If we use different representation than (2), i.e.,

$$(x,y,z,t)=\left(\begin{array}{ccccc}1& 0& e^{-t}x& 0& z\\ 0& e^t& 0& 0& x\\ 0& 0& e^{-t}& 0& y\\ 0& 0& 0& 1& t\\ 0& 0& 0& 0& 1\end{array}\right),$$

then by matrix multiplication we (again) obtain the multiplication rule (1), but also obtain a straightforward presentation of the metric:

$$\left(\begin{array}{ccccc}1& 0& -x& 0& xy-z\\ 0& e^{-t}& 0& 0& -e^{-t}x\\ 0& 0& e^t& 0& -e^{t}y\\ 0& 0& 0& 1& -t\\ 0& 0& 0& 0& 1\end{array}\right)\left(\begin{array}{c}dz\\ dx\\ dy\\ dt\\ 0\end{array}\right)=\left(\begin{array}{c}dz-xdy\\ e^{-t}dx\\ e^{t}dy\\ dt\\ 0\end{array}\right).$$

2.3. Levi–Civita Connection

The Levi–Civita connection is given by:

$$\begin{array}{cccc}{\nabla }_{e_1}{e}_1=e_4,\hfill & {\nabla }_{e_1}{e}_2=\frac{1}{2}{e}_3,\hfill & {\nabla }_{e_1}{e}_3=-\frac{1}{2}{e}_2,\hfill & {\nabla }_{e_1}{e}_4=-e_1,\hfill \\ {\nabla }_{e_2}{e}_1=-\frac{1}{2}{e}_3,\hfill & {\nabla }_{e_2}{e}_2=-e_4,\hfill & {\nabla }_{e_2}{e}_3=\frac{1}{2}{e}_1,\hfill & {\nabla }_{e_2}{e}_4=e_2,\hfill \\ {\nabla }_{e_3}{e}_1=-\frac{1}{2}{e}_2,\hfill & {\nabla }_{e_3}{e}_2=\frac{1}{2}{e}_1,\hfill & {\nabla }_{e_3}{e}_3=0,\hfill & {\nabla }_{e_3}{e}_4=0,\hfill \\ {\nabla }_{e_4}{e}_1=0,\hfill & {\nabla }_{e_4}{e}_2=0,\hfill & {\nabla }_{e_4}{e}_3=0,\hfill & {\nabla }_{e_4}{e}_4=0.\hfill \end{array}$$

(4)

Basis vector fields satisfy the following commutation relations:

$$[e_1,e_3]=[e_2,e_3]=[e_3,e_4]=0,[e_1,e_2]=e_3,[e_1,e_4]=-e_1,[e_2,e_4]=e_2.$$

(5)

2.4. Lie algebra

The Lie algebra $\mathfrak{g}$ of $G_8$ is given by:

$$\left\{\left.\left(\begin{array}{cccc}0& 0& t_1& t_3\\ 0& t_4& 0& t_1\\ 0& 0& -t_4& t_2\\ 0& 0& 0& 0\end{array}\right):t_1,t_2,t_3,t_4\in \mathbb{R}\right\}\right..$$

If we take the following basis:

$$e_1=\left(\begin{array}{cccc}0& 0& 1& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right),e_2=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 0\end{array}\right),e_3=\left(\begin{array}{cccc}0& 0& 0& 1\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right),e_4=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& -1& 0\\ 0& 0& 0& 0\end{array}\right),$$

we can verify commutation relations (5).

2.5. Riemannian and Sectional Curvatures

The non-vanishing components of the Riemannian tensor are:

$$\begin{array}{cccccccc}\hfill R(e_1,e_2)e_1& =-\frac{1}{4}{e}_2,\hfill & \hfill R(e_1,e_2)e_3& =-\frac{1}{2}{e}_4,\hfill & \hfill R(e_1,e_2)e_4& =-\frac{1}{2}{e}_3,\hfill & \hfill R(e_3,e_1)e_1& =\frac{1}{4}{e}_3,\hfill \\ \hfill R(e_4,e_1)e_1& =-e_4,\hfill & \hfill R(e_2,e_3)e_3& =\frac{1}{4}{e}_2,\hfill & \hfill R(e_2,e_4)e_4& =-e_2,\hfill & \hfill R(e_3,e_4)e_1& =-\frac{1}{2}{e}_2.\hfill \end{array}$$

Hence, we obtain the following sectional curvatures:

$$K_{12}=K_{13}=K_{23}=\frac{1}{4},K_{24}=K_{14}=-1,K_{34}=0.$$

and the scalar curvature $K=-\frac{5}{2}.$

3. Translation Curves in ${\mathrm{Sol}}_{\mathbf{1}}^{\mathbf{4}}$

3.1. Translation Curves in ${\mathrm{Sol}}_1^4$

For a given starting unit vector $(\alpha ,\beta ,\gamma ,\delta )=(\dot{x}\left(0\right),\dot{y}\left(0\right),\dot{z}\left(0\right),\dot{t}\left(0\right))$ at the origin $\left(x\right(0),y(0),z(0),t(0\left)\right)=(0,0,0,0),$ we define its image in a point $\left(x\right(s),y(s),z(s),t(s\left)\right)$ by the translation T such that:

$$\left(\begin{array}{cccc}1& 0& e^{-t}x& z\\ 0& e^t& 0& x\\ 0& 0& e^{-t}& y\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}0& 0& \alpha & \gamma \\ 0& \delta & 0& \alpha \\ 0& 0& -\delta & \beta \\ 0& 0& 0& 0\end{array}\right)=\left(\begin{array}{cccc}0& 0& \alpha -\delta x{e}^{-t}& \gamma +\beta e^{-t}x\\ 0& \delta e^t& 0& \alpha e^t\\ 0& 0& -\delta e^{-t}& \beta e^{-t}\\ 0& 0& 0& 0\end{array}\right).$$

Hence, we obtain differential equations for the curve starting at the origin in direction $(\alpha ,\beta ,\gamma ,\delta )$:

$$\begin{array}{cc}\hfill \dot{x}\left(s\right)& =\alpha e^{t\left(s\right)}\hfill \\ \hfill \dot{y}\left(s\right)& =\beta e^{-t\left(s\right)}\hfill \\ \hfill \dot{z}\left(s\right)& =\gamma +\beta x\left(s\right)e^{-t\left(s\right)},\hfill \\ \hfill \dot{t}\left(s\right)& =\delta .\hfill \end{array}$$

(6)

Remark 2.

Using the representation from Remark 1, we could explicitly obtain components of the translated initial vector:

$$\left(\begin{array}{ccccc}1& 0& e^{-t}x& 0& z\\ 0& e^t& 0& 0& x\\ 0& 0& e^{-t}& 0& y\\ 0& 0& 0& 1& t\\ 0& 0& 0& 0& 1\end{array}\right)\left(\begin{array}{c}\gamma \\ \alpha \\ \beta \\ \delta \\ 0\end{array}\right)=\left(\begin{array}{c}\gamma +\beta x{e}^{-t}\\ \alpha e^t\\ \beta e^{-t}\\ \delta \\ 0\end{array}\right).$$

Next, we solve the system (6). From the fourth equation we have $t\left(s\right)=\delta s.$ Substituting this into other equations of (6), after integration we obtain the following result.

Theorem 1.

Translation curves in ${\mathrm{Sol}}_1^4$ space starting at the origin are given by the following parametric equations:

(a) 

$$\begin{array}{ccc}\hfill x\left(s\right)& =& \frac{\alpha }{\delta }\left(e^{\delta s}-1\right),y\left(s\right)=-\frac{\beta }{\delta }\left(e^{-\delta s}-1\right),\hfill \\ \hfill z\left(s\right)& =& \left(\frac{\alpha \beta +\gamma \delta }{\delta }\right)s+\frac{\alpha \beta }{{\delta }^2}\left(e^{-\delta s}-1\right),t\left(s\right)=\delta s,\mathrm{for}\delta \ne 0\in \mathbb{R},\hfill \end{array}$$

(b) 

$$x\left(s\right)=\alpha s,y\left(s\right)=\beta s,z\left(s\right)=\gamma s+\frac{\alpha \beta }{2}{s}^2,t\left(s\right)=0,\mathrm{for}\delta =0,$$

where $\alpha ,\beta ,\gamma \in \mathbb{R},$ such that ${\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2=1$.

Remark 3.

Note that in the second case ($\delta =0$) we consider translation curves in hyperspace isometric to the ${\mathrm{Nil}}_3$ space. The obtained translation curves in this case coincide with the translation curves described in [7].

3.2. Curvature Properties of Translation Curves

Definition 1.

For a curve c in ${\mathrm{Sol}}_1^4$ space parameterized by arc length s, we say that c is a Frenet curve of osculating order r ($r=1,\dots ,4$) if there exist orthonormal vector fields $E_1$, $E_2$, $E_3$, and $E_4$ along c, such that:

$$\begin{array}{cc}\hfill & \dot{c}=E_1,{\nabla }_{\dot{c}}{E}_1=\kappa E_2,{\nabla }_{\dot{c}}{E}_2=-\kappa E_1+\tau E_3,\hfill \\ \hfill & {\nabla }_{\dot{c}}{E}_3=-\tau E_2+\sigma E_4,{\nabla }_{\dot{c}}{E}_4=-\sigma E_3,\hfill \end{array}$$

where κ, τ, and σ are positive $C^{\infty }$ functions of s.

Vector fields $E_1$, $E_2$, $E_3$, and $E_4$ are called the tangent, the normal, the binormal, and the trinormal vector field of the curve c, respectively. Functions $\kappa \left(s\right)$, $\tau \left(s\right)$, and $\sigma \left(s\right)$ are called the first, the second and the third curvature of c, respectively.

A geodesic is regarded as a Frenet curve of osculating order 1.

A helix of order 2 is a Frenet curve of osculating order 2 with constant curvature $\kappa$, i.e., it is a circle.

A helix of order 3 is a Frenet curve of osculating order 3 with constant curvatures $\kappa$ and $\tau$, i.e., it is a circular helix.

A helix of order 4 is a Frenet curve of osculating order 4 such that all curvatures $\kappa ,\tau ,$ and $\sigma$ are constant (e.g., see [12]).

Next, we determine the curvatures of translation curves.

We start with the unit velocity vector $E_1=\dot{c}=\alpha e_1+\beta e_2+\gamma e_3+\delta e_4.$ Using (4), we have:

$${\nabla }_{\dot{c}}{E}_1=(\beta \gamma -\alpha \delta )e_1+(\beta \delta -\alpha \gamma )e_2+({\alpha }^2-{\beta }^2)e_4.$$

By ${\nabla }_{\dot{c}}{E}_1=\kappa E_2$, we have:

$${\kappa }^2={({\alpha }^2-{\beta }^2)}^2+({\alpha }^2+{\beta }^2)({\gamma }^2+{\delta }^2)-4\alpha \beta \gamma \delta =const.$$

and:

$$E_2=\frac{1}{\kappa }\left\{(\beta \gamma -\alpha \delta )e_1+(\beta \delta -\alpha \gamma )e_2+({\alpha }^2-{\beta }^2)e_4\right\}.$$

(7)

Remark 4.

Notice that κ is zero in at least two cases. The first case is $\alpha =\beta =0$ and the second case is $\alpha =\pm \beta$ and $\gamma =\pm \delta$. The translation curves obtained for $\alpha =\beta =0$ coincide with geodesics we will describe later in Case 1 of the proof of Theorem 3.

Next, using ${\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2=1,$ we can rewrite the first curvature as:

$${\kappa }^2=({\alpha }^2+{\beta }^2)-4\alpha \beta (\alpha \beta +\gamma \delta ).$$

Furthermore, using (4) and (7), we calculate:

$$\begin{array}{ccc}\hfill {\nabla }_{\dot{c}}{E}_2& =& \frac{1}{\kappa }\{\left[\frac{\gamma }{2}\left(\beta \delta -\alpha \gamma \right)-\alpha ({\alpha }^2-{\beta }^2)\right]e_1+\left[\beta ({\alpha }^2-{\beta }^2)-\frac{\gamma }{2}(\beta \gamma -\alpha \delta )\right]e_2+\hfill \\ & & +\left[\alpha \beta \delta -\frac{\gamma }{2}({\alpha }^2+{\beta }^2)\right]e_3+\left[2\alpha \beta \gamma -\delta ({\alpha }^2+{\beta }^2)\right]e_4\}.\hfill \end{array}$$

Then, from ${\nabla }_{\dot{c}}{E}_2=-\kappa E_1+\tau E_3$, we have:

$$\begin{array}{ccc}\hfill \tau E_3& =& \frac{1}{\kappa }\{\left[\frac{\gamma }{2}(\beta \delta -\alpha \gamma )-\alpha ({\alpha }^2-{\beta }^2)+{\kappa }^2\alpha \right]e_1+\left[\beta ({\alpha }^2-{\beta }^2)-\frac{\gamma }{2}(\beta \gamma -\alpha \delta )+{\kappa }^2\beta \right]e_2+\hfill \\ & & +\left[\alpha \beta \delta -\frac{\gamma }{2}({\alpha }^2+{\beta }^2)+{\kappa }^2\gamma \right]e_3+\left[2\alpha \beta \gamma -\delta ({\alpha }^2+{\beta }^2)+{\kappa }^2\delta \right]e_4\}\hfill \\ & =& \frac{1}{\kappa }\{\left[\frac{\gamma }{2}(\beta \delta -\alpha \gamma )+2\alpha {\beta }^2-4{\alpha }^2\beta (\alpha \beta +\gamma \delta )\right]e_1+\hfill \\ & & +\left[-\frac{\gamma }{2}(\beta \gamma -\alpha \delta )+2{\alpha }^2\beta -4\alpha {\beta }^2(\alpha \beta +\gamma \delta )\right]e_2+\hfill \\ & & +\left[\alpha \beta \delta +\frac{\gamma }{2}({\alpha }^2+{\beta }^2)-4\alpha \beta \gamma (\alpha \beta +\gamma \delta )\right]e_3+\hfill \\ & & +\left[2\alpha \beta \gamma -4\alpha \beta \delta (\alpha \beta +\gamma \delta )\right]e_4\}.\hfill \end{array}$$

Hence, we obtain that the second curvature is constant.

$${\tau }^2=\frac{1}{{\kappa }^2}\left\{({\alpha }^2+{\beta }^2)\left(4{\alpha }^2{\beta }^2+\frac{c^2}{4}+3\alpha \beta \gamma \delta \right)-16{\alpha }^2{\beta }^2{(\alpha \beta +\gamma \delta )}^2+\alpha \beta \delta (\alpha \beta \delta -{\gamma }^3)\right\}.$$

Next, we find:

$$\begin{array}{cc}\hfill {\nabla }_{\dot{c}}{E}_3& =\frac{1}{\kappa \tau }\left\{\right[4\alpha \beta (\alpha \beta +\gamma \delta )(\alpha \delta -\beta \gamma )+\frac{1}{2}\alpha \beta (\beta \delta -2\alpha \gamma )+\hfill \\ \hfill & +\frac{\gamma }{4}\left(\alpha (\alpha \beta +\gamma \delta )+\beta ({\beta }^2-{\gamma }^2)\right)]e_1+\hfill \\ \hfill & +\left[4\alpha \beta (\alpha \beta +\gamma \delta )(\alpha \gamma -\beta \delta )+\frac{1}{2}\alpha \beta (2\beta \gamma -\alpha \delta )-\frac{\gamma }{4}(\beta (\alpha \beta +\gamma \delta )+\alpha ({\alpha }^2-{\gamma }^2))\right]e_2+\hfill \\ \hfill & +\left[({\alpha }^2-{\beta }^2)\left(\alpha \beta +\frac{1}{4}\gamma \delta \right)\right]e_3+\hfill \\ \hfill & +\left[({\beta }^2-{\alpha }^2)\left(4\alpha \beta (\alpha \beta +\gamma \delta )+\frac{1}{2}{\gamma }^2\right)\right]e_4\}.\hfill \end{array}$$

Then, from ${\nabla }_{\dot{c}}{E}_3=-\tau E_2+\sigma E_4$ we have:

$$\begin{array}{cc}\hfill \sigma E_4& =\frac{1}{\kappa \tau }\{\left[4\alpha \beta (\alpha \beta +\gamma \delta )(\alpha \delta -\beta \gamma )+\frac{1}{2}\alpha \beta (\beta \delta -2\alpha \gamma )+\frac{\gamma }{4}\left(\alpha (\alpha \beta +\gamma \delta )+\beta ({\beta }^2-{\gamma }^2)\right)-{\tau }^2(\beta \gamma -\alpha \delta )\right]e_1+\hfill \\ \hfill & +\left[4\alpha \beta (\alpha \beta +\gamma \delta )(\alpha \gamma -\beta \delta )+\frac{1}{2}\alpha \beta (2\beta \gamma -\alpha \delta )-\frac{\gamma }{4}\left(\beta (\alpha \beta +\gamma \delta )+\alpha ({\alpha }^2-{\gamma }^2)\right)-{\tau }^2(\beta \delta -\alpha \gamma )\right]e_2+\hfill \\ \hfill & +\left[({\alpha }^2-{\beta }^2)\left(\alpha \beta +\frac{1}{4}\gamma \delta \right)\right]e_3+\hfill \\ \hfill & +\left[({\beta }^2-{\alpha }^2)\left(4\alpha \beta (\alpha \beta +\gamma \delta )+\frac{1}{2}{\gamma }^2\right)-{\tau }^2({\alpha }^2-{\beta }^2)\right]e_4\}.\hfill \end{array}$$

(8)

Even though we are unable to express $\sigma E_4$ without $\kappa$ and $\tau$, we can conclude that $\sigma =const.$ From (8) we obtain that ${\sigma }^2$ can be expressed in the following form:

$${\sigma }^2=\frac{1}{{\kappa }^2{\tau }^2}\left\{{\left(\dots \right)}^2+{\left(\dots \right)}^2+{\left[({\alpha }^2-{\beta }^2)(\alpha \beta +\frac{1}{4}\gamma \delta )\right]}^2+{\left(\dots \right)}^2\right\}.$$

Thus, we conclude that the third squared term in the braces in general is not zero. Hence, $\sigma =const\ne 0$.

Thus, we have proved the following theorem.

Theorem 2.

Translation curves in ${\mathrm{Sol}}_1^4$ space are helices of order 4.

Remark 5.

Note that translation curves in ${\mathrm{Sol}}_0^4$ space are helices of order 3, i.e., circular helices [8].

3.3. Translation Spheres in ${\mathrm{Sol}}_1^4$

Let us assume that the initial unit vector of translation curve (7) is given by:

$$\alpha =sin\vartheta cos\phi cos\psi ,\beta =sin\vartheta cos\phi sin\psi ,\gamma =sin\vartheta sin\phi ,\delta =cos\vartheta .$$

Then, we can define the sphere of radius R with its center at the origin. Namely, the unit velocity translation curves ending in parameter R describe the translation spheres.

Proposition 1.

The translation sphere of radius R in ${\mathrm{Sol}}_1^4$ space is given by the following parametric equations:

$$\begin{array}{cc}\hfill x(\vartheta ,\phi ,\psi )& =tan\vartheta cos\phi cos\psi \left(e^{Rcos\vartheta }-1\right),\hfill \\ \hfill y(\vartheta ,\phi ,\psi )& =-tan\vartheta cos\phi sin\psi \left(e^{-Rcos\vartheta }-1\right),\hfill \\ \hfill z(\vartheta ,\phi ,\psi )& =Rtan\vartheta \left(\frac{1}{2}sin\vartheta {cos}^2\phi sin2\psi +cos\vartheta sin\phi \right)+\frac{1}{2}{tan}^2\vartheta {cos}^2\phi sin2\psi \left(e^{-Rcos\vartheta }-1\right),\hfill \\ \hfill t(\vartheta ,\phi ,\psi )& =Rcos\vartheta ,\hfill \end{array}$$

where $\vartheta ,\phi ,\psi \in [0,2\pi )$, and $R\in {\mathbb{R}}^{+}.$

4. Geodesics in ${\mathrm{Sol}}_{\mathbf{1}}^{\mathbf{4}}$

Here we consider geodesics in ${\mathrm{Sol}}_1^4.$ Let $c\left(s\right)=\left(x\right(s),y(s),z(s),t(s\left)\right)$ be a curve in ${\mathrm{Sol}}_1^4$ parameterized by arc length. Its unit tangent vector field is:

$$\begin{array}{cc}\hfill \dot{c}\left(s\right)=& \ddot{x}\left(s\right)\frac{\partial }{\partial x}+\dot{y}\left(s\right)\frac{\partial }{\partial y}+\dot{z}\left(s\right)\frac{\partial }{\partial z}+\dot{t}\left(s\right)\frac{\partial }{\partial t}\hfill \\ \hfill =& e^{-t\left(s\right)}\dot{x}\left(s\right)e_1+e^{t\left(s\right)}\dot{y}\left(s\right)e_2+(\dot{z}\left(s\right)-x\left(s\right)\dot{y}\left(s\right))e_3+\dot{t}\left(s\right)e_4.\hfill \end{array}$$

The arc-length condition is:

$$e^{-2t\left(s\right)}\dot{x}{\left(s\right)}^2+e^{2t\left(s\right)}\dot{y}{\left(s\right)}^2+{(\dot{z}\left(s\right)-x\left(s\right)\dot{y}\left(s\right))}^2+\dot{t}{\left(s\right)}^2=1.$$

Next, using (4), we find the acceleration:

$$\begin{array}{ccc}\hfill {\nabla }_{\dot{c}\left(s\right)}\dot{c}\left(s\right)& =& \left[\left(\ddot{x}\left(s\right)-2\dot{x}\left(s\right)\dot{t}\left(s\right)\right)e^{-t}+\dot{y}\left(s\right)(\dot{z}\left(s\right)-x\left(s\right)\dot{y}\left(s\right))e^{t\left(s\right)}\right]e_1+\hfill \\ & & +\left[\left(\ddot{y}\left(s\right)+2\dot{y}\left(s\right)\dot{t}\left(s\right)\right)e^t-\dot{x}\left(s\right)(\dot{z}\left(s\right)-x\left(s\right)\dot{y}\left(s\right))e^{-t\left(s\right)}\right]e_2+\hfill \\ & & +\left[\ddot{z}\left(s\right)-\dot{x}\left(s\right)\dot{y}\left(s\right)-x\left(s\right)\ddot{y}\left(s\right)\right]e_3+\hfill \\ & & +\left[\ddot{t}\left(s\right)-\dot{y}{\left(s\right)}^2{e}^{2t\left(s\right)}\dot{x}{\left(s\right)}^2{e}^{-2t\left(s\right)}\right]e_4.\hfill \end{array}$$

From the geodesic equation ${\nabla }_{\dot{c}\left(s\right)}\dot{c}\left(s\right)=0$, we obtain the following system:

$$\begin{array}{ccc}\hfill \ddot{x}\left(s\right)-2\dot{x}\left(s\right)\dot{t}\left(s\right)& =& -\dot{y}\left(s\right)(\dot{z}\left(s\right)-x\left(s\right)\dot{y}\left(s\right))e^{2t\left(s\right)},\hfill \\ \hfill \ddot{y}\left(s\right)+2\dot{y}\left(s\right)\dot{t}\left(s\right)& =& \dot{x}\left(s\right)(\dot{z}\left(s\right)-x\left(s\right)\dot{y}\left(s\right))e^{-2t\left(s\right)},\hfill \\ \hfill \ddot{z}\left(s\right)& =& \dot{x}\left(s\right)\dot{y}\left(s\right)+x\left(s\right)\ddot{y}\left(s\right),\hfill \\ \hfill \ddot{t}\left(s\right)& =& \dot{y}{\left(s\right)}^2{e}^{2t\left(s\right)}-\dot{x}{\left(s\right)}^2{e}^{-2t\left(s\right)}.\hfill \end{array}$$

(9)

Let the initial conditions be:

$$(x\left(0\right),y\left(0\right),z\left(0\right),t\left(0\right))=(x_0,y_0,z_0,t_0)\mathrm{and}(\dot{x}\left(0\right),\dot{y}\left(0\right),\dot{z}\left(0\right),\dot{t}\left(0\right))=(\alpha ,\beta ,\gamma ,\delta ),$$

(10)

where $\alpha ,\beta ,\gamma ,\delta \in \mathbb{R}.$

We state the following theorem.

Theorem 3.

Geodesics in ${\mathrm{Sol}}_1^4$ geometry are solutions of the system of differential equations (9). In particular, some analytical solutions of System (9) with respect to initial conditions (10) are curves:

1. 

$c_1\left(s\right)=(x_0,y_0,\gamma s+z_0,\delta s+t_0);$

2. 

$c_2\left(s\right)=(tanh\left(\alpha s\right),y_0,z_0,-ln\left[cosh\left(\alpha s\right)\right]);$

3. 

$c_3\left(s\right)=(\beta e^{2t_0}s+x_0,\beta s+y_0,x_0\beta s+\frac{1}{2}{e}^{2t_0}{\beta }^2{s}^2,t_0);$

4. 

$c_4\left(s\right)=\left(-\frac{\beta }{C}{e}^{Cs+2t_0},\frac{\beta }{C}{e}^{Cs},Cs-\frac{{\beta }^2}{2C^2}{e}^{2(Cs+t_0)},t_0\right),\mathrm{for}C=\gamma -x_0\beta \ne 0,$where $x_0,y_0,z_0,t_0,\alpha ,\beta ,\gamma ,\delta \in \mathbb{R}.$

Proof. 

Integrating the third equation of (9), with respect to (10), we have:

$$\dot{z}-x\dot{y}=C\mathrm{where}C=\gamma -x_0\beta .$$

(11)

Substituting (11) into the first and the second equations of (9), we obtain:

$$\begin{array}{ccc}\hfill \ddot{x}-2\dot{x}\dot{t}& =& -C\dot{y}{e}^{2t},\hfill \\ \hfill \ddot{y}+2\dot{y}\dot{t}& =& C\dot{x}{e}^{-2t}.\hfill \end{array}$$

(12)

Adding these equations after respective multiplication by $\dot{y},\dot{x}$ gives:

$$\begin{array}{cc}\hfill \ddot{x}\dot{y}+\dot{x}\ddot{y}& =C\left({\dot{x}}^2{e}^{-2t}-{\dot{y}}^2{e}^{2t}\right),\hfill \end{array}$$

(13)

which through the fourth equation of (9) reduces to:

$$\begin{array}{cc}\hfill 1-1{\partial }_s\left(\dot{x}\dot{y}\right)& =-C\ddot{t}.\hfill \end{array}$$

(14)

Unfortunately, solving this system presents a considerable challenge in the general case. We will look for geodesic lines in characteristic hypersurfaces of ${\mathrm{Sol}}_1^4$ space.

Case 1: $x=x_0$

From the first equation of (12), for $C\ne 0$, we have $y=y_0$ (and hence $\beta =0).$ Then, (11) and (14) imply $z\left(s\right)=\gamma s+z_0$ and $t\left(s\right)=\delta s+t_0.$ Hence,

$$c_1\left(s\right)=(x_0,y_0,\gamma s+z_0,\delta s+t_0).$$

If $C=0,$ then from the second equation of (12) and the fourth equation of (9) we obtain the system $\ddot{y}+2\dot{y}\dot{t}=0$, $\ddot{t}={\dot{y}}^2{e}^{2t}$. Integrating the first equation, we have $\dot{y}=\beta e^{-2t}.$ Substituting this in the second equation, we obtain $\ddot{t}-{\beta }^2·e^{-2t}=0.$ This equation only has a complex solution.

Case 2: $y=y_0$

From the second equation of (12), for $C\ne 0$, we have $x=x_0.$ Then, (11) and (14) again imply $z\left(s\right)=\gamma s+z_0$, and $t\left(s\right)=\delta s+t_0.$ If $C=0,$ then from the first equation of (12) and the fourth equation of (9) we obtain the system $\ddot{x}-2\dot{x}\dot{t}=0$, $\ddot{t}=-{\dot{x}}^2{e}^{-2t}$. Integrating the first equation, we have $\dot{x}=\alpha e^{2t}.$ Substituting in the second equation we obtain $\ddot{t}+{\alpha }^2·e^{2t}=0.$ The solution of this equation is $t\left(s\right)=-ln\left[cosh\left(\alpha s\right)\right].$ Hence, $\dot{x}\left(s\right)=\frac{\alpha }{{cosh}^2\left(\alpha s\right)}$ and $x\left(s\right)=tanh\left(\alpha s\right).$ From the second equation of (9) we have $z\left(s\right)=z_0.$ Thus, we obtain:

$$c_2\left(s\right)=(tanh\left(\alpha s\right),y_0,z_0,-ln\left[cosh\left(\alpha s\right)\right]).$$

Case 3: $z=z_0$

From (11) it follows that $x\dot{y}=const.=C.$ If $C=0$, then we have $x=0$ or $\dot{y}=0$. Both cases are already considered in Case 1 and Case 2, respectively. If $C\ne 0$ we differ two subcases. In the first subcase we assume that both factors are constants, i.e., $x=x_0$ and $\dot{y}=const.=\beta .$ In this subcase, the first equation of (9) leads to a contradiction. In the second subcase we assume that both factors are not constants. Then we can repeat the procedure from (12) and obtain (13). Unfortunately, we could not find a general solution for the obtained system.

Case 4: $t=t_0$

Once more, we consider two cases regarding $C=\gamma -x_0\beta$. For $C=0$, the system (9) is rather simple:

$$\ddot{x}=0,\ddot{y}=0,\dot{z}=x\dot{y},e^{2t_0}{\dot{y}}^2-e^{-2t_0}{\dot{x}}^2=0.$$

Its solution is the following curve:

$$c_3\left(s\right)=\left(\alpha s+x_0,\beta s+y_0,\gamma s+\frac{\alpha \beta }{2}{s}^2+z_0,t_0\right),\mathrm{where}\alpha =\beta e^{2t_0}\mathrm{and}\gamma =x_0\beta .$$

If $C\ne 0$, then from the fourth equation of (9), we obtain $\dot{y}=e^{-2t_0}\dot{x}.$ Substitution of this in the first equation of (9), followed by integration, obtains:

$$x\left(s\right)=\frac{\alpha }{C}{e}^{Cs}\mathrm{and}\mathrm{hence},y\left(s\right)=\frac{\beta }{C}{e}^{Cs},\mathrm{where}\beta =\alpha e^{-2t_0}.$$

From the third equation of (9), we have $z\left(s\right)=Cs+\frac{\alpha \beta }{2C^2}{e}^{2Cs}.$ Hence,

$$c_4\left(s\right)=\left(\frac{\alpha }{C}{e}^{Cs},\frac{\beta }{C}{e}^{Cs},Cs+\frac{\alpha \beta }{2C^2}{e}^{2Cs},t_0\right),\mathrm{where}C=\gamma -x_0\beta ,a=-\beta e^{2t_0}.$$

Note that for $t=0$, the system (9) is:

$$\ddot{x}=-C\dot{y},\ddot{y}=C\dot{x},C=\dot{z}-x\dot{y},{\dot{y}}^2={\dot{x}}^2.$$

The first two equations suggest a solution in sine and cosine functions, but then the fourth equation leads to a contradiction. Therefore, the only possible solution, for $C=0$, has been presented already. □