Slant Helices of (k,m)-Type According to the ED-Frame in Minkowski 4-Space

Abstract

In differential geometry, relations between curves are a large and important area of study for many researchers. Frame areas are an important tool when studying curves, specially the Frenet–Serret frame along a space curve and the Darboux frame along a surface curve in differential geometry. In this paper, we obtain slant helices of $k$-type according to the extended Darboux frame (or, for brevity, ED-frame) field by using the ED-frame field of the first kind (or, for brevity, EDFFK), which is formed with an anti-symmetric matrix for ${\epsilon }_1={\epsilon }_2={\epsilon }_3={\epsilon }_4\in \left\{-1,1\right\}$ and the ED-frame field of the second kind (or, for brevity, EDFSK), which is formed with an anti-symmetric matrix for ${\epsilon }_1={\epsilon }_2={\epsilon }_3={\epsilon }_4\in \left\{-1,1\right\}$ in four-dimensional Minkowski space ${\mathbb{E}}_1^4$. In addition, we present some characterizations of slant helices and determine (k,m)-type slant helices for the EDFFK and EDFSK in Minkowski 4-space.

1. Introduction

In classical differential geometry, curve theory is the most important area of work. Special curves and their characterizations have been studied for a long time and are still being studied. In three-dimensional Euclidean space, the Darboux frame is the velocity of the curve and is formed by the normal vector of the surface, whereas the Frenet–Serret frame is created from the acceleration and velocity of the curve. The Darboux frame field along a surface curve lying on a surface is indicated by {T, D, N} in differential geometry, where T is the unit tangent vector, N is the normal to the restricted surface of the curve, and D = T × N.

The derivatives of the vector fields of this frame are expressed in vector fields containing some real-valued functions. The curvature and torsion for the Frenet–Serret frame and the Darboux frame are called geodesic torsion, geodesic curvature, and normal curvature. Generalizations of the Frenet–Serret frame in higher-dimensional spaces are well known in the literature [1]. Spacelike normal curves in ${\mathbb{E}}_1^4$ whose Frenet frame contains only non-null vector fields, as well as the timelike normal curves in ${\mathbb{E}}_1^4$, in terms of their curvature functions and some special spacelike curves in Minkowski space-time ${\mathbb{E}}_1^4$ were constructed, respectively, by [2] and [3]. An extended Darboux frame field along a non-null curve lying on an orientable non-null hypersurface in Minkowski space-time was presented by Duldul [4]. The definition of the generalized spacelike Mannheim curve in Minkowski space-time was presented by the authors of [5]. Later, definitions of new types of slant helices were presented in Minkowski space-time [6] and four-dimensional Euclidian spaces [7].

In this paper, as given in the Euclidean 4-space, we construct $k$-type helices and $\left(k,m\right)$-type slant helices according to the extended Darboux frame field EDFFK and EDFSK in four-dimensional Minkowski space ${\mathbb{E}}_1^4$.

2. Geometric Preliminaries

Minkowski space-time ${\mathbb{E}}_1^4$ is the real vector space ${ℝ}^4$ provided with the indefinite flat metric given by

$$\langle ,\rangle =-d{a}_1^2+d{a}_2^2+d{a}_3^2+d{a}_4^2,$$

where $\left(a_1,a_2,a_3,a_4\right)$ is a rectangular coordinate system of ${\mathbb{E}}_1^4$. We call $\left({\mathbb{E}}^4,\langle ,\rangle \right)$ a Minkowski 4-space and denote it by ${\mathbb{E}}_1^4$. We say that a vector $a$ in ${\mathbb{E}}_1^4\backslash \left\{0\right\}$ is a spacelike vector, a lightlike vector, or a timelike vector if $\langle a,a\rangle$ is positive, zero, or negative, respectively. In particular, the vector $a=0$ is a spacelike vector. The norm of a vector $a\in {\mathbb{E}}_1^4$ is defined by $\Vert a\Vert =\sqrt{\left|\langle a,a\rangle \right|}$, and a vector $a$ satisfying $\langle a,a\rangle =\mp 1$ is called a unit vector. For any two vectors $a;b$ in ${\mathbb{E}}_1^4$, if $\langle a,b\rangle =0$, then the vectors $a$ and $b$ are said to be orthogonal vectors.

Let $\alpha :I\subset R\to {\mathbb{E}}_1^4$ be an arbitrary curve in ${\mathbb{E}}_1^4$; if all of the velocity vectors of $\alpha$ are spacelike, timelike, and null or lightlike vectors, the curve $\alpha$ is called a spacelike, a timelike, or a null or lightlike curve, respectively [1].

A hypersurface in the Minkowski 4-space is called a spacelike hypersurface if the induced metric on the hypersurface is a positive definite Riemannian metric, and a Lorentzian metric induced on the hypersurface is called a timelike hypersurface. The normal vector of the spacelike hypersurface is a timelike vector and the normal vector of the timelike hypersurface is a spacelike vector. Let $a=\left(a_1,a_2,a_3,a_4\right), b=\left(b_1,b_2,b_3,b_4\right), c=\left(c_1,c_2,c_3,c_4\right)\in {ℝ}_1^4;$ the vector product of $a,b,$ and $c$ is defined with the determinant

$$\begin{array}{l}a\wedge b\wedge c\hfill \end{array}=-\left|\begin{array}{llll}-e_1\hfill & e_2\hfill & e_3\hfill & e_4\hfill \\ a_1\hfill & a_2\hfill & a_3\hfill & a_4\hfill \\ b_1\hfill & b_2\hfill & b_3\hfill & b_4\hfill \\ c_1\hfill & c_2\hfill & c_3\hfill & c_4\hfill \end{array}\right|,$$

where $e_1,e_2,e_3,$ and $e_4$ are mutually orthogonal vectors (standard basis of ${ℝ}_{}^4$) satisfying the equations [1]:

$$\begin{array}{ll}{e}_2\wedge e_3\wedge e_4=e_1,\hfill & e_3\wedge e_4\wedge e_1=e_2, e_4\wedge e_1\wedge e_2=-e_3, e_1\wedge e_2\wedge e_3=e_4.\hfill \\ \hfill & \hfill \end{array}$$

Let $\mathfrak{M}$ be an oriented non-null hypersurface in ${\mathbb{E}}_1^4$ and let $\alpha$ be a non-null regular Frenet curve with speed $v=\Vert {\alpha }^{\prime }\Vert$ on $\mathfrak{M}$. Let $\left\{t,n,b_{\mathbf{1}},b_{\mathbf{2}}\right\}$ be the moving Frenet frame along the curve $\alpha$. Then, the Frenet formulas of $\alpha$ are:

$$\begin{gathered} t^{\prime }={\epsilon }_{n}v{k}_{1}n, \\ {n}^{\prime }=-{\epsilon }_{t}v{k}_{1}t+{\epsilon }_{b_1}v{k}_2{b}_{\mathbf{1}}, \\ {b}_{\mathbf{1}}^{\prime }=-{\epsilon }_{n}v{k}_{2}n-{\epsilon }_t{\epsilon }_n{\epsilon }_{b_1}v{k}_3{b}_{\mathbf{2}}, \\ {b}_{\mathbf{2}}^{\prime }=-{\epsilon }_{b_1}v{k}_3{b}_{\mathbf{1}} \end{gathered}$$

where ${\epsilon }_t=\langle t,t\rangle$, ${\epsilon }_n=\langle n,n\rangle$, ${\epsilon }_{b_1}=\langle b_1,b_1\rangle$, and ${\epsilon }_{b_2}=\langle b_2,b_2\rangle ,$ whereby ${\epsilon }_t, {\epsilon }_n, {\epsilon }_{b_1}, {\epsilon }_{b_2}\in \left\{-1,1\right\}$ and ${\epsilon }_t{\epsilon }_n{\epsilon }_{b_1}{\epsilon }_{b_2}=-1$. The vectors ${\alpha }^{\prime }, {\alpha }^{″}, {\alpha }^{‴},$ and ${\alpha }^{\left(4\right)}$ of a non-null regular curve $\alpha$ are given by

$$\begin{gathered} {\alpha }^{\prime }=vt, \\ {\alpha }^{″}=v^{\prime }t+{\epsilon }_n{v}^2{k}_{1}n, \\ {\alpha }^{‴}=\left(v^{″}-{\epsilon }_t{\epsilon }_n{v}^3{k}_1^2\right)t+{\epsilon }_n\left(3v{v}^{\prime }{k}_1+v^2{k}_1^{\prime }\right)n+{\epsilon }_n{\epsilon }_{b_1}{v}^3{k}_1{k}_2{b}_{\mathbf{1}}, \\ {\alpha }^{\left(4\right)}=\left(\dots \right)t+\left(\dots \right)n+\left(\dots \right)b_{\mathbf{1}}+\left(-{\epsilon }_t{v}^4{k}_1{k}_2{k}_3\right)b_{\mathbf{2}}. \end{gathered}$$

Then, for the Frenet vectors $t,n,b_{\mathbf{1}},b_{\mathbf{2}}$ and the curvatures $k_1,k_2,k_3$ of $\alpha$, we have

$$\begin{gathered} t=\frac{{\alpha }^{\prime }}{\Vert {\alpha }^{\prime }\Vert }, n=\frac{b_1\otimes b_2\otimes {\alpha }^{\prime }}{\Vert b_1\otimes b_2\otimes {\alpha }^{\prime }\Vert }, \\ {b}_{\mathbf{1}}=-{\epsilon }_n\frac{b_2\otimes {\alpha }^{\prime }\otimes {\alpha }^{″}}{\Vert b_2\otimes {\alpha }^{\prime }\otimes {\alpha }^{″}\Vert }, b_{\mathbf{2}}={\epsilon }_{b_1}\frac{{\alpha }^{\prime }\otimes {\alpha }^{″}\otimes {\alpha }^{‴}}{\Vert {\alpha }^{\prime }\otimes {\alpha }^{″}\otimes {\alpha }^{‴}\Vert }, \\ {k}_1=\frac{\langle n,{\alpha }^{″}\rangle }{\Vert {\alpha }^{\prime }{\Vert }^2}, k_2={\epsilon }_n\frac{\langle b_{\mathbf{1}},{\alpha }^{‴}\rangle }{\Vert {\alpha }^{\prime }{\Vert }^3{k}_1}, k_3=-{\epsilon }_t{\epsilon }_{b_2}\frac{\langle b_{\mathbf{2}},{\alpha }^{\left(4\right)}\rangle }{\Vert {\alpha }^{\prime }{\Vert }^4{k}_1{k}_2} \end{gathered}$$

Since the curve $\alpha$ lies on $\mathfrak{M}$, if we denote the unit normal vector field of $\mathfrak{M}$ restricted to $\alpha$ with $N$, and we also have the ED-frame field $\left\{T,E,D,N\right\}$ other than the Frenet frame $\left\{t,n,b_{\mathbf{1}},b_{\mathbf{2}}\right\}$ along $\alpha$, where

$$T=\frac{{\alpha }^{\prime }}{\Vert {\alpha }^{\prime }\Vert }=t,$$

If $\left\{N,T,{\alpha }^{″}\right\}$ is linearly independent:

$$E=\frac{{\alpha }^{″}-\langle {\alpha }^{″},N\rangle N}{\Vert {\alpha }^{″}-\langle {\alpha }^{″},N\rangle N\Vert }$$

$$D=-N\otimes T\otimes E$$

Then, we have the following differential equations for the ED-frame field of the first kind (EDFFK):

$${\left[\begin{array}{l}T\hfill \\ E\hfill \\ D\hfill \\ N\hfill \end{array}\right]}^{\prime }=\left[\begin{array}{llll}0\hfill & {\epsilon }_2{\kappa }_g^1\hfill & 0\hfill & {\epsilon }_4{\kappa }_n\hfill \\ -{\epsilon }_1{\kappa }_g^1\hfill & 0\hfill & {\epsilon }_3{\kappa }_g^2\hfill & {\epsilon }_4{\tau }_g^1\hfill \\ 0\hfill & -{\epsilon }_2{\kappa }_g^2\hfill & 0\hfill & {\epsilon }_4{\tau }_g^2\hfill \\ -{\epsilon }_1{\kappa }_n\hfill & -{\epsilon }_2{\tau }_g^1\hfill & -{\epsilon }_3{\tau }_g^2\hfill & 0\hfill \end{array}\right]\left[\begin{array}{l}T\hfill \\ E\hfill \\ D\hfill \\ N\hfill \end{array}\right]$$

(2.1)

and for the ED-frame field of the second kind (EDFSK):

$${\left[\begin{array}{l}T\hfill \\ E\hfill \\ D\hfill \\ N\hfill \end{array}\right]}^{\prime }=\left[\begin{array}{llll}0\hfill & 0\hfill & 0\hfill & {\epsilon }_4{\kappa }_n\hfill \\ 0\hfill & 0\hfill & {\epsilon }_3{\kappa }_g^2\hfill & {\epsilon }_4{\tau }_g^1\hfill \\ 0\hfill & -{\epsilon }_2{\kappa }_g^2\hfill & 0\hfill & 0\hfill \\ -{\epsilon }_1{\kappa }_n\hfill & -{\epsilon }_2{\tau }_g^1\hfill & 0\hfill & 0\hfill \end{array}\right]\left[\begin{array}{l}T\hfill \\ E\hfill \\ D\hfill \\ N\hfill \end{array}\right]$$

(2.2)

where ${\kappa }_g^i$ and ${\tau }_g^i$ are the geodesic curvature and the geodesic torsion of order $i,\left(i=1,2\right)$, respectively, and ${\epsilon }_1=\langle T,T\rangle , {\epsilon }_2=\langle E,E\rangle , {\epsilon }_3=\langle D,D\rangle , {\epsilon }_4=\langle N,N\rangle$ whereby ${\epsilon }_1,{\epsilon }_2,{\epsilon }_3,{\epsilon }_4\in \left\{-1,1\right\}$. In addition, when ${\epsilon }_i=-1$, then ${\epsilon }_j=1$ for all $j\ne i$, $1\le i,j\le 4$ and ${\epsilon }_1{\epsilon }_2{\epsilon }_3{\epsilon }_4=-1$ [2].

3. Differential Geometry of the ED-Frame in Minkowski 4-Space

In this section, we define some special curves according to the ED-frame of the first kind (EDFFK) and for the ED-frame field of the second kind (EDFSK) in Minkowski 4-space and obtain the Frenet vectors and the curvatures of the curve depending on the invariants of EDFFK and for EDFSK.

Definition 3.1.

Let$\alpha$be a curve in${\mathbb{E}}_1^4$with EDFFK$\left\{T,E,D,N\right\}$. If there exists a non-zero constant vector field$U$in$E_1^4$such that$\langle T,U\rangle =constant$,$\langle E,U\rangle =constant$,$\langle D,U\rangle =constant,$and$\langle N,U\rangle =constant$, then$\alpha$is said to be a$k$-type slant helix and$U$is called the slope axis of$\alpha .$

Theorem 3.1.

Let$\alpha$be a curve with Frenet formulas in EDFFK of the Minkowski space${\mathbb{E}}_1^4$. If the non-null regular$\alpha$is a$1$-type helix (or general helix), then we have

$${\epsilon }_2{\kappa }_g^1\langle E,U\rangle +{\epsilon }_4{\kappa }_n\langle N,U\rangle =0$$

(3.1)

Proof. 

Let the curve $\alpha$ be a $1$-type helix in ${\mathbb{E}}_1^4$, then for a constant field $U$, we obtain

$$\langle T,U\rangle =c$$

(3.2)

which is a constant. Differentiating (3.2) with respect to $s$, we get

$$\langle T^{\prime },U\rangle =0$$

From the Frenet equations in EDFFK (2.1), we have

$$\langle {\epsilon }_2{\kappa }_g^{1}E+{\epsilon }_4{\kappa }_{n}N,U\rangle =0$$

and it follows that (3.1) is true, which completes the proof. □

Theorem 3.2.

Let$\alpha$ be a curve with Frenet formulas in EDFFK of the Minkowski space ${\mathbb{E}}_1^4$. Hence, if the curve $\alpha$ is a $2$-type helix, then we have

$$-{\epsilon }_1{\kappa }_g^1\langle T,U\rangle +{\epsilon }_3{\kappa }_g^2\langle D,U\rangle +{\epsilon }_4{\tau }_g^1\langle N,U\rangle =0$$

(3.3)

Proof. 

Let the curve $\alpha$ be a $2$-type helix. Consider a constant field $U$ such that $\langle E,U\rangle =c_1$ is a constant. Differentiating this equation with respect to $s$, we get

$$\langle E^{\prime },U\rangle =0$$

and using the Frenet equations in EDFFK (2.1), we have Equation (3.3). □

Theorem 3.3.

Let$\alpha$be a curve with the Frenet formulas in EDFFK of the Minkowski space${\mathbb{E}}_1^4$. Then, if the curve$\alpha$is a$3$-type helix, we have the following equation

$$-{\epsilon }_2{\kappa }_g^2\langle E,U\rangle +{\epsilon }_4{\tau }_g^2\langle N,U\rangle =0$$

(3.4)

Proof. 

Let the curve $\alpha$ be a $3$-type helix. Consider a constant field $U$ such that

$$\langle D,U\rangle =c_2$$

(3.5)

is a constant. Differentiating with respect to $s$, we get

$$\langle D^{\prime },U\rangle =0$$

and using the Frenet equations in EDFFK (2.1), we can write (3.4). □

Theorem 3.4.

Let$\alpha$be a curve with the Frenet formulas in EDFFK of the Minkowski space${\mathbb{E}}_1^4$. If the curve$\alpha$is a$4$-type helix, in that case, we have

$$-{\epsilon }_1{\kappa }_n\langle T,U\rangle -{\epsilon }_2{\tau }_g^1\langle E,U\rangle -{\epsilon }_3{\tau }_g^2\langle D,U\rangle =0$$

(3.6)

Proof. 

Let the curve $\alpha$ be a $4$-type helix; then, for a constant field $U$ such that

$$\langle N,U\rangle =c_3$$

(3.7)

$c_3$is a constant. By differentiating (3.7) with respect to $s$, we get

$$\langle N^{\prime },U\rangle =0$$

By using the Frenet equations in EDFFK (2.1), we find (3.6). □

Theorem 3.5.

Let$\alpha$be a curve with the Frenet formulas inEDFSK of the Minkowski space${\mathbb{E}}_1^4$. If the curve$\alpha$is a$1$-type helix (or general helix), then we have

$$\langle E,U\rangle =\frac{-{\epsilon }_1}{{\epsilon }_2}\frac{{\kappa }_n}{{\tau }_g^1}c$$

(3.8)

where$c$is a constant,${\kappa }_n\ne 0$, and${\tau }_g^1\ne 0$.

Proof. 

Let the curve $\alpha$ be a $1$-type helix in ${\mathbb{E}}_1^4$; then, for a constant field $U$, we get

$$\langle T,U\rangle =c$$

(3.9)

which is a constant, and differentiating this equation with respect to $s$, we have

$$\langle T^{\prime },U\rangle =0$$

From the Frenet equations in EDFSK (2.2), we have

$$\langle {\epsilon }_4{\kappa }_{n}N,U\rangle =0$$

and it follows that

$${\epsilon }_4{\kappa }_n\langle N,U\rangle =0$$

If ${\kappa }_n\ne 0$, where ${\epsilon }_4$ is a constant, we can write

$$\langle N,U\rangle =0$$

(3.10)

and differentiating (3.10) with respect to $s$, we obtain

$$\langle N^{\prime },U\rangle =0.$$

Using the Frenet equations in EDFSK and (3.9), we find (3.8). □

Theorem 3.6.

Let α be a curve with the Frenet formulas inEDFSK of the Minkowski space${\mathbb{E}}_1^4$. Hence, if the curve α is a$2$-type helix, we have

$${\epsilon }_3{\kappa }_g^2\langle D,U\rangle +{\epsilon }_4{\tau }_g^1\langle N,U\rangle =0$$

(3.11)

Proof. 

Let the curve $\alpha$ be a $2$-type helix. Then, for a constant field $U$, $\langle E,U\rangle =c_1$ is a constant. Differentiating this equation with respect to $s$, we get

$$\langle E^{\prime },U\rangle =0$$

and using the Frenet equations in EDFSK (2.2), we have (3.11). □

Theorem 3.7.

Let$\alpha$be a curve with the Frenet formulas in EDFSK of the Minkowski space${\mathbb{E}}_1^4$. Then, if the curve$\alpha$is a$3$-type helix, then we have the following equation:

$$\langle N,U\rangle =-\frac{{\epsilon }_3}{{\epsilon }_4}\frac{{\kappa }_g^2}{{\tau }_g^1}{c}_2$$

(3.12)

where$c_2$is a constant,${\kappa }_g^2\ne 0,$and${\tau }_g^1\ne 0$.

Proof. 

Let the curve $\alpha$ be a $3$-type helix; thus, for a constant field $U$,

$$\langle D,U\rangle =c_2$$

(3.13)

is a constant. Differentiating (3.13) with respect to $s$, we get

$$\langle D^{\prime },U\rangle =0$$

Using the Frenet equations in EDFSK (2.2), we can write

$$-{\epsilon }_2{\kappa }_g^2\langle E,U\rangle =0$$

If ${\kappa }_g^2\ne 0$, where ${\epsilon }_2$ is a constant, we get

$$\langle E,U\rangle =0$$

(3.14)

and differentiating (3.14) with respect to $s$, we obtain

$$\langle E^{\prime },U\rangle =0$$

Using the Frenet equations in EDFSK and (3.13), if ${\tau }_g^1\ne 0$, we find (3.12). □

Theorem 3.8.

Let$\alpha$be a curve with Frenet formulas in EDFSK of the Minkowski space${\mathbb{E}}_1^4$. Then, if the curve$\alpha$is a$4$-type helix, we have

$${\epsilon }_1{\kappa }_n\langle T,U\rangle +{\epsilon }_2{\tau }_g^1\langle E,U\rangle =0$$

(3.15)

Proof. 

Let the curve $\alpha$ be a $4$-type helix; then, for a constant field $U$,

$$\langle N,U\rangle =c_3$$

(3.16)

is a constant. By differentiating (3.16) with respect to $s$, we have

$$\langle N^{\prime },U\rangle =0$$

Using the Frenet equations in EDFSK (2.2), we find (3.15). □

4. $\mathbf{\left(}\mathit{k}\mathbf{,}\mathit{m}\mathbf{\right)}$-Type Slant Helices in ${\mathbb{E}}_{\mathbf{1}}^{\mathbf{4}}$

In this section, we will define $\left(k,m\right)$-type slant helices in ${\mathbb{E}}_1^4,$ as in [7].

Definition 4.1.

Let$\alpha$be a curve in${\mathbb{E}}_1^4$with EDFFK (or EDFSK)$\left\{T,E,D,N\right\}$ $\alpha$is called a$\left(k,m\right)$-type slant helix if there exists a non-zero constant vector field$U\in {\mathbb{E}}_1^4$that satisfies$\langle U_k,U\rangle =c_k, \langle U_m,U\rangle =c_m \left(c_k, c_m are constants\right)$($or \langle T,U\rangle =c_1, \langle E,U\rangle =c_2, \langle D,U\rangle =c_3, \langle N,U\rangle =c_4 \left(c_1,c_2,c_3,c_4 are constants\right)$) for$1\le k,m\le 4$,$k\ne m$. The constant vector$U$is on the axis of$\alpha$.

Theorem 4.1.

If the curve$\alpha$ is a $\left(1,2\right)$-type slant helix in ${\mathbb{E}}_1^4$, then we have

$$\langle D,U\rangle =\frac{{\epsilon }_1{\kappa }_g^1{c}_1}{{\epsilon }_3{\kappa }_g^2}+\frac{{\epsilon }_1{\kappa }_g^1{c}_2{\tau }_g^1}{{\epsilon }_3{\kappa }_g^2{\kappa }_n}$$

where$c_1,c_2\mathit{are}\mathit{constants}$.

Proof. 

Let the curve $\alpha$ be a $\left(1,2\right)$-type slant helix in ${\mathbb{E}}_1^4$; then, for a constant field $U$, we have

$$\langle T,U\rangle =c_1$$

(4.1)

which is a constant, and

$$\langle E,U\rangle =c_2$$

(4.2)

which is a constant. Differentiating (4.1) and (4.2) with respect to $s$, we have that

$$\langle T^{\prime },U\rangle =0$$

and

$$\langle E^{\prime },U\rangle =0$$

By using the Frenet equations in EDFFK ((4.1) and (4.2)) the following equations can be obtained:

$${\epsilon }_2{\kappa }_g^1{c}_2+{\epsilon }_4{\kappa }_n\langle N,U\rangle =0$$

(4.3)

$$-{\epsilon }_1{\kappa }_g^1{c}_1+{\epsilon }_3{\kappa }_g^2\langle D,U\rangle +{\epsilon }_4{\tau }_g^1\langle N,U\rangle =0$$

(4.4)

From (4.3), we have that

$$\langle N,U\rangle =-\frac{{\epsilon }_2}{{\epsilon }_4}\frac{{\kappa }_g^1{c}_2}{{\kappa }_n}$$

(4.5)

Substituting (4.5) into (4.4), we find

$$\langle D,U\rangle =\frac{{\epsilon }_1{\kappa }_g^1{c}_1}{{\epsilon }_3{\kappa }_g^2}+\frac{{\epsilon }_1{\kappa }_g^1{c}_2{\tau }_g^1}{{\epsilon }_3{\kappa }_g^2{\kappa }_n}$$

(4.6)

which completes the proof. □

Theorem 4.2.

If the curve$\alpha$ is a $\left(1,3\right)$-type slant helix in ${\mathbb{E}}_1^4$, we have

$${\epsilon }_4\left({\kappa }_g^2{\kappa }_n+{\kappa }_g^1{\tau }_g^2\right)\langle N,U\rangle =0$$

(4.7)

Proof. 

Let the curve $\alpha$ be a $\left(1,3\right)$-type slant helix in ${\mathbb{E}}_1^4$; then, for a constant field $U$, we have

$$\langle T,U\rangle =c_1$$

(4.8)

which is a constant, and

$$\langle D,U\rangle =c_3$$

(4.9)

which is a constant. Differentiating (4.8) and (4.9) with respect to $s$, we get

$$\langle T^{\prime },U\rangle =0$$

and

$$\langle D^{\prime },U\rangle =0$$

Using the Frenet equations in EDFFK, the following equations can be obtained:

$${\epsilon }_2{\kappa }_g^1\langle E,U\rangle +{\epsilon }_4{\kappa }_n\langle N,U\rangle =0$$

(4.10)

$$-{\epsilon }_2{\kappa }_g^2\langle E,U\rangle +{\epsilon }_4{\tau }_g^2\langle N,U\rangle =0$$

(4.11)

Substituting (4.10) into (4.11), we find the following:

$${\epsilon }_4\left({\kappa }_g^2{\kappa }_n+{\kappa }_g^1{\tau }_g^2\right)\langle N,U\rangle =0$$

(4.12)

which completes the proof. □

Theorem 4.3.

If the curve$\alpha$is a$\left(1,4\right)$-type slant helix in${\mathbb{E}}_1^4$, then there exists a constant such that

$$\langle D,U\rangle =-\frac{{\epsilon }_1}{{\epsilon }_3}\frac{{\kappa }_n}{{\tau }_g^2}{c}_1+\frac{{\epsilon }_4}{{\epsilon }_3}\frac{{\kappa }_n}{{\kappa }_g^1}\frac{{\tau }_g^1}{{\tau }_g^2}{c}_4$$

where$c_1,c_4\mathit{are}\mathit{constants}$.

Proof. 

Let the curve $\alpha$ be a $\left(1,4\right)$-type slant helix in ${\mathbb{E}}_1^4$; then, for a constant field $U$, we can write

$$\langle T,U\rangle =c_1$$

(4.13)

which is a constant, and

$$\langle N,U\rangle =c_4$$

(4.14)

which is a constant. Differentiating (4.13) and (4.14) with respect to $s$, we get

$$\langle T^{\prime },U\rangle =0$$

and

$$\langle N^{\prime },U\rangle =0$$

By using the Frenet equations in EDFFK ((4.13) and (4.14)), we have the following equations:

$$\langle E,U\rangle =-\frac{{\epsilon }_4{\kappa }_n}{{\epsilon }_2{\kappa }_g^1}{c}_4,$$

(4.15)

and

$$-{\epsilon }_1{\kappa }_n{c}_1-{\epsilon }_2{\tau }_g^1\langle E,U\rangle -{\epsilon }_3{\tau }_g^2\langle D,U\rangle =0$$

(4.16)

By setting (4.15) into (4.16), we obtain the following:

$$\langle D,U\rangle =-\frac{{\epsilon }_1}{{\epsilon }_3}\frac{{\kappa }_n}{{\tau }_g^2}{c}_1+\frac{{\epsilon }_4}{{\epsilon }_3}\frac{{\kappa }_n}{{\kappa }_g^1}\frac{{\tau }_g^1}{{\tau }_g^2}{c}_4$$

(4.17)

which completes the proof. □

Theorem 4.4.

If the curve$\alpha$is a$\left(2,3\right)$-type slant helix in${\mathbb{E}}_1^4$, then there exist constants$c_2, c_3,$such that

$$\langle T,U\rangle =\frac{{\epsilon }_2}{{\epsilon }_1}\frac{{\kappa }_g^2}{{\kappa }_g^1}\frac{{\tau }_g^1}{{\tau }_g^2}{c}_2+\frac{{\epsilon }_2}{{\epsilon }_1}\frac{{\kappa }_g^2}{{\kappa }_g^1}{c}_3.$$

(4.18)

Proof. 

Let the curve $\alpha$ be a $\left(2,3\right)$-type slant helix in ${\mathbb{E}}_1^4$. Thus, for a constant field $U$, we can write that

$$\langle E,U\rangle =c_2$$

(4.19)

is a constant and that

$$\langle D,U\rangle =c_3$$

(4.20)

is a constant. Differentiating (4.19) and (4.20) with respect to $s$, we find the following equations:

$$\langle E^{\prime },U\rangle =0$$

and

$$\langle D^{\prime },U\rangle =0$$

Using the Frenet equations in EDFFK ((4.19) and (4.20)), we have

$$-{\epsilon }_1{\kappa }_g^1{c}_2+{\epsilon }_4{\tau }_g^2\langle N,U\rangle =0,$$

(4.21)

$$-{\epsilon }_2{\kappa }_g^2\langle T,U\rangle +{\epsilon }_3{\kappa }_g^2{c}_3+{\epsilon }_4{\tau }_g^1\langle N,U\rangle =0$$

(4.22)

From (4.21), we get

$$\langle N,U\rangle =\frac{{\epsilon }_2}{{\epsilon }_4}\frac{{\kappa }_g^2}{{\tau }_g^2}{c}_2,$$

(4.23)

and by setting (4.23) in (4.22), we obtain (4.18). □

Theorem 4.5.

If the curve$\alpha$is a$\left(2,4\right)$-type slant helix in${\mathbb{E}}_1^4$, then there exists a constant such that

$$\langle T,U\rangle =\frac{{\epsilon }_4{\tau }_g^1{c}_4{\epsilon }_3{\tau }_g^2-{\epsilon }_3{\kappa }_g^2{\epsilon }_2{\tau }_g^1{c}_2}{{\epsilon }_1{\kappa }_g^1{\epsilon }_3{\tau }_g^2+{\epsilon }_3{\kappa }_g^2{\epsilon }_1{\kappa }_n}$$

and

$$\langle D,U\rangle =\frac{-{\kappa }_g^1{\epsilon }_2{\tau }_g^1{c}_2-{\epsilon }_3{\tau }_g^2{\kappa }_n}{{\kappa }_g^1{\epsilon }_3{\tau }_g^2+{\epsilon }_3{\kappa }_g^2{\kappa }_n}$$

Proof. 

Let the curve $\alpha$ be a $\left(2,4\right)$-type slant helix in ${\mathbb{E}}_1^4$; then, for a constant field $U$, we can write the following equations:

$$\langle E,U\rangle =c_2$$

(4.24)

where $c_2$is a constant, and

$$\langle N,U\rangle =c_4,$$

(4.25)

where $c_4$is a constant. By differentiating (4.24) and (4.25) with respect to $s$, we get the following equations:

$$\langle E^{\prime },U\rangle =0$$

and

$$\langle N^{\prime },U\rangle =0$$

By using the Frenet equations in EDFFK ((4.24) and (4.25)), we have the following:

$$-{\epsilon }_1{\kappa }_g^1\langle T,U\rangle +{\epsilon }_3{\kappa }_g^2\langle D,U\rangle =-{\epsilon }_4{\tau }_g^1{c}_4$$

(4.26)

$$-{\epsilon }_1{\kappa }_n\langle T,U\rangle -{\epsilon }_3{\tau }_g^2\langle D,U\rangle ={\epsilon }_2{\tau }_g^1{c}_2$$

(4.27)

Substituting (4.26) in (4.27), we obtain equations in this theorem. □

Theorem 4.6.

If the curve$\alpha$is a$\left(3,4\right)$-type slant helix in${\mathbb{E}}_1^4$, then we have

$$\langle T,U\rangle =-\frac{{\epsilon }_3}{{\epsilon }_1}\frac{{\tau }_g^2}{{\kappa }_n}{c}_3-\frac{{\epsilon }_4}{{\epsilon }_1}\frac{{\tau }_g^1}{{\kappa }_n}\frac{{\tau }_g^2}{{\kappa }_g^2}{c}_4$$

(4.28)

where$c_3,c_4$are constants.

Proof. 

Let the curve $\alpha$ be a $\left(3,4\right)$-type slant helix in ${\mathbb{E}}_1^4$; then, for a constant field $U$, we can write that

$$\langle D,U\rangle =c_3$$

(4.29)

is a constant and that

$$\langle N,U\rangle =c_4$$

(4.30)

is a constant. By differentiating (4.29) and (4.30) with respect to $s$, we get

$$\langle D^{\prime },U\rangle =0$$

and

$$\langle N^{\prime },U\rangle =0.$$

By using the Frenet formulas in EDFFK ((4.29) and (4.30)), we get the following:

$$-{\epsilon }_2{\kappa }_g^2\langle E,U\rangle +{\epsilon }_4{\tau }_g^2{c}_4=0,$$

(4.31)

$$-{\epsilon }_1{\kappa }_n\langle T,U\rangle -{\epsilon }_2{\tau }_g^1\langle E,U\rangle -{\epsilon }_3{\tau }_g^2{c}_3=0,$$

(4.32)

From (4.31), we have the following equation:

$$\langle E,U\rangle =\frac{{\epsilon }_4}{{\epsilon }_2}\frac{{\tau }_g^2}{{\kappa }_g^2}{c}_4$$

(4.33)

and by setting (4.33) into (4.32), we obtain (4.28). □

Theorem 4.7.

If the curve$\alpha$is a$\left(1,2\right)$-type slant helix in${\mathbb{E}}_1^4$, then we have

$${\kappa }_n=0,{\tau }_g^1=0\mathrm{and}{\kappa }_g^2=0,$$

$${\kappa }_n\ne 0,{\tau }_g^1=-\frac{{\epsilon }_1}{{\epsilon }_2}\frac{c_1}{c_2}{\kappa }_n,$$

where$c_1,c_2\mathit{are}\mathit{constants}$.

Proof. 

Let the curve $\alpha$ be a $\left(1,2\right)$-type slant helix in ${\mathbb{E}}_1^4$; then, for a constant field $U$, we can write that

$$\langle T,U\rangle =c_1$$

(4.34)

is a constant and that

$$\langle E,U\rangle =c_2$$

(4.35)

is a constant. Differentiating (4.34) and (4.35) with respect to $\sigma$, we have that

$$\langle T^{\prime },U\rangle =0$$

and

$$\langle E^{\prime },U\rangle =0$$

Using the Frenet equations in EDFSK ((4.34) and (4.35)) satisfies the following equalities:

$${\epsilon }_4{\kappa }_n\langle N,U\rangle =0$$

(4.36)

$${\epsilon }_3{\kappa }_g^2\langle D,U\rangle +{\epsilon }_4{\tau }_g^1\langle N,U\rangle =0$$

(4.37)

where ${\epsilon }_4$ is a constant, and we evaluate the terms ${\kappa }_n$ and $\langle N,U\rangle$ in (4.36):

• For ${\kappa }_n=0$ and $\langle N,U\rangle =0$

  $$\langle N^{\prime },U\rangle =0$$

  (4.38)
  Using the Frenet equations in EDFSK, we can write

  $${\epsilon }_1{\kappa }_n{c}_1+{\epsilon }_2{\tau }_g^1{c}_2=0$$

  (4.39)
  We obtain ${\tau }_g^1=0$.
  Under these conditions, we evaluate the terms ${\kappa }_g^2$ and $\langle D,U\rangle$ in (4.37):
  • For ${\kappa }_g^2=0$ and $\langle D,U\rangle =0$

    $$\langle D^{\prime },U\rangle =0$$

    (4.40)
  Using the Frenet equations in EDFSK, we find

  $$-{\epsilon }_2{\kappa }_g^2\langle E,U\rangle =0$$

  (4.41)
  From (4.35), we get

  $${\epsilon }_2{\kappa }_g^2{c}_2=0,$$

  (4.42)

  where ${\epsilon }_2,c_2$ are constants, and we obtain ${\kappa }_g^2=0$.

  ii.

  For ${\kappa }_g^2\ne 0$ and $\langle D,U\rangle =0$,

  $$\langle D^{\prime },U\rangle =0,$$

  (4.43)

  and by using the Frenet equations in EDFSK, we find

  $$-{\epsilon }_2{\kappa }_g^2\langle E,U\rangle =0.$$

  (4.44)

  From (4.35), we get

  $${\epsilon }_2{\kappa }_g^2{c}_2=0$$

  (4.45)

  where ${\epsilon }_2,c_2$ are constants, and we obtain ${\kappa }_g^2=0$. This is a contradiction. Then, it should be different from zero.

  iii.

  For ${\kappa }_g^2=0$ and $\langle D,U\rangle \ne 0$,

  $\langle D,U\rangle$ is a constant; therefore, the same results as in case (i) are obtained.
• For ${\kappa }_n\ne 0$ and $\langle N,U\rangle =0$

  $$\langle N^{\prime },U\rangle =0,$$

  (4.46)

  and by using (4.34) and (4.35), and the Frenet equations in EDFSK, we can write

  $${\tau }_g^1=-\frac{{\epsilon }_1}{{\epsilon }_2}\frac{c_1}{c_2}{\kappa }_n$$

  (4.47)
  Then, we find that ${\tau }_g^1$ is a constant.
  Under these conditions, the same results are obtained with cases (i), (ii), and (iii).
• For ${\kappa }_n=0$ and $\langle N,U\rangle \ne 0$
  $\langle N,U\rangle =0$ is a constant, so $\langle N^{\prime },U\rangle =0;$ therefore, the same results as in case (1) are obtained.
  This completes the proof. □

Theorem 4.8.

If the curve$\alpha$is a$\left(1,3\right)$-type slant helix in${\mathbb{E}}_1^4$, then we have

$${\kappa }_n=0,{\tau }_g^1=0\mathrm{and}{\kappa }_g^2=0,$$

$${\kappa }_n\ne 0,{\tau }_g^1\ne 0\mathrm{and}{\kappa }_g^2=0$$

Proof. 

Let the curve $\alpha$ be a $\left(1,3\right)$-type slant helix in ${\mathbb{E}}_1^4$; then, for a constant field $U$, we can write that

$$\langle T,U\rangle =c_1$$

(4.48)

is a constant and that

$$\langle D,U\rangle =c_3$$

(4.49)

is a constant. Differentiating (4.48) and (4.49) with respect to $\sigma$, we have that

$$\langle T^{\prime },U\rangle =0$$

and

$$\langle D^{\prime },U\rangle =0$$

Using the Frenet equations in EDFSK satisfies the following equalities:

$${\epsilon }_4{\kappa }_n\langle N,U\rangle =0,$$

(4.50)

$${\epsilon }_2{\kappa }_g^2\langle E,U\rangle =0$$

(4.51)

where ${\epsilon }_2,{\epsilon }_4$ are constants. We shall evaluate the terms ${\kappa }_n$ and $\langle N,U\rangle$ in (4.50):

• For ${\kappa }_n=0$ and $\langle N,U\rangle =0$

  $$\langle N^{\prime },U\rangle =0.$$

  (4.52)
  Using (4.48) and the Frenet equations in EDFSK, we can write

  $${\epsilon }_1{\kappa }_n{c}_1+{\epsilon }_2{\tau }_g^1\langle E,U\rangle =0$$

  (4.53)
  We obtain ${\epsilon }_2{\tau }_g^1\langle E,U\rangle =0$.
  Under these conditions, we shall evaluate the terms ${\tau }_g^1$ and $\langle E,U\rangle$ in (4.53):
  • For ${\tau }_g^1=0$ and $\langle E,U\rangle =0$

    $$\langle E^{\prime },U\rangle =0.$$

    (4.54)
  By using the Frenet equations in EDFSK, we find

  $${\epsilon }_3{\kappa }_g^2\langle D,U\rangle +{\epsilon }_4{\tau }_g^1\langle N,U\rangle =0$$

  (4.55)
  From (4.49), we get

  $${\epsilon }_3{\kappa }_g^2{c}_3=0,$$

  (4.56)

  where ${\epsilon }_3,c_3$ are constants, and we obtain ${\kappa }_g^2=0$.

  ii.

  For ${\tau }_g^1\ne 0$, $\langle E,U\rangle =0$

  $$\langle E^{\prime },U\rangle =0$$

  (4.57)

  Using the Frenet equations in EDFSK, we find

  $${\epsilon }_3{\kappa }_g^2\langle D,U\rangle +{\epsilon }_4{\tau }_g^1\langle N,U\rangle =0,$$

  (4.58)

  and from (4.49), we obtain

  $${\epsilon }_3{\kappa }_g^2{c}_3=0$$

  (4.59)

  where ${\epsilon }_2,c_2$ are constants; thus, we find ${\kappa }_g^2=0$.

  iii.

  For ${\kappa }_g^2=0$ and $\langle D,U\rangle \ne 0$,

  $\langle D,U\rangle$ is a constant; therefore, the same results as in case (i) are obtained.
• For ${\kappa }_n\ne 0$ and $\langle N,U\rangle =0$

  $$\langle N^{\prime },U\rangle =0$$

  (4.60)
  By using (4.48) and the Frenet equations in EDFSK, we can write

  $$\langle E,U\rangle =-\frac{{\epsilon }_1}{{\epsilon }_2}\frac{{\kappa }_n{c}_1}{{\tau }_g^1}$$

  (4.61)

  where ${\tau }_g^1\ne 0$. By setting (4.61) into (4.51), we obtain

  $$\frac{{\kappa }_n{\kappa }_g^2{c}_1}{{\tau }_g^1}=0.$$

  (4.62)
  Hence, we get ${\kappa }_g^2=0$.
• For ${\kappa }_n=0$ and $\langle N,U\rangle \ne 0$
  $\langle N,U\rangle =0$ is a constant, so $\langle N^{\prime },U\rangle =0$ therefore, the same results as in case (1) are obtained.
  This completes the proof. □

Theorem 4.9.

If the curve$\alpha$is a$\left(1,4\right)$-type slant helix in${\mathbb{E}}_1^4$, then we get

$$\langle D,U\rangle =-\frac{{\epsilon }_4{\tau }_g^1{c}_4}{{\epsilon }_3{\kappa }_g^2},$$

(4.63)

where${\kappa }_g^2\ne 0$and${\kappa }_n=0$.

Proof. 

Let the curve $\alpha$ be a $\left(1,4\right)$-type slant helix in ${\mathbb{E}}_1^4$; then, for a constant field $U$, we can write that

$$\langle T,U\rangle =c_1$$

(4.64)

is a constant and that

$$\langle N,U\rangle =c_4$$

(4.65)

is a constant. Differentiating (4.64) and (4.65) with respect to $\sigma$, we have that

$$\langle T^{\prime },U\rangle =0$$

and

$$\langle N^{\prime },U\rangle =0.$$

Using (4.64), (4.65), and the Frenet equations in EDFSK satisfies the following equalities:

$${\epsilon }_4{\kappa }_n{c}_4=0$$

(4.66)

$$\langle E,U\rangle =-\frac{{\epsilon }_1{\kappa }_n{c}_1}{{\epsilon }_2{\tau }_g^1}$$

(4.67)

where ${\epsilon }_1,{\epsilon }_2,{\epsilon }_4$ are constants, and ${\tau }_g^1\ne 0$.

From (4.66), for ${\kappa }_n=0,$ we find $\langle E,U\rangle =0$. So, $\langle E^{\prime },U\rangle =0$. Using the Frenet equations in EDFSK, we get

$$\langle D,U\rangle =-\frac{{\epsilon }_4{\tau }_g^1{c}_4}{{\epsilon }_3{\kappa }_g^2}.$$

(4.68)

This completes the proof. □

Theorem 4.10.

If the curve$\alpha$is a$\left(2,3\right)$-type slant helix in${\mathbb{E}}_1^4$, then we have

$${\kappa }_n=0,\mathrm{For}{\tau }_g^1=0$$

$$\langle T,U\rangle =-\frac{{\epsilon }_2}{{\epsilon }_1}\frac{{\tau }_g^1}{{\kappa }_n}{c}_2,\mathrm{For}{\tau }_g^1\ne 0.$$

Proof. 

Let the curve $\alpha$ be a $\left(2,3\right)$-type slant helix in ${\mathbb{E}}_1^4$; then, for a constant field $U$, we can write that

$$\langle E,U\rangle =c_2$$

(4.69)

is a constant and that

$$\langle D,U\rangle =c_3$$

(4.70)

is a constant. Differentiating (4.69) and (4.70) with respect to $\sigma$, we have that

$$\langle E^{\prime },U\rangle =0$$

and

$$\langle D^{\prime },U\rangle =0$$

Using (4.69), (4.70), and the Frenet equations in EDFSK satisfies the following equalities:

$${\epsilon }_3{\kappa }_g^2{c}_3+{\epsilon }_4{\tau }_g^1\langle N,U\rangle =0$$

(4.71)

$${\epsilon }_2{\kappa }_g^2{c}_2=0,$$

(4.72)

where ${\epsilon }_2,{\epsilon }_3,{\epsilon }_4$ are constants. From (4.72), we get

$${\kappa }_g^2=0$$

(4.73)

By setting (4.69) into (4.70), we have the following equation:

$${\epsilon }_4{\tau }_g^1\langle N,U\rangle =0.$$

(4.74)

We shall evaluate terms ${\tau }_g^1$ and $\langle N,U\rangle$ in (4.74):

• For ${\tau }_g^1=0$ and $\langle N,U\rangle =0$,

  $$\langle N^{\prime },U\rangle =0,$$

  (4.75)

  and by using the Frenet equations in EDFSK, we can write

  $${\epsilon }_1{\kappa }_n\langle T,U\rangle =0.$$

  (4.76)
  Under these conditions, we find ${\kappa }_n=0$.
• For ${\tau }_g^1\ne 0$ and $\langle N,U\rangle =0$,

  $$\langle N^{\prime },U\rangle =0,$$

  (4.77)

  and by using (4.69), (4.70), and the Frenet equations in EDFSK, we can write

  $$\langle T,U\rangle =-\frac{{\epsilon }_2}{{\epsilon }_1}\frac{{\tau }_g^1}{{\kappa }_n}{c}_2$$

  (4.78)
• For ${\tau }_g^1=0$ and $\langle N,U\rangle \ne 0$,
  $\langle N,U\rangle =0$ is a constant, so $\langle N^{\prime },U\rangle =0;$ therefore, the same results as in case (2) are obtained. This completes the proof. □

Theorem 4.11.

If the curve$\alpha$is a$\left(2,4\right)$-type slant helix in${\mathbb{E}}_1^4$, then we have

$$\langle D,U\rangle =-\frac{{\epsilon }_4}{{\epsilon }_3}\frac{{\tau }_g^1}{{\kappa }_g^2}{c}_4,$$

(4.79)

$$\langle T,U\rangle =-\frac{{\epsilon }_2}{{\epsilon }_1}\frac{{\tau }_g^1}{{\kappa }_n}{c}_2,$$

(4.80)

where$c_2,c_4$are constants.

Proof. 

Let the curve $\alpha$ be a $\left(2,4\right)$-type slant helix in ${\mathbb{E}}_1^4$; then, for a constant field $U$, we can write that

$$\langle E,U\rangle =c_2$$

(4.81)

is a constant and that

$$\langle N,U\rangle =c_4$$

(4.82)

is a constant. Differentiating (4.81) and (4.82) with respect to $\sigma$, we have that

$$\langle E^{\prime },U\rangle =0$$

and

$$\langle N^{\prime },U\rangle =0.$$

Equations (4.81) and (4.82), and the Frenet equations in EDFSK involve

$${\epsilon }_3{\kappa }_g^2\langle D,U\rangle +{\epsilon }_4{\tau }_g^1{c}_4=0$$

(4.83)

$${\epsilon }_1{\kappa }_n\langle T,U\rangle +{\epsilon }_2{\tau }_g^1{c}_2=0,$$

(4.84)

where ${\epsilon }_1,{\epsilon }_2,{\epsilon }_3,{\epsilon }_4$ are constants. From (4.83) and (4.84), we obtain

$$\langle D,U\rangle =-\frac{{\epsilon }_4}{{\epsilon }_3}\frac{{\tau }_g^1}{{\kappa }_g^2}{c}_4,$$

(4.85)

$$\langle T,U\rangle =-\frac{{\epsilon }_2}{{\epsilon }_1}\frac{{\tau }_g^1}{{\kappa }_n}{c}_2$$

(4.86)

This completes the proof. □

Theorem 4.12.

If the curve$\alpha$is a$\left(3,4\right)$-type slant helix in${\mathbb{E}}_1^4$, then we have

$${\tau }_g^1=0,\mathrm{For}{\kappa }_g^2=0$$

$${\tau }_g^1=-\frac{{\epsilon }_3}{{\epsilon }_4}\frac{c_3}{c_4}{\kappa }_g^2,\mathrm{For}{\kappa }_g^2\ne 0$$

$$\langle T,U\rangle =-\frac{{\epsilon }_2}{{\epsilon }_1}\frac{{\tau }_g^1\langle E,U\rangle }{{\kappa }_n},\mathrm{For}{\kappa }_g^2=0,$$

where$c_3,c_4$are constants.

Proof. 

Let the curve $\alpha$ be a $\left(3,4\right)$-type slant helix in ${\mathbb{E}}_1^4$; then, for a constant field $U$, we can write that

$$\langle D,U\rangle =c_3$$

(4.87)

is a constant and that

$$\langle N,U\rangle =c_4$$

(4.88)

is a constant. Differentiating (4.87) and (4.88) with respect to $\sigma$, we have that

$$\langle D^{\prime },U\rangle =0$$

and

$$\langle N^{\prime },U\rangle =0.$$

Using the Frenet equations in EDFSK satisfies the following equations:

$${\epsilon }_2{\kappa }_g^2\langle E,U\rangle =0,$$

(4.89)

$${\epsilon }_1{\kappa }_n\langle T,U\rangle +{\epsilon }_2{\tau }_g^1\langle E,U\rangle =0,$$

(4.90)

where ${\epsilon }_1,{\epsilon }_2$ are constants. We shall evaulate terms ${\kappa }_g^2$ and $\langle E,U\rangle$ in (4.89):

• For ${\kappa }_g^2=0$ and $\langle E,U\rangle =0$,

  $$\langle E^{\prime },U\rangle =0,$$

  and using (4.87), (4.88), and the Frenet equations in EDFSK, we can write

  $${\epsilon }_3{\kappa }_g^2{c}_3+{\epsilon }_4{\tau }_g^1{c}_4=0.$$
  Under these conditions, we find ${\tau }_g^1=0$. Therefore, we shall evaluate terms ${\kappa }_n$ and $\langle T,U\rangle$ in (4.90):
  • For ${\kappa }_n=0$ and $\langle T,U\rangle =0$,

    $$\langle T^{\prime },U\rangle =0.$$
  By using (4.88) and the Frenet equations in EDFSK, we can write

  $${\epsilon }_4{\kappa }_n{c}_4=0.$$
  So, we find ${\kappa }_n=0$.

  ii.

  For ${\kappa }_n\ne 0$ and $\langle T,U\rangle =0$,

  $$\langle T^{\prime },U\rangle =0.$$

  By using (4.88) and the Frenet equations in EDFSK, we can write

  $${\epsilon }_4{\kappa }_n{c}_4=0.$$
  So, we find ${\kappa }_n=0$. This is a contradiction. Thus, it cannot be different from zero.

  iii.

  For ${\kappa }_n=0$ and $\langle T,U\rangle \ne 0$,

  $\langle T,U\rangle =0$ is a constant, so $\langle T^{\prime },U\rangle =0;$ therefore, the same results as in case (i) are obtained.
• For ${\kappa }_g^2\ne 0$ and $\langle E,U\rangle =0$,

  $$\langle E^{\prime },U\rangle =0$$
  By using (4.87), (4.88), and the Frenet equations in EDFSK, we can write

  $${\epsilon }_3{\kappa }_g^2{c}_3+{\epsilon }_4{\tau }_g^1{c}_4=0.$$
  Under these conditions, we find ${\tau }_g^1=-\frac{{\epsilon }_3}{{\epsilon }_4}\frac{c_3}{c_4}{\kappa }_g^2$.
• For ${\kappa }_g^2=0$ and $\langle E,U\rangle \ne 0$,
  $\langle E,U\rangle$ is a constant, so $\langle E^{\prime },U\rangle =0;$ therefore, the same results as in case (1) are obtained. Hence, from (4.90), we get

  $$\langle T,U\rangle =-\frac{{\epsilon }_2}{{\epsilon }_1}\frac{{\tau }_g^1\langle E,U\rangle }{{\kappa }_n}$$
  This completes the proof. □

5. Conclusions

In this paper, we investigate curvatures and torsions according to the extended Darboux frame field, which is formed with anti-symmetric matrixes for ${\epsilon }_1={\epsilon }_2={\epsilon }_3={\epsilon }_4\in \left\{-1,1\right\},$ and slant helices are given in four-dimensional Euclidean space; we relate these results and classify (k,m)-type slant helices in four-dimensional Minkowski space ${\mathbb{E}}_1^4$.