# Generalized Mannheim Curves in Minkowski space-time $E_{1}^{4}$

Soley Ersoya , Murat Tosuna , Hiroo Matsudab
<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
a Department of Mathematics, Sakarya University, Sakarya, TURKEY
b Department Mathematics, Kanazawa Medical University, Uchinada, Ishikawa,
920-02, JAPAN

###### Abstract

In this paper, the definition of generalized spacelike Mannheim curve in
Minkowski space-time $E_{1}^{4}$ is given. The necessary and sufficient
conditions for the generalized spacelike Mannheim curve are obtained. Also,
some characterizations of Mannheim curve are given.

Mathematics Subject Classification (2010): 53B30, 53A35, 53A04.

Keywords: Mannheim curve, Minkowski space-time

##  1 Introduction

The curves are a fundamental structure of differential geometry. An increasing
interest of the theory of curves makes a development of special curves to be
examined. A way to classification and characterization of curves is the
relationship between the Frenet vectors of the curves. For example, Saint
Venant proposed the question whether upon the surface generated by the
principal normal of a curve, a second curve can exist which has for its
principal normal of the given curve in 1845. This question was answered by
Bertrand in 1850. He showed that a necessary and sufficient condition for the
existence of such a second curve is that a linear relationship with constant
coefficients exists between the first and second curvatures of the given
original curve. The pairs of curves of this kind have been called Bertrand
partner curves or more commonly Bertrand curves [10], [14], [2]. There are
many works related with Bertrand curves in the Euclidean space and Minkowski
space, [15]–[3]. Also, generalized Bertrand curves in Euclidean 4- space are
defined and characterized in [6]. Another kind of associated curve have been
called Mannheim curve and Mannheim partner curve. The notion of Mannheim
curves was discovered by A. Mannheim in 1878. These curves in Euclidean
3-space are characterized in terms of the curvature and torsion as follows: A
space curve is a Mannheim curve if and only if its curvature $\kappa$ and
torsion $\tau$ satisfy the relation

$\kappa\left(s\right)=\alpha\left({{\kappa^{2}}\left(s\right)+{\tau^{2}}\left(s\right)}\right)$

for some constant $\alpha$. The articles concerning Mannheim curves are rather
few. In [13], a remarkable class of Mannheim curves is studied. General
Mannheim curves in the Euclidean 3-space are obtained in [11]. Mannheim
partner curves in Euclidean 3-space and Minkowski 3-space are studied and the
necessary and sufficient conditions for the Mannheim partner curves are
obtained in [5], [8]. Recently, Mannheim curves are generalized and some
characterizations and examples of generalized Mannheim curves in Euclidean
4-space ${E^{4}}$ are given by [7].

In this paper, we study the generalized spacelike Mannheim partner curves in
$4-$dimensional Minkowski space-time. We will give the necessary and
sufficient conditions for the generalized spacelike Mannheim partner curves.

##  2 Preliminaries

The basic concepts of the theory of curves in Minkowski space-time ${E^{4}}$
are briefly presented in this section. A more complete elementary treatment
can be found in [1]. Minkowski space-time ${E_{1}^{4}}$ is an Euclidean space
provided with the standard flat metric given by

$\left\langle{\,\,,\,\,}\right\rangle=-dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2}+dx_{4}^{2}$

where $\left({{x_{1}},\,{x_{2}},\,{x_{3}},\,{x_{4}}}\right)$ is a rectangular
coordinate system in ${E^{4}}$.

Since $\left\langle{\;,\;}\right\rangle$ is an indefinite metric, recall that
a vector ${\bf{v}}\in E_{1}^{4}$ can have one of the three causal characters;
it can be spacelike if $\left\langle{{\bf{v}},{\bf{v}}}\right\rangle>0$ or
${\bf{v}}={\bf{0}}$, timelike if
$\left\langle{{\bf{v}},{\bf{v}}}\right\rangle<0$ and null (lightlike) if
$\left\langle{{\bf{v}},{\bf{v}}}\right\rangle=0$ and ${\bf{v}}\neq{\bf{0}}$ .
Similarly, an arbitrary curve ${\bf{c}}={\bf{c}}\left(s\right)$ in ${E^{4}}$
can locally be spacelike, timelike or null (lightlike) if all of its velocity
vectors ${\bf{c^{\prime}}}\left(s\right)$ are, respectively, spacelike,
timelike or null. The norm of ${\bf{v}}\in E_{1}^{4}$ is given by
$\left\|{\bf{v}}\right\|=\sqrt{\left|{\left\langle{{\bf{v}},{\bf{v}}}\right\rangle}\right|}$.
If
$\left\|{{\bf{c^{\prime}}}\left(s\right)}\right\|=\sqrt{\left|{\left\langle{{\bf{c^{\prime}}}\left(s\right),{\bf{c^{\prime}}}\left(s\right)}\right\rangle}\right|}\neq
0$ for all $s\in L$, then $C$ is a regular curve in $E_{1}^{4}$. A spacelike
(timelike) regular curve $C$ is parameterized by arc-length parameter $s$
which is given by ${\bf{c}}:L\to E_{1}^{4}$, then the tangent vector
${\bf{c^{\prime}}}\left(s\right)$ along $C$ has unit length, that is,

$\left\langle{{\bf{c}}\left(s\right),{\bf{c}}\left(s\right)}\right\rangle=1\,,\,\quad\left({\left\langle{{\bf{c}}\left(s\right),{\bf{c}}\left(s\right)}\right\rangle=-1}\right)$

for all $s\in L$ .

Hereafter, curves are considered spacelike and regular ${C^{\infty}}$ curves
in $E_{1}^{4}$. Let
${{\bf{e}}_{1}}\left(s\right)={\bf{c^{\prime}}}\left(s\right)$ for all $s\in
L$, then the vector field ${{\bf{e}}_{1}}\left(s\right)$ is spacelike and it
is called spacelike unit tangent vector field on $C$.

The spacelike curve $C$ is called special spacelike Frenet curve if there
exist three smooth functions ${k_{1}}$, ${k_{2}}$, ${k_{3}}$ on $C$ and smooth
non-null frame field
$\left\\{{{{\bf{e}}_{1}},\,{{\bf{e}}_{2}},\,{{\bf{e}}_{3}},\,{{\bf{e}}_{4}}}\right\\}$
along the curve $C$. Also, the functions ${k_{1}},\,{k_{2}}$, and ${k_{3}}$
are called the first, the second, and the third curvature function on $C$,
respectively. For the ${C^{\infty}}$ special spacelike Frenet curve $C$, the
following Frenet formula is hold

$\left[\begin{array}[]{l}{{{\bf{e^{\prime}}}}_{1}}\\\
{{{\bf{e^{\prime}}}}_{2}}\\\ {{{\bf{e^{\prime}}}}_{3}}\\\
{{{\bf{e^{\prime}}}}_{4}}\\\
\end{array}\right]=\,\left[\begin{array}[]{l}\,\,\,0\,\,\,\,\,\,\,\,\,\,{k_{1}}\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,0\\\
{\mu_{1}}{k_{1}}\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,{k_{2}}\,\,\,\,\,\,\,0\\\
\,\,\,0\,\,\,\,\,\,\,{\mu_{2}}{k_{2}}\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,{k_{3}}\\\
\,\,\,0\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,{\mu_{3}}{k_{3}}\,\,\,\,0\\\
\end{array}\right]\left[\begin{array}[]{l}{{\bf{e}}_{1}}\\\
{{\bf{e}}_{\rm{2}}}\\\ {{\bf{e}}_{\rm{3}}}\\\ {{\bf{e}}_{4}}\\\
\end{array}\right]$

where ${\mu_{i}}=\mp 1,\,\,1\leq i\leq 3$, [1].
Due to characters of Frenet vectors of the spacelike curve $C$,
${\mu_{i}}\,\,\left({1\leq i\leq 3}\right)$ are defined as in the following
three subcases; Case 1: If ${{\bf{e}}_{4}}$ is timelike, then
${\mu_{i}},\,\,1\leq i\leq 3$ are

${\mu_{1}}={\mu_{2}}=-1\,\,,\,\,\,{\mu_{3}}=1$

where ${{\bf{e}}_{1}},\,{{\bf{e}}_{2}},\,{{\bf{e}}_{3}}$ and ${{\bf{e}}_{4}}$
are mutually orthogonal vector fields satisfying equations

$\left\langle{{{\bf{e}}_{1}}\,,\,{{\bf{e}}_{1}}}\right\rangle=\left\langle{{{\bf{e}}_{2}}\,,\,{{\bf{e}}_{2}}}\right\rangle=\left\langle{{{\bf{e}}_{3}}\,,\,{{\bf{e}}_{3}}}\right\rangle=1\,\,,\,\,\left\langle{{{\bf{e}}_{4}}\,,\,{{\bf{e}}_{4}}}\right\rangle=-1.$

Case 2: If ${{\bf{e}}_{3}}$ is timelike, then ${\mu_{i}},\,\,1\leq i\leq 3$
are

${\mu_{1}}=-1\,\,,\,\,{\mu_{2}}={\mu_{3}}=1$

where ${{\bf{e}}_{1}},\,{{\bf{e}}_{2}},\,{{\bf{e}}_{3}}$ and ${{\bf{e}}_{4}}$
are mutually orthogonal vector fields satisfying equations

$\left\langle{{{\bf{e}}_{1}}\,,\,{{\bf{e}}_{1}}}\right\rangle=\left\langle{{{\bf{e}}_{2}}\,,\,{{\bf{e}}_{2}}}\right\rangle=\left\langle{{{\bf{e}}_{4}}\,,\,{{\bf{e}}_{4}}}\right\rangle=1\,\,,\,\,\left\langle{{{\bf{e}}_{3}}\,,\,{{\bf{e}}_{3}}}\right\rangle=-1.$

Case 3: If ${{\bf{e}}_{2}}$ is timelike, then ${\mu_{i}},\,\,1\leq i\leq 3$
are

${\mu_{1}}={\mu_{2}}=1\,\,,\,\,{\mu_{3}}=-1$

where ${{\bf{e}}_{1}},\,{{\bf{e}}_{2}},\,{{\bf{e}}_{3}}$ and ${{\bf{e}}_{4}}$
are mutually orthogonal vector fields satisfying equations

$\left\langle{{{\bf{e}}_{1}}\,,\,{{\bf{e}}_{1}}}\right\rangle=\left\langle{{{\bf{e}}_{3}}\,,\,{{\bf{e}}_{3}}}\right\rangle=\left\langle{{{\bf{e}}_{4}}\,,\,{{\bf{e}}_{4}}}\right\rangle=1\,\,,\,\,\,\left\langle{{{\bf{e}}_{2}}\,,\,{{\bf{e}}_{2}}}\right\rangle=-1.$

For $s\in L$, the non-null frame field
$\left\\{{{{\bf{e}}_{1}},\,{{\bf{e}}_{2}},\,{{\bf{e}}_{3}},\,{{\bf{e}}_{4}}}\right\\}$
and curvature functions ${k_{1}}$ and ${k_{2}}$ are determined as follows

$\begin{array}[]{l}{1^{st}}\,\,\,\,\,\,{\rm{step}}\,\,\,\,\,\,{{\bf{e}}_{1}}\left(s\right)={\bf{c^{\prime}}}\left(s\right)\\\
{2^{nd}}\,\,\,\,{\rm{step}}\,\,\,\,\,\,\,{k_{1}}\left(s\right)=\left\|{{{{\bf{e^{\prime}}}}_{1}}\left(s\right)}\right\|>0\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\bf{e}}_{2}}\left(s\right)=\frac{1}{{{k_{1}}\left(s\right)}}{{{\bf{e^{\prime}}}}_{1}}\left(s\right)\\\
{3^{rd}}\,\,\,\,{\rm{step}}\,\,\,\,\,\,\,\,{k_{2}}\left(s\right)=\left\|{{{{\bf{e^{\prime}}}}_{2}}\left(s\right)-{\mu_{1}}{k_{1}}\left(s\right){{\bf{e}}_{1}}\left(s\right)}\right\|>0\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\bf{e}}_{3}}\left(s\right)=\frac{1}{{{k_{2}}\left(s\right)}}\left({{{{\bf{e^{\prime}}}}_{2}}\left(s\right)-{\mu_{1}}{k_{1}}\left(s\right){{\bf{e}}_{1}}\left(s\right)}\right)\\\
{4^{th}}\,\,\,{\rm{step}}\,\,\,\,\,\,\,\,\,\,{e_{4}}\left(s\right)=\varepsilon\frac{1}{{\left\|{{{{\bf{e^{\prime}}}}_{3}}\left(s\right)-{\mu_{2}}{k_{2}}\left(s\right){{\bf{e}}_{2}}\left(s\right)}\right\|}}\left({{{{\bf{e^{\prime}}}}_{3}}\left(s\right)-{\mu_{2}}{k_{2}}\left(s\right){{\bf{e}}_{2}}\left(s\right)}\right)\\\
\end{array}$

where $\varepsilon$ is taken $-1$ or $+1$ to make $+1$ the determinant of
$\left\\{{{{\bf{e}}_{1}},\,{{\bf{e}}_{2}},\,{{\bf{e}}_{3}},\,{{\bf{e}}_{4}}}\right\\}$,
that is, the non-null orthonormal frame field is of positive orientation. The
function ${k_{3}}$ is determined by

${k_{3}}\left(s\right)=\left\langle{{{{\bf{e^{\prime}}}}_{3}}\left(s\right)\,,\,{{\bf{e}}_{4}}\left(s\right)}\right\rangle\neq
0.$

So the function ${k_{3}}$ never vanishes.

In order to make sure that the spacelike curve $C$ is a special spacelike
Frenet curve, above steps must be checked, from ${1^{st}}$ step to ${4^{th}}$
step, for $s\in L$.

At each point of spacelike curve $C$, a line ${\ell_{1}}$ in the direction of
${{\bf{e}}_{2}}$ is called the first normal line, a line ${\ell_{2}}$ in the
direction of ${{\bf{e}}_{3}}$ is called the second normal line and a line
${\ell_{3}}$ in the direction of ${{\bf{e}}_{4}}$ is called the third normal
line.

Note that, according to three different case of spacelike curve $C$,
${\ell_{3}},\,{\ell_{2}}$ and ${\ell_{1}}$ can be timelike, respectively,
which are called second binormal, first binormal and principal normal line at
each point of the spacelike curve $C$.

##  3 Generalized spacelike Mannheim curves in $E_{1}^{4}$

In ${E^{4}}$ the Bertrand curves and Mannheim curves are generalized by [6]
and [7], respectively. In these regards, we have investigate generalization of
spacelike Mannheim curves Minkowski space in $E_{1}^{4}$.

###### Definition 3.1

A special spacelike curve $C$ in $E_{1}^{4}$ is a generalized spacelike
Mannheim curve if there exists a special spacelike Frenet curve ${C^{*}}$ in
$E_{1}^{4}$ such that the first normal line at each of $C$ is included in the
plane generated by the second normal line and the third normal line of
${C^{*}}$ at the corresponding point under $\phi$. Here $\phi$ is a bijection
from $C$ to ${C^{*}}$. The curve ${C^{*}}$ is called the generalized spacelike
Mannheim mate curve of $C$.

By the definition, a generalized Mannheim mate curve ${C^{*}}$ is given by

$\begin{array}[]{l}{{\bf{c}}^{*}}\left(s\right)={\bf{c}}\left(s\right)+\alpha\left(s\right){{\bf{e}}_{2}}\left(s\right),\,\,s\in
L\end{array}$ (3.1)

where $\alpha$ is a smooth function on $L$. Generally, the parameter $s$ isn’t
an arc-length of ${C^{*}}$. Let ${s^{*}}$ be the arc-length of ${C^{*}}$
defined by

${s^{*}}=\int\limits_{0}^{s}{\left\|{\frac{{d{{\bf{c}}^{*}}\left(s\right)}}{{ds}}}\right\|ds.}$

If a smooth function $f:L\to L$ is given by $f\left(s\right)={s^{*}}$, then

$\begin{array}[]{l}\,\,\,\,\frac{{d{{\bf{c}}^{*}}\left(s\right)}}{{ds}}=\,\,\,{{\bf{e}}_{1}}\left(s\right)+\alpha^{\prime}\left(s\right){{\bf{e}}_{2}}\left(s\right)+\alpha\left(s\right){\mu_{1}}{k_{1}}\left(s\right){{\bf{e}}_{1}}\left(s\right)+\alpha\left(s\right){k_{2}}\left(s\right){{\bf{e}}_{3}}\left(s\right)\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\,=\,\,\,\left({1+{\mu_{1}}\alpha\left(s\right){k_{1}}\left(s\right)}\right){{\bf{e}}_{1}}\left(s\right)+\alpha^{\prime}\left(s\right){{\bf{e}}_{2}}\left(s\right)+\alpha\left(s\right){k_{2}}\left(s\right){{\bf{e}}_{3}}\left(s\right).\\\
\end{array}$

for $\forall s\in L$. Thus, we have

$\begin{array}[]{l}f^{\prime}\left(s\right)=\frac{{d{s^{*}}}}{{ds}}=\left\|{\frac{{d{{\bf{c}}^{\bf{*}}}\left(s\right)}}{{ds}}}\right\|=\sqrt{\left|{{{\left({1+{\mu_{1}}\alpha\left(s\right){k_{1}}\left(s\right)}\right)}^{2}}+\varepsilon_{2}{{\left({\alpha^{\prime}\left(s\right)}\right)}^{2}}+\varepsilon_{3}{{\left({\alpha\left(s\right){k_{2}}\left(s\right)}\right)}^{2}}}\right|}\end{array}$

where
$\varepsilon_{i}=\left\\{\begin{array}[]{l}-1\,\,,\,\,\,{{\bf{e}}_{i}}\,\,{\rm{is}}\,{\rm{timelike}}\\\
\,\,\,\,1\,\,,\,\,\,\,{{\bf{e}}_{i}}\,{\rm{is}}\,{\rm{spacelike}}\\\
\end{array}\right.$, for $2\leq i\leq 4.$
This means that, in the Case 1, ${{\bf{e}}_{4}}$ is timelike and

$\begin{array}[]{l}f^{\prime}\left(s\right)=\sqrt{\left|{{{\left({1-\alpha\left(s\right){k_{1}}\left(s\right)}\right)}^{2}}+{{\left({\alpha^{\prime}\left(s\right)}\right)}^{2}}+{{\left({\alpha\left(s\right){k_{2}}\left(s\right)}\right)}^{2}}}\right|}\end{array}$

or in the Case 2, ${{\bf{e}}_{3}}$ is timelike and

$\begin{array}[]{l}f^{\prime}\left(s\right)=\sqrt{\left|{{{\left({1-\alpha\left(s\right){k_{1}}\left(s\right)}\right)}^{2}}+{{\left({\alpha^{\prime}\left(s\right)}\right)}^{2}}-{{\left({\alpha\left(s\right){k_{2}}\left(s\right)}\right)}^{2}}}\right|}\end{array}$

or in the Case 3, ${{\bf{e}}_{2}}$ is timelike and

$\begin{array}[]{l}f^{\prime}\left(s\right)=\sqrt{\left|{{{\left({1+\alpha\left(s\right){k_{1}}\left(s\right)}\right)}^{2}}-{{\left({\alpha^{\prime}\left(s\right)}\right)}^{2}}+{{\left({\alpha\left(s\right){k_{2}}\left(s\right)}\right)}^{2}}}\right|}.\end{array}$

The spacelike curve ${C^{*}}$ with arc-length parameter ${s^{*}}$ is

$\begin{array}[]{l}{{\bf{c}}^{*}}:\,{L^{*}}\to E_{1}^{4}\\\
\,\,\,\,\,\,\,\,\,\,\,{s^{*}}\,\,\to\,{{\bf{c}}^{*}}\left({{s^{*}}}\right).\\\
\end{array}$

For a bijection $\phi:\,C\to{C^{*}}$ defined by
$\phi\left({{\bf{c}}\left(s\right)}\right)={{\bf{c}}^{*}}\left({f\left(s\right)}\right),$
the reparametrization of ${C^{*}}$ is

$\begin{array}[]{l}{{\bf{c}}^{*}}\left({f\left(s\right)}\right)={\bf{c}}\left(s\right)+\alpha\left(s\right){{\bf{e}}_{2}}\left(s\right)\end{array}$

where $\alpha$ is a smooth function on $L$.

###### Theorem 3.1

If a special spacelike Frenet curve $C$ in $E_{1}^{4}$ is a generalized
spacelike Mannheim curve, then the first curvature function ${k_{1}}$ and the
second curvature function ${k_{2}}$ of $C$ satisfy the equality

$\begin{array}[]{l}{k_{1}}\left(s\right)=-\alpha\left({{\mu_{1}}k_{1}^{2}\left(s\right)+{\mu_{2}}k_{2}^{2}\left(s\right)}\right)\,\,,\,\,s\in
L\end{array}$ (3.2)

where $\alpha$ is a constant number and ${\mu_{1}}={\mu_{2}}=-1$ when
${{\bf{e}}_{4}}$ is timelike or ${\mu_{1}}=-1\,,\,\,{\mu_{2}}=1$ when
${{\bf{e}}_{3}}$ is timelike or ${\mu_{1}}={\mu_{2}}=1$ when ${{\bf{e}}_{2}}$
is timelike.

Proof. Let $C$ be a generalized spacelike Mannheim curve and ${C^{*}}$ be the
generalized spacelike Mannheim mate curve of $C$ with the diagram;

$\begin{array}[]{l}\,\,\,\,\,\,\,\,\,\,\,\,\mathop{\bf{c}}\limits_{\cdot\,\,\cdot}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\mathop{\bf{c}}\limits_{\cdot\,\,\cdot}^{*}}\\\
f:\,\,\,\,L\,\,\,\,\,\,\to\,\,\,\,\,{L^{*}}\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\\\
\phi\,:\,\,\,E_{1}^{4}\,\,\,\to\,\,\,E_{1}^{4}.\\\ \end{array}$

A smooth function $f$ is defined by
$f\left(s\right)=\int{\left\|{\frac{{d{{\bf{c}}^{*}}\left(s\right)}}{{ds}}}\right\|}ds={s^{*}}$
and ${s^{*}}$ is the arc-length parameter of ${C^{*}}$. Also $\phi$ is a
bijection which is defined by
$\phi\left({{\bf{c}}\left(s\right)}\right)={{\bf{c}}^{*}}\left({f\left(s\right)}\right).$
Thus, the spacelike curve ${C^{*}}$ is reparametrized by

$\begin{array}[]{l}{{\bf{c}}^{*}}\left({f\left(s\right)}\right)={\bf{c}}\left(s\right)+\alpha\left(s\right){{\bf{e}}_{2}}\left(s\right)\end{array}$
(3.3)

where $\alpha$ is a smooth function. By differentiating both sides of (3.3)
with respect to $s$

$\begin{array}[]{l}f^{\prime}\left(s\right){\bf{e}}_{1}^{*}\left({f\left(s\right)}\right)=\left({1+{\mu_{1}}\alpha\left(s\right){k_{1}}\left(s\right)}\right){{\bf{e}}_{1}}+\alpha^{\prime}\left(s\right){{\bf{e}}_{2}}\left(s\right)\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\alpha\left(s\right){k_{2}}\left(s\right){{\bf{e}}_{3}}\left(s\right)\\\
\end{array}$ (3.4)

is obtained.

On the other hand, since the first normal line at the each point of $C$ is
lying in the plane generated by the second normal line and the third normal
line of ${C^{*}}$ at the corresponding points under bijection $\phi$, the
vector field ${{\bf{e}}_{2}}\left(s\right)$ is given by

$\begin{array}[]{l}{{\bf{e}}_{2}}\left(s\right)=g\left(s\right){\bf{e}}_{3}^{*}\left({f\left(s\right)}\right)+h\left(s\right){\bf{e}}_{4}^{*}\left({f\left(s\right)}\right)\end{array}$

where $g$ and $h$ are some smooth functions on $L$. If we take into
consideration

$\begin{array}[]{l}\left\langle{{\bf{e}}_{1}^{*}\left({f\left(s\right)}\right),\,g\left(s\right){\bf{e}}_{3}^{*}\left({f\left(s\right)}\right)+h\left(s\right){\bf{e}}_{4}^{*}\left({f\left(s\right)}\right)}\right\rangle=0\end{array}$

and the equation (3.4), then we have $\alpha^{\prime}\left(s\right)=0$. So we
rewrite the equation (3.4) as

$\begin{array}[]{l}f^{\prime}\left(s\right){\bf{e}}_{1}^{*}\left({f\left(s\right)}\right)=\left({1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}\right){{\bf{e}}_{1}}\left(s\right)+\alpha{k_{2}}\left(s\right){{\bf{e}}_{3}}\left(s\right),\end{array}$
(3.5)

that is,

$\begin{array}[]{l}{\bf{e}}_{1}^{*}\left({f\left(s\right)}\right)=\frac{{\left({1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}\right)}}{{f^{\prime}\left(s\right)}}{{\bf{e}}_{1}}\left(s\right)+\frac{{\alpha{k_{2}}\left(s\right)}}{{f^{\prime}\left(s\right)}}{{\bf{e}}_{3}}\left(s\right)\end{array}$

where

$\begin{array}[]{l}f^{\prime}\left(s\right)=\sqrt{\left|{{{\left({1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}\right)}^{2}}+\varepsilon_{3}{{\left({\alpha{k_{2}}\left(s\right)}\right)}^{2}}}\right|}\,,\,\,\varepsilon_{3}=\left\\{\begin{array}[]{l}-1\,\,,\,\,{{\bf{e}}_{3}}\,\,{\rm{is\,\,timelike}}{\rm{,}}\\\
\,\,\,\,1\,\,\,,\,\,{{\bf{e}}_{3}}\,\,{\rm{is\,\,spacelike}}{\rm{.}}\\\
\end{array}\right.\end{array}$

By taking differentiation both sides of the equations (3.5) with respect to
$s$,

$\begin{array}[]{l}f^{\prime}\left(s\right)k_{1}^{*}\left({f\left(s\right)}\right){\bf{e}}_{2}^{*}\left({f\left(s\right)}\right)={\left({\frac{{1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}}{{f^{\prime}\left(s\right)}}}\right)^{\prime}}{{\bf{e}}_{1}}\left(s\right)\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left({\frac{{\left({1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}\right){k_{1}}\left(s\right)+{\mu_{2}}\alpha{{\left({{k_{2}}\left(s\right)}\right)}^{2}}}}{{f^{\prime}\left(s\right)}}}\right){{\bf{e}}_{2}}\left(s\right)\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+{\left({\frac{{\alpha{k_{2}}\left(s\right)}}{{f^{\prime}\left(s\right)}}}\right)^{\prime}}{{\bf{e}}_{3}}\left(s\right)+\left({\frac{{\alpha{k_{2}}\left(s\right){k_{3}}\left(s\right)}}{{f^{\prime}\left(s\right)}}}\right){{\bf{e}}_{4}}\left(s\right)\\\
\end{array}$ (3.6)

is obtained for $s\in L$. Since

$\begin{array}[]{l}\left\langle{{\bf{e}}_{2}^{*}\left({f\left(s\right)}\right),\,g\left(s\right){\bf{e}}_{3}^{*}\left({f\left(s\right)}\right)+h\left(s\right){\bf{e}}_{4}^{*}\left({f\left(s\right)}\right)}\right\rangle=0,\end{array}$

then in the equation (3.6) the coefficient of ${{\bf{e}}_{2}}\left(s\right)$
vanishes, that is,

$\begin{array}[]{l}\left({1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}\right){k_{1}}\left(s\right)+{\mu_{2}}\alpha{\left({{k_{2}}\left(s\right)}\right)^{2}}=0.\end{array}$

Thus,
${k_{1}}\left(s\right)=-\alpha\left({{\mu_{1}}k_{1}^{2}\left(s\right)+{\mu_{2}}k_{2}^{2}\left(s\right)}\right)$
is satisfied. This completes the proof.

If we investigate the special cases separately, then we have

in the Case 1;

$\begin{array}[]{l}{k_{1}}\left(s\right)=\alpha\left({k_{1}^{2}\left(s\right)+k_{2}^{2}\left(s\right)}\right),\end{array}$

in the Case 2;

$\begin{array}[]{l}{k_{1}}\left(s\right)=\alpha\left({k_{1}^{2}\left(s\right)-k_{2}^{2}\left(s\right)}\right),\end{array}$

in the Case 3;

$\begin{array}[]{l}{k_{1}}\left(s\right)=-\alpha\left({k_{1}^{2}\left(s\right)+k_{2}^{2}\left(s\right)}\right).\end{array}$

###### Theorem 3.2

Let $C$ be a special spacelike Frenet curve in $E_{1}^{4}$ whose curvature
functions ${k_{1}}$ and ${k_{2}}$ are non-constant functions and satisfy the
equality
${k_{1}}\left(s\right)=-\alpha\left({{\mu_{1}}k_{1}^{2}\left(s\right)+{\mu_{2}}k_{2}^{2}\left(s\right)}\right)$,
where $\alpha$ is non-zero constant, for all $s\in L$. If the spacelike curve
${C^{*}}$ given by

$\begin{array}[]{l}{{\bf{c}}^{*}}\left(s\right)={\bf{c}}\left(s\right)+\alpha{{\bf{e}}_{2}}\left(s\right)\end{array}$

is a special spacelike Frenet curve, then ${C^{*}}$ is a generalized spacelike
Mannheim mate curve of $C$.

Proof . The arc-length parameter of ${C^{*}}$ is defined by

$\begin{array}[]{l}{s^{*}}=\int\limits_{0}^{s}{\left\|{\frac{{d{{\bf{c}}^{*}}\left(s\right)}}{{ds}}}\right\|}ds\end{array}$

for all $s\in L$. Under the assumptation of

$\begin{array}[]{l}{k_{1}}\left(s\right)=-\alpha\left({{\mu_{1}}k_{1}^{2}\left(s\right)+{\mu_{2}}k_{2}^{2}\left(s\right)}\right)\end{array}$

and after calculations for all cases, separately, we obtain

in the Case 1;
$\,\,\,\,f^{\prime}\left(s\right)=\sqrt{\left|{1-\alpha{k_{1}}\left(s\right)}\right|},$

in the Case 2;
$\,\,\,\,f^{\prime}\left(s\right)=\sqrt{\left|{1-\alpha{k_{1}}\left(s\right)}\right|},$

in the Case 3;
$\,\,\,\,f^{\prime}\left(s\right)=\sqrt{\left|{1+\alpha{k_{1}}\left(s\right)}\right|}.$

Thus, we can generalize

$\begin{array}[]{l}f^{\prime}\left(s\right)=\sqrt{\left|{1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}\right|}\end{array}$

for all $s\in L$.

By differentiating the equation
${{\bf{c}}^{*}}\left({f\left(s\right)}\right)={\bf{c}}\left(s\right)+\alpha{{\bf{e}}_{2}}\left(s\right)$
with respect to $s$, it is seen that

$\begin{array}[]{l}f^{\prime}\left(s\right){\bf{e}}_{1}^{*}\left({f\left(s\right)}\right)=\left({1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}\right){{\bf{e}}_{1}}\left(s\right)+\alpha{k_{2}}\left(s\right){{\bf{e}}_{3}}\left(s\right).\end{array}$

So, it is seen that

$\begin{array}[]{l}{\bf{e}}_{1}^{*}\left({f\left(s\right)}\right)=\left({\frac{{1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}}{{\sqrt{\left|{1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}\right|}}}{{\bf{e}}_{1}}\left(s\right)+\frac{{\alpha{k_{2}}\left(s\right)}}{{\sqrt{\left|{1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}\right|}}}{{\bf{e}}_{3}}\left(s\right)}\right)\end{array}$
(3.7)

for $s\in L$.

The differentiation of the last equation with respect to $s$ is

$\begin{array}[]{l}f^{\prime}\left(s\right)k_{1}^{*}\left({f\left(s\right)}\right){\bf{e}}_{2}^{*}\left({f\left(s\right)}\right)={\left({\sqrt{\left|{1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}\right|}}\right)^{\prime}}{{\bf{e}}_{1}}\left(s\right)\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left({\frac{{\left({1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}\right){k_{1}}\left(s\right)+{\mu_{2}}\alpha
k_{2}^{2}\left(s\right)}}{{\sqrt{\left|{1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}\right|}}}}\right){{\bf{e}}_{2}}\left(s\right)\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+{\left({\frac{{\alpha{k_{2}}\left(s\right)}}{{\sqrt{\left|{1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}\right|}}}}\right)^{\prime}}{{\bf{e}}_{3}}\left(s\right)+\left({\frac{{\alpha{k_{2}}\left(s\right){k_{3}}\left(s\right)}}{{\sqrt{\left|{1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}\right|}}}}\right){{\bf{e}}_{4}}\left(s\right).\\\
\end{array}$ (3.8)

According to our assumption,

$\begin{array}[]{l}\frac{{\left({1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}\right){k_{1}}\left(s\right)+{\mu_{2}}\alpha
k_{2}^{2}\left(s\right)}}{{\sqrt{\left|{1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}\right|}}}=0\end{array}$

is hold. Thus, the coefficient of ${{\bf{e}}_{2}}\left(s\right)$ in the
equation (3.8) is zero. It is seen from the equation (3.8),
${\bf{e}}_{2}^{*}\left({f\left(s\right)}\right)$ is given by linear
combination of ${{\bf{e}}_{1}}\left(s\right),\;\,{{\bf{e}}_{3}}\left(s\right)$
and ${{\bf{e}}_{4}}\left(s\right)$. Also, from equation (3.7),
${\bf{e}}_{1}^{*}\left({f\left(s\right)}\right)$ is a linear combination of
${{\bf{e}}_{1}}\left(s\right)$ and ${{\bf{e}}_{3}}\left(s\right).$ Moreover,
${C^{*}}$ is a special spacelike Frenet curve that the vector
${{\bf{e}}_{2}}\left(s\right)$ is given by linear combination of
${\bf{e}}_{3}^{*}\left({f\left(s\right)}\right)$ and
${\bf{e}}_{4}^{*}\left({f\left(s\right)}\right)$.

Therefore, the first normal line $C$ lies in the plane generated by the second
normal line and third normal line of ${C^{*}}$ at the corresponding points
under a bijection $\phi$ which is defined by
$\phi\left({{\bf{c}}\left(s\right)}\right)={{\bf{c}}^{*}}\left({f\left(s\right)}\right)$.
Thus, the proof of the theorem is completed.

###### Remark 3.1

In 4-dimensional Minkowski space for a special spacelike Frenet curve $C$ with
curvature functions ${k_{1}}$ and ${k_{2}}$ satisfying

$\begin{array}[]{l}{k_{1}}\left(s\right)=-\alpha\left({{\mu_{1}}k_{1}^{2}\left(s\right)+{\mu_{2}}{\mu_{3}}k_{2}^{2}\left(s\right)}\right),\end{array}$

it is not clear that a smooth spacelike curve ${C^{*}}$ given by (3.1) is a
special Frenet curve. So, it is unknown whether the reverse of Theorem 3.1 is
true or false.

###### Theorem 3.3

Let $C$ be a spacelike special curve in $E_{1}^{4}$ with non-zero third
curvature function ${k_{3}}$. If there exists a spacelike special Frenet curve
${C^{*}}$ in $E_{1}^{4}$ such that the first normal line of $C$ is linearly
dependent with the third normal line of ${C^{*}}$ at the corresponding points
$\bf{c}\left(s\right)$ and ${\bf{c}^{*}}\left(s\right)$, respectively, under a
bijection $\phi:C\to{C^{*}}$, then the curvatures ${k_{1}}$ and ${k_{2}}$ of
$C$ are constant functions.

Proof. Let $C$ be a spacelike Frenet curve in $E_{1}^{4}$ with the Frenet
frame field
$\left\\{{{{\bf{e}}_{1}},\,{{\bf{e}}_{2}},\,{{\bf{e}}_{3}},\,{{\bf{e}}_{4}}}\right\\}$
and curvature functions ${k_{1}},\,{k_{2}}$ and ${k_{3}}$. Also, we assume
that ${C^{*}}$ be a spacelike special Frenet curve in $E_{1}^{4}$ with the
Frenet frame field
$\left\\{{{\bf{e}}_{1}^{*},\,{\bf{e}}_{2}^{*},\,{\bf{e}}_{3}^{*},\,{\bf{e}}_{4}^{*}}\right\\}$
and curvature functions $k_{1}^{*},\,k_{2}^{*}\,$ and $k_{3}^{*}$.

Let the first normal line of $C$ be linearly dependent with the third normal
line of ${C^{*}}$ at the corresponding points $C$ and ${C^{*}}$, respectively.
Then the parametrization of ${C^{*}}$ is

$\begin{array}[]{l}{{\bf{c}}^{*}}\left({f\left(s\right)}\right)={\bf{c}}\left(s\right)+\alpha\left(s\right){{\bf{e}}_{2}}\left(s\right)\end{array}$
(3.9)

for all $s\in L$. If ${s^{*}}$ is the arc-length parameter of ${C^{*}}$, then

$\begin{array}[]{l}{s^{*}}=\int\limits_{0}^{s}{\sqrt{\left|{{{\left({1+{\mu_{1}}\alpha{k_{1}}}\right)}^{2}}+\varepsilon_{2}\left({\alpha^{\prime}\left(s\right)}\right)+\varepsilon_{3}{{\left({\alpha\left(s\right){k_{2}}\left(s\right)}\right)}^{2}}}\right|}}ds\end{array}$
(3.10)

where

$\begin{array}[]{l}\varepsilon_{i}=\left\\{\begin{array}[]{l}-1\,\,,\,\,{{\bf{e}}_{i}}\,\,{\rm{is}}\,\,{\rm{timelike}}\\\
\,\,\,1\,\,\,,\,\,{{\bf{e}}_{i}}\,\,{\rm{is}}\,\,{\rm{spacelike}}\\\
\end{array}\right.,\,\,\,{\rm{for}}\,\,\,\,\,{\rm{}}2\leq i\leq 4\end{array}$

and

$\begin{array}[]{l}f:\,L\to{L^{*}}\\\
\,\,\,\,\,\,\,s\,\,\to\,f\left(s\right)={s^{*}}.\\\ \end{array}$

Moreover, $\phi:C\to{C^{*}}$ is a bijection given by
$\phi\left({{\bf{c}}\left(s\right)}\right)={{\bf{c}}^{*}}\left({f\left(s\right)}\right)$.

By differentiating the equation (3.9) with respect to $s$ and using Frenet
formulas, we have

$\begin{array}[]{l}f^{\prime}\left(s\right){\bf{e}}_{1}^{*}\left({f\left(s\right)}\right)=\left({1+{\mu_{1}}\alpha\left(s\right){k_{1}}\left(s\right)}\right){{\bf{e}}_{1}}\left(s\right)+\alpha^{\prime}\left(s\right){{\bf{e}}_{2}}\left(s\right)\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\alpha\left(s\right){k_{2}}\left(s\right){{\bf{e}}_{3}}\left(s\right).\\\
\end{array}$ (3.11)

Since
${\bf{e}}_{4}^{*}\left({f\left(s\right)}\right)=\mp{{\bf{e}}_{2}}\left(s\right)$,
then

$\begin{array}[]{l}\left\langle{f^{\prime}\left(s\right){\bf{e}}_{1}^{*}\left({f\left(s\right)}\right),\,{\bf{e}}_{4}^{*}\left({f\left(s\right)}\right)}\right\rangle=\left\langle{\left({1+{\mu_{1}}\alpha\left(s\right){k_{1}}\left(s\right)}\right){{\bf{e}}_{1}}\left(s\right)+\alpha^{\prime}\left(s\right){{\bf{e}}_{2}}\left(s\right)}\right.\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left.{+\alpha\left(s\right){k_{2}}\left(s\right){{\bf{e}}_{3}}\left(s\right),\,\mp{{\bf{e}}_{2}}\left(s\right)}\right\rangle,\\\
\end{array}$

that is,

$\begin{array}[]{l}0=\mp\alpha^{\prime}\left(s\right).\end{array}$

It is easily seen that $\alpha$ is a constant number from the last equation.
Thus, hereafter we can denote $\alpha\left(s\right)=\alpha$, for all $s\in L.$

From the equation (3.10), we get

$\begin{array}[]{l}f^{\prime}\left(s\right)=\sqrt{\left|{{{\left({1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}\right)}^{2}}+\varepsilon_{3}{{\left({\alpha{k_{2}}\left(s\right)}\right)}^{2}}}\right|}>0\end{array}$

where

$\begin{array}[]{l}\varepsilon_{3}=\left\\{\begin{array}[]{l}-1\,\,,\,\,{{\bf{e}}_{i}}\,\,{\rm{is}}\,\,{\rm{timelike}}\\\
\,\,\,\,1\,\,,\,\,{{\bf{e}}_{i}}\,\,{\rm{is}}\,\,{\rm{spacelike}}\\\
\end{array}\right.\,\,\,\,,\,{\rm{for}}\,\,\,\,\,2\leq i\leq 4.\end{array}$

Then, we rewrite the equation (3.11) as follows;

$\begin{array}[]{l}{\bf{e}}_{1}^{*}\left({f\left(s\right)}\right)=\left({\frac{{1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}}{{f^{\prime}\left(s\right)}}}\right){{\bf{e}}_{1}}\left(s\right)+\left({\frac{{\alpha{k_{2}}\left(s\right)}}{{f^{\prime}\left(s\right)}}}\right){{\bf{e}}_{3}}\left(s\right).\end{array}$

The differentiation of the last equation with respect to $s$ is

$\begin{array}[]{l}f^{\prime}\left(s\right)k_{1}^{*}\left({f\left(s\right)}\right){\bf{e}}_{2}^{*}\left({f\left(s\right)}\right)={\left({\frac{{1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}}{{f^{\prime}\left(s\right)}}}\right)^{\prime}}{{\bf{e}}_{1}}\left(s\right)\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left({\frac{{{k_{1}}\left(s\right)+{\mu_{1}}\alpha
k_{1}^{2}\left(s\right)+{\mu_{2}}\alpha
k_{2}^{2}\left(s\right)}}{{f^{\prime}\left(s\right)}}}\right){{\bf{e}}_{2}}\left(s\right)\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+{\left({\frac{{\alpha{k_{2}}\left(s\right)}}{{f^{\prime}\left(s\right)}}}\right)^{\prime}}{{\bf{e}}_{3}}\left(s\right)+\left({\frac{{\alpha{k_{2}}\left(s\right){k_{3}}\left(s\right)}}{{f^{\prime}\left(s\right)}}}\right){{\bf{e}}_{4}}\left(s\right).\\\
\end{array}$ (3.12)

Since
$\left\langle{f^{\prime}\left(s\right)k_{1}^{*}\left({f\left(s\right)}\right){\bf{e}}_{2}^{*}\left({f\left(s\right)}\right),\,{\bf{e}}_{4}^{*}\left({f\left(s\right)}\right)}\right\rangle=0$
and
${\bf{e}}_{4}^{*}\left({f\left(s\right)}\right)=\mp{{\bf{e}}_{2}}\left(s\right)$
for all $s\in L$, we obtain

$\begin{array}[]{l}{k_{1}}\left(s\right)+{\mu_{1}}\alpha
k_{1}^{2}\left(s\right)+{\mu_{2}}\alpha k_{2}^{2}\left(s\right)=0\end{array}$

is satisfied. Then,

$\begin{array}[]{l}\alpha=-\frac{{{k_{1}}\left(s\right)}}{{{\mu_{1}}k_{1}^{2}\left(s\right)+{\mu_{2}}k_{2}^{2}\left(s\right)}}\end{array}$
(3.13)

is a non-zero constant number. Thus, from the equation (3.12), it is seen that

$\begin{array}[]{l}{\bf{e}}_{2}^{*}\left({f\left(s\right)}\right)=\frac{1}{{f^{\prime}\left(s\right)K\left(s\right)}}{\left({\frac{{1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}}{{f^{\prime}\left(s\right)}}}\right)^{\prime}}{{\bf{e}}_{1}}\left(s\right)+\frac{1}{{f^{\prime}\left(s\right)K\left(s\right)}}\left({\frac{{\alpha{k_{2}}\left(s\right)}}{{f^{\prime}\left(s\right)}}}\right){{\bf{e}}_{3}}\left(s\right)\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\frac{1}{{f^{\prime}\left(s\right)K\left(s\right)}}\left({\frac{{\alpha{k_{2}}\left(s\right){k_{3}}\left(s\right)}}{{f^{\prime}\left(s\right)}}}\right){{\bf{e}}_{4}}\left(s\right)\\\
\end{array}$

where $K\left(s\right)=k_{1}^{*}\left({f\left(s\right)}\right)$ for all $s\in
L$. By differentiating the last equation with respect to $s$, we obtain

$\begin{array}[]{l}f^{\prime}\left(s\right)\left[{{\mu_{1}}k_{1}^{*}\left({f\left(s\right)}\right){\bf{e}}_{1}^{*}\left({f\left(s\right)}\right)+k_{2}^{*}\left({f\left(s\right)}\right){{\bf{e}}_{3}^{*}}\left({f\left(s\right)}\right)}\right]={\left({\frac{1}{{f^{\prime}\left(s\right)K\left(s\right)}}{{\left({\frac{{1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}}{{f^{\prime}\left(s\right)}}}\right)}^{\prime}}}\right)^{\prime}}{{\bf{e}}_{1}}\left(s\right)\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left({\frac{k_{1}{\left(s\right)}}{{f^{\prime}\left(s\right)K\left(s\right)}}{{\left({\frac{{1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}}{{f^{\prime}\left(s\right)}}}\right)}^{\prime}}+\frac{{{\mu_{2}}{k_{2}}\left(s\right)}}{{f^{\prime}\left(s\right)K\left(s\right)}}{{\left({\frac{{\alpha{k_{2}}\left(s\right)}}{{f^{\prime}\left(s\right)}}}\right)}^{\prime}}}\right){{\bf{e}}_{2}}\left(s\right)\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left({{{\left({\frac{1}{{f^{\prime}\left(s\right)K\left(s\right)}}{{\left({\frac{{\alpha{k_{2}}\left(s\right)}}{{f^{\prime}\left(s\right)}}}\right)}^{\prime}}}\right)}^{\prime}}+\frac{{{\mu_{3}}{k_{3}}\left(s\right)}}{{f^{\prime}\left(s\right)K\left(s\right)}}\left({\frac{{\alpha{k_{2}}\left(s\right){k_{3}}\left(s\right)}}{{f^{\prime}\left(s\right)}}}\right)}\right){{\bf{e}}_{3}}\left(s\right)\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left({{{\left({\frac{1}{{f^{\prime}\left(s\right)K\left(s\right)}}{{\left({\frac{{\alpha{k_{2}}\left(s\right){k_{3}}\left(s\right)}}{{f^{\prime}\left(s\right)}}}\right)}^{\prime}}}\right)}^{\prime}}+\frac{{{k_{3}}\left(s\right)}}{{f^{\prime}\left(s\right)K\left(s\right)}}{{\left({\frac{{\alpha{k_{2}}\left(s\right)}}{{f^{\prime}\left(s\right)}}}\right)}^{\prime}}}\right){{\bf{e}}_{4}}\left(s\right)\\\
\end{array}$

for all $s\in L$. If we take into consideration

$\begin{array}[]{l}\left\langle{f^{\prime}\left(s\right)\left({{\mu_{1}}k_{1}^{*}\left({f\left(s\right)}\right){\bf{e}}_{1}^{*}\left({f\left(s\right)}\right)+k_{2}^{*}\left({f\left(s\right)}\right){{\bf{e}}_{3}^{*}}\left({f\left(s\right)}\right)}\right)\,,\,\,{\bf{e}}_{4}^{*}\left({f\left(s\right)}\right)}\right\rangle=0\end{array}$

and

$\begin{array}[]{l}{\bf{e}}_{4}^{*}\left({f\left(s\right)}\right)=\mp{{\bf{e}}_{2}}\left(s\right),\end{array}$

then

$\begin{array}[]{l}{\mu_{1}}\alpha{k_{1}}\left(s\right){k_{1}}^{\prime}\left(s\right){f^{\prime}\left(s\right)}-{k_{1}}\left(s\right)\left({1+{\mu_{1}}\alpha{k_{1}}\left(s\right)}\right)f^{\prime\prime}\left(s\right)\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+{\mu_{2}}\alpha{k_{2}}\left(s\right){k_{2}}^{\prime}\left(s\right){f^{\prime}\left(s\right)}-{\mu_{2}}\alpha
k_{2}^{2}\left(s\right)f^{\prime\prime}\left(s\right)=0.\\\ \end{array}$

If we arrange the last equation, then we find

$\begin{array}[]{l}\alpha\left({{\mu_{1}}{k_{1}}\left(s\right){{k^{\prime}}_{1}}\left(s\right)+{\mu_{2}}{k_{2}}\left(s\right){{k^{\prime}}_{2}}\left(s\right)}\right)f^{\prime}\left(s\right)\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\left({{k_{1}}+\alpha\left({{\mu_{1}}k_{1}^{2}\left(s\right)+{\mu_{2}}k_{2}^{2}\left(s\right)}\right)}\right)f^{\prime\prime}\left(s\right)=0.\\\
\end{array}$ (3.14)

Moreover, the differentiation of the equation (3.13) with respect to $s$ is

$\begin{array}[]{l}{k^{\prime}_{1}}\left(s\right)+2\alpha\left({{\mu_{1}}{k_{1}}\left(s\right){{k^{\prime}}_{1}}\left(s\right)+{\mu_{2}}{k_{2}}\left(s\right){{k^{\prime}}_{2}}\left(s\right)}\right)=0.\end{array}$

From the above equation, we see

$\begin{array}[]{l}-\frac{{{{k^{\prime}}_{1}}\left(s\right)}}{2}=\alpha\left({{\mu_{1}}{k_{1}}\left(s\right){{k^{\prime}}_{1}}\left(s\right)+{\mu_{2}}{k_{2}}\left(s\right){{k^{\prime}}_{2}}\left(s\right)}\right).\end{array}$
(3.15)

If we substitute the equations (3.13) and (3.15) into the equation (3.14), we
obtain

$\begin{array}[]{l}-\frac{{{{k^{\prime}}_{1}}\left(s\right)}}{2}=0.\end{array}$

Finally, we find that the first curvature function is constant (that is,
positive constant).

Thus, from the equation (3.15) it is seen that the second curvature function
${k_{2}}$ is positive constant, too. This completes the proof.

In [9], a formula of parametric equation of Mannheim curve is given in
${E^{3}}$. Moreover, the parametric equation of generalized Mannheim curve in
${E^{4}}$ is obtained in [7]. The following theorem gives a parametric
representation of a generalized spacelike Mannheim curve with timelike second
binormal vector in $E_{1}^{4}$.

###### Theorem 3.4

Let be a spacelike special curve defined by

$\begin{array}[]{l}{\bf{c}}\left(u\right)=\left[{\begin{array}[]{*{20}{c}}{\alpha\int{f\left(u\right)\sinh
udu}}\\\ {\alpha\int{f\left(u\right)\cosh udu}}\\\
{\alpha\int{f\left(u\right)g\left(u\right)du}}\\\
{\alpha\int{f\left(u\right)h\left(u\right)du}}\\\
\end{array}}\right]\end{array}$

for $u\in I\subset\mathbb{R}$. Here $\alpha$ is a non-zero constant number,
$g:I\to\mathbb{R}$ and $h:I\to\mathbb{R}$ are any smooth functions and the
positive valued smooth function $f:I\to\mathbb{R}$ is given by

$\begin{array}[]{l}f\left(u\right)=\left({1+g^{2}\left(u\right)+h^{2}\left(u\right)}\right)^{{{-3}\mathord{\left/{\vphantom{{-3}2}}\right.\kern-1.2pt}2}}\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left|{-1-g^{2}\left(u\right)-h^{2}\left(u\right)+\dot{g}^{2}\left(u\right)+\dot{h}^{2}\left(u\right)+\left({\dot{g}\left(u\right)h\left(u\right)-g\left(u\right)\dot{h}\left(u\right)}\right)^{2}}\right|^{{{-5}\mathord{\left/{\vphantom{{-5}2}}\right.\kern-1.2pt}2}}\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\left|{\left({-1-g^{2}\left(u\right)-h^{2}\left(u\right)+\dot{g}^{2}\left(u\right)+\dot{h}^{2}\left(u\right)+\left({\dot{g}\left(u\right)h\left(u\right)-g\left(u\right)\dot{h}\left(u\right)}\right)^{2}}\right)^{3}}\right.\\\
\quad\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\left({1+g^{2}\left(u\right)+h^{2}\left(u\right)}\right)^{3}\left[{\left({g\left(u\right)-\ddot{g}\left(u\right)}\right)^{2}+\left({h\left(u\right)-\ddot{h}\left(u\right)}\right)^{2}}\right.\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left.{\,-\left({\left({g\left(u\right)\dot{h}\left(u\right)-\dot{g}\left(u\right)h\left(u\right)}\right)+\left({\dot{g}\left(u\right)\ddot{h}\left(u\right)-\ddot{g}\left(u\right)\dot{h}\left(u\right)}\right)}\right)^{2}+\left({g\left(u\right)\ddot{h}\left(u\right)-\ddot{g}\left(u\right)h}\right)\left(u\right)^{2}}\right|\\\
\quad\quad\quad\quad\quad\quad\;\;\quad\quad\quad\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\\\
\end{array}$

for $u\in I$. Then the curvature functions ${k_{1}}$ and ${k_{2}}$ of $C$
satisfy

$\begin{array}[]{l}{k_{1}}\left(u\right)=\alpha\left({k_{1}^{2}\left(u\right)+k_{2}^{2}\left(u\right)}\right)\end{array}$

at the each point ${\bf{c}}\left(u\right)$ of $C$.

Proof. Let $C$ be a spacelike special curve defined by

$\begin{array}[]{l}{\bf{c}}\left(u\right)=\left[{\begin{array}[]{*{20}{c}}{\alpha\int{f\left(u\right)\sinh
udu}}\\\ {\alpha\int{f\left(u\right)\cosh udu}}\\\
{\alpha\int{f\left(u\right)g\left(u\right)du}}\\\
{\alpha\int{f\left(u\right)h\left(u\right)du}}\\\
\end{array}}\right]\quad,\quad u\in I\subset\mathbb{R}\end{array}$

where $\alpha$ is a non-zero constant number, $g$ and $h$ are any smooth
functions. $f$ is a positive valued smooth function. Thus, we obtain

$\begin{array}[]{l}{\bf{\dot{c}}}\left(u\right)=\left[{\begin{array}[]{*{20}{c}}{\alpha
f\left(u\right)\sinh u}\\\ {\alpha f\left(u\right)\cosh u}\\\ {\alpha
f\left(u\right)g\left(u\right)}\\\ {\alpha f\left(u\right)h\left(u\right)}\\\
\end{array}}\right]\quad,\quad u\in I\subset\mathbb{R}\end{array}$ (3.16)

where the subscript dot (.) denotes the differentiation with respect to $u$.

The arc-length parameter $s$ of $C$ is given by

$\begin{array}[]{l}s=\psi\left(u\right)=\int\limits_{{u_{0}}}^{u}{\left\|{{\bf{\dot{c}}}\left(u\right)}\right\|}du\end{array}$

where $\left\|{{\bf{\dot{c}}}\left(u\right)}\right\|=\alpha
f\left(u\right)\sqrt{1+{g^{2}}\left(u\right)+{h^{2}}\left(u\right)}.$

If $\varphi$ denotes the inverse function of $\psi:I\to L\subset\mathbb{R}$,
then $u=\varphi\left(s\right)$ and

$\begin{array}[]{l}\varphi^{\prime}\left(s\right)={\left\|{{{\left.{\frac{{d{\bf{c}}\left(u\right)}}{{du}}}\right|}_{u=\varphi\left(s\right)}}}\right\|^{-1}}\quad,\quad
s\in I\end{array}$

where the prime $\left({}^{\prime}\right)$ denotes the differentiation with
respect to $s$.

The unit tangent vector ${{\bf{e}}_{1}}\left(s\right)$ of the curve $C$ at the
each point ${\bf{c}}\left({\varphi\left(s\right)}\right)$ is given by

$\begin{array}[]{l}{{\bf{e}}_{1}}\left(s\right)={\left({1+{g^{2}}\left({\varphi\left(s\right)}\right)+{h^{2}}\left({\varphi\left(s\right)}\right)}\right)^{-{1\mathord{\left/{\vphantom{12}}\right.\kern-1.2pt}2}}}\left[{\begin{array}[]{*{20}{c}}{\sinh\left({\varphi\left(s\right)}\right)}\\\
{\cosh\left({\varphi\left(s\right)}\right)}\\\
{g\left({\varphi\left(s\right)}\right)}\\\
{h\left({\varphi\left(s\right)}\right)}\\\ \end{array}}\right]\end{array}$
(3.17)

for all $s\in L$. Some simplifying assumptions are made for the sake of
brevity as follows;

$\begin{array}[]{l}\sinh:=\sinh\left({\varphi\left(s\right)}\right)\quad,\quad\quad\cosh:=\cosh\left({\varphi\left(s\right)}\right)\\\
f:=f\left({\varphi\left(s\right)}\right)\quad\quad\quad\,\,,\quad\quad
g:=g\left({\varphi\left(s\right)}\right)\quad\quad\quad,\quad
h:=h\left({\varphi\left(s\right)}\right),\\\
\dot{g}:=\dot{g}\left({\varphi\left(s\right)}\right)={\left.{\frac{{dg\left(u\right)}}{{du}}}\right|_{u=\varphi\left(s\right)}}\quad,\quad\dot{h}:=\dot{h}\left({\varphi\left(s\right)}\right)={\left.{\frac{{dh\left(u\right)}}{{du}}}\right|_{u=\varphi\left(s\right)}},\\\
\ddot{g}:=\ddot{g}\left({\varphi\left(s\right)}\right)={\left.{\frac{{{d^{2}}g\left(u\right)}}{{d{u^{2}}}}}\right|_{u=\varphi\left(s\right)}}\quad,\quad\ddot{h}:=\ddot{h}\left({\varphi\left(s\right)}\right)={\left.{\frac{{{d^{2}}h\left(u\right)}}{{d{u^{2}}}}}\right|_{u=\varphi\left(s\right)}},\\\
\varphi^{\prime}:=\varphi^{\prime}\left(s\right)={\left.{\frac{{d\varphi}}{{ds}}}\right|_{s}},\\\
A:=1+{g^{2}}+{h^{2}}\quad\,\,\,,\quad\quad B:=g\dot{g}+h\dot{h}\quad,\quad
C:={{\dot{g}}^{2}}+{{\dot{h}}^{2}},\\\
D:=g\ddot{g}+h\ddot{h}\quad\quad\quad,\quad\quad
E:=\dot{g}\ddot{g}+\dot{h}\ddot{h}\quad,\quad
F:={{\ddot{g}}^{2}}+{{\ddot{h}}^{2}}.\\\ \end{array}$

Then, we have

$\begin{array}[]{l}\dot{A}=2B\quad,\quad\dot{B}=C+D\quad,\quad\dot{C}=2E\quad,\quad\varphi^{\prime}={\alpha^{-1}}{f^{-1}}{A^{{{-1}\mathord{\left/{\vphantom{{-1}2}}\right.\kern-1.2pt}2}}}.\end{array}$

So, we rewrite the equation (3.17) as

$\begin{array}[]{l}{{\bf{e}}_{1}}:={{\bf{e}}_{1}}\left(s\right)={A^{-{1\mathord{\left/{\vphantom{12}}\right.\kern-1.2pt}2}}}\left[{\begin{array}[]{*{20}{c}}{\sinh}\\\
{\cosh}\\\ g\\\ h\\\ \end{array}}\right].\end{array}$ (3.18)

By differentiating the last equation with respect to $s$, we find

$\begin{array}[]{l}{{\bf{e^{\prime}}}_{1}}=\varphi^{\prime}\left[{\begin{array}[]{*{20}{c}}{-\frac{1}{2}{A^{-{3\mathord{\left/{\vphantom{32}}\right.\kern-1.2pt}2}}}\dot{A}\sinh+{A^{-{1\mathord{\left/{\vphantom{12}}\right.\kern-1.2pt}2}}}\cosh}\\\
{-\frac{1}{2}{A^{-{3\mathord{\left/{\vphantom{32}}\right.\kern-1.2pt}2}}}\dot{A}\cosh+{A^{-{1\mathord{\left/{\vphantom{12}}\right.\kern-1.2pt}2}}}\sinh}\\\
{-\frac{1}{2}{A^{-{3\mathord{\left/{\vphantom{32}}\right.\kern-1.2pt}2}}}\dot{A}g+{A^{-{1\mathord{\left/{\vphantom{12}}\right.\kern-1.2pt}2}}}\dot{g}}\\\
{-\frac{1}{2}{A^{-{3\mathord{\left/{\vphantom{32}}\right.\kern-1.2pt}2}}}\dot{A}h+{A^{-{1\mathord{\left/{\vphantom{12}}\right.\kern-1.2pt}2}}}\dot{h}}\\\
\end{array}}\right],\end{array}$

that is,

$\begin{array}[]{l}{{\bf{e^{\prime}}}_{1}}=-\varphi^{\prime}{A^{-{1\mathord{\left/{\vphantom{12}}\right.\kern-1.2pt}2}}}\left[{\begin{array}[]{*{20}{c}}{{A^{-1}}B\sinh-\cosh}\\\
{{A^{-1}}B\cosh-\sinh}\\\ {{A^{-1}}Bg-\dot{g}}\\\ {{A^{-1}}Bh-\dot{h}}\\\
\end{array}}\right].\end{array}$ (3.19)

From the last equation, we obtain

$\begin{array}[]{l}{k_{1}}:={k_{1}}\left(s\right)=\left\|{{{{\bf{e^{\prime}}}}_{1}}\left(s\right)}\right\|=\varphi^{\prime}{A^{-1}}{\left|{-A+AC-{B^{2}}}\right|^{{1\mathord{\left/{\vphantom{12}}\right.\kern-1.2pt}2}}}.\end{array}$
(3.20)

By the fact that
${{\bf{e}}_{2}}\left(s\right)={\left({{k_{1}}\left(s\right)}\right)^{-1}}{{\bf{e^{\prime}}}_{1}}\left(s\right)$,
we have

$\begin{array}[]{l}{{\bf{e}}_{2}}:={{\bf{e}}_{2}}\left(s\right)=-{A^{{1\mathord{\left/{\vphantom{12}}\right.\kern-1.2pt}2}}}{\left|{-A+AC-{B^{2}}}\right|^{{{-1}\mathord{\left/{\vphantom{{-1}2}}\right.\kern-1.2pt}2}}}\left[{\begin{array}[]{*{20}{c}}{{A^{-1}}B\sinh-\cosh}\\\
{{A^{-1}}B\cosh-\sinh}\\\ {{A^{-1}}Bg-\dot{g}}\\\ {{A^{-1}}Bh-\dot{h}}\\\
\end{array}}\right].\end{array}$

In order to get second curvature function ${k_{2}}$, we need to calculate
${k_{2}}\left(s\right)=\left\|{{{{\bf{e^{\prime}}}}_{2}}\left(s\right)-{\mu_{1}}{k_{1}}\left(s\right){{\bf{e}}_{1}}\left(s\right)}\right\|$.
It is seen from the above equation
$\left\langle{{{\bf{e}}_{2}}\left(s\right),{{\bf{e}}_{2}}\left(s\right)}\right\rangle=1$,
that is, ${{\bf{e}}_{2}}$ is spacelike. Thus, ${\mu_{1}}$ is equal to $-1$ and
${k_{2}}\left(s\right)=\left\|{{{{\bf{e^{\prime}}}}_{2}}\left(s\right)+{k_{1}}\left(s\right){{\bf{e}}_{1}}\left(s\right)}\right\|$.
After a long process of calculation, we have

$\begin{array}[]{l}{{\bf{e^{\prime}}}_{2}}+{k_{1}}{{\bf{e}}_{1}}=\varphi^{\prime}{A^{{{-3}\mathord{\left/{\vphantom{{-3}2}}\right.\kern-1.2pt}2}}}{\left|{-A+AC-{B^{2}}}\right|^{{{-3}\mathord{\left/{\vphantom{{-3}2}}\right.\kern-1.2pt}2}}}\left[{\begin{array}[]{*{20}{c}}{\left({P+Q}\right)\sinh-R\cosh}\\\
{\left({P+Q}\right)\cosh-R\sinh}\\\ {Pg-R\dot{g}+Q\ddot{g}}\\\
{Ph-R\dot{h}+Q\ddot{h}}\\\ \end{array}}\right]\end{array}$ (3.21)

where

$\begin{array}[]{l}P={\left({-A+AC-{B^{2}}}\right)^{2}}+\left({-A+AC-{B^{2}}}\right)\left({{B^{2}}-AC-
AD}\right)\\\ \,\,\,\,\,\,\,\,\,\,\,\,\,+AB\left({-B+AE-BD}\right),\\\
Q={A^{2}}\left({-A+AC-{B^{2}}}\right),\\\ R={A^{2}}\left({-B+AE-BD}\right).\\\
\end{array}$ (3.22)

If we simplify $P$ then we have

$\begin{array}[]{l}P={A^{2}}\left({1-C+BE+D-CD}\right).\end{array}$

Thus, we rewrite the equations (3.22) and (3.23) as

$\begin{array}[]{l}{{\bf{e^{\prime}}}_{2}}+{k_{1}}{{\bf{e}}_{1}}=\varphi^{\prime}{A^{{1\mathord{\left/{\vphantom{12}}\right.\kern-1.2pt}2}}}{\left|{-A+AC-{B^{2}}}\right|^{{{-3}\mathord{\left/{\vphantom{{-3}2}}\right.\kern-1.2pt}2}}}\left[{\begin{array}[]{*{20}{c}}{\left({\tilde{P}+\tilde{Q}}\right)\sinh-\tilde{R}\cosh}\\\
{\left({\tilde{P}+\tilde{Q}}\right)\cosh-\tilde{R}\sinh}\\\
{\tilde{P}g-\tilde{R}\dot{g}+\tilde{Q}\ddot{g}}\\\
{\tilde{P}h-\tilde{R}\dot{h}+\tilde{Q}\ddot{h}}\\\
\end{array}}\right]\end{array}$ (3.23)

where

$\begin{array}[]{l}\tilde{P}=1-C+BE+D-CD,\\\ \tilde{Q}=-A+AC-{B^{2}},\\\
\tilde{R}=-B+AE-BD.\\\ \end{array}$ (3.24)

Consequently, from the equations (3.24) and (3.25), we find

$\begin{array}[]{l}{\left\|{{{{\bf{e^{\prime}}}}_{2}}+{k_{1}}{{\bf{e}}_{1}}}\right\|^{2}}={\left({\varphi^{\prime}}\right)^{2}}A{\left|{-A+AC-{B^{2}}}\right|^{-3}}\,\left|{{{\left({\tilde{P}+\tilde{Q}}\right)}^{2}}-{{\tilde{R}}^{2}}}\right.\\\
\quad\quad\quad\quad\quad\,+{{\tilde{P}}^{2}}\left({{g^{2}}+{h^{2}}}\right)+{{\tilde{R}}^{2}}\left({{{\dot{g}}^{2}}+{{\dot{h}}^{2}}}\right)+{{\tilde{Q}}^{2}}\left({{{\ddot{g}}^{2}}+{{\ddot{h}}^{2}}}\right)\\\
\quad\quad\quad\quad\quad\,\left.{-2\tilde{P}\tilde{R}\left({g\dot{g}+h\dot{h}}\right)-2\tilde{R}\tilde{Q}\left({\dot{g}\ddot{g}+\dot{h}\ddot{h}}\right)+2\tilde{P}\tilde{Q}\left({g\ddot{g}+h\ddot{h}}\right)}\right|.\\\
\end{array}$

If we substitute the abbreviations into the last equation, we get

$\begin{array}[]{l}{\left\|{{{{\bf{e^{\prime}}}}_{2}}+{k_{1}}{{\bf{e}}_{1}}}\right\|^{2}}={\left({\varphi^{\prime}}\right)^{2}}A{\left|{-A+AC-{B^{2}}}\right|^{-3}}\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left|{{{\tilde{P}}^{2}}A+2\tilde{P}\tilde{Q}+{{\tilde{Q}}^{2}}-{{\tilde{R}}^{2}}+{{\tilde{R}}^{2}}C}\right.\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left.{+{{\tilde{Q}}^{2}}F-2\tilde{P}\tilde{R}B-2\tilde{R}\tilde{Q}E+2\tilde{P}\tilde{Q}D}\right|.\\\
\end{array}$

After substituting the equation (3.24) into the last equation and simplifying
it, we have

$\begin{array}[]{l}k_{2}^{2}={\left\|{{{{\bf{e^{\prime}}}}_{2}}+{k_{1}}{{\bf{e}}_{1}}}\right\|^{2}}\\\
\quad\,={\left({\varphi^{\prime}}\right)^{2}}A{\left|{-A+AC-{B^{2}}}\right|^{-2}}\,\,\left|{\left({-A+AC-{B^{2}}}\right)\left({1+F}\right)}\right.\\\
\,\,\,\,\,\,\,\,\,\,\left.{+\left({1-C}\right){{\left({1+D}\right)}^{2}}+2BE\left({1+D}\right)-A{E^{2}}}\right|\,.\\\
\end{array}$

Moreover, from the equation (3.20) it is seen that

$\begin{array}[]{l}k_{1}^{2}={\left({\varphi^{\prime}}\right)^{2}}{A^{-2}}\left|{-A+AC-{B^{2}}}\right|.\end{array}$

The last two equation gives us

$\begin{array}[]{l}k_{1}^{2}+k_{2}^{2}={\left({\varphi^{\prime}}\right)^{2}}{A^{-2}}{\left|{-A+AC-{B^{2}}}\right|^{-2}}\left|{{{\left({-A+AC-{B^{2}}}\right)}^{3}}}\right.\\\
\quad\,\,\,\,\,\,\,\,\,\,\,\,\,\left.{+{A^{3}}\left({\left({-A+AC-{B^{2}}}\right)\left({1+F}\right)+\left({1-C}\right){{\left({1+D}\right)}^{2}}+2BE\left({1+D}\right)-A{E^{2}}}\right)}\right|.\\\
\end{array}$

By the fact
$\varphi^{\prime}={\alpha^{-1}}{f^{-1}}{A^{{{-1}\mathord{\left/{\vphantom{{-1}2}}\right.\kern-1.2pt}2}}}$,
we obtain

$\begin{array}[]{l}k_{1}^{2}+k_{2}^{2}={\alpha^{-2}}{f^{-2}}{A^{-3}}{\left|{-A+AC-{B^{2}}}\right|^{-2}}\left|{{{\left({-A+AC-{B^{2}}}\right)}^{3}}}\right.\\\
\quad\quad\quad\;\left.{\,\,+{A^{3}}\left({\left({-A+AC-{B^{2}}}\right)\left({1+F}\right)+\left({1-C}\right){{\left({1+D}\right)}^{2}}+2BE\left({1+D}\right)-A{E^{2}}}\right)}\right|.\\\
\end{array}$ (3.25)

and

$\begin{array}[]{l}{k_{1}}={\alpha^{-1}}{f^{-1}}{A^{-{3\mathord{\left/{\vphantom{32}}\right.\kern-1.2pt}2}}}{\left({-A+AC-{B^{2}}}\right)^{{1\mathord{\left/{\vphantom{12}}\right.\kern-1.2pt}2}}}.\end{array}$
(3.26)

According to our assumption,

$\begin{array}[]{l}f={\left({1+{g^{2}}+{h^{2}}}\right)^{{{-3}\mathord{\left/{\vphantom{{-3}2}}\right.\kern-1.2pt}2}}}{\left|{-1-{g^{2}}-{h^{2}}+{{\dot{g}}^{2}}+{{\dot{h}}^{2}}+{{\left({\dot{g}h-g\dot{h}}\right)}^{2}}}\right|^{{{-5}\mathord{\left/{\vphantom{{-5}2}}\right.\kern-1.2pt}2}}}\\\
\,\,\,\,\,\,\,\,\,\,\,\left|{{{\left({-1-{g^{2}}-{h^{2}}+{{\dot{g}}^{2}}+{{\dot{h}}^{2}}+{{\left({\dot{g}h-g\dot{h}}\right)}^{2}}}\right)}^{3}}}\right.\\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,-{\left({1+{g^{2}}+{h^{2}}}\right)^{3}}\,\left({{{\left({g-\ddot{g}}\right)}^{2}}+{{\left({h-\ddot{h}}\right)}^{2}}}\right.\\\
\quad\left.{\left.{\,\,\,\,\,\,\,\,\,-{{\left({\left({g\dot{h}-\dot{g}h}\right)+\left({\dot{g}\ddot{h}-\ddot{g}\dot{h}}\right)}\right)}^{2}}+{{\left({g\ddot{h}-\ddot{g}h}\right)}^{2}}}\right)}\right|,\\\
\end{array}$

we obtain

$\begin{array}[]{l}f={A^{-{3\mathord{\left/{\vphantom{32}}\right.\kern-1.2pt}2}}}{\left|{-A+AC-{B^{2}}}\right|^{{{-5}\mathord{\left/{\vphantom{{-5}2}}\right.\kern-1.2pt}2}}}\left|{{{\left({-A+AC-{B^{2}}}\right)}^{3}}}\right.\\\
\,\,\,\,\,\,\,\,\left.{\,+{A^{3}}\left({\left({1+F}\right)+\left({1-C}\right){{\left({1+D}\right)}^{2}}+2BE\left({1+D}\right)-A{E^{2}}}\right)}\right|.\\\
\end{array}$

If we substitute the above equations (3.25) and (3.26), we obtain

$\begin{array}[]{l}{k_{1}}=\alpha\left({k_{1}^{2}+k_{2}^{2}}\right).\end{array}$

The proof is completed.

In the above equation ${\mu_{1}}={\mu_{2}}=-1$ which is the special Case 1.
This formula is the parametric equation of generalized spacelike Mannheim
curve with timelike second binormal vector in the Minkowski space-time
$E_{1}^{4}$.

## References

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# Colour versus Shape Goal Misgeneralization in Reinforcement Learning: A Case
Study

Karolis Ramanauskas
Department of Computer Science
University of Bath
Bath, United Kingdom
<EMAIL_ADDRESS>
Özgür Şimşek
Department of Computer Science
University of Bath
Bath, United Kingdom
<EMAIL_ADDRESS>

###### Abstract

We explore colour versus shape goal misgeneralization originally demonstrated
by Di Langosco et al. (2022) in the Procgen Maze environment, where, given an
ambiguous choice, the agents seem to prefer generalization based on colour
rather than shape. After training over 1,000 agents in a simplified version of
the environment and evaluating them on over 10 million episodes, we conclude
that the behaviour can be attributed to the agents learning to detect the goal
object through a specific colour channel. This choice is arbitrary.
Additionally, we show how, due to underspecification, the preferences can
change when retraining the agents using exactly the same procedure except for
using a different random seed for the training run. Finally, we demonstrate
the existence of outliers in out-of-distribution behaviour based on training
random seed alone.

##  1 Introduction

Goal misgeneralization in reinforcement learning is a type of failure observed
when an agent is deployed in an environment that is different from the
environment in which it was trained. Specifically, goal misgeneralization is
said to occur when an agent pursues a goal competently but that goal is not
the one intended by the system designer (Di Langosco et al., 2022; Shah et
al., 2022). For example, an agent trained to collect gold coins may fail to
collect them when the coins are placed differently in the environment but
still efficiently navigate to certain locations while successfully avoiding
enemies. Goal misgeneralization is highly relevant to AI safety. Advanced AI
systems that competently pursue unintended goals can pose significant risks to
humanity (Russell, 2019).

We focus on colour versus shape misgeneralization, following an experiment
conducted by Di Langosco et al. (2022), where the agent was trained (from
pixel observations) in different mazes to reach a yellow, line-shaped object.
When the agent was tested in mazes that did not contain a yellow line but
instead contained both a red line and a yellow gem, the agent generally
pursued the yellow gem rather than the red line. It may be argued that this
behaviour is expected given that colour is a feature directly observed by the
agent while shape is a latent one that has to be learned (Scimeca et al.,
2021). Is that what happened? Or are there other factors at play? Should we
expect more complex AI systems, such as self-driving cars, to also exhibit
such behaviour, perhaps by acting differently when driving in an area with red
school buses instead of yellow ones?

Our contributions are as follows:

  * •

We reproduce the goal misgeneralization observed by Di Langosco et al. (2022)
in a simpler environment (Section 2) that allows extensive behavioural
experimentation.

  * •

We show that a change as simple as the random seed used in training the agent
can change the preference of the agent from colour to shape (Subsection 3.1).

  * •

We show that this behaviour stems from the arbitrary and underspecified choice
of RGB colour channel through which the agent learns to detect the goal object
(Subsection 3.2).

  * •

We demonstrate the existence of outliers, on a scale of 1 in 500, in the ways
that agents learn to solve a task (Subsection 3.3). We discuss the
implications of such outliers for large-scale AI models.

##  2 Methodology

#### Original environment.

We started by trying to reproduce the results by Di Langosco et al. (2022) by
using the same code111https://github.com/jbkjr/objective-robustness-failures
and hyper-parameters. The agent uses PPO (Schulman et al., 2017) and the
IMPALA network architecture (Espeholt et al., 2018) with no recurrent
elements. The environment is based on the Procgen Maze (Cobbe et al., 2020)
but is slightly modified by replacing the original goal object (cheese) by a
yellow line. The only reward signal is $+$10 received upon reaching the goal,
which terminates the episode. An episode terminates also after 500 steps even
if the goal has not been reached. The maze layout, maze size (ranging from
3$\times$3 to 25$\times$25), goal position, and background textures are
randomized throughout training. LABEL:fig:downsampling shows some examples.
The different instances of the environment are called levels. The actions are
the four cardinal direction moves.222Procgen environments have 15 actions but
the remaining 11 actions do nothing in Maze. The agents quickly learn to use
only the four active actions.

After training, the agent is tested in unseen levels, where the yellow line is
replaced with two new objects in random locations – a yellow gem and a red
line. The goal is now ambiguous. And yet the agent reliably goes for the
yellow gem, revealing that it learned to detect the object by colour rather
than shape. This was observed in multiple training runs using different random
seeds. If the designer of the system intended the goal to be a line, goal
misgeneralization occurred.

#### Simplifying the environment.

Observations are downsampled from the 512$\times$512 pixel human view to
64$\times$64 before being given to the agent. After some experimentation, we
noticed that the downsampling procedure made line objects invisible about 50%
of the time but the gem objects only about 20% of the time. Additionally, the
yellow line and the yellow gem are slightly different shades of yellow. We
present full details and images in Appendix A. We simplified the environment
as follows to eliminate these differences between the goal objects (see
LABEL:fig:simple-maze):

  * •

We created new line and gem assets of all eight combinations of RGB colours in
their pure form, i.e. either 0 or 255 for each of the three channels, as shown
in Appendix B.

  * •

We always used a 5$\times$5 maze with no outside padding.

  * •

We revised the maze generation algorithm to not have dead ends. This prevents
cases where reaching one object without going over the other object is
impossible.

  * •

We make either object in the test environment end the episode but with
different rewards.

The simplified environment has the welcome side effect that a single training
run takes about 40 GPU minutes instead of 40 GPU hours, mainly due to fewer
agent-environment interactions needed to train the agent. A speed up of 60
times helps run more comprehensive experiments on these agents. The last two
changes allow for the systematic testing of many trained agents.

With these changes, we obtained the same result as those reported by Di
Langosco et al. (2022) on the first agent that we trained. Specifically, an
agent we trained to seek out a yellow line preferred a yellow gem to a red
line in the test environment. We then performed a large number of further
experiments, which we explain in the next section.333Code available here:
https://github.com/KarolisRam/colour-shape-goal-misgeneralization

#### Measuring preferences and capabilities.

We train 100 agents using different random seeds. This impacts the weight
initializations of the neural networks and the training environment levels.
Each agent is tested on the same set of 1,000 held-out levels. Most results
are shown as scatter plots of preferences and capabilities because goal
misgeneralization is about the interaction between preferences and
capabilities. We care the most about cases where different preferences can
appear while capabilities remain intact. For example, a training procedure
that produces two agents that both act capably but differ in their out-of-
distribution goals would be of considerable concern if observed in advanced AI
systems.

To measure capabilities, we use the mean episode length. Even an agent acting
randomly usually gets to a goal object before the episode times out. On
average, a trained agent takes 4 steps to reach a goal while a random agent
takes 96 steps. In all plots, the axis for mean episode length ranges between
0 and 100, representing the range between a competent agent and a random one.
Any marks appearing on the border show an episode length of 100 or more,
indicating performance worse than the one obtained by acting randomly.444The
mean is not a perfect measure of capabilities. For example, some agents are
fully capable on more than half of the test levels while getting stuck in the
remaining levels. This could give a mean of 200 and a median of 4. We observed
that this rarely happens in practice.

We measure preferences by the proportion of test runs in which the agent
picked up a particular object before the other one. We established a baseline
for preferences as follows. When an agent is competently navigating to a
specific object, it may step on the other object along the way, without
seeking to do so. We therefore tested an agent in an environment with a yellow
line and an invisible object, observing that the agent stepped on the
invisible object approximately 20% of the time on its way to the yellow line.
Thus, we consider picking an object 80% of the time as the baseline for the
full preference for that object.

##  3 Results

###  3.1 The first 100 training runs

After confirming that goal misgeneralization happens in our simplified Maze
environment, we ran this same training experiment 100 times, varying only the
random seed used to initialize the weights of the network and the environment
levels. The goal in these experiments is a yellow line. We trained each agent
for 10 million steps to ensure all of them converged. We then tested each
trained agent on the same set of 1,000 test levels and measured the preference
for the red line versus the yellow gem and the mean episode length. We show
the results as a scatter plot on the top left of LABEL:fig:train-yellow-line-
test-red-green-blue-lines-vs-yellow-gem. We see that most agents keep their
capabilities and most strongly prefer the yellow gem. However, we see several
agents having no preference for either object yet keeping their capabilities
intact.

We ran two further experiments, where we replaced the red line first with a
green line, and then with a blue line. The results are shown on the top row of
LABEL:fig:train-yellow-line-test-red-green-blue-lines-vs-yellow-gem. In the
case of the green line, agents mostly keep their capabilities, but many agents
prefer the green line to the yellow gem. However, other agents still preferred
the yellow gem to the green line. Why did a different colour induce a
preference for the shape? Our next clue is the capability loss in some agents
when the line was blue.

Agents do not see things in the same way that humans do. Image observations
usually come as three 2D matrices of numbers between 0 and 255, representing
the three colour channels: red, green and blue. Yellow is a combination of red
and green colours and a lack of blue. This is still somewhat similar to human
trichromacy (Horiguchi et al., 2013), but not a perfect analogy, as humans do
not have conscious access to the three separate colours. Humans also cannot
distinguish monochromatic yellow from yellow produced by mixing red and green
light (Chittka, 2022).555An analogy in music would be hearing note D exactly
like notes C and E played together, which we do not. In the case of the
agents, they would hear D (yellow) as C + E (red + green), and not as a single
tone (colour). A closer analogy might be the more separate red and yellow
fields in the retina of the pigeon, which are used to view objects at
different distances (Bloch and Martinoya, 1983). The agent can detect the line
using any of the three channels. We show what the different lines look like to
the agent in LABEL:fig:rgb-splits.

The question remains of why the red line of the original experiment did not
exhibit the same range of behaviours as the green line. We believe this is
because the different backgrounds are not uniformly distributed among the
three colour channels, which we show in detail in Appendix C. We hypothesize
that in the case of the green line versus the yellow gem, the agents that
learned to detect the yellow line primarily via the green channel prefer the
green line. In contrast, the agents that used the red channel to detect the
yellow line preferred the yellow gem. If this is true, the goal
misgeneralization in this case happens mostly through colour.

###  3.2 Removing background textures

To show an even simpler case of goal misgeneralization, we changed the
background to pure black. The black line is entirely invisible in this case
while the yellow line is visible only in the red and green channels. The lines
on a black background and their RGB splits are shown in LABEL:fig:rgb-splits.

We train 100 new agents and test each one on 1,000 levels in 14 different
experimental setups, all of which are shown in LABEL:fig:train-yellow-line-
black-background-01 and LABEL:fig:train-yellow-line-black-background-02 in
Appendix F. Here we focus on the most interesting results. The red and the
green line versus yellow gem look much more similar than in the experiments
with textures, as shown in the first two plots on the bottom row of
LABEL:fig:train-yellow-line-test-red-green-blue-lines-vs-yellow-gem. This
gives some credence to the hypothesis that backgrounds were causing that
difference. It also shows how a different type of goal misgeneralization can
occur by simply retraining the agent.

Next, we test the agents on the red line versus the green line and show a
roughly even split between agents preferring each line, as shown in
LABEL:fig:red-line-green-line-and-OLS. Many agents do not show a strong
preference, indicating learning through both red and green channels. One
interesting observation is that the agents that prefer the red line show some
capability loss, while the ones that prefer the green line do not. This
asymmetry may be due to the colours of the maze walls or maybe even the mouse.

#### Colour versus shape.

Do agents that most strongly prefer the green line versus the red line also
prefer the yellow gem versus the red line? If so, this further proves goal
misgeneralization happening mainly through the agent’s arbitrary choice of
colour channel to learn through. We perform an OLS regression on the
preferences of each agent and show the results in LABEL:fig:red-line-green-
line-and-OLS. The relationship is reasonably strong (R2 of 0.56) but less
robust than we expected. Something else seems to be happening, especially when
considering the outliers on the chart. We will call agents by the training
seeds used in their training runs. For example, Agent-8894 is the agent
closest to the bottom right corner on the right plot in LABEL:fig:red-line-
green-line-and-OLS. It prefers green line over red line over yellow gem. It
would be informative to understand how exactly this particular agent learned
to perform the task. Perhaps it learned to detect the shape and prefers green
colour, but not as strongly as it prefers the line shape.

It might be that these observations are relevant only for toy problems because
real-world images rarely have pure single or double-channel colours. However,
it would be interesting to test how different self-driving car models perceive
red, yellow, and green colours on a black background. It might be an important
thing to distinguish well while driving.666This idea is not too far fetched.
We have already seen them get confused by stickers on road signs or ghost
white lines on the road.

###  3.3 Outliers

After noticing outliers in many of the scatter plots, we decided to
investigate them further. We trained 512 agents to reach a white line on black
backgrounds. The line is visible in all three channels, which adds more
possibilities to learn the same underspecified task in different ways. As
usual, only the random seed was changed between the 512 training runs. We then
measured the preferences and capabilities of each agent in 11 different two-
object variations, 1,000 levels for each variation. This produced a dataset
with 512 $\times$ 22 data points. We used several clustering and outlier
detection tools and also simple sorting by primary and derivative features of
this dataset. We found various interesting outliers, many of them unique among
the 512 agents. For example, Agent-439 was the only one out of 512 to strongly
prefer white gem over the yellow line while being highly capable (bottom left
dot in the bottom left plot of LABEL:fig:train-white-line-black-background).
Meanwhile, Agent-2875 performed worse than a random agent when both lines were
single channel (red, green or blue). It also really liked purple lines. A
longer list of outlier behaviours can be found in Appendix D.

Could similar outliers exist in frontier AI models, and how would we know?
What does this mean for model behaviour attribution, when rare outcomes can
happen based on the training random seed alone, that is, luck? This might even
have some parallels in human behaviour, where we sometimes observe extreme
deviations from typical patterns, and individuals may receive diagnoses such
as OCD, ADHD, or autism to describe their unique characteristics and
challenges. In the Maze agents, the random seed changes the neural network
weight initialization and the training levels, which could loosely represent
the genes and the environment, both of which interact to produce the outcomes.
We try to detect these conditions as early as possible in humans to be able to
intervene if needed. Should we be trying to look for signs of outlier
behaviour in frontier AI models early in their training?

##  4 Related work

#### Behaviour attribution.

Hilton et al. (2020) use various interpretability techniques to understand the
behaviour of Procgen Coinrun agents. McGrath et al. (2022) analyzed the chess
concepts learned by AlphaZero (Silver et al., 2018) through behavioural
analysis and other methods. Atrey et al. (2019) show how saliency maps are not
enough for attributing agent behaviour and that counterfactual experiments
should be performed.

#### Goal misgeneralization.

Di Langosco et al. (2022) introduced initial empirical examples, while Shah et
al. (2022) added more, including hypothetical future ones and showed how the
problem is different from misspecification. Turner et al. (2023) explored an
object versus location misgeneralization in a similar Procgen Maze
environment. Some image classification failures can also be interpreted as
goal misgeneralization (Shah et al., 2022). Examples include classifying a
wolf as a husky if there is snow in the background (Ribeiro et al., 2016) or
classifying skin lesions as malignant if they were photographed next to a
ruler (Narla et al., 2018).

#### Underspecification.

D’Amour et al. (2022) shows how retraining the same machine learning model
using a different random seed can lead to the same test set performance but
different out-of-distribution performance. They do this in multiple domains
and input modalities. We show this in a deep reinforcement learning setting
with goal misgeneralization. Sellam et al. (2021) demonstrate a similar
underspecification phenomenon in larger scale language models, including
visible outliers on their Winogender bias correlation plot. This indicates the
possibility of different random seed based goal misgeneralization and outlier
behaviours in frontier AI models.

##  5 Conclusion

We have shown how part of the colour versus shape goal misgeneralization
behaviour in Procgen Maze can be attributed to the RGB colour encoding scheme.
We also showed that this behaviour can change based on the random seed used
for training. These observations raise the question of how far such findings
of model behaviour attribution transfer beyond the toy examples. Since we see
underspecification happening in larger models, we expect to also see empirical
examples of goal misgeneralization being discovered in such models. It is
possible that outliers in out-of-distribution behaviour exist in large-scale
models based on the training random seed alone. However, it would be
infeasible to run such experiments on frontier AI models. Thus, we should also
be careful when making claims about behaviour attribution – they might be true
for that particular checkpoint of the trained model but might not hold anymore
if it was simply retrained using a different random seed.

A potential limitation of our work is that the simplified environment might
not be diverse enough and no meaningful generalizations are learned.
Specifically, we would expect more complex vision-based deep reinforcement
learning agents to show colour versus shape preference, even if it did not
happen in the case of Procgen Maze. In the future, we hope to see more work of
this kind – taking a specific example of interesting behaviour and trying to
deeply understand how it happens.

## Acknowledgments and Disclosure of Funding

This work was supported by the UKRI Centre for Doctoral Training in
Accountable, Responsible and Transparent AI (ART-AI) [EP/S023437/1] and the
University of Bath. This research made use of Hex, the GPU Cloud in the
Department of Computer Science at the University of Bath. We thank the
reviewers for the useful feedback. We also thank Rachael Bedford, Dan Beechey,
Christoph Bumiller, Sophia Jones, Gabija Juce, Anssi Kanervisto, Stephanie
Milani, Owen Parsons and Tom Smith for helpful discussions and reading earlier
versions of this work. We thank members of the Bath Reinforcement Learning Lab
and the SANDPIT Lab for their feedback.

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##  Appendix A Disappearing objects

Here we explain the issue of disappearing objects in the original environment
used to demonstrate colour versus shape goal misgeneralization by Di Langosco
et al. (2022). Information is invariably lost when downsampling a pixel-based
environment from human view (512$\times$512 pixels) to agent view
(64$\times$64 pixels). In the case of a pixel-based maze with 25$\times$25
grid squares, each square will get downsampled to rectangles with side lengths
of 2–3 pixels. In a sample of 100 randomly selected levels, about half of the
time, this will make the line goal object completely invisible to the agent.
The rest of the time, the gem and the line objects are 1–4 coloured pixels, as
seen in LABEL:fig:downsampling. The gem object is invisible only 20% of the
time. This makes the result that the agent prefers the yellow gem relatively
weak – half the time, the agent does not even see the red line. This might
mean that the agent learns some algorithm, such as “follow the left wall”,
which helps it find the goal when it is invisible, as well as learning to
navigate to the yellow line. In addition, the yellow line and the yellow gem
are slightly different shades of yellow.

##  Appendix B Original and new environment assets

##  Appendix C Procgen Maze background texture analysis

As seen below, Procgen Maze environment texture colours are not entirely
random and have some biases towards specific colours. This can make objects
more or less visible in certain colour channels, influencing final learned
behaviour.

##  Appendix D List of outliers

Below is a list of outliers we found among the 512 agents trained to reach a
white line on a black background (unless otherwise indicated).

  * •

Agent-2875 – completely breaks to worse than random performance when the lines
are single channel (red, green or blue). Really likes purple lines.

  * •

Agent-56 and Agent-3824 – do not prefer the original white line to a yellow
line, seemingly ignoring the blue channel (3rd row, left plot of
LABEL:fig:train-white-line-black-background, two bottom dots).

  * •

Agent-1278 – very high preference for the blue line and breaks on the red line
versus the green line.

  * •

Agent-410 – very high preference for the blue line, but does not break on the
red line versus the green line.

  * •

Agent-2505 – no colour preference, highly capable in each setting.

  * •

Agent-9986 – no colour preference, much less capable on single channel
colours.

  * •

Agent-9510 – loses a large portion of capabilities when two white lines are
present (3rd row, middle plot of LABEL:fig:train-white-line-black-background,
right-most dot).

  * •

Agent-6406 – breaks transitivity, slightly preferring cyan over yellow over
purple over cyan. However, the preferences are between 45–50%, so this is
almost surely just a statistical artefact.

  * •

Agent-439 – only one out of 512 to strongly prefer white gem over yellow line
while being highly capable (bottom left plot of LABEL:fig:train-white-line-
black-background, bottom left dot).

  * •

Agent-8894 – (trained on yellow line with black backgrounds) prefers the green
line over the red line over the yellow gem (right plot in LABEL:fig:red-line-
green-line-and-OLS, top left dot). It seems to have learned about shape and
colour and prefers green, but not as strongly as the line shape.

  * •

Agent-1983 – (trained on red line with black backgrounds) slightly prefers
blue line to red gem (top middle plot of LABEL:fig:train-red-line-black-
background-01, top left dot). Also capably pursues a blue line when given a
choice between a blue line and a green line (top right plot of
LABEL:fig:train-red-line-black-background-02, bottom left dot). This one is
peculiar – why did it learn to seek a blue line while training to reach a red
line?

##  Appendix E Other experiments

We trained over 1,000 agents but showed the results for only 700 of them. Here
we briefly describe our findings with the remaining 300 agents. When we
replaced the textures with a grey background instead of the black one, making
lines visible in all three channels, the asymmetry between the red line and
the green line versus the yellow gem disappeared as expected. However, we
observed much more capability loss in this setting. We tried that by training
to reach a yellow line (100 agents) and a red line (100 agents).

When we trained another 100 agents to reach a red line on a black background,
we expected there to be no goal misgeneralization because there was only a
single channel through which to detect the object. This was not the case. When
tested on a yellow line versus a red gem, there was a roughly even split
between agents that pursued each object. We do not know how to explain this
and leave it for future research.

We include scatter plots of all these experiments in Appendix F. We believe
that interesting further insights can be gained by analysing the plots in
detail. For example, there are some agents that perform much better than
random (38 steps on average instead of 96) when both objects are invisible
(bottom left plot of LABEL:fig:train-yellow-line-black-background-02) – what
algorithm could they be using to do that?

##  Appendix F All results

Below we provide full results of all the simplified Maze agents we trained.
Some plots are duplicates of plots presented in the main paper; we present
them here for easier comparison. All scenarios had 100 agents, each tested on
1,000 episodes, except the agents trained on the white line with black
backgrounds, of which there were 512.