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# Classification Theorems of Ruled Surfaces in Minkowski Three-Space

1
Department of Mathematics Education and RINS, Gyeongsang National University, Jinju 52828, Korea
2
Department of Mathematics, Kyungpook National University, Daegu 41566, Korea
*
Author to whom correspondence should be addressed.
Mathematics 2018, 6(12), 318; https://doi.org/10.3390/math6120318
Received: 16 November 2018 / Revised: 5 December 2018 / Accepted: 8 December 2018 / Published: 11 December 2018

## Abstract

By generalizing the notion of the pointwise 1-type Gauss map, the generalized 1-type Gauss map has been recently introduced. Without any assumption, we classified all possible ruled surfaces with the generalized 1-type Gauss map in a 3-dimensional Minkowski space. In particular, null scrolls do not have the proper generalized 1-type Gauss map. In fact, it is harmonic.

## 1. Introduction

Thanks to Nash’s imbedding theorem, Riemannian manifolds can be regarded as submanifolds of Euclidean space. The notion of finite-type immersion has been used in studying submanifolds of Euclidean space, which was initiated by B.-Y. Chen by generalizing the eigenvalue problem of the immersion [1]. An isometric immersion x of a Riemannian manifold M into a Euclidean space $E m$ is said to be of finite-type if it has the spectral decomposition as:
$x = x 0 + x 1 + ⋯ + x k ,$
where $x 0$ is a constant vector and $Δ x i = λ i x i$ for some positive integer k and $λ i ∈ R$, $i = 1 , … , k$. Here, $Δ$ denotes the Laplacian operator defined on M. If $λ 1$, …, $λ k$ are mutually different, M is said to be of k-type. Naturally, we may assume that a finite-type immersion x of a Riemannian manifold into a Euclidean space is of k-type for some positive integer k.
The notion of finite-type immersion of the submanifold into Euclidean space was extended to the study of finite-type immersion or smooth maps defined on submanifolds of a pseudo-Euclidean space $E s m$ with the indefinite metric of index $s ≥ 1$. In this sense, it is very natural for geometers to have interest in the finite-type Gauss map of submanifolds of a pseudo-Euclidean space [2,3,4].
We now focus on surfaces of the Minkowski space $E 1 3$. Let M be a surface in the 3-dimensional Minkowski space $E 1 3$ with a non-degenerate induced metric. From now on, a surface M in $E 1 3$ means non-degenerate, i.e., its induced metric is non-degenerate unless otherwise stated. The map G of a surface M into a semi-Riemannian space form $Q 2 ( ϵ )$ by parallel translation of a unit normal vector of M to the origin is called the Gauss map of M, where $ϵ ( = ± 1 )$ denotes the sign of the vector field G. A helicoid or a right cone in $E 3$ has the unique form of Gauss map G, which looks like the 1-type Gauss map in the usual sense [5,6]. However, it is quite different from the 1-type Gauss map, and thus, the authors defined the following definition.
Definition 1.
([7]) The Gauss map G of a surface M in $E 1 3$ is of pointwise 1-type if the Gauss map G of M satisfies:
$Δ G = f ( G + C )$
for some non-zero smooth function f and a constant vector $C .$ Especially, the Gauss map G is called pointwise 1-type of the first kind if $C$ is a zero vector. Otherwise, it is said to be of pointwise 1-type of the second kind.
Some other surfaces of $E 3$ such as conical surfaces have an interesting type of Gauss map. A surface in $E 1 3$ parameterized by:
$x ( s , t ) = p + t β ( s ) ,$
where p is a point and $β ( s )$ a unit speed curve is called a conical surface. The typical conical surfaces are a right (circular) cone and a plane.
Example 1.
([8]) Let Mbe a surface in $E 3$ parameterized by:
$x ( s , t ) = ( t cos 2 s , t sin s cos s , t sin s ) .$
Then, the Gauss map G can be obtained by:
$G = 1 1 + cos 2 s ( − sin 3 s , ( 2 − cos 2 s ) cos s , − cos 2 s ) .$
Its Laplacian turns out to be:
$Δ G = f G + g C$
for some non-zero smooth functions$f , g$and a constant vector$C .$The surfaceMis a kind of conical surface generated by a spherical curve$β ( s ) = ( cos 2 s , sin s cos s , sin s )$on the unit sphere$S 2 ( 1 )$centered at the origin.
Based on such an example, by generalizing the notion of the pointwise 1-type Gauss map, the so-called generalized 1-type Gauss map was introduced.
Definition 2.
([8]) The Gauss map Gof a surfaceMin $E 1 3$ is said to be of generalized 1-type if the Gauss map G satisfies:
$Δ G = f G + g C$
for some non-zero smooth functions $f , g$ and a constant vector $C .$ If $f ≠ g$, Gis said to be of proper generalized 1-type.
Definition 3.
A conical surface with the generalized 1-type Gauss map is called a conical surface of G-type.
Remark 1.
([8]) We can construct a conical surface of G-type with the functionsf, gand the vector $C$ if we solve the differential Equation (1).
Here, we provide an example of a cylindrical ruled surface in the 3-dimensional Minkowski space $E 1 3$ with the generalized 1-type Gauss map.
Example 2.
Let M be a ruled surface in the Minkowski 3-space $E 1 3$ parameterized by:
$x ( s , t ) = 1 2 s s 2 − 1 − ln ( s + s 2 − 1 ) , 1 2 s 2 , t , s ≥ 1 .$
Then, the Gauss map G is given by:
$G = ( − s , − s 2 − 1 , 0 ) .$
By a direct computation, we see that its Laplacian satisfies:
$Δ G = s − s 2 − 1 ( s 2 − 1 ) 3 2 G + s ( s − s 2 − 1 ) ( s 2 − 1 ) 3 2 ( 1 , − 1 , 0 ) ,$
which indicates that M has the generalized 1-type Gauss map.

## 2. Preliminaries

Let M be a non-degenerate surface in the Minkowski 3-space $E 1 3$ with the Lorentz metric $d s 2 = − d x 1 2 + d x 2 2 + d x 3 2 ,$ where $( x 1 , x 2 , x 3 )$ denotes the standard coordinate system in $E 1 3 .$ From now on, a surface in $E 1 3$ means non-degenerate unless otherwise stated. A curve in $E 1 3$ is said to be space-like, time-like, or null if its tangent vector field is space-like, time-like, or null, respectively. Then, the Laplacian $Δ$ is given by:
$Δ = − 1 | G | ∑ i , j = 1 2 ∂ ∂ x ¯ i ( | G | g i j ∂ ∂ x ¯ j ) ,$
where $( g i j ) = ( g i j ) − 1 ,$ $G$ is the determinant of the matrix $( g i j )$ consisting of the components of the first fundamental form and ${ x ¯ i }$ are the local coordinate system of M.
A ruled surface M in the Minkowski 3-space $E 1 3$ is defined as follows: Let I and J be some open intervals in the real line R. Let $α = α ( s )$ be a curve in $E 1 3$ defined on I and $β = β ( s )$ a transversal vector field with $α ′ ( s )$ along $α$. From now on, $′$ denotes the differentiation with respect to the parameter s unless otherwise stated. The surface M with a parametrization given by:
$x ( s , t ) = α ( s ) + t β ( s ) , s ∈ I , t ∈ J$
is called a ruled surface. In this case, the curve $α = α ( s )$ is called a base curve and $β = β ( s )$ a director vector field or a ruling. A ruled surface M is said to be cylindrical if $β$ is constant. Otherwise, it is said to be non-cylindrical.
If we consider the causal character of the base and director vector field, we can divide a few different types of ruled surfaces in $E 1 3$: If the base curve $α$ is space-like or time-like, the director vector field $β$ can be chosen to be orthogonal to $α$. The ruled surface M is said to be of type $M +$ or $M −$, respectively, depending on $α$ being space-like or time-like, respectively. Furthermore, the ruled surface of type $M +$ can be divided into three types $M + 1 ,$ $M + 2$, and $M + 3$. If $β$ is space-like, it is said to be of type $M + 1$ or $M + 2$ if $β ′$ is non-null or null, respectively. When $β$ is time-like, $β ′$ must be space-like because of the character of the causal vectors, which we call $M + 3$. On the other hand, when $α$ is time-like, $β$ is always space-like. Accordingly, it is also said to be of type $M − 1$ or $M − 2$ if $β ′$ is non-null or null, respectively. The ruled surface of type $M + 1$ or $M + 2$ (resp. $M + 3 , M − 1$ or $M − 2$) is clearly space-like (resp. time-like).
If the base curve $α$ is null, the ruling $β$ along $α$ must be null since M is non-degenerate. Such a ruled surface M is called a null scroll. Other cases, such as $α$ is non-null and $β$ is null, or $α$ is null and $β$ is non-null, are determined to be one of the types $M ± 1 , M ± 2$, and $M + 3$, or a null scroll by an appropriate change of the base curve [9].
Consider a null scroll: Let $α = α ( s )$ be a null curve in $E 1 3$ with Cartan frame ${ A , B , C }$, that is $A , B , C$ are vector fields along $α$ in $E 1 3$ satisfying the following conditions:
$〈 A , A 〉 = 〈 B , B 〉 = 0 , 〈 A , B 〉 = 1 , 〈 A , C 〉 = 〈 B , C 〉 = 0 , 〈 C , C 〉 = 1 ,$
$α ′ = A , C ′ = − a A − k ( s ) B ,$
where a is a constant and $k ( s )$ a nowhere vanishing function. A null scroll parameterized by $x = x ( s , t ) = α ( s ) + t B ( s )$ is called a B-scroll, which has constant mean curvature $H = a$ and constant Gaussian curvature $K = a 2$. Furthermore, its Laplacian $Δ G$ of the Gauss map G is given by:
$Δ G = − 2 a 2 G ,$
from which we see that a B-scroll is minimal if and only if it is flat [2,10].
Throughout the paper, all surfaces in $E 1 3$ are smooth and connected unless otherwise stated.

## 3. Cylindrical Ruled Surfaces in $E 1 3$ with the Generalized 1-Type Gauss Map

Let M be a cylindrical ruled surface of type $M + 1 , M − 1$ or $M + 3$ in $E 1 3$. Then, M is parameterized by a base curve $α$ and a unit constant vector $β$ such that:
$x ( s , t ) = α ( s ) + t β$
satisfying $〈 α ′ , α ′ 〉 = ε 1 ( = ± 1 ) ,$ $〈 α ′ , β 〉 = 0$, and $〈 β , β 〉 = ε 2 ( = ± 1 ) .$
We now suppose that M has generalized 1-type Gauss map G. Then, the Gauss map G satisfies Condition (1). We put the constant vector $C = ( c 1 , c 2 , c 3 )$ in (1) for some constants $c 1 , c 2$, and $c 3 .$
Suppose that $f = g .$ In this case, the Gauss map G is of pointwise 1-type. A classification of cylindrical ruled surfaces with the pointwise 1-type Gauss map in $E 1 3$ was described in [11].
If M is of type $M + 1$, then M is an open part of a Euclidean plane or a cylinder over a curve of infinite-type satisfying:
$c 2 f − 1 3 − ln | c 2 f − 1 3 + 1 | = ± c 3 ( s + k )$
if $C$ is null, or
$c 2 f − 1 3 + 1 2 + ( − c 1 2 + c 2 2 ) − ln c 2 f − 1 3 + 1 + c 2 f − 1 3 + 1 2 + ( − c 1 2 + c 2 2 ) + ln | − c 1 2 + c 2 2 | = ± c 3 ( s + k )$
if $C$ is non-null, where c is some non-zero constant and k is a constant.
If M is of type $M − 1$, M is an open part of a Minkowski plane or a cylinder over a curve of infinite-type satisfying:
$c 2 f − 1 3 + ln | c 2 f − 1 3 − 1 | = ± c 3 ( s + k )$
or:
$c 2 f − 1 3 − 1 2 − ( − c 1 2 + c 2 2 ) + ln c 2 f − 1 3 − 1 + c 2 f − 1 3 − 1 2 + | − c 1 2 + c 2 2 | − ln | − c 1 2 + c 2 2 | = ± c 3 ( s + k )$
depending on the constant vector, $C$, being null or non-null, respectively, for some non-zero constant c and some constant k.
If M is of type $M + 3$, M is an open part of either a Minkowski plane or a cylinder over a curve of infinite-type satisfying:
$c 2 2 + c 3 2 − c 2 f − 1 3 − 1 2 − sin − 1 c 2 f − 1 3 − 1 c 2 2 + c 3 2 = ± c 3 ( s + k ) ,$
where c is a non-zero constant and k a constant.
We now assume that $f ≠ g .$ Here, we consider two cases.
Case 1. Let M be a cylindrical ruled surface of type $M + 1$ or $M − 1 ,$ i.e., $ε 2 = 1 .$ Without loss of generality, the base curve $α$ can be put as $α ( s ) = ( α 1 ( s ) , α 2 ( s ) , 0 )$ parameterized by arc length s and the director vector field $β$ as a unit constant vector $β = ( 0 , 0 , 1 ) .$ Then, the Gauss map G of M and the Laplacian $Δ G$ of the Gauss map are respectively obtained by:
$G = ( − α 2 ′ ( s ) , − α 1 ′ ( s ) , 0 ) and Δ G = ( ε 1 α 2 ‴ ( s ) , ε 1 α 1 ‴ ( s ) , 0 ) .$
With the help of (1) and (7), it immediately follows:
$C = ( c 1 , c 2 , 0 )$
for some constants $c 1$ and $c 2$. We also have:
$ε 1 α 2 ‴ = − f α 2 ′ + g c 1 , ε 1 α 1 ‴ = − f α 1 ′ + g c 2 .$
Firstly, we consider the case that M is of type $M + 1 .$ Since $α$ is space-like, we may put:
$α 1 ′ ( s ) = sinh ϕ ( s ) and α 2 ′ ( s ) = cosh ϕ ( s )$
for some function $ϕ ( s )$ of s. Then, (8) can be written in the form:
$( ϕ ′ ) 2 cosh ϕ + ϕ ″ sinh ϕ = − f cosh ϕ + g c 1 , ( ϕ ′ ) 2 sinh ϕ + ϕ ″ cosh ϕ = − f sinh ϕ + g c 2 .$
This implies that:
$( ϕ ′ ) 2 = − f + g ( c 1 cosh ϕ − c 2 sinh ϕ )$
and:
$ϕ ″ = g ( − c 1 sinh ϕ + c 2 cosh ϕ ) .$
In fact, $ϕ ′$ is the signed curvature of the base curve $α = α ( s ) .$
Suppose $ϕ$ is a constant, i.e., $ϕ ′ = 0$. Then, $α$ is part of a straight line. In this case, M is an open part of a Euclidean plane.
Now, we suppose that $ϕ ′ ≠ 0 .$ From (8), we see that the functions f and g depend only on the parameter $s ,$ i.e., $f ( s , t ) = f ( s )$ and $g ( s , t ) = g ( s ) .$ Taking the derivative of Equation (9) and using (10), we get:
$3 ϕ ′ ϕ ″ = − f ′ + g ′ ( c 1 cosh ϕ − c 2 sinh ϕ ) .$
With the help of (9), it follows that:
$3 2 ( ϕ ′ ) 2 ′ = − f ′ + g ′ g ( ϕ ′ ) 2 + f .$
Solving the above differential equation, we have:
$ϕ ′ ( s ) 2 = k 1 g 2 3 + 2 3 g 2 3 ∫ g − 2 3 f − f ′ f + g ′ g d s , k 1 ( ≠ 0 ) ∈ R .$
We put:
$ϕ ′ ( s ) = ± p ( s ) ,$
where $p ( s ) = | k 1 g 2 3 + 2 3 g 2 3 ∫ g − 2 3 f − f ′ f + g ′ g d s |$. This means that the function $ϕ$ is determined by the functions $f , g$ and a constant vector satisfying (1). Therefore, the cylindrical ruled surface M satisfying (1) is determined by a base curve $α$ such that:
$α ( s ) = ∫ sinh ϕ ( s ) d s , ∫ cosh ϕ ( s ) d s , 0$
and the director vector field $β ( s ) = ( 0 , 0 , 1 )$.
In this case, if f and g are constant, the signed curvature $ϕ ′$ of a base curve $α$ is non-zero constant, and the Gauss map G is of the usual 1-type. Hence, M is an open part of a hyperbolic cylinder or a circular cylinder [12].
Suppose that one of the functions f and g is not constant. Then, M is an open part of a cylinder over the base curve of infinite-type satisfying (11). For a curve of finite-type in a plane of $E 1 3$, see [12] for the details.
Next, we consider the case that M is of type $M − 1 .$ Since $α$ is time-like, we may put:
$α 1 ′ ( s ) = cosh ϕ ( s ) and α 2 ′ ( s ) = sinh ϕ ( s )$
for some function $ϕ ( s )$ of s.
As was given in the previous case of type $M + 1$, if the signed curvature $ϕ ′$ of the base curve $α$ is zero, M is part of a Minkowski plane.
We now assume that $ϕ ′ ≠ 0$. Quite similarly as above, we have:
$ϕ ′ ( s ) 2 = k 2 g 2 3 + 2 3 g 2 3 ∫ g − 2 3 f f ′ f − g ′ g d s , k 2 ( ≠ 0 ) ∈ R ,$
or, we put:
$ϕ ′ ( s ) = ± q ( s ) ,$
where $q ( s ) = | k 2 g 2 3 + 2 3 g 2 3 ∫ g − 2 3 f f ′ f − g ′ g d s |$.
Case 2. Let M be a cylindrical ruled surface of type $M + 3 .$ In this case, without loss of generality, we may choose the base curve $α$ to be $α ( s ) = ( 0 , α 2 ( s ) , α 3 ( s ) )$ parameterized by arc length s and the director vector field $β$ as $β = ( 1 , 0 , 0 )$. Then, the Gauss map G of M and the Laplacian $Δ G$ of the Gauss map are obtained respectively by:
$G = ( 0 , α 3 ′ , − α 2 ′ ) and Δ G = ( 0 , − α 3 ‴ , α 2 ‴ ) .$
The relationship (13) and the condition (1) imply that the constant vector C has the form:
$C = ( 0 , c 2 , c 3 )$
for some constants $c 2$ and $c 3$.
If f and g are both constant, the Gauss map is of 1-type in the usual sense, and thus, M is an open part of a circular cylinder [1].
We now assume that the functions f and g are not both constant. Then, with the help of (1) and (13), we get:
$− α 3 ‴ = f α 3 ′ + g c 2 , α 2 ‴ = − f α 2 ′ + g c 3 .$
Since $α$ is parameterized by the arc length s, we may put:
$α 2 ′ ( s ) = cos ϕ ( s ) and α 3 ′ ( s ) = sin ϕ ( s )$
for some function $ϕ ( s )$ of s. Hence, (14) can be expressed as:
$( ϕ ′ ) 2 sin ϕ − ϕ ″ cos ϕ = f sin ϕ + g c 2 , ( ϕ ′ ) 2 cos ϕ + ϕ ″ sin ϕ = f cos ϕ − g c 3 .$
It follows:
$( ϕ ′ ) 2 = f + g ( c 2 sin ϕ − c 3 cos ϕ ) .$
Thus, M is a cylinder over the base curve $α$ given by:
$α ( s ) = 0 , ∫ cos ∫ r ( s ) d s d s , ∫ sin ∫ r ( s ) d s d s$
and the ruling $β ( s ) = ( 1 , 0 , 0 )$, where $r ( s ) = | f ( s ) + g ( s ) c 2 sin ϕ ( s ) − c 3 cos ϕ ( s ) |$.
Consequently, we have:
Theorem 1.
(Classification of cylindrical ruled surfaces in $E 1 3$) Let M be a cylindrical ruled surface with the generalized 1-type Gauss map in the Minkowski 3-space $E 1 3$. Then, M is an open part of a Euclidean plane, a Minkowski plane, a circular cylinder, a hyperbolic cylinder, or a cylinder over a base curve of infinite-type satisfying (2)–(6), (11), (12), or (15).

## 4. Non-Cylindrical Ruled Surfaces with the Generalized 1-Type Gauss Map

In this section, we classify all non-cylindrical ruled surfaces with the generalized 1-type Gauss map in $E 1 3 .$
We start with the case that the surface M is non-cylindrical of type $M + 1 , M + 3$, or $M − 1 .$ Then, M is parameterized by, up to a rigid motion,
$x ( s , t ) = α ( s ) + t β ( s )$
such that $〈 α ′ , β 〉 = 0 , 〈 β , β 〉 = ε 2 ( = ± 1 )$, and $〈 β ′ , β ′ 〉 = ε 3 ( = ± 1 ) .$ Then, ${ β , β ′ , β × β ′ }$ is an orthonormal frame along the base curve $α$. For later use, we define the smooth functions $q , u , Q$, and R as follows:
$q = ∥ x s ∥ 2 = ε 4 〈 x s , x s 〉 , u = 〈 α ′ , β ′ 〉 , Q = 〈 α ′ , β × β ′ 〉 , R = 〈 β ″ , β × β ′ 〉 ,$
where $ε 4$ is the sign of the coordinate vector field $x s = ∂ x / ∂ s .$ The vector fields $α ′$, $β ″$, $α ′ × β$, and $β × β ″$ are represented in terms of the orthonormal frame ${ β , β ′ , β × β ′ }$ along the base curve $α$ as:
$α ′ = ε 3 u β ′ − ε 2 ε 3 Q β × β ′ , β ″ = − ε 2 ε 3 β − ε 2 ε 3 R β × β ′ , α ′ × β = ε 3 Q β ′ − ε 3 u β × β ′ , β × β ″ = − ε 3 R β ′ .$
Therefore, the smooth function q is given by:
$q = ε 4 ( ε 3 t 2 + 2 u t + ε 3 u 2 − ε 2 ε 3 Q 2 ) .$
Note that t is chosen so that q takes positive values.
Furthermore, the Gauss map G of M is given by:
$G = q − 1 / 2 ε 3 Q β ′ − ( ε 3 u + t ) β × β ′ .$
By using the determinants of the first fundamental form and the second fundamental form, the mean curvature H and the Gaussian curvature K of M are obtained by, respectively,
$H = 1 2 ε 2 q − 3 / 2 R t 2 + ( 2 ε 3 u R + Q ′ ) t + u 2 R + ε 3 u Q ′ − ε 3 u ′ Q − ε 2 Q 2 R , K = q − 2 Q 2 .$
Applying the Gauss and Weingarten formulas, the Laplacian of the Gauss map G of M in $E 1 3$ is represented by:
$Δ G = 2 grad H + 〈 G , G 〉 ( t r A G 2 ) G ,$
where $A G$ denotes the shape operator of the surface M in $E 1 3$ and $grad H$ is the gradient of H. Using (18), we get:
$2 grad H = 2 〈 e 1 , e 1 〉 e 1 ( H ) e 1 + 2 〈 e 2 , e 2 〉 e 2 ( H ) e 2 = 2 ε 4 e 1 ( H ) e 1 + 2 ε 2 e 2 ( H ) e 2 = q − 7 / 2 { − ε 2 ( ε 3 u + t ) A 1 β ′ − ε 4 q B 1 β + ε 3 Q A 1 β × β ′ } ,$
where $e 1 = x s | | x s | | , e 2 = x t | | x t | |$,
$A 1 = 3 ( u ′ t + ε 3 u u ′ − ε 2 ε 3 Q Q ′ ) { R t 2 + ( 2 ε 3 u R + Q ′ ) t + u 2 R + ε 3 u Q ′ − ε 3 u ′ Q − ε 2 Q 2 R } − ( ε 3 t 2 + 2 u t + ε 3 u 2 − ε 2 ε 3 Q 2 ) { R ′ t 2 + ( 2 ε 3 u ′ R + 2 ε 3 u R ′ + Q ″ ) t + 2 u u ′ R + u 2 R ′ + ε 3 u Q ″ − ε 3 u ″ Q − 2 ε 2 Q Q ′ R − ε 2 Q 2 R ′ } , B 1 = ε 3 R t 3 + ( 3 u R + 2 ε 3 Q ′ ) t 2 + ( 3 ε 3 u 2 R + 4 u Q ′ − 3 u ′ Q − ε 2 ε 3 Q 2 R ) t + u 3 R + 2 ε 3 u 2 Q ′ − ε 2 u Q 2 R − 3 ε 3 u u ′ Q + ε 2 ε 3 Q 2 Q ′ .$
The straightforward computation gives:
$tr A G 2 = − ε 2 ε 4 q − 3 D 1 ,$
where:
$D 1 = − ε 4 ( u ′ t + ε 3 u u ′ − ε 2 ε 3 Q Q ′ ) 2 + ε 3 q { ( ε 2 Q R + ε 3 u ′ ) 2 − ε 2 ( Q ′ + ε 3 u R + R t ) 2 − 2 ε 3 Q 2 } .$
Thus, the Laplacian $Δ G$ of the Gauss map G of M is obtained by:
$Δ G = q − 7 / 2 [ − ε 4 q B 1 β + { − ε 2 ( ε 3 u + t ) A 1 + ε 3 Q D 1 } β ′ + { ε 3 Q A 1 − ( ε 3 u + t ) D 1 } β × β ′ ] .$
Now, suppose that the Gauss map G of M is of generalized 1-type. Hence, from (1), (17) and (20), we get:
$q − 7 / 2 [ − ε 4 q B 1 β + { − ε 2 ( ε 3 u + t ) A 1 + ε 3 Q D 1 } β ′ + { ( ε 3 Q A 1 − ( ε 3 u + t ) D 1 } β × β ′ ] = f q − 1 / 2 ε 3 Q β ′ − ( ε 3 u + t ) β × β ′ + g C .$
If we take the indefinite scalar product to Equation (21) with $β , β ′$ and $β × β ′$, respectively, then we obtain respectively,
$− ε 2 ε 4 q − 5 / 2 B 1 = g 〈 C , β 〉 ,$
$q − 7 / 2 { − ε 2 ε 3 ( ε 3 u + t ) A 1 + Q D 1 } = f q − 1 / 2 Q + g 〈 C , β ′ 〉 ,$
$q − 7 / 2 { − ε 2 Q A 1 + ε 2 ε 3 ( ε 3 u + t ) D 1 } = f q − 1 / 2 ε 2 ε 3 ( ε 3 u + t ) + g 〈 C , β × β ′ 〉 .$
On the other hand, the constant vector $C$ can be written as;
$C = c 1 β + c 2 β ′ + c 3 β × β ′ ,$
where $c 1 = ε 2 〈 C , β 〉$, $c 2 = ε 3 〈 C , β ′ 〉$, and $c 3 = − ε 2 ε 3 〈 C , β × β ′ 〉 .$ Differentiating the functions $c 1 , c 2$, and $c 3$ with respect to s, we have:
$c 1 ′ − ε 2 ε 3 c 2 = 0 , c 1 + c 2 ′ − ε 3 R c 3 = 0 , ε 2 ε 3 R c 2 − c 3 ′ = 0 .$
Furthermore, Equations (22)–(24) are expressed as follows:
$− ε 4 q − 5 / 2 B 1 = g c 1 ,$
$q − 7 / 2 { − ε 2 ( ε 3 u + t ) A 1 + ε 3 Q D 1 } = f q − 1 / 2 ε 3 Q + g c 2 ,$
$q − 7 / 2 { − ε 3 Q A 1 + ( ε 3 u + t ) D 1 } = f q − 1 / 2 ( ε 3 u + t ) − g c 3 .$
Combining Equations (26)–(28), we have:
${ − ε 2 ( ε 3 u + t ) A 1 + ε 3 Q D 1 } c 1 + q ε 4 B 1 c 2 = q 3 f ε 3 Q c 1 ,$
${ − ε 3 Q A 1 + ( ε 3 u + t ) D 1 } c 1 − q ε 4 B 1 c 3 = q 3 f ( ε 3 u + t ) c 1 .$
Hence, Equations (29) and (30) yield that:
$− ε 2 ε 3 A 1 c 1 + B 1 { c 2 ( ε 3 u + t ) + ε 3 Q c 3 } = 0 .$
First of all, we prove:
Theorem 2.
Let M be a non-cylindrical ruled surface of type $M + 1 , M + 3$, or $M − 1$ parameterized by the base curve α and the director vector field β in $E 1 3$ with the generalized 1-type Gauss map. If β, $β ′$, and $β ″$ are coplanar along α, then M is an open part of a plane, the helicoid of the first kind, the helicoid of the second kind or the helicoid of the third kind.
Proof.
If the constant vector $C$ is zero, then we can pass this case to that of the pointwise 1-type Gauss map of the first kind. Thus, according to the classification theorem in [4], M is an open part of the helicoid of the first kind, the helicoid of the second kind, or the helicoid of the third kind.
Now, we assume that the constant vector $C$ is non-zero. If the function Q is identically zero on $M ,$ then M is an open part of a plane because of (18).
We now consider the case of the function Q being not identically zero. Consider a non-empty open subset $U = { s ∈ dom ( α ) | Q ( s ) ≠ 0 }$ of dom($α$). Since $β$, $β ′$, and $β ″$ are coplanar along $α$, R vanishes. Thus, $c 3$ is a constant, and $c 1 ″ = − ε 2 ε 3 c 1$ from (25). Since the left-hand side of (31) is a polynomial in t with functions of s as the coefficients, all of the coefficients that are functions of s must be zero. From the leading coefficient, we have:
$ε 2 ε 3 c 1 Q ″ + 2 c 2 Q ′ = 0 .$
Observing the coefficient of the term involving $t 2$ of (31), with the help of (32), we get:
$ε 2 ε 3 c 1 ( 3 u ′ Q ′ + u ″ Q ) + 3 c 2 u ′ Q − 2 c 3 Q Q ′ = 0 .$
Examining the coefficient of the linear term in t of (31) and using (32) and (33), we also get:
$Q { c 1 ε 2 ( u ′ ) 2 + ( Q ′ ) 2 + ε 2 ε 3 c 2 Q Q ′ − ε 3 c 3 u ′ Q } = 0 .$
On U,
$c 1 ε 2 ( u ′ ) 2 + ( Q ′ ) 2 + ε 2 ε 3 c 2 Q Q ′ − ε 3 c 3 u ′ Q = 0 .$
Similarly, from the constant term with respect to t of (31), we have:
$ε 3 c 1 ( − 3 u ′ Q ′ + u ″ Q ) + ε 2 c 3 Q Q ′ = 0$
by using (32)–(34). Combining (33) and (35), we obtain:
$2 ε 3 c 1 u ′ Q ′ + ε 2 c 2 u ′ Q − ε 2 c 3 Q Q ′ = 0 .$
Now, suppose that $u ′ ( s ) ≠ 0$ at some point $s ∈ U$ and then $u ′ ≠ 0$ on an open interval $U 1 ⊂ U$. Equation (34) yields:
$ε 3 c 3 Q = 1 u ′ { c 1 ε 2 ( u ′ ) 2 + ( Q ′ ) 2 + ε 2 ε 3 c 2 Q Q ′ } .$
Substituting (37) into (36), we get:
${ ( u ′ ) 2 − ε 2 ( Q ′ ) 2 } ( ε 3 c 1 Q ′ + ε 2 c 2 Q ) = 0 ,$
or, using $c 2 = ε 2 ε 3 c 1 ′$ in (25),
${ ( u ′ ) 2 − ε 2 ( Q ′ ) 2 } ( c 1 Q ) ′ = 0 .$
Suppose that $( u ′ ) 2 − ε 2 ( Q ′ ) 2 ( s 0 ) ≠ 0$ for some $s 0 ∈ U 1$. Then, $c 1 Q$ is constant on a component $U 2$ containing $s 0$ of $U 1$.
If $c 1 = 0$ on $U 2 ,$ we easily see that $c 2 = 0$ by (25). Hence, (34) yields that $c 3 u ′ Q = 0$, and so, $c 3 = 0$. Since $C$ is a constant vector, $C$ is zero on M. This contradicts our assumption. Thus, $c 1 ≠ 0$ on $U 2$. From the equation $c 1 ″ + ε 2 ε 3 c 1 = 0 ,$ we get:
$c 1 = k 1 cos ( s + s 1 ) or c 1 = k 2 cosh ( s + s 2 )$
for some non-zero constants $k i$ and $s i ∈ R ( i = 1 , 2 ) .$ Since $c 1 Q$ is constant, $k 1$ and $k 2$ must be zero. Hence, $c 1 = 0 ,$ a contradiction. Thus, $( u ′ ) 2 − ε 2 ( Q ′ ) 2 = 0$ on $U 1 ,$ from which we get $ε 2 = 1$ and $u ′ = ± Q ′ .$ If $u ′ ≠ − Q ′ ,$ then $u ′ = Q ′$ on an open subset $U 3$ in $U 1 .$ Hence, (34) implies that $Q ′ ( 2 ε 3 c 1 Q ′ + c 2 Q − c 3 Q ) = 0 .$ On $U 3$, we get $c 3 Q = 2 ε 3 c 1 Q ′ + c 2 Q .$ Putting it into (35), we have:
$ε 3 c 1 ( Q ′ ) 2 − ε 3 c 1 Q Q ″ − c 2 Q Q ′ = 0 .$
Combining (32) and (38), $c 1 Q$ is constant on $U 3 .$ Similarly as above, we can derive that $C$ is zero on M, which is a contradiction. Therefore, we have $u ′ = − Q ′$ on $U 1$. Similarly, as we just did to the case under the assumption $u ′ ≠ − Q ′$, it is also proven that the constant vector $C$ becomes zero. It is also a contradiction, and so, $U 1 = ∅$. Thus, $u ′ = 0$ and $Q ′ = 0 .$ From (18), the mean curvature H vanishes. In this case, the Gauss map G is of pointwise 1-type of the first kind. Hence, the open set U is empty. Therefore, we see that if the director vector field $β$, $β ′$, and $β ″$ are coplanar, the function Q vanishes on M. Hence, M is an open part of a plane because of (18). □
From now on, we assume that R is non-vanishing, i.e., $β ∧ β ′ ∧ β ″ ≠ 0$ everywhere on $M .$
If $f = g ,$ the Gauss map of the non-cylindrical ruled surface of type $M + 1 , M − 1$ or $M + 3$ in $E 1 3$ is of pointwise 1-type. According to the classification theorem given in [5,13], M is part of a circular cone or a hyperbolic cone.
Now, we suppose that $f ≠ g$ and the constant vector $C$ is non-zero unless otherwise stated. Similarly as before, we develop our argument with (31). The left-hand side of (31) is a polynomial in t with functions of s as the coefficients, and thus, they are zero. From the leading coefficient of the left-hand side of (31), we obtain:
$ε 2 c 1 R ′ + ε 3 c 2 R = 0 .$
With the help of (25), $c 1 R$ is constant. If we examine the coefficient of the term of $t 3$ of the left-hand side of (31), we get:
$c 1 ( − ε 2 ε 3 u ′ R + ε 2 Q ″ ) + 2 c 2 ε 3 Q ′ + c 3 Q R = 0 .$
From the coefficient of the term involving $t 2$ in (31), using (25) and (40), we also get:
$c 1 ( − 3 ε 2 ε 3 u ′ Q ′ + Q Q ′ R − ε 2 ε 3 u ″ Q − Q 2 R ′ ) − 3 c 2 u ′ Q + 2 c 3 Q Q ′ = 0 .$
Furthermore, considering the coefficient of the linear term in t of (31) and making use of Equations (25), (40), and (41), we obtain:
$Q { c 1 ( ε 2 ( u ′ ) 2 + ( Q ′ ) 2 ) + c 2 ε 2 ε 3 Q Q ′ − c 3 ε 3 u ′ Q } = 0 .$
Now, we consider the open set $V = { s ∈ dom ( α ) | Q ( s ) ≠ 0 }$. Suppose $V ≠ ∅ .$ From (42),
$c 1 ( ε 2 ( u ′ ) 2 + ( Q ′ ) 2 ) + c 2 ε 2 ε 3 Q Q ′ − c 3 ε 3 u ′ Q = 0 .$
Similarly as above, observing the constant term in t of the left-hand side of (31) with the help of (25) and (39), and using (40), (41) and (43), we have:
$Q 2 ( 2 c 1 ε 3 u ′ Q ′ + c 2 ε 2 u ′ Q − c 3 ε 2 Q Q ′ ) = 0 .$
Since $Q ≠ 0$ on V, one can have:
$2 c 1 ε 3 u ′ Q ′ + c 2 ε 2 u ′ Q − c 3 ε 2 Q Q ′ = 0 .$
Our making use of the first and the second equations in (25), (40) reduces to:
$c 1 ε 2 u ′ R − ε 2 ε 3 ( c 1 Q ) ″ − c 1 Q = 0 .$
Suppose that $u ′ ( s ) ≠ 0$ for some $s ∈ V$. Then, $u ′ ≠ 0$ on an open subset $V 1 ⊂ V$. From (43), on $V 1$:
$c 3 Q = 1 u ′ { ε 2 ε 3 c 1 ( u ′ ) 2 + ε 3 c 1 ( Q ′ ) 2 + ε 2 c 2 Q Q ′ } .$
Putting (46) into (44), we have ${ ( u ′ ) 2 − ε 2 ( Q ′ ) 2 } ( ε 3 c 1 Q ′ + ε 2 c 2 Q ) = 0 .$ With the help of $c 1 ′ = ε 2 ε 3 c 2$, it becomes:
${ ( u ′ ) 2 − ε 2 ( Q ′ ) 2 } ( c 1 Q ) ′ = 0 .$
Suppose that $( u ′ ) 2 − ε 2 ( Q ′ ) 2 ( s ) ≠ 0$ on $V 1$. Then, $c 1 Q$ is constant on a component $V 2$ of $V 1$. Hence, (45) yields that:
$c 1 Q = ε 2 c 1 u ′ R .$
If $c 1 ≡ 0$ on $V 2$, (25) gives that $c 2 = 0$ and $c 3 R = 0$. Since $R ≠ 0$, $c 3 = 0$. Hence, the constant vector $C$ is zero, a contradiction. Therefore, $c 1 ≠ 0$ on $V 2$. From (47), $Q = ε 2 u ′ R$. Moreover, $u ′$ is a non-zero constant because $c 1 Q$ and $c 1 R$ are constants. Thus, (41) and (44) can be reduced to as follows:
$c 1 Q ′ R − c 1 Q R ′ + 2 c 3 Q ′ = 0 ,$
$ε 3 c 1 u ′ Q ′ − ε 2 c 3 Q Q ′ = 0 .$
Upon our putting $Q = ε 2 u ′ R$ into (48), $c 3 Q ′ = 0$ is derived. By (49), $c 1 u ′ Q ′ = 0 .$ Hence, $Q ′ = 0$. It follows that Q and R are non-zero constants on $V 2$.
On the other hand, since the torsion of the director vector field $β$ viewed as a curve in $E 1 3$ is zero, $β$ is part of a plane curve. Moreover, $β$ has constant curvature $ε 2 − ε 2 ε 3 R 2$. Hence, $β$ is a circle or a hyperbola on the unit pseudo-sphere or the hyperbolic space of radius 1 in $E 1 3 .$ Without loss of generality, we may put:
$β ( s ) = 1 p ( R , cos p s , sin p s ) or β ( s ) = 1 p ( sinh p s , cosh p s , R ) ,$
where $p 2 = ε 2 ( 1 − ε 3 R 2 )$ and $p > 0$. Then, the function $u = 〈 α ′ , β ′ 〉$ is given by:
$u = − α 2 ′ ( s ) sin p s + α 3 ′ ( s ) cos p s or u = − α 1 ′ ( s ) cosh p s + α 2 ′ ( s ) sinh p s ,$
where $α ′ ( s ) = ( α 1 ′ ( s ) , α 2 ′ ( s ) , α 3 ′ ( s ) ) .$ Therefore, we have:
$u ′ = − ( α 2 ″ + p α 3 ′ ) sin p s − ( p α 2 ′ − α 3 ″ ) cos p s or u ′ = ( − α 1 ″ + p α 2 ′ ) cosh p s − ( p α 1 ′ − α 2 ″ ) sinh p s .$
Since $u ′$ is a constant, $u ′$ must be zero. It is a contradiction on $V 1$, and so:
$( u ′ ) 2 = ε 2 ( Q ′ ) 2$
on $V 1$. It immediately follows that:
$ε 2 = 1$
on $V 1$. Therefore, we get $u ′ = ± Q ′ .$ Suppose $u ′ ≠ − Q ′$ on $V 1$. Then, $u ′ = Q ′$ and (43) can be written as:
$Q ′ ( 2 ε 3 c 1 Q ′ + c 2 Q − c 3 Q ) = 0 .$
Since $Q ′ ≠ 0$ on V,
$c 3 Q = 2 ε 3 c 1 Q ′ + c 2 Q .$
Putting (50) into (40) and (41), respectively, we obtain:
$ε 3 c 1 Q ′ R + c 2 Q R + 2 ε 3 c 2 Q ′ + c 1 Q ″ = 0 ,$
$ε 3 c 1 ( Q ′ ) 2 + c 1 Q Q ′ R − ε 3 c 1 Q Q ″ − c 1 Q 2 R ′ − c 2 Q Q ′ = 0 .$
Putting together Equations (51) and (52) with the help of (39), we get:
$( ε 3 c 1 Q ′ + c 2 Q ) ( Q ′ + 2 ε 3 Q R ) = 0 .$
Suppose $( ε 3 c 1 Q ′ + c 2 Q ) ( s ) ≠ 0$ on $V 1$. Then, $Q ′ = − 2 ε 3 Q R .$ If we make use of it, we can derive $R ( ε 3 c 1 Q ′ + c 2 Q ) = 0$ from (51). Since R is non-vanishing, $ε 3 c 1 Q ′ + c 2 Q = 0$, a contradiction. Thus:
$ε 3 c 1 Q ′ + c 2 Q = 0 ,$
that is, $c 1 Q$ is constant on each component of $V 1$. From (45), $c 1 Q = c 1 u ′ R .$ Similarly as before, it is seen that $c 1 ≠ 0$ and $u ′$ is a non-zero constant. Hence, $Q = u ′ R .$ If we use the fact that $c 1 Q$ and $Q ′$ are constant, $c 2 Q ′ = 0$ is derived from (51). Therefore, $c 2 = 0$ on each component of $V 1 .$ By (53), $c 1 = 0$ on each component of $V 1$. Hence, (50) implies that $c 3 = 0$ on each component of $V 1$. The vector $C$ is constant and thus zero on M, a contradiction. Thus, we obtain $u ′ = − Q ′$ on $V 1$. Equation (43) with $u ′ = − Q ′$ gives that:
$c 3 Q = − 2 ε 3 c 1 Q ′ − c 2 Q .$
Putting (54) together with $u ′ = − Q ′$ into (40), we have:
$c 1 Q ″ = ε 3 c 1 Q ′ R + c 2 Q R − 2 ε 3 c 2 Q ′ .$
Furthermore, Equations (39), (41), (54) and (55) give:
$( ε 3 c 1 Q ′ + c 2 Q ) ( Q ′ − 2 ε 3 Q R ) = 0$
on $V 1$. Suppose $ε 3 c 1 Q ′ + c 2 Q ≠ 0$. Then, $Q ′ = 2 ε 3 Q R$, and thus, $Q ″ = 2 ε 3 Q ′ R + 2 ε 3 Q R ′$. Putting it into (55) with the help of (39), we get:
$R ( ε 3 c 1 Q ′ + c 2 Q ) = 0 ,$
from which $ε 3 c 1 Q ′ + c 2 Q = 0$, a contradiction. Therefore, we get:
$ε 3 c 1 Q ′ + c 2 Q = 0$
on $V 1$. Thus, $c 1 Q$ is constant on each component of $V 1$. Similarly developing the argument as before, we see that the constant vector $C$ is zero, which contradicts our assumption. Consequently, the open subset $V 1$ is empty, i.e., the functions u and Q are constant on each component of V. Since $Q = u ′ R$, Q vanishes on V. Thus, the open subset V is empty, and hence, Q vanishes on M. Thus, (18) shows that the Gaussian curvature K automatically vanishes on $M .$
Thus, we obtain:
Theorem 3.
Let M be a non-cylindrical ruled surface of type $M + 1 , M + 3$, or $M − 1$ parameterized by the non-null base curve α and the director vector field β in $E 1 3$ with the generalized 1-type Gauss map. If β, $β ′$, and $β ″$ are not coplanar along α, then M is flat.
Combining Definition 3, Theorems 2 and 3, and the classification theorem of flat surfaces with the generalized 1-type Gauss map in Minkowski 3-space in [8], we have the following:
Theorem 4.
Let M be a non-cylindrical ruled surface of type $M + 1 , M + 3$, or $M − 1$ in $E 1 3$ with the generalized 1-type Gauss map. Then, M is locally part of a plane, the helicoid of the first kind, the helicoid of the second kind, the helicoid of the third kind, a circular cone, a hyperbolic cone, or a conical surface of G-type.
We now consider the case that the ruled surface M is non-cylindrical of type $M + 2 , M − 2 .$ Then, up to a rigid motion, a parametrization of M is given by:
$x ( s , t ) = α ( s ) + t β ( s )$
satisfying $〈 α ′ , β 〉 = 0 , 〈 α ′ , α ′ 〉 = ε 1 ( = ± 1 ) , 〈 β , β 〉 = 1$, and $〈 β ′ , β ′ 〉 = 0$ with $β ′ ≠ 0$.
Again, we put the smooth functions q and u as follows:
$q = ∥ x s ∥ 2 = | 〈 x s , x s 〉 | , u = 〈 α ′ , β ′ 〉 .$
We see that the null vector fields $β ′$ and $β × β ′$ are orthogonal, and they are parallel. It is easily derived as $β ′ = β × β ′$. Moreover, we may assume that $β ( 0 ) = ( 0 , 0 , 1 )$ and $β$ can be taken by:
$β ( s ) = ( a s , a s , 1 )$
for a non-zero constant $a .$ Then, ${ α ′ , β , α ′ × β }$ forms an orthonormal frame along the base curve $α$. With respect to this frame, we can put:
$β ′ = ε 1 u ( α ′ − α ′ × β ) and α ″ = − u β + u ′ u α ′ × β .$
Note that the function u is non-vanishing.
On the other hand, we can compute the Gauss map G of M such as:
$G = q − 1 / 2 ( α ′ × β − t β ′ ) .$
We also easily get the mean curvature H and the Gaussian curvature K of M by the usual procedure, respectively,
$H = 1 2 q − 3 / 2 u ′ t − ε 1 u ′ u and K = q − 2 u 2 .$
Upon our using (19), the Laplacian of the Gauss map G of M is expressed as:
$Δ G = q − 7 / 2 A 2 α ′ + B 2 β + D 2 α ′ × β$
with respect to the orthonormal frame ${ α ′ , β , α ′ × β } ,$ where we put:
$A 2 = 3 ε 1 ( u ′ ) 2 u t + ε 4 ε 1 q − u ″ u + ( u ′ ) 2 u 2 + u u ″ t 2 + ε 1 ( u ′ ) 2 u t + q ( u ′ ) 2 u t − 3 ε 1 u ( u ′ ) 2 t 3 + ε 4 ε 1 u ( u ′ ) 2 t 3 + 2 ε 4 ε 1 q u 3 t , B 2 = ε 4 q u ′ ( 4 ε 1 − u t ) , D 2 = 3 ε 1 u ( u ′ ) 2 t 3 − 3 ( u ′ ) 2 t 2 − ε 4 q ε 1 u u ″ t 2 − u ″ t + ( u ′ ) 2 u t − ε 1 q ( u ′ ) 2 u 2 − q ( u ′ ) 2 u t − ε 4 ( u ′ ) 2 t 2 − 2 ε 4 q u 2 − ε 4 ε 1 u ( u ′ ) 2 t 3 − 2 ε 4 ε 1 q u 3 t .$
We now suppose that the Gauss map G of M is of generalized 1-type satisfying Condition (1). Then, from (56), (57), and (59), we get:
$q − 7 / 2 A 2 α ′ + B 2 β + D 2 α ′ × β = f q − 1 / 2 { ( 1 + ε 1 u t ) α ′ × β − ε 1 u t α ′ } + g C .$
If the constant vector $C$ is zero, the Gauss map G is nothing but of pointwise 1-type of the first kind. By the result of [4], M is part of the conjugate of Enneper’s surface of the second kind.
From now on, for a while, we assume that $C$ is a non-zero constant vector. Taking the indefinite scalar product to Equation (60) with the orthonormal vector fields $α ′ , β$, and $α ′ × β ,$ respectively, we obtain:
$ε 1 q − 7 / 2 A 2 = − f q − 1 / 2 u t + g 〈 C , α ′ 〉 ,$
$q − 7 / 2 B 2 = g 〈 C , β 〉 ,$
$ε 1 q − 7 / 2 D 2 = f q − 1 / 2 ( ε 1 + u t ) − g 〈 C , α ′ × β 〉 .$
In terms of the orthonormal frame ${ α ′ , β , α ′ × β }$, the constant vector $C$ can be written as:
$C = c 1 α ′ + c 2 β + c 3 α ′ × β ,$
where we have put $c 1 = ε 1 〈 C , α ′ 〉$, $c 2 = 〈 C , β 〉$, and $c 3 = − ε 1 〈 C , α ′ × β 〉 .$ Then, Equations (61)–(63) are expressed as follows:
$ε 1 q − 7 / 2 A 2 = − f q − 1 / 2 u t + ε 1 g c 1 ,$
$q − 7 / 2 B 2 = g c 2 ,$
$ε 1 q − 7 / 2 D 2 = f q − 1 / 2 ( ε 1 + u t ) + ε 1 g c 3 .$
Differentiating the functions $c 1 , c 2$, and $c 3$ with respect to the parameter s, we get:
$c 1 ′ = − ε 1 u c 2 − u ′ u c 3 , c 2 ′ = u c 1 + u c 3 , c 3 ′ = − u ′ u c 1 + ε 1 u c 2 .$
Combining Equations (64)–(66), we obtain:
$c 2 ( ε 1 + u t ) A 2 − { ε 1 c 1 + ( c 1 + c 3 ) u t } B 2 + c 2 u t D 2 = 0 .$
As before, from (68), we obtain the following:
$c 2 ( 2 u u ″ − 3 ( u ′ ) 2 ) + ( c 1 + c 3 ) u 2 u ′ = 0 ,$
$7 c 2 ( u ′ ) 2 − 5 c 1 u 2 u ′ − 7 c 3 u 2 u ′ = 0 ,$
$c 2 ( 7 ( u ′ ) 2 − 3 u u ″ ) − 11 c 1 u 2 u ′ − 4 c 3 u 2 u ′ = 0 ,$
$c 2 ( u u ″ − ( u ′ ) 2 ) + 4 c 1 u 2 u ′ = 0 .$
Combining Equations (69) and (71), we get:
$5 c 2 ( u u ″ − ( u ′ ) 2 ) − 7 c 1 u 2 u ′ = 0 .$
From (72) and (73), we get $c 1 u ′ = 0 .$ Hence, Equations (70) and (72) become:
$u ′ ( c 2 u ′ − c 3 u 2 ) = 0 ,$
$c 2 ( u u ″ − ( u ′ ) 2 ) = 0 .$
Now, suppose that $u ′ ( s 0 ) ≠ 0$ at some point $s 0 ∈ dom ( α )$. Then, there exists an open interval J such that $u ′ ≠ 0$ on J. Then, $c 1 = 0$ on $J .$ Hence, (67) reduces to:
$ε 1 u 2 c 2 + u ′ c 3 = 0 , c 2 ′ = u c 3 , c 3 ′ = ε 1 u c 2 .$
From the above relationships, we see that $c 2 ′$ is constant on J. In this case, if $c 2 = 0 ,$ then $c 3 = 0 .$ Hence, $C$ is zero on J. Thus, the constant vector $C$ is zero on M. This contradicts our assumption. Therefore, $c 2$ is non-zero. Solving the differential Equation (74) with the help of $c 2 ′ = u c 3$ in (76), we get $u = k c 2$ for some non-zero constant $k .$ Moreover, since $c 2 ′$ is constant, $u ″ = 0$. Thus, Equation (75) implies that $u ′ = 0 ,$ which is a contradiction. Therefore, there does not exist such a point $s 0 ∈ dom ( α )$ such that $u ′ ( s 0 ) ≠ 0 .$ Hence, u is constant on M. With the help of (58), the mean curvature H of M vanishes on $M .$ It is easily seen from (19) that the Gauss map G of M is of pointwise 1-type of the first kind, which means (1) is satisfied with $C = 0$. Thus, this case does not occur.
As a consequence, we give the following classification:
Theorem 5.
Let M be a non-cylindrical ruled surface of type $M + 2$ or $M − 2$ in $E 1 3$ with the generalized 1-type Gauss map G. Then, the Gauss map G is of pointwise 1-type of the first kind and M is an open part of the conjugate of Enneper’s surface of the second kind.
Remark 2.
There do not exist non-cylindrical ruled surfaces of type $M + 2$ or $M − 2$ in $E 1 3$ with the proper generalized 1-type Gauss map G.

## 5. Null Scrolls in the Minkowski 3-Space $E 1 3$

In this section, we examine the null scrolls with the generalized 1-type Gauss map in the Minkowski 3-space $E 1 3 .$ In particular, we focus on proving the following theorem.
Theorem 6.
Let M be a null scroll in the Minkowski 3-space $E 1 3$. Then, M has generalized 1-type Gauss map G if and only if M is part of a Minkowski plane or a B-scroll.
Proof.
Suppose that a null scroll M has the generalized 1-type Gauss map. Let $α = α ( s )$ be a null curve in $E 1 3$ and $β = β ( s )$ a null vector field along $α$ such that $〈 α ′ , β 〉 = 1 .$ Then, the null scroll M is parameterized by:
$x ( s , t ) = α ( s ) + t β ( s )$
and we have the natural coordinate frame ${ x s , x t }$ given by:
$x s = α ′ + t β ′ and x t = β .$
We put the smooth functions $u , v , Q$, and R by:
$u = 〈 α ′ , β ′ 〉 , v = 〈 β ′ , β ′ 〉 , Q = 〈 α ′ , β ′ × β 〉 , R = 〈 α ′ , β ″ × β 〉 .$
Then, ${ α ′ , β , α ′ × β }$ is a pseudo-orthonormal frame along $α$.
Straightforward computation gives the Gauss map G of M and the Laplacian $Δ G$ of G by:
$G = α ′ × β + t β ′ × β and Δ G = − 2 β ″ × β + 2 ( u + t v ) β ′ × β .$
With respect to the pseudo-orthonormal frame ${ α ′ , β , α ′ × β }$, the vector fields $β ′$, $β ′ × β$, and $β ″ × β$ are represented as:
$β ′ = u β − Q α ′ × β , β ′ × β = Q β and β ″ × β = R β − v α ′ × β .$
Thus, the Gauss map G and its Laplacian $Δ G$ are expressed by:
$G = α ′ × β + t Q β and Δ G = − 2 ( R − u Q − t v Q ) β + 2 v α ′ × β .$
Since M has the generalized 1-type Gauss map, the Gauss map G satisfies:
$Δ G = f G + g C$
for some non-zero smooth functions $f , g$ and a constant vector $C$. From (79), we get:
$− 2 ( R − u Q − t v Q ) β + 2 v α ′ × β = f ( α ′ × β + t Q β ) + g C .$
If the constant vector $C$ is zero, M is an open part of a Minkowski plane or a B-scroll according to the classification theorem in [4].
We now consider the case that the constant vector $C$ is non-zero. If we take the indefinite inner product to Equation (81) with $α ′ , β$, and $α ′ × β ,$ respectively, we get:
$− 2 ( R − u Q − t v Q ) = f t Q + g c 2 , g c 1 = 0 , 2 v = f + g c 3 ,$
where we have put
$c 1 = 〈 C , β 〉 , c 2 = 〈 C , α ′ 〉 and c 3 = 〈 C , α ′ × β 〉 .$
Since $g ≠ 0$, Equation (82) gives $〈 C , β ′ 〉 = 0 .$ Together with (78), we see that $c 3 Q = 0 .$ Suppose that $Q ( s ) ≠ 0$ on an open interval $I ˜ ⊂ dom ( α )$. Then, $c 3 = 0$ on $I ˜$. Therefore, the constant vector $C$ can be written as $C = c 2 β$ on $I ˜$. If we differentiate $C = c 2 β$ with respect to s, $c 2 ′ β + c 2 β ′ = 0$, and thus, $c 2 v = 0 .$ On the other hand, from (77) and (78), we have $v = Q 2 .$ Hence, v is non-zero on $I ˜$, and so, $c 2 = 0 .$ It contradicts that $C$ is a non-zero vector. In the sequel, Q vanishes identically. Then, $β ′ = u β$, which implies $R = 0$. Thus, the Gauss map G is reduced to $G = α ′ × β$, which depends only on the parameter s, from which the shape operator S of M is easily derived as:
$S = 0 0 0 0 or S = 0 0 k ( s ) 0$
for some non-vanishing function k. Therefore, the null scroll M is part of a Minkowski plane or a flat B-scroll described in Section 2 determined by $A = α ′ , B = β , C = G$ satisfying $C ′ = − k ( s ) B$. Thus, null scrolls in $E 1 3$ with the generalized 1-type Gauss map satisfying (80) are part of Minkowski planes or B-scrolls whether $C$ is zero or not.
The converse is obvious. This completes the proof. □
Corollary 1.
There do not exist null scrolls in $E 1 3$ with the proper generalized 1-type Gauss map.
Open problem: Classify ruled submanifolds with the generalized 1-type Gauss map in Minkowski space.

## Author Contributions

M.C. computed the problems. Y.H.K. have the inspiration for the problem and polished the paper.

## Funding

Y.H. Kim was supported by the National Research Foundation of Korea (NRF) grant funded by the Korean Government (MSIP) (2016R1A2B1006974).

## Acknowledgments

The authors would like to express their deep thanks to the referees for their valuable suggestions to improve the paper.

## Conflicts of Interest

The authors declare no conflict of interest.

## References

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