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Article

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.
Keywords: ruled surface; null scroll; Minkowski space; pointwise 1-type Gauss map; generalized 1-type Gauss map; conical surface of G-type ruled surface; null scroll; Minkowski space; pointwise 1-type Gauss map; generalized 1-type Gauss map; conical surface of G-type

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

  1. Chen, B.-Y. Total Mean Curvature and Submanifolds of Finite Type, 2nd ed.; World Scientific: Hackensack, NJ, USA, 2015. [Google Scholar]
  2. Kim, D.-S.; Kim, Y.H.; Yoon, D.W. Characterization of generalized B-scrolls and cylinders over finite type curves. Indian J. Pure Appl. Math. 2003, 33, 1523–1532. [Google Scholar]
  3. Kim, Y.H.; Yoon, D.W. Ruled surfaces with finite type Gauss map in Minkowski spaces. Soochow J. Math. 2000, 26, 85–96. [Google Scholar]
  4. Kim, Y.H.; Yoon, D.W. Ruled surfaces with pointwise 1-type Gauss map. J. Geom. Phys. 2000, 34, 191–205. [Google Scholar] [CrossRef]
  5. Choi, M.; Kim, D.-S.; Kim, Y.H.; Yoon, D.W. Circular cone and its Gauss map. Colloq. Math. 2012, 129, 203–210. [Google Scholar] [CrossRef]
  6. Choi, M.; Kim, Y.H. Characterization of the helicoid as ruled surfaces with pointwise 1-type Gauss map. Bull. Korean Math. Soc. 2001, 38, 753–761. [Google Scholar]
  7. Chen, B.-Y.; Choi, M.; Kim, Y.H. Surfaces of revolution with pointwise 1-type Gauss map. J. Korean Math. Soc. 2005, 42, 447–455. [Google Scholar] [CrossRef]
  8. Yoon, D.W.; Kim, D.-S.; Kim, Y.H.; Lee, J.W. Hypersurfaces with generalized 1-type Gauss map. Mathematics 2018, 6, 130. [Google Scholar] [CrossRef]
  9. Kim, D.-S.; Kim, Y.H.; Yoon, D.W. Finite type ruled surfaces in Lorentz-Minkowski space. Taiwanese J. Math. 2007, 11, 1–13. [Google Scholar] [CrossRef]
  10. Graves, L.K. Codimension one isometric immersions between Lorentz spaces. Trans. Am. Math. Soc. 1979, 252, 367–392. [Google Scholar] [CrossRef]
  11. Choi, M.; Kim, Y.H.; Yoon, D.W. Classification of ruled surfaces with pointwise 1-type Gauss map in Minkowski 3-space. Taiwanese J. Math. 2011, 15, 1141–1161. [Google Scholar] [CrossRef]
  12. Chung, H.-S.; Kim, D.-S. Finite type curves in Lorentz-Minkowski plane. Honam Math. J. 1995, 17, 41–47. [Google Scholar]
  13. Jung, S.M.; Kim, Y.H. Gauss map and its applications on ruled submanifolds in Minkowski space. Symmetry 2018, 10, 218. [Google Scholar] [CrossRef]
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