Classification Theorems of Ruled Surfaces in Minkowski Three-Space

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.


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: where x 0 is a constant vector and ∆x i = λ i x i for some positive integer k and λ i ∈ R, i = 1, . . ., k.Here, ∆ 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 m s 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 3 1 .Let M be a surface in the 3-dimensional Minkowski space E 3  1 with a non-degenerate induced metric.From now on, a surface M in E 3 1 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 ( ) by parallel translation of a unit normal vector of M to the origin is called the Gauss map of M, where (= ±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 3  1 is of pointwise 1-type if the Gauss map G of M satisfies: 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 3  1 parameterized by: x(s, t) = p + tβ(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 M be a surface in E 3 parameterized by: x(s, t) = (t cos 2 s, t sin s cos s, t sin s).
Its Laplacian turns out to be: ∆G = f G + gC for some non-zero smooth functions f , g and a constant vector C. The surface M is 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 G of a surface M in E 3  1 is said to be of generalized 1-type if the Gauss map G satisfies: for some non-zero smooth functions f , g and a constant vector C. If f = g, G is 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 functions f , g and 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 3 1 with the generalized 1-type Gauss map.

Example 2.
Let M be a ruled surface in the Minkowski 3-space E 3 1 parameterized by: Then, the Gauss map G is given by: By a direct computation, we see that its Laplacian satisfies: (1, −1, 0), which indicates that M has the generalized 1-type Gauss map.

Preliminaries
Let M be a non-degenerate surface in the Minkowski 3-space E 3 1 with the Lorentz metric , where (x 1 , x 2 , x 3 ) denotes the standard coordinate system in E 3 1 .From now on, a surface in E 3  1 means non-degenerate unless otherwise stated.A curve in E 3 1 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 ∆ is given by: where (g ij ) = (g ij ) −1 , G is the determinant of the matrix (g ij ) consisting of the components of the first fundamental form and { xi } are the local coordinate system of M.
A ruled surface M in the Minkowski 3-space E 3 1 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 3  1 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) + tβ(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 3  1 : 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 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 3 1 with Cartan frame {A, B, C}, that is A, B, C are vector fields along α in E 3  1 satisfying the following conditions: where a is a constant and k(s) a nowhere vanishing function.A null scroll parameterized by x = x(s, t) = α(s) + tB(s) is called a B-scroll, which has constant mean curvature H = a and constant Gaussian curvature K = a 2 .Furthermore, its Laplacian ∆G of the Gauss map G is given by: 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 3 1 are smooth and connected unless otherwise stated.

Cylindrical Ruled
Surfaces in E 3 1 with the Generalized 1-Type Gauss Map Let M be a cylindrical ruled surface of type Then, M is parameterized by a base curve α and a unit constant vector β such that: 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 3  1 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: if C is null, or if C is non-null, where c is some non-zero constant and k is a constant.
is an open part of a Minkowski plane or a cylinder over a curve of infinite-type satisfying: or: depending on the constant vector, C, being null or non-null, respectively, for some non-zero constant c and some constant k.
M is an open part of either a Minkowski plane or a cylinder over a curve of infinite-type satisfying: 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 ∆G of the Gauss map are respectively obtained by: With the help of ( 1) and ( 7), it immediately follows: for some constants c 1 and c 2 .We also have: Firstly, we consider the case that M is of type M 1 + .Since α is space-like, we may put: for some function φ(s) of s.Then, ( 8) can be written in the form: This implies that: (φ and: 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: With the help of ( 9), it follows that: Solving the above differential equation, we have: We put: 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: 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 3  1 , 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: 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: or, we put: 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 ∆G of the Gauss map are obtained respectively by: G = (0, α 3 , −α 2 ) and ∆G = (0, −α 3 , α 2 ). ( The relationship (13) and the condition (1) imply that the constant vector C has the form: 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: Since α is parameterized by the arc length s, we may put: for some function φ(s) of s.Hence, (14) can be expressed as: It follows: Thus, M is a cylinder over the base curve α given by: α(s) = 0, cos r(s)ds ds, sin r(s)ds ds and the ruling β(s) = (1, 0, 0), where r(s Consequently, we have: Let M be a cylindrical ruled surface with the generalized 1-type Gauss map in the Minkowski 3-space E 3 1 .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).

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 3  1 .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) + tβ(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: 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: Therefore, the smooth function q is given by: Note that t is chosen so that q takes positive values.Furthermore, the Gauss map G of M is given by: 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, Applying the Gauss and Weingarten formulas, the Laplacian of the Gauss map G of M in E 3 1 is represented by: where A G denotes the shape operator of the surface M in E where The straightforward computation gives: where: Thus, the Laplacian ∆G of the Gauss map G of M is obtained by: Now, suppose that the Gauss map G of M is of generalized 1-type.Hence, from (1), ( 17) and (20), we get: If we take the indefinite scalar product to Equation (21) with β, β and β × β , respectively, then we obtain respectively, On the other hand, the constant vector C can be written as; where Differentiating the functions c 1 , c 2 , and c 3 with respect to s, we have: Furthermore, Equations ( 22)-( 24) are expressed as follows: Combining Equations ( 26)-(28), we have: Hence, Equations ( 29) and (30) yield that: First of all, we prove: 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 3  1 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: Observing the coefficient of the term involving t 2 of (31), with the help of (32), we get: Examining the coefficient of the linear term in t of (31) and using (32) and (33), we also get: Similarly, from the constant term with respect to t of (31), we have: by using (32)-(34).Combining (33) and (35), we obtain: 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: Substituting ( 37) into (36), we get: 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: 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 Putting it into (35), we have: 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 1 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: 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: From the coefficient of the term involving t 2 in (31), using ( 25) and ( 40), we also get: Furthermore, considering the coefficient of the linear term in t of (31) and making use of Equations ( 25), (40), and (41), we obtain: Now, we consider the open set V = {s ∈ dom(α)|Q(s) = 0}.Suppose V = ∅.From (42), 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: Since Q = 0 on V, one can have: Our making use of the first and the second equations in (25), (40) reduces to: Suppose that u (s) = 0 for some s ∈ V.Then, u = 0 on an open subset V 1 ⊂ V. From (43), on V 1 : Putting ( 46) into (44), we have {(u Hence, (45) yields that: 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: On the other hand, since the torsion of the director vector field β viewed as a curve in E 3 1 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 3  1 .Without loss of generality, we may put: where and p > 0.Then, the function u = α , β is given by: where α (s) = (α 1 (s), α 2 (s), α 3 (s)).Therefore, we have: Since u is a constant, u must be zero.It is a contradiction on V 1 , and so: It immediately follows that: Then, u = Q and (43) can be written as: Putting (50) into (40) and (41), respectively, we obtain: Putting together Equations ( 51) and ( 52) with the help of (39), we get: If we make use of it, we can derive a contradiction.Thus: 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: Putting (54) together with u = −Q into (40), we have: Furthermore, Equations (39), ( 41), ( 54) and (55) give: Then, Q = 2ε 3 QR, and thus, Q = 2ε 3 Q R + 2ε 3 QR .Putting it into (55) with the help of (39), we get: a contradiction.Therefore, we get: 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.
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 3  1 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 3 1 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.
Again, we put the smooth functions q and u as follows: 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) = (as, as, 1) (65) Differentiating the functions c 1 , c 2 , and c 3 with respect to the parameter s, we get: Combining Equations ( 64)-(66), we obtain: As before, from (68), we obtain the following: Combining Equations ( 69) and (71), we get: From ( 72) and (73), we get c 1 u = 0. Hence, Equations (70) and (72) become: 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: 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 = uc 3 in (76), we get u = kc 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: In this section, we examine the null scrolls with the generalized 1-type Gauss map in the Minkowski 3-space E 3  1 .In particular, we focus on proving the following theorem.
Theorem 6.Let M be a null scroll in the Minkowski 3-space E 3 1 .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 3  1 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) + tβ(s) and we have the natural coordinate frame {x s , x t } given by: x s = α + tβ and x t = β.
We put the smooth functions u, v, Q, and R by: With respect to the pseudo-orthonormal frame {α , β, α × β}, the vector fields β , β × β, and β × β are represented as: Thus, the Gauss map G and its Laplacian ∆G are expressed by: Since M has the generalized 1-type Gauss map, the Gauss map G satisfies: for some non-zero smooth functions f , g and a constant vector C. From (79), we get: 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: 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 Ĩ ⊂ dom(α).Then, c 3 = 0 on Ĩ.Therefore, the constant vector C can be written as C = c 2 β on Ĩ.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 Ĩ, and so, c 2 = 0.It contradicts that C is a non-zero vector.In the sequel, Q vanishes identically.Then, β = uβ, 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:

Theorem 5 .Remark 2 . 5 .
Let M be a non-cylindrical ruled surface of type M 2 + or M 2− in E 3 1 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.There do not exist non-cylindrical ruled surfaces of type M 2 + or M 2 − in E 3 1 with the proper generalized 1-type Gauss map G. Null Scrolls in the Minkowski 3-Space E 3 1 )Then, {α , β, α × β} is a pseudo-orthonormal frame along α.Straightforward computation gives the Gauss map G of M and the Laplacian ∆G of G by:G = α × β + tβ × β and ∆G = −2β × β + 2(u + tv)β × β.

Corollary 1 .
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 E3  1 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.There do not exist null scrolls in E 3 1 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).