Classification Theorems of Ruled Surfaces in Minkowski Three-Space

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 ${\mathbb{E}}^m$ is said to be of finite-type if it has the spectral decomposition as:

$$x=x_0+x_1+\cdots +x_k,$$

where $x_0$ is a constant vector and $\Delta x_i={\lambda }_i{x}_i$ for some positive integer k and ${\lambda }_i\in \mathbb{R}$, $i=1,\dots ,k$. Here, $\Delta$ denotes the Laplacian operator defined on M. If ${\lambda }_1$, …, ${\lambda }_k$ 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 ${\mathbb{E}}_s^m$ with the indefinite metric of index $s\ge 1$. 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 ${\mathbb{E}}_1^3$. Let M be a surface in the 3-dimensional Minkowski space ${\mathbb{E}}_1^3$ 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 $Q^2\left(ϵ\right)$ by parallel translation of a unit normal vector of M to the origin is called the Gauss map of M, where $ϵ(=\pm 1)$ denotes the sign of the vector field G. A helicoid or a right cone in ${\mathbb{E}}^3$ 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 ${\mathbb{E}}_1^3$ is of pointwise 1-type if the Gauss map G of M satisfies:

$$\Delta G=f(G+\mathbf{C})$$

for some non-zero smooth function f and a constant vector $\mathbf{C}.$ Especially, the Gauss map G is called pointwise 1-type of the first kind if $\mathbf{C}$ is a zero vector. Otherwise, it is said to be of pointwise 1-type of the second kind.

Some other surfaces of ${\mathbb{E}}^3$ such as conical surfaces have an interesting type of Gauss map. A surface in ${\mathbb{E}}_1^3$ parameterized by:

$$x(s,t)=p+t\beta \left(s\right),$$

where p is a point and $\beta \left(s\right)$ 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 ${\mathbb{E}}^3$ parameterized by:

$$x(s,t)=(t{cos}^{2}s,tsinscoss,tsins).$$

Then, the Gauss map G can be obtained by:

$$G=\frac{1}{\sqrt{1+{cos}^{2}s}}(-{sin}^{3}s,(2-{cos}^{2}s)coss,-{cos}^{2}s).$$

Its Laplacian turns out to be:

$$\Delta G=fG+g\mathbf{C}$$

for some non-zero smooth functions$f,g$and a constant vector$\mathbf{C}.$The surfaceMis a kind of conical surface generated by a spherical curve$\beta \left(s\right)=({cos}^{2}s,sinscoss,sins)$on the unit sphere${\mathbb{S}}^2\left(1\right)$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 ${\mathbb{E}}_1^3$ is said to be of generalized 1-type if the Gauss map G satisfies:

$$\Delta G=fG+g\mathbf{C}$$

(1)

for some non-zero smooth functions $f,g$ and a constant vector $\mathbf{C}.$ If $f\ne g$, 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 $\mathbf{C}$ if we solve the differential Equation (1).

Here, we provide an example of a cylindrical ruled surface in the 3-dimensional Minkowski space ${\mathbb{E}}_1^3$ with the generalized 1-type Gauss map.

Example 2.

Let M be a ruled surface in the Minkowski 3-space ${\mathbb{E}}_1^3$ parameterized by:

$$x(s,t)=\left(\frac{1}{2}\left(s\sqrt{s^2-1}-ln(s+\sqrt{s^2-1})\right),\frac{1}{2}{s}^2,t\right),s\ge 1.$$

Then, the Gauss map G is given by:

$$G=(-s,-\sqrt{s^2-1},0).$$

By a direct computation, we see that its Laplacian satisfies:

$$\Delta G=\frac{s-\sqrt{s^2-1}}{{(s^2-1)}^{\frac{3}{2}}}G+\frac{s(s-\sqrt{s^2-1})}{{(s^2-1)}^{\frac{3}{2}}}(1,-1,0),$$

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 ${\mathbb{E}}_1^3$ 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 ${\mathbb{E}}_1^3.$ From now on, a surface in ${\mathbb{E}}_1^3$ means non-degenerate unless otherwise stated. A curve in ${\mathbb{E}}_1^3$ 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 $\Delta$ is given by:

$$\Delta =-\frac{1}{\sqrt{\left|\mathcal{G}\right|}}\sum _{i,j=1}^2\frac{\partial }{\partial {\overline{x}}_i}\left(\sqrt{\left|\mathcal{G}\right|}{g}^{ij}\frac{\partial }{\partial {\overline{x}}_j}\right),$$

where $(g^{ij})={(g_{ij})}^{-1},$ $\mathcal{G}$ is the determinant of the matrix $(g_{ij})$ consisting of the components of the first fundamental form and $\left\{{\overline{x}}_i\right\}$ are the local coordinate system of M.

A ruled surface M in the Minkowski 3-space ${\mathbb{E}}_1^3$ is defined as follows: Let I and J be some open intervals in the real line R. Let $\alpha =\alpha \left(s\right)$ be a curve in ${\mathbb{E}}_1^3$ defined on I and $\beta =\beta \left(s\right)$ a transversal vector field with ${\alpha }^{\prime }\left(s\right)$ along $\alpha$. From now on, ${}^{\prime }$ denotes the differentiation with respect to the parameter s unless otherwise stated. The surface M with a parametrization given by:

$$x(s,t)=\alpha \left(s\right)+t\beta \left(s\right),s\in I,t\in J$$

is called a ruled surface. In this case, the curve $\alpha =\alpha \left(s\right)$ is called a base curve and $\beta =\beta \left(s\right)$ a director vector field or a ruling. A ruled surface M is said to be cylindrical if $\beta$ 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 ${\mathbb{E}}_1^3$: If the base curve $\alpha$ is space-like or time-like, the director vector field $\beta$ can be chosen to be orthogonal to $\alpha$. The ruled surface M is said to be of type $M_{+}$ or $M_{-}$, respectively, depending on $\alpha$ 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 $\beta$ is space-like, it is said to be of type $M_{+}^1$ or $M_{+}^2$ if ${\beta }^{\prime }$ is non-null or null, respectively. When $\beta$ is time-like, ${\beta }^{\prime }$ must be space-like because of the character of the causal vectors, which we call $M_{+}^3$. On the other hand, when $\alpha$ is time-like, $\beta$ is always space-like. Accordingly, it is also said to be of type $M_{-}^1$ or $M_{-}^2$ if ${\beta }^{\prime }$ 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 $\alpha$ is null, the ruling $\beta$ along $\alpha$ must be null since M is non-degenerate. Such a ruled surface M is called a null scroll. Other cases, such as $\alpha$ is non-null and $\beta$ is null, or $\alpha$ is null and $\beta$ is non-null, are determined to be one of the types $M_{\pm }^1,M_{\pm }^2$, and $M_{+}^3$, or a null scroll by an appropriate change of the base curve [9].

Consider a null scroll: Let $\alpha =\alpha \left(s\right)$ be a null curve in ${\mathbb{E}}_1^3$ with Cartan frame $\{A,B,C\}$, that is $A,B,C$ are vector fields along $\alpha$ in ${\mathbb{E}}_1^3$ satisfying the following conditions:

$$⟨A,A⟩=⟨B,B⟩=0,⟨A,B⟩=1,⟨A,C⟩=⟨B,C⟩=0,⟨C,C⟩=1,$$

$${\alpha }^{\prime }=A,C^{\prime }=-aA-k\left(s\right)B,$$

where a is a constant and $k\left(s\right)$ a nowhere vanishing function. A null scroll parameterized by $x=x(s,t)=\alpha \left(s\right)+tB\left(s\right)$ is called a B-scroll, which has constant mean curvature $H=a$ and constant Gaussian curvature $K=a^2$. Furthermore, its Laplacian $\Delta G$ of the Gauss map G is given by:

$$\Delta G=-2a^{2}G,$$

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 ${\mathbb{E}}_1^3$ are smooth and connected unless otherwise stated.

3. Cylindrical Ruled Surfaces in ${\mathbb{E}}_1^3$ with the Generalized 1-Type Gauss Map

Let M be a cylindrical ruled surface of type $M_{+}^1,M_{-}^1$ or $M_{+}^3$ in ${\mathbb{E}}_1^3$. Then, M is parameterized by a base curve $\alpha$ and a unit constant vector $\beta$ such that:

$$x(s,t)=\alpha \left(s\right)+t\beta$$

satisfying $⟨{\alpha }^{\prime },{\alpha }^{\prime }⟩={\epsilon }_1(=\pm 1),$ $⟨{\alpha }^{\prime },\beta ⟩=0$, and $⟨\beta ,\beta ⟩={\epsilon }_2(=\pm 1).$

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 $\mathbf{C}=(c_1,c_2,c_3)$ 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 ${\mathbb{E}}_1^3$ 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:

$$c^2{f}^{-\frac{1}{3}}-ln|c^2{f}^{-\frac{1}{3}}+1|=\pm c^3(s+k)$$

(2)

if $\mathbf{C}$ is null, or

$$\begin{array}{c}\sqrt{{\left(c^2{f}^{-\frac{1}{3}}+1\right)}^2+(-c_1^2+c_2^2)}-ln\left(c^2{f}^{-\frac{1}{3}}+1+\sqrt{{\left(c^2{f}^{-\frac{1}{3}}+1\right)}^2+(-c_1^2+c_2^2)}\right)\hfill \\ +ln\sqrt{|-c_1^2+c_2^2|}=\pm c^3(s+k)\hfill \end{array}$$

(3)

if $\mathbf{C}$ 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:

$$c^2{f}^{-\frac{1}{3}}+ln|c^2{f}^{-\frac{1}{3}}-1|=\pm c^3(s+k)$$

(4)

or:

$$\begin{array}{c}\sqrt{{\left(c^2{f}^{-\frac{1}{3}}-1\right)}^2-(-c_1^2+c_2^2)}+ln\left(c^2{f}^{-\frac{1}{3}}-1+\sqrt{{\left(c^2{f}^{-\frac{1}{3}}-1\right)}^2+|-c_1^2+c_2^2|}\right)\hfill \\ -ln\sqrt{|-c_1^2+c_2^2|}=\pm c^3(s+k)\hfill \end{array}$$

(5)

depending on the constant vector, $\mathbf{C}$, 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:

$$\sqrt{c_2^2+c_3^2-{\left(c^2{f}^{-\frac{1}{3}}-1\right)}^2}-{sin}^{-1}\left(\frac{c^2{f}^{-\frac{1}{3}}-1}{\sqrt{c_2^2+c_3^2}}\right)=\pm c^3(s+k),$$

(6)

where c is a non-zero constant and k a constant.

We now assume that $f\ne g.$ Here, we consider two cases.

Case 1. Let M be a cylindrical ruled surface of type $M_{+}^1$ or $M_{-}^1,$ i.e., ${\epsilon }_2=1.$ Without loss of generality, the base curve $\alpha$ can be put as $\alpha \left(s\right)=({\alpha }_1\left(s\right),{\alpha }_2\left(s\right),0)$ parameterized by arc length s and the director vector field $\beta$ as a unit constant vector $\beta =(0,0,1).$ Then, the Gauss map G of M and the Laplacian $\Delta G$ of the Gauss map are respectively obtained by:

$$G=(-{\alpha }_2^{\prime }\left(s\right),-{\alpha }_1^{\prime }\left(s\right),0)\mathrm{and}\Delta G=({\epsilon }_1{\alpha }_2^{‴}\left(s\right),{\epsilon }_1{\alpha }_1^{‴}\left(s\right),0).$$

(7)

With the help of (1) and (7), it immediately follows:

$$\mathbf{C}=(c_1,c_2,0)$$

for some constants $c_1$ and $c_2$. We also have:

$$\begin{array}{cc}\hfill {\epsilon }_1{\alpha }_2^{‴}& =-f{\alpha }_2^{\prime }+g{c}_1,\hfill \\ \hfill {\epsilon }_1{\alpha }_1^{‴}& =-f{\alpha }_1^{\prime }+g{c}_2.\hfill \end{array}$$

(8)

Firstly, we consider the case that M is of type $M_{+}^1.$ Since $\alpha$ is space-like, we may put:

$${\alpha }_1^{\prime }\left(s\right)=sinh\varphi \left(s\right)\mathrm{and}{\alpha }_2^{\prime }\left(s\right)=cosh\varphi \left(s\right)$$

for some function $\varphi \left(s\right)$ of s. Then, (8) can be written in the form:

$$\begin{array}{c}{\left({\varphi }^{\prime }\right)}^{2}cosh\varphi +{\varphi }^{″}sinh\varphi =-fcosh\varphi +g{c}_1,\hfill \\ {\left({\varphi }^{\prime }\right)}^{2}sinh\varphi +{\varphi }^{″}cosh\varphi =-fsinh\varphi +g{c}_2.\hfill \end{array}$$

This implies that:

$${\left({\varphi }^{\prime }\right)}^2=-f+g(c_{1}cosh\varphi -c_{2}sinh\varphi )$$

(9)

and:

$${\varphi }^{″}=g(-c_{1}sinh\varphi +c_{2}cosh\varphi ).$$

(10)

In fact, ${\varphi }^{\prime }$ is the signed curvature of the base curve $\alpha =\alpha \left(s\right).$

Suppose $\varphi$ is a constant, i.e., ${\varphi }^{\prime }=0$. Then, $\alpha$ is part of a straight line. In this case, M is an open part of a Euclidean plane.

Now, we suppose that ${\varphi }^{\prime }\ne 0.$ From (8), we see that the functions f and g depend only on the parameter $s,$ i.e., $f(s,t)=f\left(s\right)$ and $g(s,t)=g\left(s\right).$ Taking the derivative of Equation (9) and using (10), we get:

$$3{\varphi }^{\prime }{\varphi }^{″}=-f^{\prime }+g^{\prime }(c_{1}cosh\varphi -c_{2}sinh\varphi ).$$

With the help of (9), it follows that:

$$\frac{3}{2}{\left({\left({\varphi }^{\prime }\right)}^2\right)}^{\prime }=-f^{\prime }+\frac{g^{\prime }}{g}\left({\left({\varphi }^{\prime }\right)}^2+f\right).$$

Solving the above differential equation, we have:

$${\varphi }^{\prime }{\left(s\right)}^2=k_1{g}^{\frac{2}{3}}+\frac{2}{3}{g}^{\frac{2}{3}}\int g^{-\frac{2}{3}}f\left(-\frac{f^{\prime }}{f}+\frac{g^{\prime }}{g}\right)ds,k_1(\ne 0)\in R.$$

(11)

We put:

$${\varphi }^{\prime }\left(s\right)=\pm \sqrt{p\left(s\right)},$$

where $p\left(s\right)=|k_1{g}^{\frac{2}{3}}+\frac{2}{3}{g}^{\frac{2}{3}}\int g^{-\frac{2}{3}}f\left(-\frac{f^{\prime }}{f}+\frac{g^{\prime }}{g}\right)ds|$. This means that the function $\varphi$ 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 $\alpha$ such that:

$$\alpha \left(s\right)=\left(\int sinh\varphi (s)ds,\int cosh\varphi (s)ds,0\right)$$

and the director vector field $\beta \left(s\right)=(0,0,1)$.

In this case, if f and g are constant, the signed curvature ${\varphi }^{\prime }$ of a base curve $\alpha$ 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 ${\mathbb{E}}_1^3$, see [12] for the details.

Next, we consider the case that M is of type $M_{-}^1.$ Since $\alpha$ is time-like, we may put:

$${\alpha }_1^{\prime }\left(s\right)=cosh\varphi \left(s\right)\mathrm{and}{\alpha }_2^{\prime }\left(s\right)=sinh\varphi \left(s\right)$$

for some function $\varphi \left(s\right)$ of s.

As was given in the previous case of type $M_{+}^1$, if the signed curvature ${\varphi }^{\prime }$ of the base curve $\alpha$ is zero, M is part of a Minkowski plane.

We now assume that ${\varphi }^{\prime }\ne 0$. Quite similarly as above, we have:

$${\varphi }^{\prime }{\left(s\right)}^2=k_2{g}^{\frac{2}{3}}+\frac{2}{3}{g}^{\frac{2}{3}}\int g^{-\frac{2}{3}}f\left(\frac{f^{\prime }}{f}-\frac{g^{\prime }}{g}\right)ds,k_2(\ne 0)\in R,$$

(12)

or, we put:

$${\varphi }^{\prime }\left(s\right)=\pm \sqrt{q\left(s\right)},$$

where $q\left(s\right)=|k_2{g}^{\frac{2}{3}}+\frac{2}{3}{g}^{\frac{2}{3}}\int g^{-\frac{2}{3}}f\left(\frac{f^{\prime }}{f}-\frac{g^{\prime }}{g}\right)ds|$.

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 $\alpha$ to be $\alpha \left(s\right)=(0,{\alpha }_2\left(s\right),{\alpha }_3\left(s\right))$ parameterized by arc length s and the director vector field $\beta$ as $\beta =(1,0,0)$. Then, the Gauss map G of M and the Laplacian $\Delta G$ of the Gauss map are obtained respectively by:

$$G=(0,{\alpha }_3^{\prime },-{\alpha }_2^{\prime })\mathrm{and}\Delta G=(0,-{\alpha }_3^{‴},{\alpha }_2^{‴}).$$

(13)

The relationship (13) and the condition (1) imply that the constant vector C has the form:

$$\mathbf{C}=(0,c_2,c_3)$$

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:

$$\begin{array}{cc}\hfill -{\alpha }_3^{‴}& =f{\alpha }_3^{\prime }+g{c}_2,\hfill \\ \hfill {\alpha }_2^{‴}& =-f{\alpha }_2^{\prime }+g{c}_3.\hfill \end{array}$$

(14)

Since $\alpha$ is parameterized by the arc length s, we may put:

$${\alpha }_2^{\prime }\left(s\right)=cos\varphi \left(s\right)\mathrm{and}{\alpha }_3^{\prime }\left(s\right)=sin\varphi \left(s\right)$$

for some function $\varphi \left(s\right)$ of s. Hence, (14) can be expressed as:

$$\begin{array}{c}{\left({\varphi }^{\prime }\right)}^{2}sin\varphi -{\varphi }^{″}cos\varphi =fsin\varphi +g{c}_2,\hfill \\ {\left({\varphi }^{\prime }\right)}^{2}cos\varphi +{\varphi }^{″}sin\varphi =fcos\varphi -g{c}_3.\hfill \end{array}$$

It follows:

$${\left({\varphi }^{\prime }\right)}^2=f+g(c_{2}sin\varphi -c_{3}cos\varphi ).$$

(15)

Thus, M is a cylinder over the base curve $\alpha$ given by:

$$\alpha \left(s\right)=\left(0,\int cos\left(\int \sqrt{r\left(s\right)}ds\right)ds,\int sin\left(\int \sqrt{r\left(s\right)}ds\right)ds\right)$$

and the ruling $\beta \left(s\right)=(1,0,0)$, where $r\left(s\right)=|f\left(s\right)+g\left(s\right)\left(c_{2}sin\varphi \left(s\right)-c_{3}cos\varphi \left(s\right)\right)|$.

Consequently, we have:

Theorem 1.

(Classification of cylindrical ruled surfaces in ${\mathbb{E}}_1^3$) Let M be a cylindrical ruled surface with the generalized 1-type Gauss map in the Minkowski 3-space ${\mathbb{E}}_1^3$. 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 ${\mathbb{E}}_1^3.$

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,

$$x(s,t)=\alpha \left(s\right)+t\beta \left(s\right)$$

such that $⟨{\alpha }^{\prime },\beta ⟩=0,⟨\beta ,\beta ⟩={\epsilon }_2(=\pm 1)$, and $⟨{\beta }^{\prime },{\beta }^{\prime }⟩={\epsilon }_3(=\pm 1).$ Then, $\{\beta ,{\beta }^{\prime },\beta \times {\beta }^{\prime }\}$ is an orthonormal frame along the base curve $\alpha$. For later use, we define the smooth functions $q,u,Q$, and R as follows:

$$q=\parallel x_s{\parallel }^2={\epsilon }_4⟨x_s,x_s⟩,u=⟨{\alpha }^{\prime },{\beta }^{\prime }⟩,Q=⟨{\alpha }^{\prime },\beta \times {\beta }^{\prime }⟩,R=⟨{\beta }^{″},\beta \times {\beta }^{\prime }⟩,$$

where ${\epsilon }_4$ is the sign of the coordinate vector field $x_s=\partial x/\partial s.$ The vector fields ${\alpha }^{\prime }$, ${\beta }^{″}$, ${\alpha }^{\prime }\times \beta$, and $\beta \times {\beta }^{″}$ are represented in terms of the orthonormal frame $\{\beta ,{\beta }^{\prime },\beta \times {\beta }^{\prime }\}$ along the base curve $\alpha$ as:

$$\begin{array}{c}{\alpha }^{\prime }={\epsilon }_{3}u{\beta }^{\prime }-{\epsilon }_2{\epsilon }_{3}Q\beta \times {\beta }^{\prime },\hfill \\ {\beta }^{″}=-{\epsilon }_2{\epsilon }_3\beta -{\epsilon }_2{\epsilon }_{3}R\beta \times {\beta }^{\prime },\hfill \\ {\alpha }^{\prime }\times \beta ={\epsilon }_{3}Q{\beta }^{\prime }-{\epsilon }_{3}u\beta \times {\beta }^{\prime },\hfill \\ \beta \times {\beta }^{″}=-{\epsilon }_{3}R{\beta }^{\prime }.\hfill \end{array}$$

(16)

Therefore, the smooth function q is given by:

$$q={\epsilon }_4({\epsilon }_3{t}^2+2ut+{\epsilon }_3{u}^2-{\epsilon }_2{\epsilon }_3{Q}^2).$$

Note that t is chosen so that q takes positive values.

Furthermore, the Gauss map G of M is given by:

$$G=q^{-1/2}\left({\epsilon }_{3}Q{\beta }^{\prime }-({\epsilon }_{3}u+t)\beta \times {\beta }^{\prime }\right).$$

(17)

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,

$$\begin{array}{cc}\hfill H& =\frac{1}{2}{\epsilon }_2{q}^{-3/2}\left(R{t}^2+(2{\epsilon }_{3}uR+Q^{\prime })t+u^{2}R+{\epsilon }_{3}u{Q}^{\prime }-{\epsilon }_3{u}^{\prime }Q-{\epsilon }_2{Q}^{2}R\right),\hfill \\ \hfill K& =q^{-2}{Q}^2.\hfill \end{array}$$

(18)

Applying the Gauss and Weingarten formulas, the Laplacian of the Gauss map G of M in ${\mathbb{E}}_1^3$ is represented by:

$$\Delta G=2\mathrm{grad}H+⟨G,G⟩\left(\mathrm{t}r{A}_G^2\right)G,$$

(19)

where $A_G$ denotes the shape operator of the surface M in ${\mathbb{E}}_1^3$ and $\mathrm{grad}H$ is the gradient of H. Using (18), we get:

$$\begin{array}{cc}\hfill 2\mathrm{grad}H& =2⟨e_1,e_1⟩e_1\left(H\right)e_1+2⟨e_2,e_2⟩e_2\left(H\right)e_2\hfill \\ & =2{\epsilon }_4{e}_1\left(H\right)e_1+2{\epsilon }_2{e}_2\left(H\right)e_2\hfill \\ & =q^{-7/2}\{-{\epsilon }_2({\epsilon }_{3}u+t)A_1{\beta }^{\prime }-{\epsilon }_{4}q{B}_1\beta +{\epsilon }_{3}Q{A}_1\beta \times {\beta }^{\prime }\},\hfill \end{array}$$

where $e_1=\frac{x_s}{\left|\right|x_s\left|\right|},e_2=\frac{x_t}{\left|\right|x_t\left|\right|}$,

$$\begin{array}{cc}\hfill A_1=& 3(u^{\prime }t+{\epsilon }_{3}u{u}^{\prime }-{\epsilon }_2{\epsilon }_{3}Q{Q}^{\prime })\{R{t}^2+(2{\epsilon }_{3}uR+Q^{\prime })t+u^{2}R+{\epsilon }_{3}u{Q}^{\prime }-{\epsilon }_3{u}^{\prime }Q-{\epsilon }_2{Q}^{2}R\}\hfill \\ & -({\epsilon }_3{t}^2+2ut+{\epsilon }_3{u}^2-{\epsilon }_2{\epsilon }_3{Q}^2)\{R^{\prime }{t}^2+(2{\epsilon }_3{u}^{\prime }R+2{\epsilon }_{3}u{R}^{\prime }+Q^{″})t+2u{u}^{\prime }R+u^2{R}^{\prime }\hfill \\ & +{\epsilon }_{3}u{Q}^{″}-{\epsilon }_3{u}^{″}Q-2{\epsilon }_{2}Q{Q}^{\prime }R-{\epsilon }_2{Q}^2{R}^{\prime }\},\hfill \\ \hfill B_1=& {\epsilon }_{3}R{t}^3+(3uR+2{\epsilon }_3{Q}^{\prime })t^2+(3{\epsilon }_3{u}^{2}R+4u{Q}^{\prime }-3u^{\prime }Q-{\epsilon }_2{\epsilon }_3{Q}^{2}R)t+u^{3}R+2{\epsilon }_3{u}^2{Q}^{\prime }\hfill \\ & -{\epsilon }_{2}u{Q}^{2}R-3{\epsilon }_{3}u{u}^{\prime }Q+{\epsilon }_2{\epsilon }_3{Q}^2{Q}^{\prime }.\hfill \end{array}$$

The straightforward computation gives:

$$\mathrm{tr}{A}_G^2=-{\epsilon }_2{\epsilon }_4{q}^{-3}{D}_1,$$

where:

$$D_1=-{\epsilon }_4{(u^{\prime }t+{\epsilon }_{3}u{u}^{\prime }-{\epsilon }_2{\epsilon }_{3}Q{Q}^{\prime })}^2+{\epsilon }_{3}q\{{({\epsilon }_{2}QR+{\epsilon }_3{u}^{\prime })}^2-{\epsilon }_2{(Q^{\prime }+{\epsilon }_{3}uR+Rt)}^2-2{\epsilon }_3{Q}^2\}.$$

Thus, the Laplacian $\Delta G$ of the Gauss map G of M is obtained by:

$$\Delta G=q^{-7/2}[-{\epsilon }_{4}q{B}_1\beta +\{-{\epsilon }_2({\epsilon }_{3}u+t)A_1+{\epsilon }_{3}Q{D}_1\}{\beta }^{\prime }+\{{\epsilon }_{3}Q{A}_1-({\epsilon }_{3}u+t)D_1\}\beta \times {\beta }^{\prime }].$$

(20)

Now, suppose that the Gauss map G of M is of generalized 1-type. Hence, from (1), (17) and (20), we get:

$$\begin{array}{cc}\hfill q^{-7/2}& [-{\epsilon }_{4}q{B}_1\beta +\{-{\epsilon }_2({\epsilon }_{3}u+t)A_1+{\epsilon }_{3}Q{D}_1\}{\beta }^{\prime }+\{({\epsilon }_{3}Q{A}_1-({\epsilon }_{3}u+t)D_1\}\beta \times {\beta }^{\prime }]\hfill \\ \hfill =& f{q}^{-1/2}\left({\epsilon }_{3}Q{\beta }^{\prime }-({\epsilon }_{3}u+t)\beta \times {\beta }^{\prime }\right)+g\mathbf{C}.\hfill \end{array}$$

(21)

If we take the indefinite scalar product to Equation (21) with $\beta ,{\beta }^{\prime }$ and $\beta \times {\beta }^{\prime }$, respectively, then we obtain respectively,

$$-{\epsilon }_2{\epsilon }_4{q}^{-5/2}{B}_1=g⟨\mathbf{C},\beta ⟩,$$

(22)

$$q^{-7/2}\{-{\epsilon }_2{\epsilon }_3({\epsilon }_{3}u+t)A_1+Q{D}_1\}=f{q}^{-1/2}Q+g⟨\mathbf{C},{\beta }^{\prime }⟩,$$

(23)

$$q^{-7/2}\{-{\epsilon }_{2}Q{A}_1+{\epsilon }_2{\epsilon }_3({\epsilon }_{3}u+t)D_1\}=f{q}^{-1/2}{\epsilon }_2{\epsilon }_3({\epsilon }_{3}u+t)+g⟨\mathbf{C},\beta \times {\beta }^{\prime }⟩.$$

(24)

On the other hand, the constant vector $\mathbf{C}$ can be written as;

$$\mathbf{C}=c_1\beta +c_2{\beta }^{\prime }+c_3\beta \times {\beta }^{\prime },$$

where $c_1={\epsilon }_2⟨\mathbf{C},\beta ⟩$, $c_2={\epsilon }_3⟨\mathbf{C},{\beta }^{\prime }⟩$, and $c_3=-{\epsilon }_2{\epsilon }_3⟨\mathbf{C},\beta \times {\beta }^{\prime }⟩.$ Differentiating the functions $c_1,c_2$, and $c_3$ with respect to s, we have:

$$\begin{array}{c}{c}_1^{\prime }-{\epsilon }_2{\epsilon }_3{c}_2=0,\hfill \\ c_1+c_2^{\prime }-{\epsilon }_{3}R{c}_3=0,\hfill \\ {\epsilon }_2{\epsilon }_{3}R{c}_2-c_3^{\prime }=0.\hfill \end{array}$$

(25)

Furthermore, Equations (22)–(24) are expressed as follows:

$$-{\epsilon }_4{q}^{-5/2}{B}_1=g{c}_1,$$

(26)

$$q^{-7/2}\{-{\epsilon }_2({\epsilon }_{3}u+t)A_1+{\epsilon }_{3}Q{D}_1\}=f{q}^{-1/2}{\epsilon }_{3}Q+g{c}_2,$$

(27)

$$q^{-7/2}\{-{\epsilon }_{3}Q{A}_1+({\epsilon }_{3}u+t)D_1\}=f{q}^{-1/2}({\epsilon }_{3}u+t)-g{c}_3.$$

(28)

Combining Equations (26)–(28), we have:

$$\{-{\epsilon }_2({\epsilon }_{3}u+t)A_1+{\epsilon }_{3}Q{D}_1\}{c}_1+q{\epsilon }_4{B}_1{c}_2=q^{3}f{\epsilon }_{3}Q{c}_1,$$

(29)

$$\{-{\epsilon }_{3}Q{A}_1+({\epsilon }_{3}u+t)D_1\}{c}_1-q{\epsilon }_4{B}_1{c}_3=q^{3}f({\epsilon }_{3}u+t)c_1.$$

(30)

Hence, Equations (29) and (30) yield that:

$$-{\epsilon }_2{\epsilon }_3{A}_1{c}_1+B_1\{c_2({\epsilon }_{3}u+t)+{\epsilon }_{3}Q{c}_3\}=0.$$

(31)

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 ${\mathbb{E}}_1^3$ with the generalized 1-type Gauss map. If β, ${\beta }^{\prime }$, and ${\beta }^{″}$ 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 $\mathbf{C}$ 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 $\mathbf{C}$ 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 $U=\{s\in \mathrm{dom}(\alpha \left)\right|Q\left(s\right)\ne 0\}$ of dom($\alpha$). Since $\beta$, ${\beta }^{\prime }$, and ${\beta }^{″}$ are coplanar along $\alpha$, R vanishes. Thus, $c_3$ is a constant, and $c_1^{″}=-{\epsilon }_2{\epsilon }_3{c}_1$ 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:

$${\epsilon }_2{\epsilon }_3{c}_1{Q}^{″}+2c_2{Q}^{\prime }=0.$$

(32)

Observing the coefficient of the term involving $t^2$ of (31), with the help of (32), we get:

$${\epsilon }_2{\epsilon }_3{c}_1(3u^{\prime }{Q}^{\prime }+u^{″}Q)+3c_2{u}^{\prime }Q-2c_{3}Q{Q}^{\prime }=0.$$

(33)

Examining the coefficient of the linear term in t of (31) and using (32) and (33), we also get:

$$Q\{c_1\left({\epsilon }_2{\left(u^{\prime }\right)}^2+{\left(Q^{\prime }\right)}^2\right)+{\epsilon }_2{\epsilon }_3{c}_{2}Q{Q}^{\prime }-{\epsilon }_3{c}_3{u}^{\prime }Q\}=0.$$

On U,

$$c_1\left({\epsilon }_2{\left(u^{\prime }\right)}^2+{\left(Q^{\prime }\right)}^2\right)+{\epsilon }_2{\epsilon }_3{c}_{2}Q{Q}^{\prime }-{\epsilon }_3{c}_3{u}^{\prime }Q=0.$$

(34)

Similarly, from the constant term with respect to t of (31), we have:

$${\epsilon }_3{c}_1(-3u^{\prime }{Q}^{\prime }+u^{″}Q)+{\epsilon }_2{c}_{3}Q{Q}^{\prime }=0$$

(35)

by using (32)–(34). Combining (33) and (35), we obtain:

$$2{\epsilon }_3{c}_1{u}^{\prime }{Q}^{\prime }+{\epsilon }_2{c}_2{u}^{\prime }Q-{\epsilon }_2{c}_{3}Q{Q}^{\prime }=0.$$

(36)

Now, suppose that $u^{\prime }\left(s\right)\ne 0$ at some point $s\in U$ and then $u^{\prime }\ne 0$ on an open interval $U_1\subset U$. Equation (34) yields:

$${\epsilon }_3{c}_{3}Q=\frac{1}{u^{\prime }}\{c_1\left({\epsilon }_2{\left(u^{\prime }\right)}^2+{\left(Q^{\prime }\right)}^2\right)+{\epsilon }_2{\epsilon }_3{c}_{2}Q{Q}^{\prime }\}.$$

(37)

Substituting (37) into (36), we get:

$$\{{\left(u^{\prime }\right)}^2-{\epsilon }_2{\left(Q^{\prime }\right)}^2\}({\epsilon }_3{c}_1{Q}^{\prime }+{\epsilon }_2{c}_{2}Q)=0,$$

or, using $c_2={\epsilon }_2{\epsilon }_3{c}_1^{\prime }$ in (25),

$$\{{\left(u^{\prime }\right)}^2-{\epsilon }_2{\left(Q^{\prime }\right)}^2\}{\left(c_{1}Q\right)}^{\prime }=0.$$

Suppose that $\left({(u^{\prime })}^2-{\epsilon }_2{(Q^{\prime })}^2\right)(s_0)\ne 0$ for some $s_0\in U_1$. 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 $c_3{u}^{\prime }Q=0$, and so, $c_3=0$. Since $\mathbf{C}$ is a constant vector, $\mathbf{C}$ is zero on M. This contradicts our assumption. Thus, $c_1\ne 0$ on $U_2$. From the equation $c_1^{″}+{\epsilon }_2{\epsilon }_3{c}_1=0,$ we get:

$$c_1=k_{1}cos(s+s_1)\mathrm{or}{c}_1=k_{2}cosh(s+s_2)$$

for some non-zero constants $k_i$ and $s_i\in \mathbb{R}(i=1,2).$ Since $c_{1}Q$ is constant, $k_1$ and $k_2$ must be zero. Hence, $c_1=0,$ a contradiction. Thus, ${\left(u^{\prime }\right)}^2-{\epsilon }_2{\left(Q^{\prime }\right)}^2=0$ on $U_1,$ from which we get ${\epsilon }_2=1$ and $u^{\prime }=\pm Q^{\prime }.$ If $u^{\prime }\ne -Q^{\prime },$ then $u^{\prime }=Q^{\prime }$ on an open subset $U_3$ in $U_1.$ Hence, (34) implies that $Q^{\prime }(2{\epsilon }_3{c}_1{Q}^{\prime }+c_{2}Q-c_{3}Q)=0.$ On $U_3$, we get $c_{3}Q=2{\epsilon }_3{c}_1{Q}^{\prime }+c_{2}Q.$ Putting it into (35), we have:

$${\epsilon }_3{c}_1{\left(Q^{\prime }\right)}^2-{\epsilon }_3{c}_{1}Q{Q}^{″}-c_{2}Q{Q}^{\prime }=0.$$

(38)

Combining (32) and (38), $c_{1}Q$ is constant on $U_3.$ Similarly as above, we can derive that $\mathbf{C}$ is zero on M, which is a contradiction. Therefore, we have $u^{\prime }=-Q^{\prime }$ on $U_1$. Similarly, as we just did to the case under the assumption $u^{\prime }\ne -Q^{\prime }$, it is also proven that the constant vector $\mathbf{C}$ becomes zero. It is also a contradiction, and so, $U_1=\varnothing$. Thus, $u^{\prime }=0$ and $Q^{\prime }=0.$ 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 $\beta$, ${\beta }^{\prime }$, and ${\beta }^{″}$ 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., $\beta \wedge {\beta }^{\prime }\wedge {\beta }^{″}\ne 0$ 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 ${\mathbb{E}}_1^3$ 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 $f\ne g$ and the constant vector $\mathbf{C}$ 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:

$${\epsilon }_2{c}_1{R}^{\prime }+{\epsilon }_3{c}_{2}R=0.$$

(39)

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:

$$c_1(-{\epsilon }_2{\epsilon }_3{u}^{\prime }R+{\epsilon }_2{Q}^{″})+2c_2{\epsilon }_3{Q}^{\prime }+c_{3}QR=0.$$

(40)

From the coefficient of the term involving $t^2$ in (31), using (25) and (40), we also get:

$$c_1(-3{\epsilon }_2{\epsilon }_3{u}^{\prime }{Q}^{\prime }+Q{Q}^{\prime }R-{\epsilon }_2{\epsilon }_3{u}^{″}Q-Q^2{R}^{\prime })-3c_2{u}^{\prime }Q+2c_{3}Q{Q}^{\prime }=0.$$

(41)

Furthermore, considering the coefficient of the linear term in t of (31) and making use of Equations (25), (40), and (41), we obtain:

$$Q\{c_1({\epsilon }_2{\left(u^{\prime }\right)}^2+{\left(Q^{\prime }\right)}^2)+c_2{\epsilon }_2{\epsilon }_{3}Q{Q}^{\prime }-c_3{\epsilon }_3{u}^{\prime }Q\}=0.$$

(42)

Now, we consider the open set $V=\{s\in \mathrm{dom}(\alpha \left)\right|Q\left(s\right)\ne 0\}$. Suppose $V\ne \varnothing .$ From (42),

$$c_1({\epsilon }_2{\left(u^{\prime }\right)}^2+{\left(Q^{\prime }\right)}^2)+c_2{\epsilon }_2{\epsilon }_{3}Q{Q}^{\prime }-c_3{\epsilon }_3{u}^{\prime }Q=0.$$

(43)

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:

$$Q^2(2c_1{\epsilon }_3{u}^{\prime }{Q}^{\prime }+c_2{\epsilon }_2{u}^{\prime }Q-c_3{\epsilon }_{2}Q{Q}^{\prime })=0.$$

Since $Q\ne 0$ on V, one can have:

$$2c_1{\epsilon }_3{u}^{\prime }{Q}^{\prime }+c_2{\epsilon }_2{u}^{\prime }Q-c_3{\epsilon }_{2}Q{Q}^{\prime }=0.$$

(44)

Our making use of the first and the second equations in (25), (40) reduces to:

$$c_1{\epsilon }_2{u}^{\prime }R-{\epsilon }_2{\epsilon }_3{\left(c_{1}Q\right)}^{″}-c_{1}Q=0.$$

(45)

Suppose that $u^{\prime }\left(s\right)\ne 0$ for some $s\in V$. Then, $u^{\prime }\ne 0$ on an open subset $V_1\subset V$. From (43), on $V_1$:

$$c_{3}Q=\frac{1}{u^{\prime }}\{{\epsilon }_2{\epsilon }_3{c}_1{\left(u^{\prime }\right)}^2+{\epsilon }_3{c}_1{\left(Q^{\prime }\right)}^2+{\epsilon }_2{c}_{2}Q{Q}^{\prime }\}.$$

(46)

Putting (46) into (44), we have $\{{\left(u^{\prime }\right)}^2-{\epsilon }_2{\left(Q^{\prime }\right)}^2\}({\epsilon }_3{c}_1{Q}^{\prime }+{\epsilon }_2{c}_{2}Q)=0.$ With the help of $c_1^{\prime }={\epsilon }_2{\epsilon }_3{c}_2$, it becomes:

$$\{{\left(u^{\prime }\right)}^2-{\epsilon }_2{\left(Q^{\prime }\right)}^2\}{\left(c_{1}Q\right)}^{\prime }=0.$$

Suppose that $\left({(u^{\prime })}^2-{\epsilon }_2{(Q^{\prime })}^2\right)(s)\ne 0$ on $V_1$. Then, $c_{1}Q$ is constant on a component $V_2$ of $V_1$. Hence, (45) yields that:

$$c_{1}Q={\epsilon }_2{c}_1{u}^{\prime }R.$$

(47)

If $c_1\equiv 0$ on $V_2$, (25) gives that $c_2=0$ and $c_{3}R=0$. Since $R\ne 0$, $c_3=0$. Hence, the constant vector $\mathbf{C}$ is zero, a contradiction. Therefore, $c_1\ne 0$ on $V_2$. From (47), $Q={\epsilon }_2{u}^{\prime }R$. Moreover, $u^{\prime }$ 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:

$$c_1{Q}^{\prime }R-c_{1}Q{R}^{\prime }+2c_3{Q}^{\prime }=0,$$

(48)

$${\epsilon }_3{c}_1{u}^{\prime }{Q}^{\prime }-{\epsilon }_2{c}_{3}Q{Q}^{\prime }=0.$$

(49)

Upon our putting $Q={\epsilon }_2{u}^{\prime }R$ into (48), $c_3{Q}^{\prime }=0$ is derived. By (49), $c_1{u}^{\prime }{Q}^{\prime }=0.$ Hence, $Q^{\prime }=0$. 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 $\beta$ viewed as a curve in ${\mathbb{E}}_1^3$ is zero, $\beta$ is part of a plane curve. Moreover, $\beta$ has constant curvature $\sqrt{{\epsilon }_2-{\epsilon }_2{\epsilon }_3{R}^2}$. Hence, $\beta$ is a circle or a hyperbola on the unit pseudo-sphere or the hyperbolic space of radius 1 in ${\mathbb{E}}_1^3.$ Without loss of generality, we may put:

$$\beta \left(s\right)=\frac{1}{p}(R,cosps,sinps)\mathrm{or}\beta \left(s\right)=\frac{1}{p}(sinhps,coshps,R),$$

where $p^2={\epsilon }_2(1-{\epsilon }_3{R}^2)$ and $p>0$. Then, the function $u=⟨{\alpha }^{\prime },{\beta }^{\prime }⟩$ is given by:

$$u=-{\alpha }_2^{\prime }\left(s\right)sinps+{\alpha }_3^{\prime }\left(s\right)cosps\mathrm{or}u=-{\alpha }_1^{\prime }\left(s\right)coshps+{\alpha }_2^{\prime }\left(s\right)sinhps,$$

where ${\alpha }^{\prime }\left(s\right)=({\alpha }_1^{\prime }\left(s\right),{\alpha }_2^{\prime }\left(s\right),{\alpha }_3^{\prime }\left(s\right)).$ Therefore, we have:

$$u^{\prime }=-({\alpha }_2^{″}+p{\alpha }_3^{\prime })sinps-(p{\alpha }_2^{\prime }-{\alpha }_3^{″})cosps\mathrm{or}{u}^{\prime }=(-{\alpha }_1^{″}+p{\alpha }_2^{\prime })coshps-(p{\alpha }_1^{\prime }-{\alpha }_2^{″})sinhps.$$

Since $u^{\prime }$ is a constant, $u^{\prime }$ must be zero. It is a contradiction on $V_1$, and so:

$${\left(u^{\prime }\right)}^2={\epsilon }_2{\left(Q^{\prime }\right)}^2$$

on $V_1$. It immediately follows that:

$${\epsilon }_2=1$$

on $V_1$. Therefore, we get $u^{\prime }=\pm Q^{\prime }.$ Suppose $u^{\prime }\ne -Q^{\prime }$ on $V_1$. Then, $u^{\prime }=Q^{\prime }$ and (43) can be written as:

$$Q^{\prime }(2{\epsilon }_3{c}_1{Q}^{\prime }+c_{2}Q-c_{3}Q)=0.$$

Since $Q^{\prime }\ne 0$ on V,

$$c_{3}Q=2{\epsilon }_3{c}_1{Q}^{\prime }+c_{2}Q.$$

(50)

Putting (50) into (40) and (41), respectively, we obtain:

$${\epsilon }_3{c}_1{Q}^{\prime }R+c_{2}QR+2{\epsilon }_3{c}_2{Q}^{\prime }+c_1{Q}^{″}=0,$$

(51)

$${\epsilon }_3{c}_1{\left(Q^{\prime }\right)}^2+c_{1}Q{Q}^{\prime }R-{\epsilon }_3{c}_{1}Q{Q}^{″}-c_1{Q}^2{R}^{\prime }-c_{2}Q{Q}^{\prime }=0.$$

(52)

Putting together Equations (51) and (52) with the help of (39), we get:

$$({\epsilon }_3{c}_1{Q}^{\prime }+c_{2}Q)(Q^{\prime }+2{\epsilon }_{3}QR)=0.$$

Suppose $({\epsilon }_3{c}_1{Q}^{\prime }+c_{2}Q)\left(s\right)\ne 0$ on $V_1$. Then, $Q^{\prime }=-2{\epsilon }_{3}QR.$ If we make use of it, we can derive $R({\epsilon }_3{c}_1{Q}^{\prime }+c_{2}Q)=0$ from (51). Since R is non-vanishing, ${\epsilon }_3{c}_1{Q}^{\prime }+c_{2}Q=0$, a contradiction. Thus:

$${\epsilon }_3{c}_1{Q}^{\prime }+c_{2}Q=0,$$

(53)

that is, $c_{1}Q$ is constant on each component of $V_1$. From (45), $c_{1}Q=c_1{u}^{\prime }R.$ Similarly as before, it is seen that $c_1\ne 0$ and $u^{\prime }$ is a non-zero constant. Hence, $Q=u^{\prime }R.$ If we use the fact that $c_{1}Q$ and $Q^{\prime }$ are constant, $c_2{Q}^{\prime }=0$ 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 $\mathbf{C}$ is constant and thus zero on M, a contradiction. Thus, we obtain $u^{\prime }=-Q^{\prime }$ on $V_1$. Equation (43) with $u^{\prime }=-Q^{\prime }$ gives that:

$$c_{3}Q=-2{\epsilon }_3{c}_1{Q}^{\prime }-c_{2}Q.$$

(54)

Putting (54) together with $u^{\prime }=-Q^{\prime }$ into (40), we have:

$$c_1{Q}^{″}={\epsilon }_3{c}_1{Q}^{\prime }R+c_{2}QR-2{\epsilon }_3{c}_2{Q}^{\prime }.$$

(55)

Furthermore, Equations (39), (41), (54) and (55) give:

$$({\epsilon }_3{c}_1{Q}^{\prime }+c_{2}Q)(Q^{\prime }-2{\epsilon }_{3}QR)=0$$

on $V_1$. Suppose ${\epsilon }_3{c}_1{Q}^{\prime }+c_{2}Q\ne 0$. Then, $Q^{\prime }=2{\epsilon }_{3}QR$, and thus, $Q^{″}=2{\epsilon }_3{Q}^{\prime }R+2{\epsilon }_{3}Q{R}^{\prime }$. Putting it into (55) with the help of (39), we get:

$$R({\epsilon }_3{c}_1{Q}^{\prime }+c_{2}Q)=0,$$

from which ${\epsilon }_3{c}_1{Q}^{\prime }+c_{2}Q=0$, a contradiction. Therefore, we get:

$${\epsilon }_3{c}_1{Q}^{\prime }+c_{2}Q=0$$

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 $\mathbf{C}$ 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 $Q=u^{\prime }R$, 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 ${\mathbb{E}}_1^3$ with the generalized 1-type Gauss map. If β, ${\beta }^{\prime }$, and ${\beta }^{″}$ 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 ${\mathbb{E}}_1^3$ 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:

$$x(s,t)=\alpha \left(s\right)+t\beta \left(s\right)$$

satisfying $⟨{\alpha }^{\prime },\beta ⟩=0,⟨{\alpha }^{\prime },{\alpha }^{\prime }⟩={\epsilon }_1(=\pm 1),⟨\beta ,\beta ⟩=1$, and $⟨{\beta }^{\prime },{\beta }^{\prime }⟩=0$ with ${\beta }^{\prime }\ne 0$.

Again, we put the smooth functions q and u as follows:

$$q=\parallel x_s{\parallel }^2=\left|⟨x_s,x_s⟩\right|,u=⟨{\alpha }^{\prime },{\beta }^{\prime }⟩.$$

We see that the null vector fields ${\beta }^{\prime }$ and $\beta \times {\beta }^{\prime }$ are orthogonal, and they are parallel. It is easily derived as ${\beta }^{\prime }=\beta \times {\beta }^{\prime }$. Moreover, we may assume that $\beta \left(0\right)=(0,0,1)$ and $\beta$ can be taken by:

$$\beta \left(s\right)=(as,as,1)$$

for a non-zero constant $a.$ Then, $\{{\alpha }^{\prime },\beta ,{\alpha }^{\prime }\times \beta \}$ forms an orthonormal frame along the base curve $\alpha$. With respect to this frame, we can put:

$${\beta }^{\prime }={\epsilon }_{1}u({\alpha }^{\prime }-{\alpha }^{\prime }\times \beta )\mathrm{and}{\alpha }^{″}=-u\beta +\frac{u^{\prime }}{u}{\alpha }^{\prime }\times \beta .$$

(56)

Note that the function u is non-vanishing.

On the other hand, we can compute the Gauss map G of M such as:

$$G=q^{-1/2}({\alpha }^{\prime }\times \beta -t{\beta }^{\prime }).$$

(57)

We also easily get the mean curvature H and the Gaussian curvature K of M by the usual procedure, respectively,

$$H=\frac{1}{2}{q}^{-3/2}\left(u^{\prime }t-{\epsilon }_1\frac{u^{\prime }}{u}\right)\mathrm{and}K=q^{-2}{u}^2.$$

(58)

Upon our using (19), the Laplacian of the Gauss map G of M is expressed as:

$$\Delta G=q^{-7/2}\left(A_2{\alpha }^{\prime }+B_2\beta +D_2{\alpha }^{\prime }\times \beta \right)$$

(59)

with respect to the orthonormal frame $\{{\alpha }^{\prime },\beta ,{\alpha }^{\prime }\times \beta \},$ where we put:

$$\begin{array}{cc}\hfill A_2=& 3{\epsilon }_1\frac{{\left(u^{\prime }\right)}^2}{u}t+{\epsilon }_4{\epsilon }_{1}q\left(-\frac{u^{″}}{u}+\frac{{\left(u^{\prime }\right)}^2}{u^2}+u{u}^{″}{t}^2+{\epsilon }_1\frac{{\left(u^{\prime }\right)}^2}{u}t\right)+q\frac{{\left(u^{\prime }\right)}^2}{u}t-3{\epsilon }_{1}u{\left(u^{\prime }\right)}^2{t}^3\hfill \\ & +{\epsilon }_4{\epsilon }_{1}u{\left(u^{\prime }\right)}^2{t}^3+2{\epsilon }_4{\epsilon }_{1}q{u}^{3}t,\hfill \\ \hfill B_2=& {\epsilon }_{4}q{u}^{\prime }(4{\epsilon }_1-ut),\hfill \\ \hfill D_2=& 3{\epsilon }_{1}u{\left(u^{\prime }\right)}^2{t}^3-3{\left(u^{\prime }\right)}^2{t}^2-{\epsilon }_{4}q\left({\epsilon }_{1}u{u}^{″}{t}^2-u^{″}t+\frac{{\left(u^{\prime }\right)}^2}{u}t\right)-{\epsilon }_{1}q\frac{{\left(u^{\prime }\right)}^2}{u^2}-q\frac{{\left(u^{\prime }\right)}^2}{u}t\hfill \\ & -{\epsilon }_4{\left(u^{\prime }\right)}^2{t}^2-2{\epsilon }_{4}q{u}^2-{\epsilon }_4{\epsilon }_{1}u{\left(u^{\prime }\right)}^2{t}^3-2{\epsilon }_4{\epsilon }_{1}q{u}^{3}t.\hfill \end{array}$$

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:

$$q^{-7/2}\left(A_2{\alpha }^{\prime }+B_2\beta +D_2{\alpha }^{\prime }\times \beta \right)=f{q}^{-1/2}\{(1+{\epsilon }_{1}ut){\alpha }^{\prime }\times \beta -{\epsilon }_{1}ut{\alpha }^{\prime }\}+g\mathbf{C}.$$

(60)

If the constant vector $\mathbf{C}$ 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 $\mathbf{C}$ is a non-zero constant vector. Taking the indefinite scalar product to Equation (60) with the orthonormal vector fields ${\alpha }^{\prime },\beta$, and ${\alpha }^{\prime }\times \beta ,$ respectively, we obtain:

$${\epsilon }_1{q}^{-7/2}{A}_2=-f{q}^{-1/2}ut+g⟨\mathbf{C},{\alpha }^{\prime }⟩,$$

(61)

$$q^{-7/2}{B}_2=g⟨\mathbf{C},\beta ⟩,$$

(62)

$${\epsilon }_1{q}^{-7/2}{D}_2=f{q}^{-1/2}({\epsilon }_1+ut)-g⟨\mathbf{C},{\alpha }^{\prime }\times \beta ⟩.$$

(63)

In terms of the orthonormal frame $\{{\alpha }^{\prime },\beta ,{\alpha }^{\prime }\times \beta \}$, the constant vector $\mathbf{C}$ can be written as:

$$\mathbf{C}=c_1{\alpha }^{\prime }+c_2\beta +c_3{\alpha }^{\prime }\times \beta ,$$

where we have put $c_1={\epsilon }_1⟨\mathbf{C},{\alpha }^{\prime }⟩$, $c_2=⟨\mathbf{C},\beta ⟩$, and $c_3=-{\epsilon }_1⟨\mathbf{C},{\alpha }^{\prime }\times \beta ⟩.$ Then, Equations (61)–(63) are expressed as follows:

$${\epsilon }_1{q}^{-7/2}{A}_2=-f{q}^{-1/2}ut+{\epsilon }_{1}g{c}_1,$$

(64)

$$q^{-7/2}{B}_2=g{c}_2,$$

(65)

$${\epsilon }_1{q}^{-7/2}{D}_2=f{q}^{-1/2}({\epsilon }_1+ut)+{\epsilon }_{1}g{c}_3.$$

(66)

Differentiating the functions $c_1,c_2$, and $c_3$ with respect to the parameter s, we get:

$$\begin{array}{c}{c}_1^{\prime }=-{\epsilon }_{1}u{c}_2-\frac{u^{\prime }}{u}{c}_3,\hfill \\ c_2^{\prime }=u{c}_1+u{c}_3,\hfill \\ c_3^{\prime }=-\frac{u^{\prime }}{u}{c}_1+{\epsilon }_{1}u{c}_2.\hfill \end{array}$$

(67)

Combining Equations (64)–(66), we obtain:

$$c_2({\epsilon }_1+ut)A_2-\{{\epsilon }_1{c}_1+(c_1+c_3)ut\}{B}_2+c_{2}ut{D}_2=0.$$

(68)

As before, from (68), we obtain the following:

$$c_2(2u{u}^{″}-3{(u^{\prime })}^2)+(c_1+c_3)u^2{u}^{\prime }=0,$$

(69)

$$7c_2{\left(u^{\prime }\right)}^2-5c_1{u}^2{u}^{\prime }-7c_3{u}^2{u}^{\prime }=0,$$

(70)

$$c_2(7{(u^{\prime })}^2-3u{u}^{″})-11c_1{u}^2{u}^{\prime }-4c_3{u}^2{u}^{\prime }=0,$$

(71)

$$c_2(u{u}^{″}-{(u^{\prime })}^2)+4c_1{u}^2{u}^{\prime }=0.$$

(72)

Combining Equations (69) and (71), we get:

$$5c_2(u{u}^{″}-{\left(u^{\prime }\right)}^2)-7c_1{u}^2{u}^{\prime }=0.$$

(73)

From (72) and (73), we get $c_1{u}^{\prime }=0.$ Hence, Equations (70) and (72) become:

$$u^{\prime }(c_2{u}^{\prime }-c_3{u}^2)=0,$$

(74)

$$c_2(u{u}^{″}-{(u^{\prime })}^2)=0.$$

(75)

Now, suppose that $u^{\prime }\left(s_0\right)\ne 0$ at some point $s_0\in \mathrm{dom}\left(\alpha \right)$. Then, there exists an open interval J such that $u^{\prime }\ne 0$ on J. Then, $c_1=0$ on $J.$ Hence, (67) reduces to:

$$\begin{array}{c}{\epsilon }_1{u}^2{c}_2+u^{\prime }{c}_3=0,\hfill \\ c_2^{\prime }=u{c}_3,\hfill \\ c_3^{\prime }={\epsilon }_{1}u{c}_2.\hfill \end{array}$$

(76)

From the above relationships, we see that $c_2^{\prime }$ is constant on J. In this case, if $c_2=0,$ then $c_3=0.$ Hence, $\mathbf{C}$ is zero on J. Thus, the constant vector $\mathbf{C}$ is zero on M. This contradicts our assumption. Therefore, $c_2$ is non-zero. Solving the differential Equation (74) with the help of $c_2^{\prime }=u{c}_3$ in (76), we get $u=k{c}_2$ for some non-zero constant $k.$ Moreover, since $c_2^{\prime }$ is constant, $u^{″}=0$. Thus, Equation (75) implies that $u^{\prime }=0,$ which is a contradiction. Therefore, there does not exist such a point $s_0\in \mathrm{dom}\left(\alpha \right)$ such that $u^{\prime }\left(s_0\right)\ne 0.$ 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 $\mathbf{C}=0$. 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 ${\mathbb{E}}_1^3$ 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 ${\mathbb{E}}_1^3$ with the proper generalized 1-type Gauss map G.

5. Null Scrolls in the Minkowski 3-Space ${\mathbb{E}}_1^3$

In this section, we examine the null scrolls with the generalized 1-type Gauss map in the Minkowski 3-space ${\mathbb{E}}_1^3.$ In particular, we focus on proving the following theorem.

Theorem 6.

Let M be a null scroll in the Minkowski 3-space ${\mathbb{E}}_1^3$. 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 $\alpha =\alpha \left(s\right)$ be a null curve in ${\mathbb{E}}_1^3$ and $\beta =\beta \left(s\right)$ a null vector field along $\alpha$ such that $⟨{\alpha }^{\prime },\beta ⟩=1.$ Then, the null scroll M is parameterized by:

$$x(s,t)=\alpha \left(s\right)+t\beta \left(s\right)$$

and we have the natural coordinate frame $\{x_s,x_t\}$ given by:

$$x_s={\alpha }^{\prime }+t{\beta }^{\prime }\mathrm{and}{x}_t=\beta .$$

We put the smooth functions $u,v,Q$, and R by:

$$u=⟨{\alpha }^{\prime },{\beta }^{\prime }⟩,v=⟨{\beta }^{\prime },{\beta }^{\prime }⟩,Q=⟨{\alpha }^{\prime },{\beta }^{\prime }\times \beta ⟩,R=⟨{\alpha }^{\prime },{\beta }^{″}\times \beta ⟩.$$

(77)

Then, $\{{\alpha }^{\prime },\beta ,{\alpha }^{\prime }\times \beta \}$ is a pseudo-orthonormal frame along $\alpha$.

Straightforward computation gives the Gauss map G of M and the Laplacian $\Delta G$ of G by:

$$G={\alpha }^{\prime }\times \beta +t{\beta }^{\prime }\times \beta \mathrm{and}\Delta G=-2{\beta }^{″}\times \beta +2(u+tv){\beta }^{\prime }\times \beta .$$

With respect to the pseudo-orthonormal frame $\{{\alpha }^{\prime },\beta ,{\alpha }^{\prime }\times \beta \}$, the vector fields ${\beta }^{\prime }$, ${\beta }^{\prime }\times \beta$, and ${\beta }^{″}\times \beta$ are represented as:

$${\beta }^{\prime }=u\beta -Q{\alpha }^{\prime }\times \beta ,{\beta }^{\prime }\times \beta =Q\beta \mathrm{and}{\beta }^{″}\times \beta =R\beta -v{\alpha }^{\prime }\times \beta .$$

(78)

Thus, the Gauss map G and its Laplacian $\Delta G$ are expressed by:

$$G={\alpha }^{\prime }\times \beta +tQ\beta \mathrm{and}\Delta G=-2(R-uQ-tvQ)\beta +2v{\alpha }^{\prime }\times \beta .$$

(79)

Since M has the generalized 1-type Gauss map, the Gauss map G satisfies:

$$\Delta G=fG+g\mathbf{C}$$

(80)

for some non-zero smooth functions $f,g$ and a constant vector $\mathbf{C}$. From (79), we get:

$$-2(R-uQ-tvQ)\beta +2v{\alpha }^{\prime }\times \beta =f({\alpha }^{\prime }\times \beta +tQ\beta )+g\mathbf{C}.$$

(81)

If the constant vector $\mathbf{C}$ 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 $\mathbf{C}$ is non-zero. If we take the indefinite inner product to Equation (81) with ${\alpha }^{\prime },\beta$, and ${\alpha }^{\prime }\times \beta ,$ respectively, we get:

$$-2(R-uQ-tvQ)=ftQ+g{c}_2,g{c}_1=0,2v=f+g{c}_3,$$

(82)

where we have put

$$c_1=⟨\mathbf{C},\beta ⟩,c_2=⟨\mathbf{C},{\alpha }^{\prime }⟩\mathrm{and}{c}_3=⟨\mathbf{C},{\alpha }^{\prime }\times \beta ⟩.$$

Since $g\ne 0$, Equation (82) gives $⟨\mathbf{C},{\beta }^{\prime }⟩=0.$ Together with (78), we see that $c_{3}Q=0.$ Suppose that $Q\left(s\right)\ne 0$ on an open interval $\tilde{I}\subset \mathrm{dom}\left(\alpha \right)$. Then, $c_3=0$ on $\tilde{I}$. Therefore, the constant vector $\mathbf{C}$ can be written as $\mathbf{C}=c_2\beta$ on $\tilde{I}$. If we differentiate $\mathbf{C}=c_2\beta$ with respect to s, $c_2^{\prime }\beta +c_2{\beta }^{\prime }=0$, 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 $\tilde{I}$, and so, $c_2=0.$ It contradicts that $\mathbf{C}$ is a non-zero vector. In the sequel, Q vanishes identically. Then, ${\beta }^{\prime }=u\beta$, which implies $R=0$. Thus, the Gauss map G is reduced to $G={\alpha }^{\prime }\times \beta$, which depends only on the parameter s, from which the shape operator S of M is easily derived as:

$$S=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\mathrm{or}S=\left(\begin{array}{cc}0& 0\\ k\left(s\right)& 0\end{array}\right)$$

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 $A={\alpha }^{\prime },B=\beta ,C=G$ satisfying $C^{\prime }=-k\left(s\right)B$. Thus, null scrolls in ${\mathbb{E}}_1^3$ with the generalized 1-type Gauss map satisfying (80) are part of Minkowski planes or B-scrolls whether $\mathbf{C}$ is zero or not.

The converse is obvious. This completes the proof. □

Corollary 1.

There do not exist null scrolls in ${\mathbb{E}}_1^3$ with the proper generalized 1-type Gauss map.

Open problem: Classify ruled submanifolds with the generalized 1-type Gauss map in Minkowski space.