Extension of Eigenvalue Problems on Gauss Map of Ruled Surfaces

Abstract

A finite-type immersion or smooth map is a nice tool to classify submanifolds of Euclidean space, which comes from the eigenvalue problem of immersion. The notion of generalized 1-type is a natural generalization of 1-type in the usual sense and pointwise 1-type. We classify ruled surfaces with a generalized 1-type Gauss map as part of a plane, a circular cylinder, a cylinder over a base curve of an infinite type, a helicoid, a right cone and a conical surface of G-type.

1. Introduction

Nash’s embedding theorem enables us to study Riemannian manifolds extensively by regarding a Riemannian manifold as a submanifold of Euclidean space with sufficiently high codimension. By means of such a setting, we can have rich geometric information from the intrinsic and extrinsic properties of submanifolds of Euclidean space. Inspired by the degree of algebraic varieties, B.-Y. Chen introduced the notion of order and type of submanifolds of Euclidean space. Furthermore, he developed the theory of finite-type submanifolds and estimated the total mean curvature of compact submanifolds of Euclidean space in the late 1970s ([1]).

In particular, the notion of finite-type immersion is a direct generalization of the eigenvalue problem relative to the immersion of a Riemannian manifold into a Euclidean space: Let $x:M\to {\mathbb{E}}^m$ be an isometric immersion of a submanifold M into the Euclidean m-space ${\mathbb{E}}^m$ and $\Delta$ the Laplace operator of M in ${\mathbb{E}}^m$. The submanifold M is said to be of finite-type if x has a spectral decomposition by $x=x_0+x_1+...+x_k$, where $x_0$ is a constant vector and $x_i$ are the vector fields satisfying $\Delta x_i={\lambda }_i{x}_i$ for some ${\lambda }_i\in \mathbb{R}$$(i=1,2,...,k)$. If the eigenvalues ${\lambda }_1,{\lambda }_2,...,{\lambda }_k$ are different, it is called k-type. Since this notion was introduced, many works have been made in this area (see [1,2]). This notion of finite-type immersion was naturally extended to that of pseudo-Riemannian manifolds in pseudo-Euclidean space and it was also applied to smooth maps, particularly the Gauss map defined on submanifolds of Euclidean space or pseudo-Euclidean space ([1,3,4,5,6]).

Regarding the Gauss map of finite-type, B.-Y. Chen and P. Piccini ([7]) studied compact surfaces with 1-type Gauss map, that is, $\Delta G=\lambda (G+\mathbb{C})$, where $\mathbb{C}$ is a constant vector and $\lambda \in \mathbb{R}$. Since then, many works regarding finite-type Gauss map have been established ([1,3,4,8,9,10,11,12,13,14,15]).

However, some surfaces have an interesting property concerning the Gauss map: The helicoid in ${\mathbb{E}}^3$ parameterized by

$$x(u,v)=(ucosv,usinv,av),a\ne 0$$

has the Gauss map and its Laplacian respectively given by

$$G=\frac{1}{\sqrt{a^2+u^2}}(asinv,-acosv,u)$$

and

$$\Delta G=\frac{2a^2}{{(a^2+u^2)}^2}G.$$

The right (or circular) cone $C_a$ with parametrization

$$x(u,v)=(ucosv,usinv,au),a\ge 0$$

has the Gauss map

$$G=\frac{1}{\sqrt{1+a^2}}(acosv,asinv,-1)$$

which satisfies

$$\Delta G=\frac{1}{u^2}(G+(0,0,\frac{1}{\sqrt{1+a^2}}))$$

(Reference [8,10]). The Gauss maps above are similar to be of 1-type, but not to be of the 1-type Gauss map in the usual sense. Based upon such cases, B.-Y. Chen and the present authors defined the notion of pointwise 1-type Gauss map ([8]).

Definition 1.

A submanifold M in ${\mathbb{E}}^m$ is said to have pointwise 1-type Gauss map if the Gauss map G of M satisfies

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

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

Let p be a point of ${\mathbb{E}}^3$ and $\beta =\beta \left(s\right)$ a unit speed curve such that p does not lie on $\beta$. A surface parameterized by

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

is called a conical surface. A typical conical surface is a right cone and a plane.

Let us consider a following example of a conical surface.

Example 1

([15]).Let M be a surface in ${\mathbb{E}}^3$ parameterized by

$$x(s,t)=(s{cos}^{2}t,ssintcost,ssint).$$

Then, the Gauss map G can be obtained by

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

After a considerably long computation, its Laplacian turns out to be

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

for some non-zero smooth functions $f,g$ and a constant vector $\mathbb{C}.$ The surface M is a kind of conical surfaces generated by a spherical curve $\beta \left(t\right)=({cos}^{2}t,sintcost,sint)$ on the unit sphere ${\mathbb{S}}^2\left(1\right)$ centered at the origin.

Inspired by such an example, we would like to generalize the notion of pointwise 1-type Gauss map as follows:

Definition 2

([15]).The Gauss map G of a submanifold M in ${\mathbb{E}}^m$ is of generalized 1-type if the Gauss map G of M satisfies

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

(1)

for some non-zero smooth functions $f,g$ and a constant vector $\mathbb{C}.$

Especially, we define a conical surface of G-type.

Definition 3.

A conical surface with generalized 1-type Gauss map is called a conical surface of G-type.

Remark 1

([15]).A conical surface of G-type is constructed by the functions f, g and the constant vector $\mathbb{C}$ by solving the differential equations generated by Equation (1).

In [15], the authors classified flat surfaces with a generalized 1-type Gauss map in ${\mathbb{E}}^3$. In fact, flat surfaces are ruled surfaces which are locally cones, cylinders or tangent developable surfaces. In the present paper, without such an assumption of flatness, we prove that non-cylindrical ruled surfaces with a generalized 1-type Gauss map are flat and thus we completely classify ruled surfaces with generalized 1-type Gauss map in ${\mathbb{E}}^3$.

2. Preliminaries

Let M be a surface of ${\mathbb{E}}^3$. The map $G:M\to {\mathbb{S}}^2\left(1\right)\subset {\mathbb{E}}^3$ which maps each point p of M to a point $G_p$ of ${\mathbb{S}}^2\left(1\right)$ by identifying the unit normal vector $N_p$ to M at the point with $G_p$ is called the Gauss map of the surface $M,$ where ${\mathbb{S}}^2\left(1\right)$ is the unit sphere in ${\mathbb{E}}^3$ centered at the origin.

For the matrix $\tilde{g}=\left({\tilde{g}}_{ij}\right)$ consisting of the components of the metric on M, we denote by ${\tilde{g}}^{-1}=\left({\tilde{g}}^{ij}\right)$ (resp. $\mathcal{G}$ ) the inverse matrix (resp. the determinant) of the matrix $\left({\tilde{g}}_{ij}\right).$ Then the Laplacian $\Delta$ on M is in turn given by

$$\Delta =-\frac{1}{\sqrt{\mathcal{G}}}\sum _{i,j}\frac{\partial }{\partial x^i}\left(\sqrt{\mathcal{G}}{\tilde{g}}^{ij}\frac{\partial }{\partial x^j}\right).$$

(2)

Let $\alpha =\alpha \left(s\right)$ be a regular curve in ${\mathbb{E}}^3$ defined on an open interval I and $\beta =\beta \left(s\right)$ a transversal vector field to ${\alpha }^{\prime }\left(s\right)$ along $\alpha .$ Then a ruled surface M can be parameterized by

$$x(s,t)=\alpha \left(s\right)+t\beta \left(s\right),s\in I,t\in \mathbb{R}$$

satisfying $⟨{\alpha }^{\prime },\beta ⟩=0$ and $⟨\beta ,\beta ⟩=1,$ where ${}^{\prime }$ denotes $d/ds$. The curve $\alpha$ is called the base curve and $\beta$ the director vector field or ruling. It is said to be cylindrical if $\beta$ is constant, or, non-cylindrical otherwise.

Throughout this paper, we assume that all the functions and vector fields are smooth and surfaces under consideration are connected unless otherwise stated.

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

In this section, we study the cylindrical ruled surfaces with the generalized 1-type Gauss map in ${\mathbb{E}}^3.$

Let M be a cylindrical ruled surface in ${\mathbb{E}}^3$. We can parameterize M with a plane curve $\alpha =\alpha \left(s\right)$ and a constant vector $\beta$ as

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

Here the plane curve $\alpha$ is assumed to be defined by $\alpha \left(s\right)=({\alpha }_1\left(s\right),{\alpha }_2\left(s\right),0)$ with the arc length s and $\beta$ a constant unit vector, namely $\beta =(0,0,1)$. In the sequel, the Gauss map G of M is given by

$$G={\alpha }^{\prime }\times \beta =({\alpha }_2^{\prime },-{\alpha }_1^{\prime },0)$$

(3)

and the Laplacian $\Delta G$ of the Gauss map G using Equation (2) is obtained by

$$\Delta G=(-{\alpha }_2^{{}^{‴}},{\alpha }_1^{{}^{‴}},0),$$

(4)

where ${}^{\prime }$ stands for $d/ds$.

From now on, ${}^{\prime }$ denotes the differentiation with respect to the parameter s relative to the base curve.

Suppose that the Gauss map G of M is of generalized 1-type, i.e., G satisfies Equation (1). We now consider two cases either $f=g$ or $f\ne g$.

Case 1. $f=g$.

In this case, the Gauss map G is of pointwise 1-type described in Definition 1. According to Classification Theorem in [10,11], we have that the ruled surface M is part of a plane, a circular cylinder or a cylinder over a base curve of an infinite-type satisfying

$${sin}^{-1}\left(\frac{c^2{f}^{-\frac{1}{3}}-1}{\sqrt{c_1^2+c_2^2}}\right)-\sqrt{c_1^2+c_2^2-{\left(c^2{f}^{-\frac{1}{3}}-1\right)}^2}=\pm c^3(s+k),$$

(5)

where $\mathbb{C}=(c_1,c_2,0)$, and $c(\ne 0)$ and k are constants.

Case 2. $f\ne g.$

By a direct computation using Equations (3) and (4), we see that the third component $c_3$ of the constant vector $\mathbb{C}$ is zero. We put $\mathbb{C}=(c_1,c_2,0)$. Then, we have the following system of ordinary differential equations

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

(6)

Since $\alpha$ is of unit speed, that is, ${\left({\alpha }_1^{\prime }\right)}^2+{\left({\alpha }_2^{\prime }\right)}^2=1$, we may put

$${\alpha }_1^{\prime }\left(s\right)=cos\theta \left(s\right)\mathrm{and}{\alpha }_2^{\prime }\left(s\right)=sin\theta \left(s\right)$$

for a smooth function $\theta =\theta \left(s\right)$ of s. One can write Equation (6) as

$$\begin{array}{c}{\left({\theta }^{\prime }\right)}^{2}sin\theta -{\theta }^{″}cos\theta =fsin\theta +g{c}_1,\hfill \\ {\left({\theta }^{\prime }\right)}^{2}cos\theta +{\theta }^{″}sin\theta =fcos\theta -g{c}_2,\hfill \end{array}$$

which give

$${\left({\theta }^{\prime }\right)}^2=f+g(c_{1}sin\theta -c_{2}cos\theta ),$$

(7)

$$-{\theta }^{″}=g(c_{1}cos\theta +c_{2}sin\theta ).$$

(8)

Taking the derivative of Equation (7), we have

$$2{\theta }^{\prime }{\theta }^{″}=f^{\prime }+g^{\prime }(c_{1}sin\theta -c_{2}cos\theta )+g(c_{1}cos\theta +c_{2}sin\theta ){\theta }^{\prime }.$$

With the help of Equations (7) and (8) it implies that

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

Solving the above differential equation, we get

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

If we put

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

(9)

where $p\left(s\right)=|k{g}^{\frac{2}{3}}\left(s\right)+\frac{2}{3}{g}^{\frac{2}{3}}\left(s\right)\int g^{-\frac{2}{3}}\left(s\right)f\left(s\right)(\frac{f^{\prime }}{f}-\frac{g^{\prime }}{g})ds|$ for some non-zero constant k, we get a base curve $\alpha$ of M as follows:

$$\alpha \left(s\right)=\left(\int cos\theta (s)ds,\int sin\theta (s)ds,0\right),$$

(10)

where $\theta \left(s\right)=\pm \int \sqrt{p\left(s\right)}ds.$ In fact, ${\theta }^{\prime }$ is the signed curvature of the curve $\alpha$ which is precisely determined by the given functions f, g and the constant vector $\mathbb{C}$.

Note that if f and g are constant, the Gauss map G is of 1-type in the usual sense. In this case, the signed curvature of $\alpha$ is non-zero constant and thus M is part of a circular cylinder.

Suppose that one of the functions f and g is not constant. Since a plane curve in ${\mathbb{E}}^3$ is of finite-type if and only if it is part of a straight line or a circle, the base curve $\alpha$ defined by Equation (10) is of an infinite-type ([2]). Thus, by putting together Cases 1 and 2, we have a classification theorem as follows:

Theorem 1.

Let M be a cylindrical ruled surface in ${\mathbb{E}}^3$ with the generalized 1-type Gauss map. Then it is an open part of a plane, a circular cylinder or a cylinder over a base curve of an infinite-type satisfying Equations (5), (9) and (10).

4. Classification Theorem

In this section, we examine non-cylindrical ruled surfaces with generalized 1-type Gauss map in ${\mathbb{E}}^3$ and obtain a classification theorem.

Let M be a non-cylindrical ruled surface in ${\mathbb{E}}^3$ parameterized by a base curve $\alpha$ and a director vector field $\beta$. Up to a rigid motion, its parametrization is given by

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

such that $⟨{\alpha }^{\prime },\beta ⟩=0,⟨\beta ,\beta ⟩=1$ and $⟨{\beta }^{\prime },{\beta }^{\prime }⟩=1$. Then, we have an orthonormal frame $\{\beta ,{\beta }^{\prime },\beta \times {\beta }^{\prime }\}$ along $\alpha$. With the frame $\{x_s,x_t\}$ , we define the smooth functions $q,u,Q$ and R as follows:

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

With such functions above, we can express the vector fields ${\alpha }^{\prime },{\beta }^{″},{\alpha }^{\prime }\times \beta ,$ $\beta \times {\beta }^{″}$ in the following:

$$\begin{array}{c}{\alpha }^{\prime }=u{\beta }^{\prime }+Q\beta \times {\beta }^{\prime },\hfill \\ {\beta }^{″}=-\beta +R\beta \times {\beta }^{\prime },\hfill \\ {\alpha }^{\prime }\times \beta =Q{\beta }^{\prime }-u\beta \times {\beta }^{\prime },\hfill \\ \beta \times {\beta }^{″}=-R{\beta }^{\prime },\hfill \end{array}$$

(11)

from which, the smooth function q and the Gauss map G are represented respectively as

$$q=t^2+2ut+u^2+Q^2$$

and

$$G=\frac{x_s\times x_t}{\left|\right|x_s\times x_t\left|\right|}=q^{-1/2}\left(Q{\beta }^{\prime }-(u+t)\beta \times {\beta }^{\prime }\right).$$

(12)

Then, by straightforward computation, the mean curvature H and the Gaussian curvature K of M are respectively represented as:

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

(13)

Remark 2.

If $R=0$, then the director vector field β is a plane curve.

By the Gauss and Weingarten formulas, the following equation is easily obtained:

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

where $\nabla H$ is the gradient of H and A denotes the shape operator of M. From Equation (13), we get

$$\begin{array}{cc}\hfill 2\nabla H& =2e_1\left(H\right)e_1+2e_2\left(H\right)e_2\hfill \\ & =q^{-3}{B}_1{e}_1+q^{-5/2}{A}_1{e}_2\hfill \\ & =q^{-7/2}\left(q{A}_1\beta +(u+t)B_1{\beta }^{\prime }+Q{B}_1\beta \times {\beta }^{\prime }\right),\hfill \end{array}$$

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

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

We also have

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

where

$$D_1={\{-R{t}^2-(2uR+Q^{\prime })t-u(uR+Q^{\prime })+Q(u^{\prime }-QR)\}}^2+2Q^2(t^2+2ut+u^2+Q^2).$$

Thus we obtain the Laplacian $\Delta G$ of the Gauss map G of M given by

$$\Delta G=q^{-7/2}[q{A}_1\beta +\left((u+t)B_1+D_{1}Q\right){\beta }^{\prime }+\left(Q{B}_1-D_1(u+t)\right)\beta \times {\beta }^{\prime }].$$

(14)

Suppose that M has generalized 1-type Gauss map G. Then, with the help of Equations (1), (12) and (14), we obtain

$$\begin{array}{c}{q}^{-7/2}[q{A}_1\beta +\{(u+t)B_1+D_{1}Q\}{\beta }^{\prime }+\{Q{B}_1-D_1(u+t)\}\beta \times {\beta }^{\prime }]\hfill \\ =f{q}^{-1/2}\{Q{\beta }^{\prime }-(u+t)\beta \times {\beta }^{\prime }\}+g\mathbb{C}.\hfill \end{array}$$

(15)

By taking the inner product to Equation (15) with $\beta ,{\beta }^{\prime }$ and $\beta \times {\beta }^{\prime }$ respectively, we get the following:

$$q^{-5/2}{A}_1=g⟨\mathbb{C},\beta ⟩,$$

(16)

$$q^{-7/2}\{(u+t)B_1+D_{1}Q\}=f{q}^{-1/2}Q+g⟨\mathbb{C},{\beta }^{\prime }⟩,$$

(17)

$$q^{-7/2}\{Q{B}_1-(u+t)D_1\}=-f{q}^{-1/2}(u+t)+g⟨\mathbb{C},\beta \times {\beta }^{\prime }⟩.$$

(18)

Combining Equations (16), (17) and (18), we have

$$q{A}_1{\omega }_2-\{(u+t)B_1+D_{1}Q\}{\omega }_1+f{q}^{3}Q{\omega }_1=0,$$

(19)

$$q{A}_1{\omega }_3-\{Q{B}_1-(u+t)D_1\}{\omega }_1-f{q}^3(u+t){\omega }_1=0,$$

(20)

$$\{(u+t)B_1+D_{1}Q\}{\omega }_3-\{Q{B}_1-(u+t)D_1\}{\omega }_2-f{q}^3\{Q{\omega }_3+(u+t){\omega }_2\}=0,$$

(21)

where we have put ${\omega }_1=⟨\mathbb{C},\beta ⟩$, ${\omega }_2=⟨\mathbb{C},{\beta }^{\prime }⟩$ and ${\omega }_3=⟨\mathbb{C},\beta \times {\beta }^{\prime }⟩.$

On the other hand, differentiating a constant vector $\mathbb{C}={\omega }_1\beta +{\omega }_2{\beta }^{\prime }+{\omega }_3\beta \times {\beta }^{\prime }$ with respect to the parameter s and using Equation (11), we get

$$\begin{array}{cc}\hfill {\omega }_1^{\prime }& -{\omega }_2=0,\hfill \\ \hfill {\omega }_3^{\prime }& +{\omega }_{2}R=0,\hfill \\ \hfill {\omega }_1& +{\omega }_2^{\prime }-{\omega }_{3}R=0.\hfill \end{array}$$

(22)

Combining Equations (19) and (20), we obtain

$$A_1\{{\omega }_2(u+t)+{\omega }_{3}Q\}-B_1{\omega }_1=0.$$

(23)

First of all, we consider the case of $R=0$.

Theorem 2.

Let M be a non-cylindrical ruled surface in ${\mathbb{E}}^3$ with generalized 1-type Gauss map. If $R=0,$ then M is part of a plane or a helicoid.

Proof. 

If the constant vector $\mathbb{C}$ is zero in the definition given by Equation (1), then the Gauss map G is of nothing but pointwise 1-type Gauss map of the first kind. By Characterization Theorem, M is part of a helicoid ([10]).

We now assume that the constant vector $\mathbb{C}$ is non-zero. In this case, we will show $Q=0$ on M and thus M is part of a plane due to Equation (13).

Suppose that the open subset $U=\{s\in \mathrm{dom}(\alpha \left)\right|Q\left(s\right)\ne 0\}$ of $\mathbb{R}$ is not empty. Then, on a component $U_C$ of U, we have from Equation (22) that ${\omega }_3$ is a constant and ${\omega }_1^{″}=-{\omega }_1$. Since the left hand side of Equation (23) is a polynomial in t with functions of s as the coefficients, the leading coefficient consisting of functions of s must vanish and ${\omega }_1^2{Q}^{\prime }$ is a constant on $U_C$ with the help of Equation (22).

Next, from the coefficient of $t^2$ in Equation (23), we obtain

$$3{\omega }_2{u}^{\prime }Q-2{\omega }_{3}Q{Q}^{\prime }+3{\omega }_1{u}^{\prime }{Q}^{\prime }+{\omega }_1{u}^{″}Q=0.$$

(24)

Similar to the above, from the coefficient of the linear term in t of Equation (23) with the help of Equation (24), we get

$${\omega }_{2}Q{Q}^{\prime }+{\omega }_3{u}^{\prime }Q-{\omega }_1{\left(u^{\prime }\right)}^2+{\omega }_1{\left(Q^{\prime }\right)}^2=0.$$

(25)

In addition, the constant term in Equation (23) relative to t is automatically zero. If we make use of Equation (24), we obtain

$$\begin{array}{cc}\hfill Q[{\omega }_1\{3u{\left(u^{\prime }\right)}^2+3u^{\prime }Q{Q}^{\prime }-3u{\left(Q^{\prime }\right)}^2& -u^{\prime \prime }{Q}^2\}-3{\omega }_{2}uQ{Q}^{\prime }\hfill \\ & -{\omega }_3(3u{u}^{\prime }Q+Q^2{Q}^{\prime })]=0.\hfill \end{array}$$

Hence, on $U_C$, we have

$$\begin{array}{cc}\hfill {\omega }_1\{3u{\left(u^{\prime }\right)}^2+3u^{\prime }Q{Q}^{\prime }-3u{\left(Q^{\prime }\right)}^2& -u^{\prime \prime }{Q}^2\}-3{\omega }_{2}uQ{Q}^{\prime }\hfill \\ & -{\omega }_3(3u{u}^{\prime }Q+Q^2{Q}^{\prime })=0.\hfill \end{array}$$

(26)

Using Equations (24) and (25), Equation (26) can be reduced to

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

(27)

Suppose that there is a point $s_0\in U_C$ such that $u^{\prime }\left(s_0\right)\ne 0$. Then, $u^{\prime }\left(s\right)\ne 0$ everywhere on an open interval I containing $s_0$. So, Equation (25) yields

$${\omega }_{3}Q=\frac{1}{u^{\prime }}\{{\omega }_1{\left(u^{\prime }\right)}^2-{\omega }_1{\left(Q^{\prime }\right)}^2-{\omega }_{2}Q{Q}^{\prime }\}.$$

(28)

Putting Equation (28) into (27), $({u^{\prime }}^2+{Q^{\prime }}^2)({\omega }_{2}Q+{\omega }_1{Q}^{\prime })=0,$ which implies ${\omega }_{2}Q+{\omega }_1{Q}^{\prime }=0.$ Since ${\omega }_2={\omega }_1^{\prime }$, we see that ${\omega }_{1}Q$ is constant on I.

If ${\omega }_1=0$ on some subinterval J in I, ${\omega }_2=0$ on J. Equation (25) gives ${\omega }_3=0$ on J. Since $\mathbb{C}$ is a constant vector, $\mathbb{C}$ is zero vector, which is a contradiction. Thus, without loss of generality we may assume that ${\omega }_1\ne 0$ everywhere on I and it is of the form ${\omega }_1=k_{1}cos(s+s_1)$ for some non-zero constant $k_1$ and $s_1\in \mathbb{R}$. Since ${\omega }_1^2{Q}^{\prime }$ is constant and ${\omega }_{1}Q$ is constant on I, ${\omega }_1$ must be zero on I, which contradicts ${\omega }_1=k_{1}cos(s+s_1)$ for some non-zero constant $k_1$. Therefore, the open interval I is empty and thus $u^{\prime }=0$ on $U_C$. If we take into account Equations (25) and (27), we get $Q^{\prime }({\omega }_{2}Q+{\omega }_1{Q}^{\prime })=0$ and ${\omega }_3{Q}^{\prime }=0,$ respectively.

Suppose that $Q^{\prime }\left(s_2\right)\ne 0$ at some point $s_2\in U_C$. Then ${\omega }_3=0$ and ${\omega }_{1}Q$ is a constant on an open interval $J_1$ containing $s_2.$ Similar to the above argument, since ${\omega }_1^2{Q}^{\prime }$ and ${\omega }_{1}Q$ are constant on $J_1$, it follows that ${\omega }_1=0.$ By Equation (22), ${\omega }_2$ is zero. Hence the constant vector $\mathbb{C}$ is zero, a contradiction. Thus $J_1$ is empty. Therefore, Q is constant on $U_C$. By continuity, Q is either a non-zero constant or zero on M. Because of Equation (13), M is minimal and it is an open part of a helicoid, which means that the Gauss map is of pointwise 1-type of the first kind. Therefore, the open subset U is empty. Consequently, Q is zero on M. Hence, M is an open part of a plane. ☐

Now, we assume that the function R is not vanishing everywhere.

If $f=g$, the Gauss map G of M is of pointwise 1-type. Thus, M is characterized as an open part of a right cone including the case that M is a plane or a helicoid depending upon whether the constant vector $\mathbb{C}$ is non-zero or zero ([9]).

From now on, we may assume the constant vector $\mathbb{C}$ is non-zero and $f\ne g$ unless otherwise stated. Similarly as before, Equation (23) yields

$${\omega }_{2}R+{\omega }_1{R}^{\prime }=0.$$

(29)

Since ${\omega }_1^{\prime }={\omega }_2$ in Equation (22), we see that ${\omega }_{1}R$ is constant. In addition, the coefficient of the term involving $t^3$ in Equation (23) must be zero.

With the help of Equation (29), we get

$$2{\omega }_2{Q}^{\prime }+{\omega }_{3}QR-{\omega }_1{u}^{\prime }R+{\omega }_1{Q}^{″}=0.$$

(30)

If we examine the coefficient of the term involving $t^2$ in Equation (23), using Equations (29) and (30) we obtain

$${\omega }_1{Q}^2{R}^{\prime }-3{\omega }_2{u}^{\prime }Q+2{\omega }_{3}Q{Q}^{\prime }-{\omega }_{1}Q{Q}^{\prime }R-3{\omega }_1{u}^{\prime }{Q}^{\prime }-{\omega }_1{u}^{″}Q=0.$$

(31)

Furthermore, from the coefficient of the linear term in t in Equation (23) with the help of Equations (29)–(31), we also get

$$Q\{{\omega }_{2}Q{Q}^{\prime }+{\omega }_3{u}^{\prime }Q-{\omega }_1{\left(u^{\prime }\right)}^2+{\omega }_1{\left(Q^{\prime }\right)}^2\}=0.$$

(32)

Suppose that the function Q is not zero, i.e., the open subset $V=\{s\in \mathrm{dom}(\alpha \left)\right|Q\left(s\right)\ne 0\}$ of dom$\left(\alpha \right)$ is not empty. Equation (32) gives that

$${\omega }_{2}Q{Q}^{\prime }+{\omega }_3{u}^{\prime }Q-{\omega }_1{\left(u^{\prime }\right)}^2+{\omega }_1{\left(Q^{\prime }\right)}^2=0.$$

(33)

Moreover, considering the constant term relative to t in Equation (23) and using Equations (29)–(31), we obtain

$$\begin{array}{cc}\hfill Q[{\omega }_1\{3u{\left(u^{\prime }\right)}^2+3u^{\prime }Q{Q}^{\prime }& -Q^2{Q}^{\prime }R-3u{\left(Q^{\prime }\right)}^2-u^{″}{Q}^2+Q^3{R}^{\prime }\}\hfill \\ & -3{\omega }_{2}uQ{Q}^{\prime }-{\omega }_3(3u{u}^{\prime }Q+Q^2{Q}^{\prime })]=0.\hfill \end{array}$$

Hence, on the open subset V in $\mathbb{R}$,

$$\begin{array}{cc}\hfill {\omega }_1\{3u{\left(u^{\prime }\right)}^2+3u^{\prime }Q{Q}^{\prime }& -Q^2{Q}^{\prime }R-3u{\left(Q^{\prime }\right)}^2-u^{″}{Q}^2+Q^3{R}^{\prime }\}\hfill \\ & -3{\omega }_{2}uQ{Q}^{\prime }-{\omega }_3(3u{u}^{\prime }Q+Q^2{Q}^{\prime })=0.\hfill \end{array}$$

(34)

Applying Equations (31) and (33) to Equation (34), we have

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

(35)

On the other hand, since ${\omega }_{3}R={\omega }_1+{{\omega }_2}^{\prime }$ in Equation (22), Equation (30) becomes

$${\left({\omega }_{1}Q\right)}^{″}+{\omega }_{1}Q-{\omega }_1{u}^{\prime }R=0.$$

(36)

Suppose that the function u is not constant, i.e., the open subset $V_1=\{s\in V|u^{\prime }\left(s\right)\ne 0\}$ is not empty. Then Equation (33) yields

$${\omega }_{3}Q=\frac{1}{u^{\prime }}\{{\omega }_1{\left(u^{\prime }\right)}^2-{\omega }_1{\left(Q^{\prime }\right)}^2-{\omega }_{2}Q{Q}^{\prime }\}.$$

(37)

Putting Equation (37) into (35), $({u^{\prime }}^2+{Q^{\prime }}^2)({\omega }_{2}Q+{\omega }_1{Q}^{\prime })=0$ and thus ${\omega }_{2}Q+{\omega }_1{Q}^{\prime }=0.$ Therefore, ${\omega }_{1}Q$ is constant on a component $\mathcal{C}$ of $V_1$. From Equation (36), we get ${\omega }_{1}Q={\omega }_1{u}^{\prime }R.$

If ${\omega }_1=0$ on an open interval $\tilde{I}$$\subset \mathcal{C}$, the constant vector $\mathbb{C}$ is zero on M, a contradiction. Thus, ${\omega }_1\ne 0$ and so $Q=u^{\prime }R$ on $\mathcal{C}.$ The fact that ${\omega }_{1}Q$ and ${\omega }_{1}R$ are constant on $\mathcal{C}$ implies that $u^{\prime }$ is a non-zero constant on $\mathcal{C}$. Then, Equations (31) and (35) are simplified as follows:

$${\omega }_1{Q}^2{R}^{\prime }+2{\omega }_{3}Q{Q}^{\prime }-{\omega }_{1}Q{Q}^{\prime }R=0,$$

(38)

$${\omega }_1{u}^{\prime }{Q}^{\prime }-{\omega }_{3}Q{Q}^{\prime }=0.$$

(39)

Putting $Q=u^{\prime }R$ into Equation (38), ${\omega }_3{Q}^{\prime }=0$ is derived. Thus, Equation (39) implies that ${\omega }_1{Q}^{\prime }=0$ and so $Q^{\prime }=0$ on $\mathcal{C}$. Hence, Q and R are both non-zero constants on $\mathcal{C}$.

On the other hand, without difficulty, we can show that the torsion of the director vector field $\beta =\beta \left(s\right)$ viewed as a curve is zero and so $\beta$ is part of a plane curve which is a small circle on the unit sphere centered at the origin with the normal curvature –1 and the geodesic curvature R on $\mathcal{C}$. Up to a rigid motion, we may put

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

on $\mathcal{C}$, where we have put $p=\sqrt{1+R^2}$. Then, $u=⟨{\alpha }^{\prime },{\beta }^{\prime }⟩=-{\alpha }_1^{\prime }sinps+{\alpha }_2^{\prime }cosps$, 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, on $\mathcal{C}$, we get

$$u^{\prime }=-({\alpha }_1^{″}+{\alpha }_2^{\prime }p)sinps+({\alpha }_2^{″}-{\alpha }_1^{\prime }p)cosps,$$

from which, we see that $u^{\prime }=0$ on $\mathcal{C}\subset V_1$, a contradiction. Hence, $V_1$ is empty and so $u^{\prime }=0$ on V. Then, Equations (30), (33) and (35) can be respectively reduced to

$$2{\omega }_2{Q}^{\prime }+{\omega }_{3}QR+{\omega }_1{Q}^{″}=0,$$

(40)

$${\omega }_{2}Q{Q}^{\prime }+{\omega }_1{\left(Q^{\prime }\right)}^2=0,$$

(41)

$${\omega }_{3}Q{Q}^{\prime }=0.$$

(42)

Suppose that $Q^{\prime }\left({\tilde{s}}_0\right)\ne 0$ at a point ${\tilde{s}}_0$ in V. From Equations (41) and (42), ${\omega }_3=0$ and ${\omega }_{1}Q$ is a constant on an open interval $\tilde{J}$$\subset V$ containing ${\tilde{s}}_0$. Hence, ${{\omega }_2}^{\prime }Q=0$ is derived from Equation (40). Therefore, ${{\omega }_2}^{\prime }=0$ on $\tilde{J}$. The third equation of (22) yields ${\omega }_1=0$. It follows that ${\omega }_2=0$. Since $\mathbb{C}$ is a constant vector, $\mathbb{C}$ is zero on $M,$ a contradiction. So, $Q^{\prime }=0$ on $V.$ Thus, Q is non-zero constant on each component of V. If we consider Equations (30) and (31), we have

$${\omega }_{3}R=0\mathrm{and}{\omega }_1{R}^{\prime }=0.$$

Since $R\ne 0,$${\omega }_3=0$ on each component of V. By Equation (29), ${\omega }_{2}R=0$, which yields that $\mathbb{C}$ is zero on $M.$ It is a contradiction. Hence, the open subset V of $\mathbb{R}$ is empty and the function Q is vanishing on $M.$ Thus, M is flat due to Equation (13). Since the ruled surface M is non-cylindrical, M is one of an open part of a tangent developable surface or a conical surface. One of the authors proved that tangential developable surfaces do not have a generalized 1-type Gauss map and a conical surface of G-type can be constructed by the given functions $f,g$ and the constant vector $\mathbb{C}$ ([15]).

Consequently, we have

Theorem 3.

Let M be a non-cylindrical ruled surface in ${\mathbb{E}}^3$ with generalized 1-type Gauss map. Then, M is an open part of a plane, a helicoid, a right cone or a conical surface of G-type.

Summing up our results, we obtain the following classification theorem.

Theorem 4.

(Classification) Let M be a ruled surface in ${\mathbb{E}}^3$ with a generalized 1-type Gauss map. Then, M is an open part of a plane, a circular cylinder, a cylinder over a base curve of an infinite-type satisfying Equations (5), (9) and (10), a helicoid, a right cone or a conical surface of G-type.