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Article

Extension of Eigenvalue Problems on Gauss Map of Ruled Surfaces

1
Department of Mathematics Education and RINS, Gyeongsang National University, Jinju 52828, Korea
2
Department of Mathematics, Kyungpook National University, Daegu 41566, Korea
*
Author to whom correspondence should be addressed.
Symmetry 2018, 10(10), 514; https://doi.org/10.3390/sym10100514
Received: 20 September 2018 / Revised: 9 October 2018 / Accepted: 12 October 2018 / Published: 16 October 2018

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 E m be an isometric immersion of a submanifold M into the Euclidean m-space E m and Δ the Laplace operator of M in 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 Δ x i = λ i x i for some λ i R ( i = 1 , 2 , . . . , k ) . If the eigenvalues λ 1 , λ 2 , . . . , λ 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, Δ G = λ ( G + C ) , where C is a constant vector and λ 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 E 3 parameterized by
x ( u , v ) = ( u cos v , u sin v , a v ) , a 0
has the Gauss map and its Laplacian respectively given by
G = 1 a 2 + u 2 ( a sin v , a cos v , u )
and
Δ G = 2 a 2 ( a 2 + u 2 ) 2 G .
The right (or circular) cone C a with parametrization
x ( u , v ) = ( u cos v , u sin v , a u ) , a 0
has the Gauss map
G = 1 1 + a 2 ( a cos v , a sin v , 1 )
which satisfies
Δ G = 1 u 2 ( G + ( 0 , 0 , 1 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 E m is said to have pointwise 1-type Gauss map if the Gauss map G of M satisfies
Δ G = f ( G + C )
for some non-zero smooth function f and a constant vector C . In particular, if 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 E 3 and β = β ( s ) a unit speed curve such that p does not lie on β . A surface parameterized by
x ( s , t ) = p + t β ( s )
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 E 3 parameterized by
x ( s , t ) = ( s cos 2 t , s sin t cos t , s sin t ) .
Then, the Gauss map G can be obtained by
G = 1 1 + cos 2 t ( sin 3 t , ( 2 cos 2 t ) cos t , cos 2 t ) .
After a considerably long computation, its Laplacian turns out to be
Δ G = f G + g C
for some non-zero smooth functions f , g and a constant vector C . The surface M is a kind of conical surfaces generated by a spherical curve β ( t ) = ( cos 2 t , sin t cos t , sin t ) on the unit sphere S 2 ( 1 ) 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 E m is of generalized 1-type if the Gauss map G of M satisfies
Δ G = f G + g C
for some non-zero smooth functions f , g and a constant vector 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 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 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 E 3 .

2. Preliminaries

Let M be a surface of E 3 . The map G : M S 2 ( 1 ) E 3 which maps each point p of M to a point G p of S 2 ( 1 ) 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 S 2 ( 1 ) is the unit sphere in E 3 centered at the origin.
For the matrix g ˜ = ( g ˜ i j ) consisting of the components of the metric on M, we denote by g ˜ 1 = ( g ˜ i j ) (resp. G ) the inverse matrix (resp. the determinant) of the matrix ( g ˜ i j ) . Then the Laplacian Δ on M is in turn given by
Δ = 1 G i , j x i G g ˜ i j x j .
Let α = α ( s ) be a regular curve in E 3 defined on an open interval I and β = β ( s ) a transversal vector field to α ( s ) along α . Then a ruled surface M can be parameterized by
x ( s , t ) = α ( s ) + t β ( s ) , s I , t R
satisfying α , β = 0 and β , β = 1 , where denotes d / d s . The curve α is called the base curve and β the director vector field or ruling. It is said to be cylindrical if β 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 E 3 with Generalized 1-Type Gauss Map

In this section, we study the cylindrical ruled surfaces with the generalized 1-type Gauss map in E 3 .
Let M be a cylindrical ruled surface in E 3 . We can parameterize M with a plane curve α = α ( s ) and a constant vector β as
x ( s , t ) = α ( s ) + t β .
Here the plane curve α is assumed to be defined by α ( s ) = ( α 1 ( s ) , α 2 ( s ) , 0 ) with the arc length s and β a constant unit vector, namely β = ( 0 , 0 , 1 ) . In the sequel, the Gauss map G of M is given by
G = α × β = ( α 2 , α 1 , 0 )
and the Laplacian Δ G of the Gauss map G using Equation (2) is obtained by
Δ G = ( α 2 , α 1 , 0 ) ,
where stands for d / d s .
From now on, 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 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 c 2 f 1 3 1 c 1 2 + c 2 2 c 1 2 + c 2 2 c 2 f 1 3 1 2 = ± c 3 ( s + k ) ,
where C = ( c 1 , c 2 , 0 ) , and c ( 0 ) and k are constants.
Case 2. f g .
By a direct computation using Equations (3) and (4), we see that the third component c 3 of the constant vector C is zero. We put C = ( c 1 , c 2 , 0 ) . Then, we have the following system of ordinary differential equations
α 2 = f α 2 + g c 1 , α 1 = f α 1 + g c 2 .
Since α is of unit speed, that is, ( α 1 ) 2 + ( α 2 ) 2 = 1 , we may put
α 1 ( s ) = cos θ ( s ) and α 2 ( s ) = sin θ ( s )
for a smooth function θ = θ ( s ) of s. One can write Equation (6) as
( θ ) 2 sin θ θ cos θ = f sin θ + g c 1 , ( θ ) 2 cos θ + θ sin θ = f cos θ g c 2 ,
which give
( θ ) 2 = f + g ( c 1 sin θ c 2 cos θ ) ,
θ = g ( c 1 cos θ + c 2 sin θ ) .
Taking the derivative of Equation (7), we have
2 θ θ = f + g ( c 1 sin θ c 2 cos θ ) + g ( c 1 cos θ + c 2 sin θ ) θ .
With the help of Equations (7) and (8) it implies that
3 2 ( θ 2 ) = f + g g ( ( θ ) 2 f ) .
Solving the above differential equation, we get
θ ( s ) 2 = k g 2 3 ( s ) + 2 3 g 2 3 ( s ) g 2 3 ( s ) f ( s ) ( f f g g ) d s , k ( 0 ) R .
If we put
θ ( s ) = ± p ( s ) ,
where p ( s ) = | k g 2 3 ( s ) + 2 3 g 2 3 ( s ) g 2 3 ( s ) f ( s ) ( f f g g ) d s | for some non-zero constant k, we get a base curve α of M as follows:
α ( s ) = cos θ ( s ) d s , sin θ ( s ) d s , 0 ,
where θ ( s ) = ± p ( s ) d s . In fact, θ is the signed curvature of the curve α which is precisely determined by the given functions f, g and the constant vector 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 α 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 E 3 is of finite-type if and only if it is part of a straight line or a circle, the base curve α 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 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 E 3 and obtain a classification theorem.
Let M be a non-cylindrical ruled surface in E 3 parameterized by a base curve α and a director vector field β . Up to a rigid motion, its parametrization is given by
x ( s , t ) = α ( s ) + t β ( s )
such that α , β = 0 , β , β = 1 and β , β = 1 . Then, we have an orthonormal frame { β , β , β × β } along α . 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 = α , β , Q = α , β × β , R = β , β × β .
With such functions above, we can express the vector fields α , β , α × β , β × β in the following:
α = u β + Q β × β , β = β + R β × β , α × β = Q β u β × β , β × β = R β ,
from which, the smooth function q and the Gauss map G are represented respectively as
q = t 2 + 2 u t + u 2 + Q 2
and
G = x s × x t | | x s × x t | | = q 1 / 2 Q β ( u + t ) β × β .
Then, by straightforward computation, the mean curvature H and the Gaussian curvature K of M are respectively represented as:
H = 1 2 q 3 / 2 ( R t 2 ( 2 u R + Q ) t + u Q Q 2 R u 2 R u Q ) , K = Q 2 q 2 .
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:
Δ G = 2 H + ( t r A 2 ) G ,
where H is the gradient of H and A denotes the shape operator of M. From Equation (13), we get
2 H = 2 e 1 ( H ) e 1 + 2 e 2 ( H ) e 2 = q 3 B 1 e 1 + q 5 / 2 A 1 e 2 = q 7 / 2 q A 1 β + ( u + t ) B 1 β + Q B 1 β × β ,
where e 1 = x s | | x s | | , e 2 = x t | | x t | | ,
A 1 = R t 3 + ( 3 u R + 2 Q ) t 2 + ( Q 2 R 3 u Q + 3 u 2 R + 4 u Q ) t + ( u Q 2 R 3 u u Q + u 3 R + 2 u 2 Q Q 2 Q ) , B 1 = 3 ( u t + u u + Q Q ) { R t 2 + ( 2 u R + Q ) t u Q + Q 2 R + u 2 R + u Q } + ( t 2 + 2 u t + u 2 + Q 2 ) { R t 2 ( 2 u R + 2 u R + Q ) t + u Q 2 Q Q R Q 2 R 2 u u R u 2 R u Q } .
We also have
tr A 2 = q 3 D 1 ,
where
D 1 = { R t 2 ( 2 u R + Q ) t u ( u R + Q ) + Q ( u Q R ) } 2 + 2 Q 2 ( t 2 + 2 u t + u 2 + Q 2 ) .
Thus we obtain the Laplacian Δ G of the Gauss map G of M given by
Δ G = q 7 / 2 [ q A 1 β + ( u + t ) B 1 + D 1 Q β + Q B 1 D 1 ( u + t ) β × β ] .
Suppose that M has generalized 1-type Gauss map G. Then, with the help of Equations (1), (12) and (14), we obtain
q 7 / 2 [ q A 1 β + { ( u + t ) B 1 + D 1 Q } β + { Q B 1 D 1 ( u + t ) } β × β ] = f q 1 / 2 { Q β ( u + t ) β × β } + g C .
By taking the inner product to Equation (15) with β , β and β × β respectively, we get the following:
q 5 / 2 A 1 = g C , β ,
q 7 / 2 { ( u + t ) B 1 + D 1 Q } = f q 1 / 2 Q + g C , β ,
q 7 / 2 { Q B 1 ( u + t ) D 1 } = f q 1 / 2 ( u + t ) + g C , β × β .
Combining Equations (16), (17) and (18), we have
q A 1 ω 2 { ( u + t ) B 1 + D 1 Q } ω 1 + f q 3 Q ω 1 = 0 ,
q A 1 ω 3 { Q B 1 ( u + t ) D 1 } ω 1 f q 3 ( u + t ) ω 1 = 0 ,
{ ( u + t ) B 1 + D 1 Q } ω 3 { Q B 1 ( u + t ) D 1 } ω 2 f q 3 { Q ω 3 + ( u + t ) ω 2 } = 0 ,
where we have put ω 1 = C , β , ω 2 = C , β and ω 3 = C , β × β .
On the other hand, differentiating a constant vector C = ω 1 β + ω 2 β + ω 3 β × β with respect to the parameter s and using Equation (11), we get
ω 1 ω 2 = 0 , ω 3 + ω 2 R = 0 , ω 1 + ω 2 ω 3 R = 0 .
Combining Equations (19) and (20), we obtain
A 1 { ω 2 ( u + t ) + ω 3 Q } B 1 ω 1 = 0 .
First of all, we consider the case of R = 0 .
Theorem 2.
Let M be a non-cylindrical ruled surface in 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 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 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 dom ( α ) | Q ( s ) 0 } of R is not empty. Then, on a component U C of U, we have from Equation (22) that ω 3 is a constant and ω 1 = ω 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 ω 1 2 Q 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 ω 2 u Q 2 ω 3 Q Q + 3 ω 1 u Q + ω 1 u Q = 0 .
Similar to the above, from the coefficient of the linear term in t of Equation (23) with the help of Equation (24), we get
ω 2 Q Q + ω 3 u Q ω 1 ( u ) 2 + ω 1 ( Q ) 2 = 0 .
In addition, the constant term in Equation (23) relative to t is automatically zero. If we make use of Equation (24), we obtain
Q [ ω 1 { 3 u ( u ) 2 + 3 u Q Q 3 u ( Q ) 2 u Q 2 } 3 ω 2 u Q Q ω 3 ( 3 u u Q + Q 2 Q ) ] = 0 .
Hence, on U C , we have
ω 1 { 3 u ( u ) 2 + 3 u Q Q 3 u ( Q ) 2 u Q 2 } 3 ω 2 u Q Q ω 3 ( 3 u u Q + Q 2 Q ) = 0 .
Using Equations (24) and (25), Equation (26) can be reduced to
2 ω 1 u Q + ω 2 u Q ω 3 Q Q = 0 .
Suppose that there is a point s 0 U C such that u ( s 0 ) 0 . Then, u ( s ) 0 everywhere on an open interval I containing s 0 . So, Equation (25) yields
ω 3 Q = 1 u { ω 1 ( u ) 2 ω 1 ( Q ) 2 ω 2 Q Q } .
Putting Equation (28) into (27), ( u 2 + Q 2 ) ( ω 2 Q + ω 1 Q ) = 0 , which implies ω 2 Q + ω 1 Q = 0 . Since ω 2 = ω 1 , we see that ω 1 Q is constant on I.
If ω 1 = 0 on some subinterval J in I, ω 2 = 0 on J. Equation (25) gives ω 3 = 0 on J. Since C is a constant vector, C is zero vector, which is a contradiction. Thus, without loss of generality we may assume that ω 1 0 everywhere on I and it is of the form ω 1 = k 1 cos ( s + s 1 ) for some non-zero constant k 1 and s 1 R . Since ω 1 2 Q is constant and ω 1 Q is constant on I, ω 1 must be zero on I, which contradicts ω 1 = k 1 cos ( s + s 1 ) for some non-zero constant k 1 . Therefore, the open interval I is empty and thus u = 0 on U C . If we take into account Equations (25) and (27), we get Q ( ω 2 Q + ω 1 Q ) = 0 and ω 3 Q = 0 , respectively.
Suppose that Q ( s 2 ) 0 at some point s 2 U C . Then ω 3 = 0 and ω 1 Q is a constant on an open interval J 1 containing s 2 . Similar to the above argument, since ω 1 2 Q and ω 1 Q are constant on J 1 , it follows that ω 1 = 0 . By Equation (22), ω 2 is zero. Hence the constant vector 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 C is non-zero or zero ([9]).
From now on, we may assume the constant vector C is non-zero and f g unless otherwise stated. Similarly as before, Equation (23) yields
ω 2 R + ω 1 R = 0 .
Since ω 1 = ω 2 in Equation (22), we see that ω 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 ω 2 Q + ω 3 Q R ω 1 u R + ω 1 Q = 0 .
If we examine the coefficient of the term involving t 2 in Equation (23), using Equations (29) and (30) we obtain
ω 1 Q 2 R 3 ω 2 u Q + 2 ω 3 Q Q ω 1 Q Q R 3 ω 1 u Q ω 1 u Q = 0 .
Furthermore, from the coefficient of the linear term in t in Equation (23) with the help of Equations (29)–(31), we also get
Q { ω 2 Q Q + ω 3 u Q ω 1 ( u ) 2 + ω 1 ( Q ) 2 } = 0 .
Suppose that the function Q is not zero, i.e., the open subset V = { s dom ( α ) | Q ( s ) 0 } of dom ( α ) is not empty. Equation (32) gives that
ω 2 Q Q + ω 3 u Q ω 1 ( u ) 2 + ω 1 ( Q ) 2 = 0 .
Moreover, considering the constant term relative to t in Equation (23) and using Equations (29)–(31), we obtain
Q [ ω 1 { 3 u ( u ) 2 + 3 u Q Q Q 2 Q R 3 u ( Q ) 2 u Q 2 + Q 3 R } 3 ω 2 u Q Q ω 3 ( 3 u u Q + Q 2 Q ) ] = 0 .
Hence, on the open subset V in R ,
ω 1 { 3 u ( u ) 2 + 3 u Q Q Q 2 Q R 3 u ( Q ) 2 u Q 2 + Q 3 R } 3 ω 2 u Q Q ω 3 ( 3 u u Q + Q 2 Q ) = 0 .
Applying Equations (31) and (33) to Equation (34), we have
2 ω 1 u Q + ω 2 u Q ω 3 Q Q = 0 .
On the other hand, since ω 3 R = ω 1 + ω 2 in Equation (22), Equation (30) becomes
( ω 1 Q ) + ω 1 Q ω 1 u R = 0 .
Suppose that the function u is not constant, i.e., the open subset V 1 = { s V | u ( s ) 0 } is not empty. Then Equation (33) yields
ω 3 Q = 1 u { ω 1 ( u ) 2 ω 1 ( Q ) 2 ω 2 Q Q } .
Putting Equation (37) into (35), ( u 2 + Q 2 ) ( ω 2 Q + ω 1 Q ) = 0 and thus ω 2 Q + ω 1 Q = 0 . Therefore, ω 1 Q is constant on a component C of V 1 . From Equation (36), we get ω 1 Q = ω 1 u R .
If ω 1 = 0 on an open interval I ˜ C , the constant vector C is zero on M, a contradiction. Thus, ω 1 0 and so Q = u R on C . The fact that ω 1 Q and ω 1 R are constant on C implies that u is a non-zero constant on C . Then, Equations (31) and (35) are simplified as follows:
ω 1 Q 2 R + 2 ω 3 Q Q ω 1 Q Q R = 0 ,
ω 1 u Q ω 3 Q Q = 0 .
Putting Q = u R into Equation (38), ω 3 Q = 0 is derived. Thus, Equation (39) implies that ω 1 Q = 0 and so Q = 0 on C . Hence, Q and R are both non-zero constants on C .
On the other hand, without difficulty, we can show that the torsion of the director vector field β = β ( s ) viewed as a curve is zero and so β 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 C . Up to a rigid motion, we may put
β ( s ) = 1 p ( cos p s , sin p s , R )
on C , where we have put p = 1 + R 2 . Then, u = α , β = α 1 sin p s + α 2 cos p s , where α ( s ) = ( α 1 ( s ) , α 2 ( s ) , α 3 ( s ) ) . Therefore, on C , we get
u = ( α 1 + α 2 p ) sin p s + ( α 2 α 1 p ) cos p s ,
from which, we see that u = 0 on C V 1 , a contradiction. Hence, V 1 is empty and so u = 0 on V. Then, Equations (30), (33) and (35) can be respectively reduced to
2 ω 2 Q + ω 3 Q R + ω 1 Q = 0 ,
ω 2 Q Q + ω 1 ( Q ) 2 = 0 ,
ω 3 Q Q = 0 .
Suppose that Q ( s ˜ 0 ) 0 at a point s ˜ 0 in V. From Equations (41) and (42), ω 3 = 0 and ω 1 Q is a constant on an open interval J ˜ V containing s ˜ 0 . Hence, ω 2 Q = 0 is derived from Equation (40). Therefore, ω 2 = 0 on J ˜ . The third equation of (22) yields ω 1 = 0 . It follows that ω 2 = 0 . Since C is a constant vector, C is zero on M , a contradiction. So, Q = 0 on V . Thus, Q is non-zero constant on each component of V. If we consider Equations (30) and (31), we have
ω 3 R = 0 and ω 1 R = 0 .
Since R 0 , ω 3 = 0 on each component of V. By Equation (29), ω 2 R = 0 , which yields that C is zero on M . It is a contradiction. Hence, the open subset V of 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 C ([15]).
Consequently, we have
Theorem 3.
Let M be a non-cylindrical ruled surface in 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 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.

Author Contributions

Y.H.K. gave the idea to establish the Classification Theorem of ruled surfaces with generalized 1-type Gauss map and M.C. computed the details. Y.H.K. checked and polished the draft.

Funding

This research was funded by the National Research Foundation of Korea (NRF) Grant funded by the Korea Government (MSIP) grant number 2016R1A2B1006974.

Acknowledgments

We would like to thank the referee for the careful review and the valuable comments to improve the paper.

Conflicts of Interest

The authors declare no conflict of interest.

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MDPI and ACS Style

Choi, M.; Kim, Y.H. Extension of Eigenvalue Problems on Gauss Map of Ruled Surfaces. Symmetry 2018, 10, 514. https://doi.org/10.3390/sym10100514

AMA Style

Choi M, Kim YH. Extension of Eigenvalue Problems on Gauss Map of Ruled Surfaces. Symmetry. 2018; 10(10):514. https://doi.org/10.3390/sym10100514

Chicago/Turabian Style

Choi, Miekyung, and Young Ho Kim. 2018. "Extension of Eigenvalue Problems on Gauss Map of Ruled Surfaces" Symmetry 10, no. 10: 514. https://doi.org/10.3390/sym10100514

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