EXTENSION OF EIGENVALUE PROBLEMS ON GAUSS MAP OF RULED SURFACES

A finite-type immersion or smooth map is a nice tool to classify submanifolds of Euclidean space, which comes from eigenvalue problem of immersion. The notion of generalized 1-type is a natural generalization of those of 1-type in the usual sense and pointwise 1-type. We classify ruled surfaces with 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.


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 finitetype submanifolds and estimated the total mean curvature of compact submanifolds of Euclidean space in the late 1970s ( [3]).
In particular, the notion of finite-type immersion is a direct generalization of 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 ∆ 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 ∆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 [3,5]).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 ( [1,2,3,10,11]).
Regarding the Gauss map of finite-type, B.-Y. Chen and P. Piccini ( [6]) initiated to study submanifolds of Euclidean space with finite-type Gauss map and classified compact surfaces with 1-type Gauss map, that is, ∆G = λ(G + C), where C is a constant vector and λ ∈ R. Since then, quite a few works on ruled surfaces and ruled submanifolds with finite-type Gauss map in Euclidean space or pseudo-Euclidean space have been established ( [1,2,3,4,7,8,9,12,13,14,15]).
However, some surfaces including a helicoid and a right cone in Euclidean 3-space have an interesting property concerning the Gauss map: The helicoid in E 3 parameterized by x(u, v) = (u cos v, u sin v, av), a = 0 has the Gauss map and its Laplacian respectively given by The right (or circular) cone C a with parametrization x(u, v) = (u cos v, u sin v, au), a ≥ 0 has the Gauss map [4,8]).The Gauss maps above are similar to 1-type, but it is not of 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 ( [4]).
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 pointewise 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 parametrized by 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 1.2.( [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 After a considerably long computation, its Laplacian turns out to be ∆G = f G + gC for some non-zero smooth functions f, g and a constant vector C. The surface M is a kind of conical 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: for some non-zero smooth functions f, g and a constant vector C.
Especially we define a conical surface with generalized 1-type Gauss map.
Definition 1.4.A conical surface with generalized 1-type Gauss map is called a conical surface of G-type.
Remark 1.5.( [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 (1.1).
In the present paper, we classify a ruled surface with generalized 1-type Gauss map in E 3 as a plane, a circular cylinder, a cylinder over a base curve of an infinite type generated by the given function f, g and the constant vector C, a helicoid, a right cone and a conical surface of G-type.

Preliminaries
Let M be a surface of the 3-dimensional Euclidean space 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 ij ) consisting of the components of the metric on M , we denote by g−1 = (g ij ) (resp.G ) the inverse matrix (resp.the determinant) of the matrix (g ij ).Then the Laplacian ∆ on M is in turn given by Now, we define a ruled surface M in the 3-dimensional Euclidean space E 3 .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 the ruled surface M can be parameterized by x(s, t) = α(s) + tβ(s), s ∈ I, t ∈ R satisfying α , β = 0 and β, β = 1, where denotes d/ds.The curve α is called the base curve and β the director vector field or ruling.In particular, M is said to be cylindrical if β is constant, or, non-cylindrical otherwise.Throughout this paper, we assume that all the geometric objects are smooth and all surfaces are connected unless otherwise stated.

Cylindrical ruled surfaces in E 3 with generalized 1-type Gauss map
In this section, we study the cylindrical ruled surfaces with generalized 1-type Gauss map in E 3 .
Let M be a cylindrical ruled surface in E 3 .Without loss of generality, we assume that M is parameterized by x(s, t) = α(s) + tβ, where α(s) = (α 1 (s), α 2 (s), 0) is a plane curve parameterized by the arc-length s and β a constant vector, namely β = (0, 0, 1).In this case, the Gauss map and the Laplacian ∆G of the Gauss map G using (2.1) is obtained by where stands for d/ds.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 (1.1).We now consider two cases for equation (1.1).
In this case, the Gauss map G is of pointwise 1-type described in Definition 1.1.According to Classification Theorem in [8] and [9], we have the ruled surface M is part of a plane, a circular cylinder or a cylinder over a base curve of an infinite-type satisfying where C = (c 1 , c 2 , 0), c( = 0) and k are constants.
Case 2. f = g.By a direct computation using (3.1) and (3.2), 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 Since α is a unit speed curve, that is, (α 1 ) 2 + (α 2 ) 2 = 1, we may put for a smooth function θ = θ(s) of s.It enables equation (3.4) to be rewritten in the form (θ ) (3.6) Taking the derivative of (3.5), we have With the help of (3.5) and (3.6) Solving the above differential equation, we get where p(s) = |kg for some non-zero constant k, we get a base curve α of M as follows: where θ(s) = ± p(s) ds.In fact, θ is the signed curvature of the base 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 the base curve α is non-zero constant.So, the cylindrical ruled surface 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 defined by (3.8) is of an infinite-type ( [5]).Thus, by putting together Cases 1 and 2, we have a classification theorem of cylindrical ruled surface with generalized 1-type Gauss map in E 3 .Theorem 3.1.Let M be a cylindrical ruled surface in E 3 .Suppose that M has 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 (3.3), (3.7) and (3.8).

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) + tβ(s) such that α , β = 0, β, β = 1 and β , β = 1.Then, we have the natural frame {x s , x t } given by x s = α (s) + tβ (s) and x t = β(s).
From this setting, we have an orthonormal frame {β, β , β × β }.For later use, we define the smooth functions q, u, Q and R as follows: In terms of the orthonormal frame {β, β , β × β }, we obtain from which, the smooth function q is given by Let H and K be the mean curvature and the Gaussian curvature of M respectively.By straightforward computation in using the first and second fundamental forms, they are given as follows: (4.3) Remark 4.1.If R ≡ 0, then the director vector field β is a plane curve.
The following formula is well known with respect to the Laplacian of the Gauss map of M in E 3 , which are easily obtained by applying the Gauss formula and the Weingarten formula: ∆G = 2gradH + (trA 2 )G, where A denotes the shape operator of the surface M .
From (4.3), we get where We also have trA 2 = q −3 D 1 , where Thus we obtain the Laplacian ∆G of the Gauss map G of M given by Suppose that M has generalized 1-type Gauss map G.Then, with the help of (1.1), (4.2) and (4.4), we obtain for some non-zero smooth functions f, g and a constant vector C.
If we take the inner product to equation (4.5) with β, β and β × β respectively, then we get the following: ) Combining equations (4.6), (4.7) and (4.8), we have ) where we have put 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 (4.1), we get Combining equations (4.9) and (4.10), we obtain First of all, we consider the case of R ≡ 0.
Theorem 4.2.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 (1.1), then the Gauss map G is nothing but of pointwise 1-type Gauss map of the first kind.By Characterization Theorem of a ruled surface with pointwise 1-type Gauss map of the first kind, M is part of a helicoid ( [8]).
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 (4.3).
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 (4.12) that ω 3 is a constant and ω 1 = −ω 1 .Observing equation (4.13), the left side is a polynomial in t with functions of s as the coefficients.Hence the leading coefficient must vanish and ω 2  1 Q is a constant on U C with the help of (4.12).
Next, by examining the coefficient of the term involving t 2 in (4.13), we obtain Similarly as above, from the coefficient of the linear term in t of (4.13) with the help of (4.14), we get Also, the constant term in (4.13) with respect to the parameter t is automatically zero.
If we make use of (4.14), we obtain Hence, on U C , we have 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, (4.15) yields Putting (4.18) into (4.17),(u If ω 1 ≡ 0 on some subinterval J in I, ω 2 = 0 on J. (4.15) 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 2 1 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 it into account of (4.15) and (4.17), we get 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 .Similarly as above, since ω 2 1 Q and ω 1 Q are constant on J 1 , it follows that ω 1 = 0.By (4.12), 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 (4.3), 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, without loss of generality we may assume that the function R is not vanishing everywhere.If f = g, the non-cylindrical ruled surface M has pointwise 1-type Gauss map which 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 ( [7]).
From now on, we may assume the constant vector C is non-zero and f = g unless otherwise stated.Similarly as before, the leading coefficient of the polynomial in the left side of equation (4.13) in t with functions of s as the coefficients is zero and we get

.19)
Since ω 1 = ω 2 in (4.12), we see that ω 1 R is constant.Also, the coefficient of the term involving t 3 in (4.13) must be zero.With the help of (4.19), we get If we examine the coefficient of the term involving t 2 in (4.13), using (4.19) and (4.20) we obtain Furthermore, from the coefficient of the linear term in t in (4.13) with the help of (4.19), (4.20) and (4.21), we also get Consider an open set V = {s ∈ dom(α)|Q(s) = 0}.Suppose that V is not empty.Equation (4.22) gives that

.25)
On the other hand, since ω 3 R = ω 1 + ω 2 in (4.12), (4.20) becomes If ω 1 ≡ 0 on an open interval Ĩ ⊂ 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, (4.21) and (4.25) are simplified as follows: (4.29) Putting Q = u R into (4.28),ω 3 Q = 0 is derived.Thus, (4.29) 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) viewing 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. Without loss of generality, we may put (4.32) Suppose that Q (s 0 ) = 0 at a point s0 in V. From (4.31) and (4.32), ω 3 = 0 and ω 1 Q is a constant on an open interval J ⊂ V containing s0 .Hence, ω 2 Q = 0 is derived from (4.30).Therefore, ω 2 = 0 on J.The third equation of (4.12) 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 (4.20) and (4.21), we have ω 3 R = 0 and ω 1 R = 0. Since R = 0, ω 3 = 0 on each component of V.By (4.19), 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 (4.3).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 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]).