Department of Mathematics, Kyungpook National University, Daegu 41566, Korea
Author to whom correspondence should be addressed.
Received: 12 May 2018 / Accepted: 11 June 2018 / Published: 13 June 2018
We study ruled submanifolds in Minkowski space in regard to the Gauss map satisfying some partial differential equation. As a generalization of usual cylinders, cones and null scrolls in a three-dimensional Minkowski space, a cylinder over a space curve, a product manifold of a right cone and a k-plane, a product manifold of a hyperbolic cone and a k-plane which look like kinds of cylinders over cones in 3-space, and the generalized B-scroll kind in Minkowski space are characterized with the partial differential equation regarding the Gauss map, where k is a positive integer.
finite-type immersion; pointwise 1-type Gauss map of the second kind; generalized B-scroll kind
According to Nash’s imbedding theorem, a Riemannian manifold can be imbedded in a Euclidean space with considerably high codimension. That naturally enables us to study Riemannian manifolds as submanifolds of a Euclidean space. In the late 1970’s, the notion of finite-type immersion of Riemannian manifolds into Euclidean space was introduced, which is a generalization of the so-called eigenvalue problem of the immersion : An isometric immersion of x of a Riemannian manifold M into a Euclidean space is said to be of finite-type if it can be expressed as
for some positive integer k, where is a constant vector and for some , . Here, denotes the Laplace operator defined on M. If , ⋯, are mutually different, M is said to be of k-type. We may assume that a finite-type immersion of x of a Riemannian manifold into a Euclidean space is of k-type for some non-negative integer k.
Let be an m-dimensional pseudo-Euclidean space of signature . The notion of finite-type immersion was extended to that of submanifolds in pseudo-Euclidean space and to that of smooth maps defined on submanifolds of Euclidean space or pseudo-Euclidean space . In particular, the study of finite-type immersions and finite-type Gauss map of submanifolds in the Minkowski m-space denoted by has been made extensively ([2,3,4,5,6,7,8,9,10,11,12,13,14,15]).
On the other hand, the Gauss map of some nice surfaces in the three-dimensional Euclidean space has an interesting property regarding the Laplacian. The helicoid in parameterized by
has the Gauss map G satisfying
The Gauss map of the right (or circular) cone in with parametrization
(cf. [16,17]). The Gauss map of those surfaces is similar to of 1-type, but obviously not of 1-type in the usual sense. We need to know what other manifolds have such a property. Based on these examples, the following definition was introduced.
(). An oriented n-dimensional submanifold M of the Euclidean space or the pseudo- Euclidean space is said to have pointwise 1-type Gauss map or the Gauss map is of pointwise 1-type if it satisfies
where f is a non-zero smooth function on M and a constant vector in the ambient space. In particular, if is zero, the Gauss map G is said to be of pointwise 1-type of the first kind. Otherwise, it is said to be of the second kind ([19,20,21,22,23,24]).
The notion of ruled submanifold is a concept of great interest in the Riemannian geometry, which has been investigated by many authors. Several results involving ruled submanifolds in manifolds equipped with remarkable geometric structures were recently obtained in [25,26,27,28].
In [19,20], the authors of the present paper et al. studied ruled submanifolds in the Euclidean space with pointwise 1-type Gauss map and proved that the ruled submanifold M in the Euclidean space is minimal if and only if the Gauss map G of M is of pointwise 1-type Gauss map of the first kind. Further, we showed that the only non-cylindrical ruled submanifold M in the Euclidean space with pointwise 1-type Gauss map of the second kind is the generalized right cone.
In , the authors of the present paper and et al. investigated the ruled submanifolds in the Lorentz-Minkowski m-space with pointwise 1-type Gauss map of the first kind and then established the equivalent conditions for the minimality of the ruled submanifold in the Lorentz-Minkowski m-space by means of the Gauss map.
In this paper, we will study ruled submanifolds in with pointwise 1-type Gauss map of the second kind and thereby complete the classification of the ruled submanifolds in with pointwise 1-type Gauss map.
A curve in is said to be space-like, time-like or null if its tangent vector field is space-like, time-like or null, respectively.
Let be an isometric immersion of an n-dimensional pseudo-Riemannian manifold M into . Throughout the present paper, a submanifold in always means pseudo-Riemannian, in other words, each tangent space of the submanifold in is non-degenerate.
Let be a local coordinate system of M in . For the components of the pseudo-Riemannian metric on M induced from that of , we denote by (respectively, ) the inverse matrix (respectively, the determinant) of the matrix of the components of the induced metric . Then, the Laplacian defined on M is given by
We now define the Gauss map G on M. Consider the map of a point p of M mapped to an oriented tangent space at p, where is the Grassmannian manifold consisting of all oriented n-planes passing through the origin. Roughly speaking it can be achieved by parallel displacement of the oriented tangent space at p to the origin of . By an isomorphism, can be identified with in a natural manner. Let us express the Gauss map rigorously. Choose an adapted local orthonormal frame in such that are tangent to M and normal to M. Define the map (), .
An indefinite scalar product on is defined by
Then, is an orthonormal basis of for some positive integer k.
Now, let us recall the notion of a ruled submanifold M in ([7,8,9,10]). A non-degenerate -dimensional submanifold M in is called a ruled submanifold if M is foliated by r-dimensional totally geodesic submanifolds of along a regular curve on M defined on an open interval I. Thus, a parametrization of a ruled submanifold M in can be given by
where ’s are some open intervals for . Without loss of generality, we may assume that for all . For each s, is open in Span, , ⋯, , which is the linear span of linearly independent vector fields , , ⋯, along the curve . Here, we assume that are either non-degenerate or degenerate for all s along . We call the rulings and the base curve of the ruled submanifold M. In particular, the ruled submanifold M is said to be cylindrical if are parallel along , or non-cylindrical otherwise.
([7,8]). (1) If the rulings of M are non-degenerate, then the base curve α can be chosen to be orthogonal to the rulings as follows: Let V be a unit vector field on M which is orthogonal to the rulings. Then α can be taken as an integral curve of V.
(2) If the rulings are degenerate, we can choose a null base curve which is transversal to the rulings: Let V be a null vector field on M which is not tangent to the rulings. An integral curve of V can be the base curve.
By solving a system of ordinary differential equations similarly set up relative to a frame along a curve in as given in , we have
(). Let be a smooth l-dimensional non-degenerate distribution in the Minkowski m-space along a curve , where and . Then, we can choose orthonormal vector fields along α which generate the orthogonal complement satisfying for .
3. Characterization of Cylinders over Spatial Base Curves
Let M be an -dimensional ruled submanifold in with non-degenerate rulings. Then, by Remark 1, the base curve can be chosen to be orthogonal to the rulings. Without loss of generality, we may assume that is a unit speed curve, that is, . From now on, the prime denotes unless otherwise stated. By Lemma 1, we may choose orthonormal vector fields along satisfying
A parametrization of M is given by
In this section, we always assume that the parametrization (3) satisfies condition (2). Then, the Gauss map G of M is given by
where and are the function and the vectors respectively, defined by
First, we consider the case of cylindrical ruled submanifolds that are one of two typical types of ruled submanifolds, which are cylindrical or non-cylindrical. Before discussing cylindrical ruled submanifolds, we cite the following lemma.
(). Suppose that a unit speed curve in the m-dimensional Minkowski space defined on an open interval I satisfies
where g is a function of the parameter s and a constant vector in . Then, the curve α lies in a 3-dimensional affine space in . In particular, if the constant vector is zero, we see that α is a plane curve.
We now prove that if an -dimensional cylindrical ruled submanifold M in has pointwise 1-type Gauss map of the second kind satisfying (1), then it is part of a -plane or a cylinder over a curve in 3-dimensional affine space.
Let M be a cylindrical -dimensional ruled submanifold in generated by non-degenerate rulings which is parameterized by (3). Without loss of generality, we may assume that generating the rulings are constant vectors.
The Laplacian of M is then naturally expressed by
where and the Gauss map G of M is given by
We now suppose that the Gauss map G is of pointwise 1-type of the second kind, that is, for some non-zero smooth function f and some non-zero constant vector . Then, the equation is written as
From Equation (6), we see that f is a function of s. We may assume that f is non-zero on the open interval . Then, differentiation of Equation (6) with respect to s gives
which implies that for some constant vector , where denotes zero vector. Namely, if we denote by the Laplacian of , we have
According to Lemma 2, we see that the curve lies in a 3-dimensional affine space in .
If a cylindrical ruled submanifold M is part of an -plane or a cylinder over a 3-dimensional affine space satisfying (8), it is obvious that the Gauss map G is of pointwise 1-type of the second kind. Thus, we have
Let M be an -dimensional cylindrical ruled submanifold of . Then, M has pointwise 1-type Gauss map of the second kind if and only if M is part of an -dimensional plane or a cylinder over a curve in a 3-dimensional affine space in satisfying (8).
Next, we consider the case that non-cylindrical ruled submanifolds have pointwise 1-type Gauss map of the second kind. Let M be an -dimensional non-cylindrical ruled submanifold parameterized by (3) in . Then, we have
for . The function q defined in the beginning of this section is given by
where and for . Note that q is a polynomial in with functions in s as coefficients.
From now on, for a polynomial in , deg denotes the degree of in unless otherwise stated.
If we adapt the proof of Proposition 3.3 of  to the case of a non-cylindrical ruled submanifold in the Minkowski m-space , we may assume that the generator vector fields of the rulings of M satisfy
on the domain I of for all if M has pointwise 1-type Gauss map of the second kind. Then, we get the components of the metric on M
It is enough for us to consider the case of . Accordingly, Equation (9) gives . By definition, we have the Laplacian of the form
First, we suppose that are non-null. Then, using the Formula (10), can be expressed as
If we use the indefinite scalar product on , we have
where we have put
Then, we get
By taking the indefinite scalar product with the vector to both sides of (11), we obtain
where we have put
Let be the orthonormal vector fields which are normal to M along If we apply Lemma 1 to the normal space of M, then there exists an orthonormal frame of the normal space satisfying
Let M be an -dimensional non-cylindrical and non-planar ruled submanifold parameterized by (3) in . Let , , ⋯, be the orthonormal generators of the rulings along the base curve α such that are non-null for all . If the Gauss map G of M satisfies for some non-zero function f and non-zero constant vector , then
Let We suppose that the interior Int of is non-empty.
If we put
Equation (13) tells us that on Int. Using on Int, equation yields that
as the coefficients of terms containing , and , respectively, for .
Now, we will proceed with the proof according to the following steps.
Step 1. on Int.
We suppose that at some point in Int. Then, in (21) and hence
by taking the indefinite scalar product with . Thus, we can see that
as the coefficients of terms containing and for , by virtue of . Equations (23), (30) and (31) indicate that is orthogonal to all vectors of (24), so the coefficient of has to be identically zero, which yields that for . It contradicts our assumption.
Therefore, we conclude that the functions are constant for all , that is,
Step 2. An expression for f on .
According to Step 1, Equation (24) is simplified as
We repeat taking the indefinite scalar product to to (32) and then we obtain
If , then and hence in (33) for all . In this case, and are given by (27) and
respectively, where we have put
Together with (27) and (34), (32) yields that the constant vector can be expressed as
and that the equations containing and are given by
respectively. Therefore, and are of the form
If , then in (37). By definitions of and , and hence for all and . Thus, we have and (36) is simplified as
for . Equation (38) implies that and for and . Thus, and hence . Also, under these conditions, by computation, we get which implies that by virtue of (27), (34) and (35). Therefore, the Gauss map G is a constant vector, a contradiction. Therefore, we see that . Then, it follows from (37) that
for some non-vanishing function h of s. Putting (39) into (36), we have
which allows us to have the following equation
for some non-vanishing function g of s. Comparing the coefficients of terms containing , and of (40) gives us three equations:
where is a polynomial in t such that deg with vector functions of s as coefficients. Considering the degrees of (57), we see that
for some constant c.
Suppose that there exist ∈ such that are not a multiple of for . By (58), we get
for some constant . By hypothesis, we can put
for some constants and polynomials in t with deg for . Then, has to be a multiple of because of (59), a contradiction. Thus, we have
by comparing the both sides of (60) for all . It contradicts that is not a polynomial. Therefore, we have
for all s and t.
Since on M, the rational function f defined by (54) becomes
Let M be an -dimensional non-cylindrical and non-planar ruled submanifold parameterized by (3) in . Let , , ⋯, be the orthonormal generators of the rulings along the base curve α such that are non-null for all . If M has pointwise 1-type Gauss map of the second kind, then we have
for all .
If M is Lorentzian, it is obvious. We now suppose that M is space-like. In this case,
In (80), we can see that the vector ∧ ∧⋯∧∧⋯∧ ∧⋯∧ of is orthogonal to and other vectors in (80) except the vectors having the same form. Note that
for and .
We multiply by (77) and then compare (78) and the equation obtained in such a way. Then, we can obtain
with the help of (17) and (79). By taking the wedge product with to (82) for some k, (80) and (82) induce the following
Again, taking the wedge product with to (83) for we get
for all and . If for some j and a, then putting (16) into (84) implies that
for . Here, we note that because of . Using (85), we can see that the value of (81) becomes zero, which means that the coefficients of in (82) must be identically zero by the orthogonality of vectors. Therefore, we have and hence on , for all j and a.
In (92), considering the constant terms with respect to t and the coefficients of terms containing , we see
for all . Since , are constant for all and . Together with Lemma 4, we have
Let M be an -dimensional non-cylindrical and non-planar ruled submanifold parameterized by (3) in with pointwise 1-type Gauss map of the second kind. Let , , ⋯, be the orthonormal generators of the rulings along the base curve α. If are non-null for all , then the functions
are constant functions on the open interval for all , where
Furthermore, for the case that are non-null for all j, we have
Let M be an -dimensional non-cylindrical and non-planar ruled submanifold parameterized by (3) in . We suppose that are non-null for all . If M has pointwise 1-type Gauss map of the second kind, we can choose an orthonormal frame of the normal space of M along α satisfying
Let M be an -dimensional non-cylindrical and non-planar ruled submanifold parameterized by (3) in . Let , , ⋯, be the orthonormal generators of the rulings along the base curve α such that are non-null for all . If M has pointwise 1-type Gauss map of the second kind, then the parametrization of M is given by
where is part of a circle or a hyperbola in with , , ⋯, orthonormal constant vectors satisfying = , a constant vector and for some open intervals and .
Suppose on the whole domain I of . In this case, we showed that M is part of an plane in . Clearly, a plane can be parameterized as (94) for some suitable constant vectors .
Now, we suppose that M is not part of an plane, that is, is not empty. Note that
for some negative constant c, which means that the function is constant and hence the functions are constant for all by virtue of (99). By continuity, the interval J is the whole domain I of . Furthermore, (97) implies
for the constant . By Lemma 2, we can see that the curve is contained in a 2-dimensional subspace of . Equation (100) gives that the curvature is non-zero constant and hence the plane curve is part of a circle or a hyperbola.
Considering Lemmas 4–6, we may put
for some constant vectors and for such that , , , ⋯, are linearly independent for each s. By applying Gram-Schmidt’s orthogonalization, we get orthonormal constant vectors , ⋯, from , ⋯, . are constant and thus are also constant for all .
We put for all . Define
where . Then is a non-zero constant since , , ⋯, are linearly independent. Take , where . After appropriate change of parameters , the parametrization of (3) for M can be reduced to
for some constant vector .
If is a circle, we can see that the trace of position vectors of is a circle on the unit sphere by virtue of the first equation of (101). □
Note that if , that is, is time-like, then we can see that is part of a hyperbola in by applying the same arguments developed in the proof of Proposition 1. Therefore, we can also obtain the parametrization (94) for M such that the curve is part of a hyperbola.
We now consider the case that some of the generators of rulings have null derivatives. Let M be an -dimensional non-cylindrical ruled submanifold parameterized by (3) in . Again, if we use Proposition 3.3 of , we may assume that for all .
Case 3. Suppose that are null for all . We then have three possible cases according to the degree of q.
Subcase 3.1. Let deg , that is, are null with for and for . Note that for all . Then M has the Gauss map of the form
for all . Equation (102) shows that the function f depends on the parameter s only. From , we can put
where are non-vanishing functions of s for all . Also, ∧ follows from and hence we have
for some functions of s and . By (18) and (104), we see
By straightforward computation, equation of (102) provides that
Now, on the non-empty open interval , the first equation of (102) implies
In this case, we recall that
because of . From the definition of , we get
with the aid of (18), (103)–(105) and (108), where we have put
Considering (107) and (109)–(111), we have the following:
as the coefficients of vectors , and , respectively, for all and . Note that are non-vanishing for all . Multiplying (114) by and adding the equations obtained in such a way together with respect to a , we obtain
By putting (112) into (113) and then considering (116), we have
Thus, is non-positive because of (108). Since are null and for , the vector can not be time-like and thus Therefore, in (113), we see that
for all . If , then and then f is vanishing because of (106), a contradiction on . Therefore, on and hence f is a non-zero constant function on M by continuity. This means that G is of 1-type in the usual sense. For ruled submanifolds with finite-type Gauss map, see .
Subcase 3.2. Let deg . In this case, for some i and the null vector fields satisfy for . We note that and for all . Thus, implies
Note that because of deg . Therefore, using the function f obtained from (117), we repeat the same process to get (55). Then, we have the following equation
where we have put
Since deg , the left side of (118) has to be vanishing, that is,
We note that of (131) is non-vanishing for all s. If this not the case, since + is a polynomial in t of degree 1 and is constant with respect to t, has to be vanishing for s and hence , a contradiction to for all i. Therefore, we have
for all . Differentiating ’ of (132) with respect to s and taking the indefinite scalar product with to the equation obtained in such a way, we get
which implies that is a non-zero constant function for all s. If not, that is, , the vector is zero, a contradiction. Therefore, (132) yields
for all . This is also a contradiction to the characters of and .
Consequently, we can conclude that there is no ruled submanifold with deg which has pointwise 1-type Gauss map of the second kind.
Subcase 3.3. Let deg . In this case, we can easily obtain the same conclusion such as Lemma 4 by referring to the case that , , ⋯, are non-null. But this is impossible according to the characters of the vectors and . Therefore, we see that no ruled submanifold in with deg has pointwise 1-type Gauss map of the second kind.
Case 4. Suppose that , ⋯, are null for and are non-null for , .
In this case, deg q = 2. If we follow a similar argument for the case that , , ⋯, are non-null, for the same reason as in Subcase 3.3, we can conclude that there is no ruled submanifold in with pointwise 1-type Gauss map of the second kind under these assumptions.
Until now, we have considered the necessary conditions for ruled submanifolds to have a pointwise 1-type Gauss map of the second kind. That is, if the ruled submanifold M in parameterized by (3) has a pointwise 1-type Gauss map of the second kind, then according to the characters of , M is part of a product manifold of a right cone (or a hyperbolic cone) and a plane, or M has a 1-type Gauss map in the usual sense. Conversely, by straightforward computations, we can see that the Gauss maps of these ruled submanifolds are of pointwise 1-type of the second kind.
Therefore, we have
Let M be an -dimensional non-cylindrical ruled submanifold with non-degenerate rulings in the Minkowski m-space . Then, M has a pointwise 1-type Gauss map G of the second kind if and only if M is one of the following:
(1) M has a 1-type Gauss map in the usual sense, i.e., the Gauss map G satisfies for some non-zero and some constant vector .
(2) M is part of a product manifold of a right cone and a plane of the form or .
(3) M is part of a product manifold of a hyperbolic cone and a plane .
(4) M is part of an -plane in .
4. Generalized Null Scrolls in
Let M be an -dimensional ruled submanifold in with degenerate rulings along a regular curve with a parametrization , where . Since is degenerate, it can be spanned by a degenerate frame such that
Without loss of generality as was shown in Lemma 1, we may assume that
Since the tangent space of M at is non-degenerate and contains the degenerate ruling , there exists a tangent vector field A to M which satisfies
Let be an integral curve of the vector field A on M. Then we can define another parametrization x of M as follows:
where . A ruled submanifold defined as above is called a generalized null scroll. We refer to two lemmas for later use.
Differentiating (137) with respect to , with the aid of (138) we get
as the coefficient of the vector for all and .
If for all h, (134) implies that is tangent to M. With the help of Lemma 8 and Theorem 3, we can see that M is minimal and hence the Gauss map G of M is harmonic. In this case, G can be chosen as the constant vector . That is, M is part of a Lorentzian -plane in .
Now, we suppose that for some . If we put
in the same manner as (15) by virtue of Lemma 1, differentiating (138) with respect to s provides
Comparing the coefficients of the vectors in (138) and (141), we obtain the following four equations:
which means that the function is constant. Since , the function is constant, so are for all . In (153), we can see that is a zero vector because of , a contradiction. Therefore, from (152), we conclude that
for some positive constant . Since , we see that
We now introduce another kind of generalized null scroll as follows:
For a null curve in , we consider a null frame along satisfying
Let be the matrix consisting of column vectors of , , , ⋯, with respect to the standard coordinate system in . Then we have
where denotes the transpose of , diag and
Consider a system of ordinary differential equations
where , and are some smooth functions of s and are constant satisfying
for some positive constant d.
For a given initial condition satisfying , there exists a unique solution to on the whole domain I of containing 0. Since T is symmetric and is skew-symmetric, and hence we have
for all . Therefore, , , , ⋯, form a null frame along a null curve in on I. Let .
We now give the following definition.
A generalized null scroll satisfying (157) parameterized by
is called the generalized B-scroll kind.
In the case of with , a generalized B-scroll kind is a so-called B-scroll.
Therefore, we can see that the parametrization of a generalized null scroll M with a pointwise 1-type Gauss map of the second kind can be given by (159). Furthermore, by combining the first two equations of (158), we can see that these ruled submanifolds satisfy
Conversely, for a generalized B-scroll kind M parameterized by (159), by computation, can be expressed as
where is the constant vector given by
It means that the Gauss map G of M is of pointwise 1-type of the second kind.
In this case, the open subset is non-empty. Comparing the vectors composing the constant vector of (133) and (135), by the orthogonality of them, we get
on W for all and .
If for all , we obtain the result that the open subset W of M is part of a Lorentzian -plane by Lemma 8 and Theorem 3.
If for some , then and for all . Then, Equation (133) is simplified as
on W. By differentiating (160) with respect to and using (161), we can obtain
as the coefficients of . Since on W, we have , a contradiction to for some h.
Therefore, we can conclude that if the open set W is non-empty, then the functions are identically zero on W for all , and hence we see that W is an open part of a Lorentzian plane in . By continuity, M is a Lorentzian -plane.
Therefore, we have
Let M be a generalized null scroll in the Minkowski m-space . Then, M has a pointwise 1-type Gauss map of the second kind if and only if M is part of a Lorentzian -plane in or a generalized B-scroll kind.
In particular, by straightforward computation, we have
Let M be a null scroll in the Minkowski 3-space . Then, M has a pointwise 1-type Gauss map of the second kind if and only if M is part of a time-like plane or a flat B-scroll.
Y.H.K. gave the idea to start the research on Gauss Map and its applications on ruled submanifolds in Minkowski Space, computed the last section and polish the draft. S.M.J. devoted to compute the details.
The second named author was supported by Kyungpook National University Bokhyeon Research Fund, 2017.
Authors would like to express their appreciation to the referees for their comments and valuable suggestions to improve the paper.
Conflicts of Interest
The authors declare no conflict of interest.
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