2. Preliminaries
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.
Remark 1 ([
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 [
30], we have
Lemma 1 ([
8]).
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
or, equivalently
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.
Lemma 2 ([
29]).
Suppose that a unit speed curve in the m-dimensional Minkowski space defined on an open interval I satisfieswhere 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
or, equivalently
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
Theorem 1. 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 [
19] 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
for
.
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
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
for all
. Then we can put
where
and
for
. From (
16), we get
where
for
. And, we may put
where
for all
.
Suppose that
M is not an
-plane, that is,
. To deal with (
13), we consider the subset
Without loss of generality, we may assume that
on
. Then, on
,
By (
9) and
, we see that
and hence
which implies that
Also, it follows from (
16) that on
,
for all
.
Lemma 3. 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 , thenon Proof. Let We suppose that the interior Int of is non-empty.
Equation (
13) tells us that
on Int
. Using
on Int
, equation
yields that
which provides
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
because of (
12), (
16) and (
18). Since
and
,
Now, we suppose that
for some
. Then, at
Equation (
11) is rewritten as
and taking the indefinite scalar product with
to (
24) gives us the following
where we have put
Substituting (
25) into (
24) and then considering the constant terms of the equation obtained in such a way, we get
By straightforward computation, it follows from (
19), (
21) and (
22) that at
where we have put
Combining (
17), (
26) and (
27), we can obtain
for all
and
. Using (
19), (
21) and (
28), we also get at
Since
, the functions
and
are identically zero on
for all
, with the help of (
19), (
21), (
22) and (
29). Thus, we have
which means that
is orthogonal to all vectors of (
24) except the constant vector
, so taking the indefinite scalar product with
to (
24) yields that
which is a contradiction.
Therefore, we have
on
. Using (
30), (
24) implies that
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,
on
.
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
and
where
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:
Combining these equations, we get
Therefore,
that is,
and hence
for all
and
. From (
39), we can obtain
and then we have
from (
32). By regarding (
43) as the polynomial in
t of degree 1, we get
which produce that
for all
j. With the help of (
17), (
27), (
34) and (
41), the equation above provides that
which means that
due to (
42), a contradiction.
Therefore, we conclude that
on
and hence the function
f is given by
from (
33).
Step 3. We find the another equation for f on .
First, we suppose that
of (
35) for all
and
. Then, we have
and
for
and
. In this case,
and
are given by
With the help of (
46), the first three equations of (
45) provide
for all
and
.
Taking the indefinite scalar product with
to (
32) gives us the equation containing
in the following
Using (
46) and (
47), Equation (
48) is rewritten as
which gives
and hence
for all
and
. Thus,
, a contradiction. Therefore, we conclude that
or
for some
k,
a and
b. Now we assume that
. Then, taking the indefinite scalar product with
to (
32), we obtain
or, equivalently,
Comparing two Equations (
44) and (
49) regarding the function
f, we get
which provides that
as the coefficients of terms containing
,
and
for
. Combining these three equations above, we have
for all
and
. If
for some
j, then we can see that
by applying the same arguments used to show that
on
. But, it contradicts
. Therefore, we have
and hence the function
f becomes
for all
and
.
Step 4. We compare the equations for f obtained in Steps 2 and 3.
Putting (
50) into (
32) and then considering the coefficients of terms containing
and
in the equation obtained in such a way, we get
for
. With the help of (
46), Equation (
51) gives that
, so we see that
if and only if
.
If
, then
because of (
46). In this case,
which means that
or
for
and
. The case of
also guarantees
, a contradiction. Therefore, we see that
and
for all
and
. By computation, equation
gives us
and equation
provides
for all
j,
k,
a and
b. In particular, by replacing
k with
j in (
53), we have
which implies that
by virtue of (
52). This is a contradiction.
According to Steps 1, 2, 3 and 4, we can conclude that the subset is empty, that is, we may assume that on M. □
By Lemma 3, we can see that the function
f of (
13) is a rational function in
t with functions in
s as coefficients of the form
If we substitute (
54) into (
11) and multiply
by the equation obtained in such a way, then we have
Next, we will show that the function q is independent of the parameter s and it is a form of perfect square expression in t of degree 2.
We suppose that
is not a polynomial in
t. Then we have
By following the same argument to prove Lemma 3.4 in [
19], (
56) implies that
Then, we deduce from (
55) the following
or,
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
which yields
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
Lemma 4. 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 havefor all . Proof. If
M is Lorentzian, it is obvious. We now suppose that
M is space-like. In this case,
for all
. Therefore, (
55) can be rewritten as
where we have put
According to (
63), we may have put
where
,
and
are vector functions of
s for
. Then, by considering the degrees of polynomials (
62) and (
64) in
t, we can see that
which implies that
for all
. From the above two equations, we have
If
, then
in (
17), that is,
for all
and
.
Now, we consider
and suppose that
. Then, on
,
and hence
With the help of (
63) and (
64), the relations
provide us with the following results
for
. Considering the orthogonality of vectors of the right sides in (
67) and (
68), we can see that the
must be zero. Therefore, it follows from (
66) that
Case 1. If
on some open interval
, then
on
. Then, (
55) is simplified as
or,
where
is a polynomial in
t with vector functions of
s as coefficients.
If
for some functions
a,
of
s and a vector
of
s, then Equation (
71) gives us
as the coefficients of terms containing
,
and
for
. Putting (
72) into (
73) and substituting the equation obtained in such a way into (
74), we get
which implies that
on
.
If is of the form for some vectors and along s, we also have the only possible case of .
Then, the condition of
renders (
70) simple as follows
Here, we may assume that
. If not, that is,
, it follows from (
75) that
because of
. By computations,
and (
76) implies that
. According to Theorem 3.4 in [
9], we can see that it is part of an
-plane in
.
Considering the constant terms with respect to
t and the coefficients of terms containing
of (
75), we have
Differentiating (
17) with respect to
s gives
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.
Case 2. We consider
and suppose that
. Then, on
,
and Equation (
68) is simplified as
With the help of (
86), taking the indefinite scalar product with
to (
87) gives
which yields
because of
,
and
. If
, then
, which implies that
by applying the same arguments used in Case 1. If
on
, then, by continuity,
on
A and hence
on
A for the same reasons as Case 1.
According to Cases 1 and 2, we can conclude that
on
and hence on
M. This proof is complete. □
From Lemma 4, the Gauss map
G is given by
and thus
yields
Taking the indefinite scalar product to (
88) with
, we obtain
Suppose that on I. Then we have two cases concerning .
If on I, then the Gauss map G is a constant vector field and hence M is an open part of an -plane in .
Now, we suppose that
is null on some interval
U. Then the normal part of
of (
18) has to be null on
U as well. Therefore, we have
for
. By (
89), we get
Since
, we see that
on
U. Then, Equation (
88) is rewritten as
which yields
for all
. By the definition of
and Lemma 4, we have
Taking the wedge product with
to (
90) for some
k, we obtain
Without loss of generality, we may assume that
is a non-zero constant. So we have
which means that
is tangent to
M, a contradiction.
Therefore, we conclude that if on I, then and M is part of an -plane in .
We now suppose that the open subset
is not empty. Then, we may put
Using
and putting (
91) into (
88), we have
In (
92), considering the constant terms with respect to
t and the coefficients of terms containing
, we see
and hence
for all
. Since
,
are constant for all
and
. Together with Lemma 4, we have
Lemma 5. 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 functionsare constant functions on the open interval for all , where Furthermore, for the case that
are non-null for all
j, we have
Lemma 6. 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 satisfyingfor all . Proof. It is sufficient to see Lemma 3.5 in [
20]. □
Proposition 1. 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 bywhere is part of a circle or a hyperbola in with , , ⋯, orthonormal constant vectors satisfying = , a constant vector and for some open intervals and . Proof. 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
Then, Equation (
93) implies
or, equivalently,
on
J. Equation (
96) yields
for some positive constant
. Therefore, from (
93) we get
for some non-zero constant
.
Meanwhile, according to Lemma 6, we can put
for all
. Since
is constant, by straightforward computation, we have
where
. By the orthogonality of the vectors
,
and
for all
and
, (
98) yields
for all
. Since
on
J and
, there exists a non-zero function
for some
. Then, (
99) implies
So we can see that
for some non-zero real number
. By (
95), we have
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 [
19], 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
Therefore,
implies
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
which implies
By straightforward computation, equation
of (
102) provides that
with the help of (
103) and (
104).
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
or,
for all
. By (
112), (
115) yields that
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 [
10].
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,
Using
and
, Equation (
119) can be expressed as
for some vector
. In [
29], it was proved that Equation (
120) implies
Therefore, (
119) is simplified as
which furnishes us with three equations as follows:
for all
. Combining (
122)–(
124), we get
for all
. By the characters of
and
, we see that the functions
of (
16) are non-vanishing for all
s and it is impossible to have
. Thus, we have
for all
.
Meanwhile, we note that
for all
. Then, we can put
for some
with
, where
are non-vanishing functions for all
. From the definition of
, we have
which implies that
are non-zero constant for all
. Indeed, we see that
for all
. By (
14) and (
126), we also obtain
for all
. Thus, the following vector and function of (
121) are induced as
and
by virtue of (
122), (
124), (
125), (
127) and (
128). Using
and (
119), and substituting (
129) and (
130) into (
118), Equation (
118) is rewritten as
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
or, equivalently,
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
Theorem 2. 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
at
.
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.
Lemma 7 ([
7]).
We may assume that for all s. Lemma 8 ([
8]).
Let M be a ruled submanifold with degenerate rulings. Then, the following are equivalent.(1) M is minimal
(2) is tangent to M.
If we put
and
, Lemma 7 implies
where
,
and
for
.
Note that
P and
Q are polynomials in
with functions in
s as coefficients. Then the Laplacian
of
M can be expressed as follows:
where
.
By definition of the indefinite scalar product
on
, we may put
Then the Gauss map
G is given by
In [
9], the authors proved the following theorem.
Theorem 3 ([
9]).
Let M be a generalized null scroll in . Then, the following are equivalent.(1) M is minimal.
(2) M has a harmonic Gauss map.
We now suppose that a generalized null scroll
M has a pointwise 1-type Gauss map
. Without loss of generality, we may assume that
. Then by straightforward computation, we get
where we have put
for
and
.
Now, we note that
Q is constant with respect to
. Then, by differentiating (
133) with respect to
, we get
Case 5. on M.
Equation (
135) implies that
Putting (
136) into (
133), we obtain the following polynomial in
t of degree 1 with functions of
s as coefficients
Comparing the constant terms with respect to
t of (
137) and using
, we have
From (
136) we see that
.
Differentiating (
137) with respect to
, with the aid of (
138) we get
which implies
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:
Substituting (
139) into (
143), we get
for some
h with
. Putting (
146) into (
144) gives
which implies
for some constant
. Therefore, we can put
for some constants
and
.
Recall that Equation (
139) is valid for all
. By replacing
h with
, ⋯,
, respectively, and comparing equations obtained in such a way, with the help of (
147) we can get
for all
and for all
. By virtue of (
147) and (
148), Equation (
142) is simplified as
for all
. Putting (
147) into (
149) and repeating the method to get (
148), we have
for all
. If we put (
147) and (
148) into (
139), then we obtain
because of
. Substituting (
139) into (
145) provides
which yields
with the help of (
151).
If
, (
138) implies that
and hence
Combining (
153) and (
154), we have
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
or, equivalently,
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
for
.
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
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.
Definition 2. A generalized null scroll satisfying (157) parameterized byis called the generalized B-scroll kind. Remark 2. 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.
Case 6.
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
or,
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
Theorem 4. 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
Corollary 1. 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.