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Kyungpook Mathematical Journal 2017; 57(1): 133-144

Published online March 23, 2017

Copyright © Kyungpook Mathematical Journal.

On the Ruled Surfaces with L1-Pointwise 1-Type Gauss Map

Young Ho Kim1
Nurettìn Cenk Turgay2

Department of Mathematics, Kyungpook National University, Daegu 41566, Korea1
Department of Mathematics, Istanbul Technical University, 34469 Maslak, Istanbul, Turkey2

Received: June 18, 2014; Accepted: October 17, 2014

In this paper, we study ruled surfaces in 3-dimensional Euclidean and Minkowski space in terms of their Gauss map. We obtain classification theorems for these type of surfaces whose Gauss map G satisfying □G = f(G+C) for a constant vector C and a smooth function f, where □ denotes the Cheng-Yau operator.

Keywords: Cheng-Yau operator, Gauss map, null scroll, pointwise 1-type, ruled surface

Let M be a hypersurface of the Euclidean space . A smooth mapping φ: M is said to be of k-type if it can be expressed as a sum of eigenvectors of Laplace operator Δ corresponding to k distinct eigenvalues of Δ ([7]). If φ is an immersion from M into is of k-type, then M itself is said to be of k-type ([3]). The study of finite type mappings was summed up in a report by B.-Y. Chen ([4]).

On the other hand, if the Gauss map G of M is of 1-type, then it satisfies

ΔG=λ(G+C)

for a constant λ ∈ ℝ and a constant vector C. In this case, M is said to have 1-type Gauss map, [8]. However, Gauss map of some important submanifolds such as a helicoid and a catenoid in and several rotational surfaces in satisfy a very similar equation to (1.1), namely

ΔG=f(G+C)

for some function fC(M) and a constant vector C, ([11, 12]). These submanifolds whose Gauss map G satisfying (1.2) is said to have pointwise 1-type Gauss map. Submanifolds with pointwise 1-type Gauss map have been worked in several papers (cf. [5, 11, 16, 17, 18, 20, 21]).

In the recent years, the definition of being k-type of an hypersurface is extended in a natural way by replacing Laplace operator Δ with a sequence operators L0, L1, L2, …, Lk such that L0 = −Δ, where Lk is the linearized operator of the first variation of the (k + 1)-th mean curvature arising from normal variations of a hypersurface M of the Euclidean space . For convenience, the notation □ is used to denote the operator L1 which is called as theCheng-Yau operator introduced in [9]. The authors Alías et al. studied an isometric immersion x: MnRn+1 satisfying Lk(x) = Ax + b for a constant matrix A and a constant vector b, where k is a positive integer.

In [15], the authors give the following definition.

Definition 1.([15])

An oriented surface M of Euclidean space is said to have □-pointwise 1-type Gauss map if its Gauss map satisfies

G=f(G+C)

for a smooth function fC(M) and a constant vector C. More precisely, a □-pointwise 1-type Gauss map is said to be of the first kind if (1.3) is satisfied for C = 0; otherwise, it is said to be of the second kind. Moreover, if (1.3) is satisfied for a constant function f, then we say M has □-(global) 1-type Gauss map.

In the same paper, authors states

Open Problem

Classify surfaces in with □-1-type Gauss map.

On the other hand, there are many studies done on rotational surfaces, ruled surfaces and translation surfaces in terms of being finite type or having pointwise 1-type Gauss map. For example, in [5] and [14], the rotational surfaces of the Euclidean 3-space and the Minkowski 3-space E13 with (Δ-)pointwise 1-type Gauss map have been studied. Also, a classification of ruled surfaces in terms of their Gauss map was studied in [6] and [16].

In this paper, we study rotational surfaces, ruled surfaces and translation surfaces in and E13 with □-pointwise 1-type Gauss map.

Let q ∈ {0, 1} and Eq3 denote the 3-dimensional semi-Euclidean space with the canonical semi-Euclidean metric tensor of index q given by

g=,=(-1)qdx12+dx22+dx32.

A non-zero vector u in the Minkowski space E13 is called space-like (resp., time-like or light-like) if 〈u, u〉 > 0 (resp., 〈u, u〉 < 0 or 〈u, u〉 = 0). Furthermore, a curve β is called space-like (resp., time-like or light-like) if its tangent vector β′ is space-like (resp., time-like or light-like) at every point.

On the other hand, a two dimensional subspace U of E13 is called non-degenerate if UU= {0} and a non-degenerate subspace U of index r is called space-like (resp., time-like) if r = 0 (resp., r = 1). Eventually, a surface M in E13 is called non-degenerate, (resp., degenerate, space-like or time-like) if its tangent space TpM is non-degenerate, (resp., degenerate, space-like or time-like) at every point pM.

The following lemmas are well-known and useful (see, for instance [13]):

Lemma 2.1

Let u, v be two orthogonal vectors in E13. If u is time-like, then v is space-like.

Lemma 2.2

Two light-like vectors are orthogonal if and only if they are linearly dependent.

Lemma 2.3

A two dimensional subspace U of E13is a time-like space if and only if it contains two linearly independent lightlike vectors.

2.1. Surfaces in 3-dimensional Euclidean and Minkowski spaces

Let M be an oriented surface in Eq3. We denote the Levi-Civita connections of E13 and M by ∇̃ and ∇, respectively and D stands for the normal connection of M. We put

ɛ={1if Mis space-like-1if Mis time-like.

Then, we have 〈N, N〉 = (−1)qɛ, where N is the unit normal vector field associated with the orientation of M. The mapping G:MEq3 which assigns every point p to N(p) is called the Gauss map of M.

The well-known Gauss and Weingarten formulas are given by

˜XY=XY+h(X,Y),˜XN=-S(X)

for tangent vector fields X, Y of M, where h is the second fundamental form and S is the shape operator of M. The covariant derivative of h is defined by

(˜Xh)(Y,Z)=DXh(Y,Z)-h(XY,Z)-h(Y,XZ)

Then, the Codazzi equation is given by

(¯Xh)(Y,Z)=(¯Yh)(X,Z)

for tangent vector fields X, Y, Z of M. Note that S and h satisfy 〈S(X), Y 〉 = 〈h(X, Y ), N〉.

The functions , H and K defined by (λ) = det(S − λI) = λ2 − 2Hλ + K are called the characteristic polynomial of S, the mean curvature of M and the Gaussian curvature of M, respectively. M is said to be minimal (resp., flat) if H (resp., K) vanishes identically. Sometimes the (complex valued) functions λ1 and λ2 satisfying i) = 0, i = 1, 2 are called the principal curvatures of M.

The Gauss equation is given by

R(e1,e2,e2,e1)=K,

where R is the curvature tensor associated with connection ∇ and e1, e2 are orthonormal vector fields on M.

We will use χ(M) to denote the space of all smooth functions from M into Eq3 and C(M) the space of all smooth functions defined on M. Let = {e1, e2, e3} be an orthonormal frame field defined on M, i.e., 〈e1, e1〉 = ɛ, 〈e2, e2〉 = 1, e3 = N and 〈ei, ej〉 = 0 for i j (i, j = 1, 2, 3). If Xχ(M) is tangent to M, its divergence divX is defined by divX = ɛ〈∇e1X, e1〉 + 〈∇e2X, e2〉. On the other hand, the gradient of a function fC(M) is given by ∇f = ɛe1(f)e1 + e2(f)e2 and the Laplace operator acting on M is given as Δ = −ɛe1e1 −∇e2e2 + e1e1 + e2e2.

2.2. Surfaces with □-pointwise 1-type Gauss map

Let M be a surface in Eq3 and P0, P1 the Newton transformations given by P0 = I, P1 = 2HIS, where I is the identity operator acting on the tangent bundle of M. Then, the second order differential operators Lk: C(M) → C(M) associated with Pk are given by Lk(f) = tr(Pk ∘∇2f), k = 1, 2. Note that we have L0 = −Δ and L1 = □, where □ is the Cheng-Yau operator introduced in [9]. As a matter of fact, it turns out to be

Lkf=div (Pk(f))

for fC(M) ([2]).

We will use following lemma and theorems in [15].

Lemma 2.4.([15])

Let M be an oriented surface in with Gaussian curvature K and mean curvature H. Then, the Gauss map G of M satisfies

G=-K-2HKG.

Theorem 2.5.([15])

An oriented surface M in has -harmonic Gauss map if and only if it is flat, i.e, its Gaussian curvature vanishes identically.

Theorem 2.6.([15])

An oriented surface M in has -pointwise 1-type Gauss map of the first kind if and only if it has constant Gaussian curvature.

Theorem 2.7.([15])

An oriented minimal surface M in has -pointwise 1-type Gauss map if and only if it is an open part of a plane.

Let M be a ruled surface in given by (2.5). Then, as a surface, we have xt = β ≠ 0 and x(s, t) = α + t̃ β̃ where = tβ, β1/2 and β̃ = β/β, β1/2. Thus, without loss of generality, we may assume 〈β, β〉 = 1. By re-defining s appropriately, we also suppose that 〈β, β′〉 = 1. Moreover, for another base curve ᾱ of M given by ᾱ(s) = α(s)+g(s)β(s) with g′(s)+〈α′(s), β(s)〉 = 0 we have 〈ᾱ, β〉 = 0. Hence, without loss of generality, we may also assume 〈α, β〉 = 0.

Because of these assumptions, we have

α=aβ+bββ

and

β=-β+cββ

for some smooth functions a = a(s), b = b(s) and c = c(s).

We choose an orthonormal frame field as

e1=1Es,e2=t,G=1E(bβ-(a+t)ββ)

where

E=b2+(a+t)2.

By a direct calculation, we obtain the connection form ω12 as

ω12=wθ1,         w=-a+tE2,

where {θ1, θ2} is the dual base of {e1, e2}. Moreover, the Gaussian curvature K and the mean curvature H are given by

K=-h22,H=h1/2,

where

h1=h(e1,e1),G=b(a-bc)-b(a+t)-c(a+t)2E3,h2=h(e1,e2),G=bE2.

On the other hand, from the Codazzi equation (2.3) and the Gauss equation (2.4) we obtain

wt=w2+K,h1,t=h2,sE+wh1,h2,t=2wh2.

By using (3.5) and (3.11), we obtain

Kt=4wK.

3.1. Ruled surfaces with □-pointwise 1-type Gauss map of the first kind

We first give the following theorem:

Theorem 3.1

Let M be a non-cylindrical ruled surface whose position vector given by (2.5). Then, the following statements are equivalent:

  • M has -pointwise 1-type Gauss map of the first kind.

  • M has -harmonic Gauss map.

  • α′ = aβfor a smooth function a.

Proof
  • ⇔ (2): Kt = 0 implies Kw = 0 because of (3.12). Thus, if Kt = 0 and K ≠ 0 at a point p of M, then there exists a neighborhood of p such that = 0 which is not possible because of (3.4). Therefore, we have if K is constant, then K = 0. Hence, from Theorem 2.5 and Theorem 2.6 we obtain (1) ⇔ (2).

  • ⇔ (3): Because of (3.5) and (3.8), M is flat if and only if b ≡ 0 which is equivalent to α′ = aβ′ because of (3.1).

Theorem 3.2

A ruled surface in has -pointwise 1-type Gauss map of the second kind if and only if M is flat.

Proof

Let M be a ruled surface in given by (2.5) with □-pointwise 1-type Gauss map of the second kind. Then, there exist a function f and a vector C = C1e1 + C2e2 + C3G such that

fC1=-KsE,fC2=-Kt,f(C3+1)=-2KH.

Note that from (3.12) and (3.13b) we obtain

fC2=-4Kw.

On the other hand, by using Gauss and Weingarten formulas, we obtain

tC=(C1,t)e1+(C2,t)e2+(C3,t)G+C1h2G-C3h2e1.

Thus, ∇tC = 0 implies

C1,t=h2C3,C2,t=0,C3,t=-h2C1.

Now, we assume towards a contradiction that M is not flat, i.e., the open subset ℳ = {pM|K(p) ≠ 0} of M is not empty. By multiplying both sides of (3.13c) by C2 and using (3.6) and (3.14), we obtain K (4w(C3 + 1) − h1C2) = 0 from which we get

4w(C3+1)=h1C2

on ℳ. By taking derivative of this equation and using (3.9), (3.10), we obtain

4(w2+K)(C3+1)-4wh2C1=(h2,sE+wh1)C2

on ℳ. Next, we multiply both sides of this equation by f and use (3.13a), (3.13c) and (3.14) to obtain K(h22h1E+3wh2s)=0 from which we get

h22h1E+3wh2s=0

By using (3.3), (3.4), (3.7) and (3.8) in (3.16), we obtain

b3(a-bc)+2b2b(a+t)+(-cb2+6ab)(a+t)2-3b(a+t)3=0

on ℳ, from which, we obtain b is a constant and

b(a-bc)=0,b(6a-bc)=0.

Note that if b = 0, then (3.1) implies α′ = aβ′ and from Theorem 3.1 we have ℳ is flat which yields a contradiction. Thus, we have b ≠ 0.

From (3.18) we have a′ = c = 0. Therefore, (3.6) and (3.7) imply that ℳ is minimal. However, Theorem 2.7 implies that ℳ is an open part of a plane which is contradiction since K ≠ 0 on ℳ. Hence we have ℳ is an empty set, i. e., M is flat.

The converse is obvious.

In this section, we suppose that M is a ruled surface whose Gauss map satisfies □G = AG for some matrix A ∈ ℝ3x3 with real entities. From this equation and (2.6) we obtain

-AG=e1(K)e1+e2(K)e2+2KHG.

By taking covariant derivative of this equation on the direction e2 we have

-˜e2(AG)=ASe2=(e2e1(K)-2KHh2)e1+e2e2(K)e2+(h2e1(K)+2e2(KH))G

as ∇e2e1 = ∇e2e2 = 0 and h(e2, e2) = 0. From this equation we obtain

h2Ae1,e2=Ktt.

Note that, by using (3.9) and (3.12), one can obtain Ktt = 20w2K + 4K2. On the other hand, by using (3.2) we obtain

Ae1,e2=1E((a+t)Aβ,β+bA(ββ),β).

From this equation and (3.19) we have

Aβ,βE5(a+t)+bA(ββ),βE5+20b(a+t)-4b3=0

which implies b = 0, i. e., M is flat. Hence, we have

Theorem 3.3

The Gauss map G of a ruled surface in satisfies G = AG for a matrix A ∈ ℝ3x3 if and only if M is flat.

Combining Theorem 3.2 and Theorem 3.3, we obtain

Theorem 3.4

Let M be a non-cylindrical ruled surface whose position vector given by (2.5). Then, the following statements are equivalent:

  • M has -pointwise 1-type Gauss map of the second kind.

  • The Gauss map G of M satisfies G = AG for a matrix A ∈ ℝ3x3.

  • M is flat.

A non-degenerate ruled surface M in E13 given by (2.5) is called a null scroll if 〈β, β〉 = 〈α, α′〉 = 0 and 〈α, β〉 ≠ 0. In this case, without loss of generality we may assume that 〈α, β〉 = 1. Furthermore, we may choose an appropriate parameter s in such a way that 〈α, β′〉 = 0, which is possible if the base curve α is chosen as a null geodesic of M.

On the other hand, if 〈β, β′〉 = 0 at an open subset ℳ of M, then there exists a function a such that β′ = aβ which implies β = (β1, β2, β3) is β = β1c0 for a constant light-like vector c0E13. Hence, we may assume β = c0 which implies that ℳ is cylindirical. Therefore, we may locally assume 〈β, β′〉 = E2 > 0.

The tangent vector fields e1 = −∂s + (t2E2/2)∂t and e2 = ∂t form a pseudo-orthonormal frame field and the unit nomal vector field is N = −E1β′ + tEβ. By a direct calculation, we obtain

Se1=-Ee1+be2,         Se2=-Ee2,˜e1e2=-tE2e2+EN,         ˜e2e2=0

for a non-vanishing function b. From (4.1a) we have H = E which implies

P1=-2EI-S.

4.1. Gauss map of null scrolls

In the next lemma, we obtain □G for a null scroll in E13.

Lemma 4.1

Let M be a null scroll in E13. Then, the Gauss map G of M satisfies

G=-2EEe2+2E3G.
Proof

Let C be a constant vector in E13. By a direct computation, we obtain

G,C=-Ee2,Ce1-Ee1,Ce2+be2,Ce2.

By considering (4.2), we get

P1(G,C)=E2e2,Ce1+E2e1,Ce2.

By using this equation and (2.5), we obtain

G,C=-e1(E2e2,Ce1+E2e1,Ce2),e2-e2(E2e2,Ce1+E2E2e1,Ce2),e1=e1(E2)e2,C+e2(E2)e1,C+2E2h(e1,e2),C=-2EEe2+2E3G,C.

Thus, we have (4.3).

Example 1

If α(s) is a null curve in E13 with the Cartan frame {A, B, C} such that 〈A, A〉 = 〈B, B〉 = 0, 〈A, B〉 = −1, 〈A, C〉 = 〈B, C〉 = 0 and 〈C, C〉 = 1 with α′ = A, A′ = k1(s)C and B′ = k2C for a constant k2 and a smooth function k1 and β(s) = B(s), then the null scroll given by (2.5) is said to be a B-scroll. It is well-known that a null scroll M is a B-scroll if and only if E is a constant (see [19]). In this case, the Gauss map of M satisfies

G=2E3G

because of (4.3) which implies M has □-1-type Gauss map of the first kind.

Next, we want to give classification of null scrolls in E13 with □-pointwise 1-type Gauss map.

Proposition 4.2

A null scroll in E13 has -pointwise 1-type Gauss map if and only if it is a B-scroll.

Proof

Let M be a null scroll in E13 with □-pointwise 1-type Gauss map. Then, the Gauss map G of M satisfies

-2EEe2+2E3G=f(G+C)

for a constant vector C and a smooth function f. From (4.5) we have

fC,e2=0.

Now, we consider the open subset ℳ = {pM|f(p) ≠ 0} of M on which 〈C, e2〉 = 0 is satisfied. From this equation we get

e1(e2,C)=0

on ℳ. By a further calculation taking into account of Gauss formula (2.1), (4.6) and (4.7), we obtain

G,C=0.

By combining (4.6) and (4.8), we obtain C = C1e2 = C1β. From which we get C = 0. Thus, (4.5) implies E is constant. Hence M is a B-scroll.

The converse is given in Example 1.

Now, we obtain the following proposition.

Proposition 4.3

Let M be a null scroll in E13. Then, its Gauss map satisfies G = AG for a constant 3 × 3-matrix A if and only if M is a B-scroll.

Proof

Suppose the the Gauss map G of M satisfies □G = AG for a constant 3 × 3-matrix A. Then, we have

-2EEe2+2E3G=AG,

from which, we get

-2EE˜e2e2+2E3˜e2G=A(˜e2G).

By using (4.1a), we obtain

Ae2=2E3e2

from which we get A(∇̃e1e2) = 2E3 ∇̃e1e2. From this equation and (4.1b) we obtain

-tE2Ae2+EAG=E3(-tE2e2+EG).

By combining this equation with (4.9) and (4.10), we obtain EE′ = 0 which implies E is constant. Hence, M is a B-scroll.

The converse is given in Example 1.

By combining Proposition 4.2 and Proposition 4.3 with the result of [1], we obtain the following theorem.

Theorem 4.4

Let M be a null scroll in E13. Then the following conditions are equivalent.

  • M has -pointwise 1-type Gauss map.

  • The Gauss map G of M satisfies ΔG = AG for a constant 3 × 3-matrix A.

  • The Gauss map G of M satisfies G = AG for a constant 3 × 3-matrix A.

  • M is a B-scroll.

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