On Non-Developable Ruled Surfaces in Euclidean 3-Spaces∗

Indian J. pure appl. Math., 38(4): 281-290, August 2007 °c Printed in India.

ON NON-DEVELOPABLE RULED SURFACES IN EUCLIDEAN 3-SPACES∗

DAE WON YOON

Department of Mathematics Education and RINS, Gyeongsang National University, Chinju 660-701, Korea e-mail: [email protected]

(Received 8 February 2005; after ﬁnal revision 4 November 2006; accepted 12 December 2006)

In this paper, we classify ruled surfaces in a 3-dimensional Euclidean space satisfying some algebraic equations in terms of the second Gaussian curvature, the mean curvature and the Gaussian curvature. Key Words: Gaussian Curvature; Second Gaussian Curvature; Mean Curvature; Minimal Surface; Ruled Surface

1. INTRODUCTION

The inner geometry of the second fundamental form has been a popular research topic for ages. We will refer the term “non-developable,” and by a non-developable surface we mean that a surface free of points of vanishing Gaussian curvature in a 3-dimensional Euclidean space. It is readily seen that the second fundamental form of a surface is non-degenerate if and only if a surface is non-developable.

On a non-developable surface M, we can consider the Gaussian curvature KII of the second fundamental form which is regarded as a new Riemannian metric. Therefore, KII can be deﬁned formally and it is the curvature of the Riemannian or pseudo-Riemannian manifold (M,II). Using classical notation, we denote the component functions of the second fundamental form by e, f and g. Thus we deﬁne the second Gaussian curvature by (cf. [2]) ¯ ¯ ¯ ¯ ¯ 1 1 1 1 ¯ ¯ 1 1 ¯ ¯ − 2 ett + fst − 2 gss 2 es fs − 2 et ¯ ¯ 0 2 et 2 gs ¯ 1 ¯ ¯ ¯ ¯ K = ¯ f − 1 g e f ¯ − ¯ 1 e e f ¯ . (1.1) II (|eg| − f 2)2 ¯ t 2 s ¯ ¯ 2 t ¯ ¯ 1 ¯ ¯ 1 ¯ 2 gt f g 2 gs f g It is well known that a minimal surface has vanishing second Gaussian curvature but that a surface with vanishing second Gaussian curvature need not be minimal.

∗This work was supported by Korea Research Foundation Grant (KRF-2004-041-C00039). 282 DAE WON YOON

For the study of the second Gaussian curvature, Koutrouﬁotis [9] has shown that a closed ovaloid √ is a sphere if KII = cK for some constant c or if KII = K, where K is the Gaussian curvature.

Koufogiorgos and Hasanis [8] proved that the sphere is the only closed ovaloid satisfying KII = H, where H is the mean curvature. Also, Kuhnel¨ [10] studied surfaces of revolution satisfying KII = H. One of the natural generalizations of surfaces of revolution is the helicoidal surfaces. In [1]

Baikoussis and Koufogiorgos proved that the helicoidal surfaces satisfying KII = H are locally characterized by constancy of the ratio of the principal curvatures. On the other hand, Blair and Koufogiorgos [2] investigated a non-developable ruled surface in a 3-dimensional Euclidean space R3 satisfying the condition

aKII + bH = constant, 2a + b 6= 0, (1.2) along each ruling. Also, they proved that a ruled surface with vanishing second Gaussian curvature is a helicoid.

Recently, the present author [15] studied a non-developable ruled surface in a 3-dimensional Euclidean space R3 satisfying the conditions

aH + bK = constant, a 6= 0, (1.3)

aKII + bK = constant, a 6= 0, (1.4) along each ruling.

In particular, if it satisﬁes the condition (1.3), then a surface is called a linear Weingarten surface (see [11]).

On the other hand, in [6] Kim and the present author investigated a non-developable ruled surface in a 3-dimensional Lorentz-Minkowski space satisfying the conditions (1.2), (1.3) and (1.4).

In this article, we will study a non-developable ruled surface in a 3-dimensional Euclidean space R3 satisfying the conditions

aH2 + 2bHK + cK2 = constant, a 6= 0, (1.5) 2 2 aH + 2bHKII + cKII = constant, a 6= −4(b + c), c 6= 0, (1.6) 2 2 aK + 2bKKII + cKII = constant, c 6= 0. (1.7)

If a surface satisfies the equations (1.5), (1.6) and (1.7), then a surface is said to be a HK-quadric surface, HKII -quadric surface and KKII -quadric surface, respectively. On the other hand, many geometers have been interested in studying submanifolds of Euclidean and pseudo-Euclidean space in terms of the so-called finite type immersion. Also, such a notion can be extended to smooth maps on submanifolds, namely the Gauss map [3, 5]. The Gauss map G on ON NON-DEVELOPABLE RULED SURFACES IN EUCLIDEAN 3-SPACES 283 a submanifold M of a Euclidean space or pseudo-Euclidean space is said to be of pointwise 1-type if ∆G = fG for some smooth function f on M where ∆ denotes the Laplace operator defined on M [5].

In [3] Choi and Kim proved the following theorem which will be useful to prove our theorems in this paper.

Theorem 1.1 — ([3]). Let M be a non-cylindrical ruled surface in a 3-dimensional Euclidean space. Then, the Gauss map is of pointwise 1-type if and only if M is an open part of a helicoid.

2. MAIN RESULTS

In this section we classify a non-developable ruled surface in a 3-dimensional Euclidean space R3 satisfying the equations (1.5), (1.6) and (1.7). It is well known that a cylindrical ruled surface is developable, i.e., the Gaussian curvature K is identically zero. Therefore, the second fundamental form II is degenerate. Thus, non-cylindrical ruled surfaces are meaningful for our study.

Theorem 2.1 — Let M be a non-developable ruled surface in a 3-dimensional Euclidean space. Then, M is a HK-quadric surface if and only if M is an open part of a helicoid. PROOF : Let M be a non-developable ruled surface in R3. Then the parametrization for M is given by x = x(s, t) = α(s) + tβ(s) such that hβ, βi = 1, hβ0, β0i = 1 and hα0, β0i = 0. In this case α is the striction curve of x, and the parameter is the arc-length on the spherical curve β. And we have the natural frame {xs, xt} given 0 0 by xs = α + tβ and xt = β. Then, the components of the ﬁrst fundamental form are given by

E = hα0, α0i + t2,F = hα0, βi,G = 1. √ We put D = EG − F 2. In terms of the orthonormal basis {β, β0, β × β0} we obtain

α0 = F β + Qβ × β0, (2.1) β00 = −β − Jβ × β0, (2.2) α0 × β = Qβ0, (2.3) where Q = hα0, β × β0i 6= 0,J = hβ00, β0 × βi. Thus, we get p D = Q2 + t2, (2.4) from which the unit normal vector N is written as 1 1 N = (α0 × β + tβ0 × β) = (Qβ0 − tβ × β0). D D 284 DAE WON YOON

This leads to the components e, f and g of the second fundamental form

1 Q e = (Q(F + QJ) − Q0t + Jt2), f = 6= 0, g = 0. D D

Therefore, using the data described above, the mean curvature H, the Gaussian curvature K and the second Gaussian curvature KII are given respectively by

1 Eg − 2F f + Ge 1 ¡ ¢ H = = Jt2 − Q0t + Q(QJ − F ) , (2.5) 2 EG − F 2 2D3

eg − f 2 Q2 K = = − (2.6) EG − F 2 D4 and

µ ¶ 1 1 1 K = ff (f − e ) − f 2(− e + f ) II f 4 t s 2 t 2 tt st

1 ¡ ¢ = Jt4 + Q(F + 2QJ)t2 − 2Q2Q0t + Q3(QJ − F ) . (2.7) 2Q2D3

We now differentiate H, K and KII with respect to t, the results are

1 H = (−Jt3 + 2Q0t2 + Q(−QJ + 3F )t − Q2Q0), (2.8) t 2D5

4Q2 K = t (2.9) t D6 and

1 (K ) = (Jt5 + Q(2QJ − F )t3 + 4Q2Q0t2 II t 2Q2D5

+Q3(5F + QJ)t − 2Q4Q0). (2.10)

Suppose that a non-developable ruled surface is HK-quadric. Then we have (1.5)

aHHt + b(HtK + HKt) + cKKt = 0, which implies we have

2 2 2 2 4 2 2 2 2 4b D A1 = a D A2 + 8acD A2A3 + 16c A3, (2.11) ON NON-DEVELOPABLE RULED SURFACES IN EUCLIDEAN 3-SPACES 285 where we put

2 3 2 0 2 4 3 4 0 A1 = 5Q Jt − 6Q Q t + (5Q J − 7Q F )t + Q Q , 2 5 0 4 2 2 02 3 2 0 0 2 A2 = J t − 3Q Jt − (4QJF − 2Q J − 2Q )t − (2Q Q J + 5QQ F )t −(Q2Q02 + 4Q3JF − Q4J 2 − 3Q2F 2)t − (Q3Q0F − Q4Q0J), 4 A3 = 4Q t. (2.12)

Therefore, by using (2.4) and (2.12) we can show that the coefﬁcient of the highest order t14 of the equation (2.11) is a2J 4 = 0.

Since a 6= 0, we infer that J = 0. Therefore, we can rewrite (2.12) in the form

2 0 2 3 4 0 A1 = −6Q Q t − 7Q F t + Q Q , 02 3 02 2 2 02 2 2 3 0 A2 = 2Q t + 5QQ F t − (Q Q − 3Q F )t − Q Q F, 4 A3 = 4Q t. (2.13)

By (2.13) the coefﬁcient of the highest order t10 of the equation (2.11) is

4a2Q04 = 0, from which Q0 = 0. Therefore, (2.13) becomes

3 A1 = −7Q F t, 2 2 A2 = 3Q F t, 4 A3 = 4Q t,

0 since Q = 0. Hence, from (2.11) we obtain F = 0 and c = 0. Consequently, KII = H = 0. In this case, the surface is minimal, that is, helicoid. ¤

Remark 2.1 : On a non-developable ruled surface in a 3-dimensional Euclidean space,

1. KII = 0 if and only if H = 0.

2. KII = H if and only if KII = H = 0. Theorem 2.2 — Let M be a non-developable ruled surface in a 3-dimensional Euclidean space.

Then, M is a HKII -quadric surface if and only if M is an open part of a helicoid.

PROOF : As is described in Theorem 2.1 we assume that a non-developable ruled surface M is parametrized by x = x(s, t) = α(s) + tβ(s) such that hβ, βi = 1, hβ0, β0i = 1 and hα0, β0i = 0. 286 DAE WON YOON

Suppose that a non-developable ruled surface is HKII -quadric. Then the equation (1.6) implies

aHHt + b(HtKII + H(KII )t) + cKII (KII )t = 0.

From (2.5), (2.7), (2.8), (2.10) and the above equation we have

4 2 aQ B1 + bQ B2 + cB3 = 0, (2.14) where we put

2 5 0 4 2 2 02 3 2 0 0 2 B1 = −J t + 3JQ t + (4QJF − 2Q J − 2Q )t + (2Q Q J − 5QQ F )t +(Q2Q02 + 4Q3JF − Q4J 2 − 3Q2F 2)t + (Q3Q0F − Q4Q0J), 0 6 2 0 0 4 3 2 02 2 2 3 B2 = Q Jt + (7Q Q J + 3QQ F )t + (8Q JF − 8Q Q + 4Q F )t +(3Q4Q0J + 18Q3Q0F )t2 + (8Q5JF − 8Q4F 2 + 4Q4Q02)t − 3Q5Q0(QJ − F ), 2 9 2 2 7 2 0 6 4 2 3 2 2 5 B3 = J t + 4Q J t + 2Q Q Jt + (6Q J + 4Q JF − Q F )t +(2Q4Q0J + 6Q3Q0F )t4 + (4Q6J 2 + 8Q5JF + 6Q4F 2 − 8Q4Q02)t3 −(2Q6Q0J + 16Q5Q0F )t2 + (4Q6Q02 + Q8J 2 + 4Q7JF − 5Q6F 2)t −2Q4Q0(Q4J − Q3F ). (2.15)

By (2.15) the coefﬁcient of the highest order of the equation (2.14) is

cJ 2 = 0.

Therefore, one ﬁnds J = 0 since c 6= 0, which implies (2.15) becomes

02 3 0 2 2 02 2 2 3 0 B1 = −2Q t − 5QQ F t + (Q Q − 3Q F )t + Q Q F, 0 4 2 02 2 2 3 3 0 2 B2 = 3QQ F t + (−8Q Q + 4Q F )t + 18Q Q F t +(−8Q4F 2 + 4Q4Q02)t + 3Q5Q0F, 2 2 5 3 0 4 4 2 4 02 3 B3 = −Q F t + 6Q Q F t + (6Q F − 8Q Q )t −16Q5Q0F t2 + (4Q6Q02 − 5Q6F 2)t + 2Q7Q0F. (2.16)

Similarly to Theorem 2.1, one easily sees that J = 0, F = 0 and Q02(a + 4b + 4c) = 0 since c 6= 0. If a 6= −4(b + c), then Q0 = 0 and the mean curvature H is identically zero by the help of (2.5). Thus, the surface is minimal. ¤

Remark 2.2 : In Theorem 2.2, if a = −4(b + c), then J = F = 0 with arbitrary Q0 implies the equation KII = 2H.

00 Without loss of generality, we may assume β(0) = (1, 0, 0). Thus, by (2.2) β = −β implies

β(s) = (d1 sin s, d2 sin s, cos s + d3 sin s) ON NON-DEVELOPABLE RULED SURFACES IN EUCLIDEAN 3-SPACES 287

2 2 2 2 2 for some constants d1, d2, d3 satisfying d1 + d2 + d3 = 1. Since hβ, βi = 1, we have d1 + d2 = 1 and d3 = 0. From this we can obtain q 2 β(s) = (d1 sin s, ± 1 − d1 sin s, cos s), where −1 ≤ d1 ≤ 1. On the other hand, by (2.1) we have µ q ¶ 2 α(s) = ∓ 1 − d1, d1, 0 f(s) + E, R where f(s) = Q(s)ds and E = (e1, e2, e3) is constant vector. Thus, the surface M has the parametrization of the form µ q q ¶ 2 2 x(s, t) = ∓ 1 − d1f(s) + td1 sin s + e1, d1f(s) ∓ t 1 − d1 sin s + e2, t cos s + e3 , (2.17) R where −1 ≤ d1 ≤ 1, f(s) = Q(s)ds and (e1, e2, e3) is constant vector. For speciﬁc function f(s) and appropriate intervals of s and t in (2.17), we have the graph shown in Figure 1.

FIG. 1. 288 DAE WON YOON

If d1 = 0 or d1 = ±1, then, the surface M is a right conoid. Therefore, a right conoid satisﬁes the equation KII = 2H.

Theorem 2.3 — Let M be a non-developable ruled surface in a 3-dimensional Euclidean space.

Then, M is a KKII -quadric surface if and only if M is an open part of a helicoid.

PROOF : As is described in Theorem 2.1 we assume that a non-developable ruled surface M is parametrized by x = x(s, t) = α(s) + tβ(s) such that hβ, βi = 1, hβ0, β0i = 1 and hα0, β0i = 0.

Suppose that a non-developable ruled surface is KKII -quadric. Then the equation (1.7) implies

aKKt + b(KtKII + K(KII )t) + cKII (KII )t = 0, which implies we obtain from (2.1), (2.3), (2.4), (2.6) and (2.7)

2 8 2 2 4 2 4 2 2 8 2 2 16a Q C1 + c D C3 + 8acQ D C1C3 − 4b Q D C2 = 0, (2.18) where

4 C1 = −4Q t, 5 2 3 2 0 2 4 3 4 0 C2 = 3Jt + (6Q J + 5QF )t − 12Q Q t + (3Q J − 9Q F )t + 2Q Q , 2 9 2 2 7 2 0 6 3 2 3 2 5 C3 = J t + 4Q J t + 2Q Q Jt + (2Q J + 4Q JF + 2Q J − QF )t +(2Q4Q0J + 6Q3Q0F )t4 + (6Q6J 2 + 8Q5JF + 6Q4F 2 − 8Q4Q02)t3 −(2Q6Q0J + 16Q5Q0F )t2 + (4Q6Q02 + Q8J 2 + 4Q7JF − 5Q6F 2)t −2Q4Q0(Q4J − Q3F ). (2.19)

In this case we can also obtain J = F = Q0 = 0 and a = 0. Consequently, the surface is minimal. ¤

Combining the results of Theorem 1.1, our Theorems 2.1, 2.2, 2.3 and main Theorems in [2, 15], we have

Theorem 2.4 — Let M be a non-developable ruled surface in a 3-dimensional Euclidean space. Then, the following are equivalent:

1. M has pointwise 1-type Gauss map.

2. M satisﬁes the equation aH2 + 2bHK + cK2 = constant, a 6= 0, b, c ∈ R, along each ruling. ON NON-DEVELOPABLE RULED SURFACES IN EUCLIDEAN 3-SPACES 289

2 2 3. M satisﬁes the equation aH + 2bHKII + cKII = constant, a 6= −4(b + c), c 6= 0, b ∈ R, along each ruling.

2 2 4. M satisﬁes the equation aK + 2bKKII + cKII = constant, c 6= 0, a, b ∈ R, along each ruling.

5. M satisﬁes the equation aKII + bK = constant, a 6= 0, b ∈ R, along each ruling.

6. M satisﬁes the equation aH + bK = constant, a 6= 0, b ∈ R, along each ruling.

7. M satisﬁes the equation aKII + bH = constant, 2a + b 6= 0, b ∈ R, along each ruling.

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