J. geom. 64 (1999) 141 - 149 0047-2468/99/020141-09 $1.50+0.20/0 9 Birlda~iuser Verlag, Basel, 1999

TRANSLATION SURFACES WITH IN 3-DIMENSIONAL SPACES

Journal of Geometry

CONSTANT

IVIEAN CURVATURE

Dedicated to Professor Udo Simon on the occation of his sixtieth birthday Huili Liu*

We give the classification of the translation surfaces with constant mean curvature or constant Gauss curvature in 3-dimensional Euclidean space E 8 and 3-dimensional Minkowski space

E~. 1991 Mathematics subject Classifications. 53 C 42, 53 C 40, 53 C 50. Keywords and phrases. Mean curvature, translation surface, spacelike surface, timelike surface.

0. INTRODUCTION Spacelike constant mean curvature hypersurfaces in arbitrary spacetime have interest in reletivity theory. They are convenient initial surfaces for the Cauchy problem and provide a time guage which is imp9rtant in the study of singularities, the positivity of mass, and gravitational radiation. The surfaces of constant mean curvature or Gaussian curvature in 3-dimensional Euclidean space E s or 3-dimensional Minkowski space E~ have been studied extensively. A representation formula for spaeelike surfaces with prescribed mean curvature has been obtained in [I]; such a representation formula for timelike surfaces has been give in [7]. In [2] and [3], the spacelike and timelike surfaces with constant mean curvature or Gaussian curvature in 3-dimensional Minkowski space E~ have been studied via the theory of finite-type harmonic maps. For the study of the surfaces theory in E a or E13, it is a very important and interesting problem to construct the constant mean curvature or Gaussian curvature surfaces. In this paper, to develop a corresponding work for the translation surfaces with zero mean curvature in 3-dimensional space in [5], we classify the translation surfaces *The author is supported by the EDU. COMM. University.

of CHINA, the NSF of Liaoning and the Northeastern

142

Liu

of constant mean curvature or Gaussian curvature in 3-dimensional Euclidean space E a and Minkowski space E~ with the classical methods. We will prove: T h e o r e m 1. Let S be a translation surface with constant Gauss curvature K in 3 - dimensional Euclidean space E ~ or 3-dimensional Minkowski space E~. Then S is congruent to a cylinder, so K = O.

T h e o r e m 2. (1) Let S be a translation surface with constant mean curvature H ~ 0 in 3-dimensional Euclidean space E 3. Then S is congruent to the following surface or a part in E~:

(a)

z=

2H

~/1 - 4 H 2 x 2 - ay,

a E R;

(2) Let S be a translation surface with constant mean curvature H ~ 0 in 3-dimensional Minkowski space E31. Then (i) if S is spaceIike, it is congruent to the following surfaces or a part in E~: (b)

v/l'a

2

z = ~ 2 - - - w - - J 1 + 4H~x2 - ay,

lal < 1

or

(c)

x = ay

v ~2+~l

v/4H2z2 - 1

or

(d)

x - "v/~/~ - 1 ~/1 + 4H2y 2 - az,

lal

> 1;

(ii) if S is timelike, it is congruent to the following surfaces or a part in E~: -~/1

(e)

z -

-

2H

a2 v/4H2x 2 - 1 - ay,

lal < z

or

(f)

z=

vz~ T l x / 1 2H

4Hex 2 -

ay,

[al > 1

or

(g)

x = ay + ~ - - a+- - x / 1 2H

+ 4H2z 2

or

(h)

x--

-x/a 2-1V/4H2y 2_l_az, 2-H

lal>l

~ / 1 - a 211/1_4H2y 2 _ a z ,

lal < l.

or

(i)

Liu

Remark.

143

For the translation' surfaces with zero m e a n curvature, we have (cf. [5]):

T h e o r e m 3. (1) Let S be a translation surface with zero mean curvature in 3-dimensional Euclidean space E ~. Then S is congruent to the following surface or a part in ES: 1 1 z = - log(cos(ax)) - - log(cos(ay)), a a

0 r a E R;

(2) Let S be a translation surface with zero mean curvature in 3-dimensional Minkowski space E~. Then (i) if S is spacelike, it is congruent to the spacelike part of the following surfaces in E~:

(a)

z = - log(cosh(ax) ) -

log(cosh(ay))

a

or

(b)

x = -log(cos(ay)) -

log(sinh(az));

a

(ii) if S is timelike, it is congruent to the timelike part of (a), (b) or the following surfaces (or a part) in E~: (c)

z = _1 log(sinh(ax)) _ _1 log(cosh(ay)) a

a

or

(d)

z = la log(cosh(ax)) - 1 log(sinh(ay))

or

(e)

z = _1 log(sinh(ax)) - -1 log(sinh(ay)) a

a

or

(f)

1 1 x = -log(cos(ay)) - -log(cosh(az)). a

a

i. PRELIMINARIES Let E 3 be the 3-dimensional Euclidean space with the metric < , >__dx 2+@2+dz

2

and E~ the 3-dimensional Minkowski space with the metric

< , > = dx 2 _~. @2 _ dz 2.

144

Liu

We denote a surface S in E ~ or E~ by

~(~,~)={x(~,v),y(~,v),z(~,~)}. Then the first fundamental form I of the surface S is defined by I = E d u 2 + 2Fdudv + Gdv 2, ,

,

E:,

,

F=
,

,

v >,

G=<%,r

,

v>,

, ru-

ar(u,v) cOu

For the surface in E a, E G - F 2 > 0; for the spacelike surface in E~, E G - F 2 > 0; for the timelike surface in E~, E G - F 2 < 0. We define the second fundamental form I I of S by I I : Ldu 2 + 2 M d u d v + N d v 2,

1

L --

M

det(r~, ' rv, ' r ~' ) ,

~/IEG- F21 1

--

I I I det (r~, %, r~v),

~/IEG- F2I N

1

--

I I t det (r~, %, %~).

~/IEa- F=I

Then the Gauss curvature K and the mean curvature H of S is given by K-

LN

-

EG-

M 2

F 2'

H=

EN - 2FM - GL 2(EG-

F 2)

The surface S in E 3 or E~ is called a translation surface, if it can be written as z=g(x)-h(y)

(see [%

2. THE

or y - - g ( z ) - h ( x )

or x = - g ( y ) - h ( z )

[5]).

PROOF

OF

THE

THEOREM

1

In Euclidean space E 3, by a transformation in E ~, the translation surface S can be written as

(2.1)

z = g(x) - h(y).

Then the Gauss curvature of S is given by (2.2)

-g"(x)h'(y) K = (1 + g,(x)2 + h'(y)2)~"

Liu

145

I f / ( is constant, from (2.2), when 9"(x) # O, we have h"'(y)(1 + g,(~)2 + h,(y)~) _ 4h,(y)h,,(y)~ = 0; when h ' ( y ) # O, we have 9"(x) (1 + g'(x) 2 + h'(y) 2) - 4g'(x )g"(x ) 2 = O. Hence g"(x) = 0 or h"(y) = O. Assume that g(x) = ax + b, a, b E R , then r ( z , y) = (x, y, ax + b - h(y)) = (0, y, b - h(y)) + x(1, 0, a).

The surface is a cylinder. In Minkowski space E~, by a transformation in E~, the translation surface S can be written as

(2.3.a)

z = g(~) - h(y)

or

x=g(v)-h(z).

(2.3.b)

Then the Gauss curvature of S is given by K=

(2.4.a)

-g"(x)h"(y) (1-r

or

-g"(y)h"(z)

K =

(2.4.b)

(h,(z)2 _ g,(y)2 _ 1)2"

If K is constant, from (2.4) we have also g" = 0 or h" = 0. The surface is a cylinder. This completes the proof of Theorem 1. Q . E . D

3. T H E P R O O F

OF T H E O R E M 2

Let S be the surface with constant mean curvature H ~ 0 in Euclidean space E a. From (2.1) we have (3.1)

H = 9"(x)(1 + h'(y) 2) - h"(y) (1 + 9'(x) 2)

2(1 + r

+ h,(y)2)~

Differentiating (3.1) with respect to x and y, we obtain: 0 = (g"(1 + h '2) - 2 J J ' h " ) (1 + g,2 + h,2)_~ , , 9 - 3 g g (g , (1 + h ,2) _ h"(1 + J 2 ) ) ( 1 + g,2 + h,2)_~

= (J"(1 + h '2) - 2JJ'h")(1 + g,2 + h,2)_ ~ - ~ H g g ' " (1+g'2 + h,~)_l

146

Liu

t h a t is

(g"(1 +

h '2) - 2g' g'h')(1 + g,2 + hn)_89 = 6Hg' g'; 0 = (2h'h'g" - 2g'g'h")(1 + g,2 + h,2)-89 -h'h"(g"(1 + h '2) - 2g' g"h")(1 + g,2 + h,2)-~

= (2h%'g" - 2g'g'h")(1 + g,2 + h,2)_89 _ 6Hg'g"h'h'(1 + g,2 + h,2)_1, t h a t is

(2h'h'g'" - 2g'g'h")(1 + g,2 + h,~)89 _ 6Hg'g'h'h" = O. We assume t h a t g'(x) • 0 and h'(y) ~ O. T h e n g" 3H = (g~,

(3.2)

h"

-

h~-~h,)(l+g'~+h'2) 89

Differentiating (3.2) with respect to x, we have

d " , , , tl + g-~)

g ,2 + h,2)~ + 3Hgtg" = O.

Hence 9"(x) ~ 0 and h'(y) ~ 0 yield H = 0. It contradicts to the a s s u m p t i o n H r 0. Therefore we o b t a i n t h a t g" = 0 or h" = 0. By a t r a n s f o r m a t i o n in E 8 we can assume t h a t h" = 0 and write h(y) = ay. From (3.1) we have

(3.3)

d'(~)(1 + ~ ) = 2H(1 + ~ + d(~)~)~.

Solving this equation we get

(3.4)

~/1 + a 2

g(x) = -

2H

~/1 - 4H~(~ + ~)~ + ~ , ~1, ~ 9 a .

T h e r e f o r e the surface is

(3.5)

z-

~/l+a 2

It is congruent to the surface (a) given by Theorem 2. Let S be the surface with constant mean curvature H ~ 0 in Minkowski space E~. From (2.3) we have: (3.6.a)

H ---- g ' ( x ) ( 1 ~ E(y) 2) - h ' ( y ) ( 1 - g~(x) 2)

for

z = g(x) - h(y);

for

x = g(y) - h(z).

211 - g'(x) ~ - h'(y)21~

(3.6.5)

H =

J ' ( ~ ) ( h ' ( z ) 2 - 1) - h " ( z ) ( 1 + g ' ( y ) 2 ) 21h,(z)2 _ g,(y)~ :- 11.~

Liu

147

We assume that 9" ~: 0 and h" ~ O. By (3.6.a), differentiating with respect to x and y we get (3.7.a)

C -g'(x)

3H = , g ~ )

h"(y)

+ h,~y))

.

1 - g'(x) 2 - h'(y)~l 89

LFrom (3.6.b), differentiating with respect to y and z we get (3.7.5)

-3H = (g"(Y)

9'(Y)g"(Y)

h'(z) h,(z)h,,(z))lh,(z)2 _ 9,(y)2 _ 1[89

(3.7) yields also H = 0. It contradicts to the assumption H r 0. For the surface (2.3.a), by a transformation in E~ we can assume that h" = 0 and write h(y) = ay. From (3.6.@ we have

(3.8.1)

g"(x)(1 - a 2) = 2H(1 - a N - g'(x)2)~

or

(3.s.2)

J'(x)(1 . a 2). . 2H(g'(x) . ~

(1

a))~.~-~

Solving these equations we obtain ~/1

(3.9)

g(x) -

-

~r

a2

~/1 + 4H2(x + cl) 2 + c2, cl,c2 9 R,

lal < 1,

the surface is spacelike and congruent to the surface (b) given by Theorem 2; (3.10)

g(x) - - , , / 1 - a S ~/4H2( x + cl)2 _ 1 + c2, cl, c2 9 R,

2H

lal <

1,

the surface is timelike and congruent to the surface (e) given by Theorem 2;

(3.11)

g(~)-

v~-

7#

11/1-4H2(x+cl)2+c2,

cl,c2 9 R,

I~1 > 1,

the surface is timelike and congruent to the surface (f) given by Theorem 2. For the surface (2.3.b), when 9" = 0 by a transformation in E~ we can write g(y) = ay. From (3.6.b) we have

(3.12.1)

- h " ( z ) ( 1 + a 2) = 2 H ( h ' ( z ) 2 - a 2 - 1)~

or

(3.12.2)

- h " ( z ) ( 1 + a 2) = 2 H ( a 2 + 1 - h'(z)2)~.

Solving these equations we obtain (3.13)

h(z) -

l v ~ - - ~ / 4 H ~ ( z + cl)~ - 1 + c~, ~1, ~ e R , 2H

148

Liu

the surface is spacelike and congruent to the surface (c) given by Theorem 2; (3.14)

h(z) - - v / 1 + a2~/4H2(z + cl) 2 + 1 + c2, cl,c2 9 R , 2H

the surface is timelike and congruent to the surface (g) given by Theorem 2. When h" = 0 we assume that h(z) = az. by (3.6.b) we have (3.15.1)

g"(y)(a 2 - 1) = 2H(a 2 - 1 - g,(y)2)+

or y'(y)(a 2 - 1) = 2H(g'(y) 2 + 1 - a2)~.

(3.15.2)

Solving these equations we obtain

(3.16)

g(Y)- 4 J 7H l~/l+4H2(y+ci)2+c~' cl,c2eR, la[>l,

the surface is spacelike and congruent to the surface (d) given by Theorem 2; (3.17)

g(Y) -

~-H- 1 j 4 H ~ ( y + cl) 2 - 1 + c~, Cl, c~ e R,

lal > 1,

the surface is timelike and congruent to the surface (h) given by Theorem 2; (3.18)

g(y)-

~/1~ - a~ v/1 _ 4H2(y + Cl)~ + c2, cl, c~ e R,

lal < 1,

the surface is timelike and congruent to the surface (i) given by Theorem 2. This completes the proof of Theorem 2. Q.E.D

REFERENCES [1] K. AKUTAGAWA AND S. NISHAKAWA,The Gauss map and spacelike surfaces with prescribed mean curvature in Minkowski 3-space, Tohoku Math. J., 42(1990), 67-82. [2] J. INOGUCHI, Spacelike surfaces and harmonic maps of finite type, preprint, Tokyo Metropoitan University. [3] J. INOGUCHI, Timelike surfaces of constant mean curvature in Minkowski 3-space, preprint, Tokyo Metropoitan University. [4] H. L. LIU, Minimal immersions of Pseudo-Riemannian manifolds, Tsukuba J. Math.,

16(1992), 1-1o. [5] H. L. LIu, Translation surfaces with dependent Gauss and mean curvature in 3-space, J. NEUT, 14(1993), 88-93.

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149

[6] H. L. LIu AND G. L. LIU, Rotation surfaces with constant mean curvature in 4dimensional Pseudo-Euclidean space, Kyushu J. Math., 48(1994), 35-42. [7] M. A. MACID, Timelike surfaces in Lorentz 3-space with prescribed mean curvature and Gauss map, Hokkaido Math. J., 19(1985) 447-464. [8] T. K. MILNOR, Harmonic maps and classical surfaces theory in Minkowski 3-space, Trans. AMS, 280(1983), 161-185. [9] B. O'NIELL, Semi-Riemannian Geometry, Academic Press, Orland, 1983.

[10] A. TI~EIBERGS,Entire spacelike hypersurfaces of constant mean curvature in Minkowski space, Invent. Math., 66(1982), 39-56. [11] A. TREIBERGS, Gauss map of spacelike constant mean curvature hypersurfaces of Minkowski space, J. Diff. Geom., 32(1990), 775-817.

Huili Liu Department of Mathematics Northeastern University Shenyang 110006 P. R. China liuhiQramm.neu.edu.cn Eingegangen am 25. Novenber 1997