New examples of surfaces in H3 with conformal normal Gauss map

In this paper we give some examples of surfaces in H3 with conformal normal Gauss map with respect to the second conformal structure and prove some global properties.

by using the second conformal structure on surfaces (see section 1 for the definition).From this, surfaces whose normal Gauss map are conformal maps have been found(see Theorem 1).
The propose of this paper is to classify locally the ruled surfaces with conformal normal Gauss map in H 3 and also give some new examples of complete properly embedded ruled surfaces.The rest of the paper is divided into three sections.The first one describes the definitions of the normal Gauss map in context of this paper and the second conformal structure on surfaces and states Theorem 1.The second section gets two local examples within the Euclidean ruled surface and in the last section some global properties of the ruled surfaces and translational surfaces are proven.

PRELIMINARIES
Take the upper half-space model of the hyperbolic 3-space H 3 = {(x 1 , x 2 , x 3 ) ∈ R 3 : x 3 > 0} with the Riemannian metric ds 2 = 1 x 2 3 (dx 2 1 + dx 2 2 + dx 2 3 ) and constant sectional curvature −1.Let be a connected 2-dimensional smooth surface and x : → H 3 be an immersion of into H 3 with local coordinates u, v.The first and the second fundamental forms of the immersion are written, respectively, as I = Edu 2 + 2Fdudv + Gdv 2 and II = Ldu 2 + 2Mdudv + N dv 2 .The unit normal vector field of x( ) in H 3 is written as where e 2 31 + e 2 32 + e 2 33 = 1.The Gauss equation is Identifying H 3 with the Lie group the multiplication is defined as matrix multiplication and the unit element is e = (0, 0, 1).The Riemannian metric is left−invariant and are the left−invariant unit orthonormal vector fields.Now, the unit normal vector field of x( ) can be written as n = e 31 X 1 + e 32 X 2 + e 33 X 3 .Left translating n to T e (H 3 ), we obtain By the stereographic projection, we get the map → C {∞} , Call g 1 (or g 2 ) the normal Gauss map of surface x( ) (Kokubu 1997).On U 1 U 2 , g 1 g 2 = 1.In this paper, we only consider g 1 and write the normal Gauss map as g : → C {∞}.Then we have Consider an immersion x : → H 3 with Gauss curvature K > −1.By the Gauss equation, we can choose a suitable orientation on x( ) such that the second fundamental form II becomes a positive definite metric on and induces a conformal structure on , which is called the second conformal structure like (Klotz 1963).
will be considered as a Riemannian surface with the second conformal structure.
THEOREM 1 (Shi 2004).Let be a connected Riemannian surface and x : → H 3 be an immersion with Gauss curvature K > −1.Assume that the set of umbilics has no interior point.Then normal Gauss map g : → C {∞} of x( ) is conformal map if and only if the Gauss curvature K and the normal Gauss map g satisfy K = − 4|g| 2 (1+|g| 2 ) 2 > −1.The Weierstrass representation formulas for these surfaces can be found in (Shi 2004).
REMARK.For the Bryant's hyperbolic Gauss map of surface in H 3 (see (Bryant 1987) for the definition), Gálvez, Martínez and Milán (Gálvez et al. 2000) proven that the hyperbolic Gauss map is conformal if and only if the surface is either flat or totally umbilic.

EXAMPLES
We consider the surfaces in H 3 as those ones in R 3 .The simplest examples of surfaces with conformal normal Gauss map are the equidistant surfaces and horospheres x 3 = const > 0, i.e. ordinary Euclidean planes.They are totally umbilics with constant Gauss curvature and constant normal Gauss map.
The following theorem gives all the ruled surfaces with conformal normal Gauss map in H 3 .
THEOREM 2. Up to an isometric transformation of H 3 , every ruled surface with conformal normal Gauss map in H 3 is locally a part of one of the following, (1) equidistant surface with respect to a vertical hyperbolic plane, An Acad Bras Cienc (2006) 78 (1) (2) REMARK.We may check that, among all isometric transformations of H 3 (Korevaar et al. 1992), the horizontal Euclidean translations and rotations (2), the hyperbolic reflections with respect to a vertical hyperplane and the vertical hyperbolic translations preserve the concept of the ruled surfaces and the conformality of the normal Gauss map of any surface with conformal normal Gauss map and the hyperbolic reflections preserve the conformality of the normal Gauss map of all totally umbilics but totally geodesics.
PROOF.Generally, considered as surfaces in R 3 , the ruled surfaces in H 3 can be represented as where D(⊂ R 2 ) is a parameter domain and α(v) and β(v) are two vector value functions into R 3 corresponding to two curves in R 3 .
First, we assume that β is locally nonconstant and without loss of generality assume that where •, • is the inner product in R 3 .Write as e = (0, 0, 1).We have x 3 = x, e and β = 0.The unit normal vector of the surface x(u, v) : D → H 3 is given by where X ∧ Y is the exterior product of the vectors X and By a straight computation, we have, By the Gauss equation K = −1 + L N−M 2 EG−F 2 and (4), we know that Expanding the above formula and noting the linearly independence of 1, u, u 2 and u 3 , we get By ( 7), if β ∧ β , e = 0 at a point p 0 , then β , β ∧ β = 0 in a neighbourhood U of p 0 and the curver β(v) is a geodesic of S 2 .Hence, β ∧ β , e is locally constant and so is globally constant.CASE 1.If β ∧ β , e = 0, then β ∧ β , e = 0. Hence β, β , β and e lie on same plane.We get β, β ∧ β = 0.So the curve β(v) is a plane curve in R 3 , i.e. a unit circle, and the plane π on which β lies passes through the origin of R 3 .By β ∧ β , e = 0, we know π is a vertical plane on which x 3 -axis lies.Revolving π around x 3 -axis, i.e. making an isometric transformation (x 1 , x 2 , x 3 ) → (x 1 cos θ − x 2 sin θ, x 1 sin θ + x 2 cos θ, x 3 ), we may assume that π is the plane , then equation systems (3) and ( 7)-( 10) may be written as Solving (11)(12)(13), we get where c 1 = 0, c 2 ,c 3 and c 4 are constants.Making a translation (x 1 , x 2 , x 3 ) → (x 1 +c 3 , x 2 +c 4 , x 3 ), we get (3) and (4) of Theorem 2.
CASE 2. If β ∧ β , e = 0, then β , β ∧ β ≡ 0. We also get that the curve β is a plane curve in R 3 and the plane π on which the curve β lies passes through the origin of R 3 but is not totally geodesic.Similar to case 1, making a rotation transformation around x 3 -axis, we may assume that x 1 -axis lies on π.Now, β(v)is a unit circle on π .β(v) = (cos v, sin v cos θ, sin v sin θ), where constant θ is the angle between plane x 1 ox 2 and π .By ( 8), we get α , β ∧ β = 0. Multiplying (10) by β ∧ β , e and using ( 9), we get Then, according to whether β , β ∧ α = 0 or not, we again get two cases from equation systems ( 3) and ( 7)−( 10), and In the case (i), taking derivative on α , β∧β = 0, we get α , β∧β = 0.So α ∧α , α = 0, i.e. the torsion of the curve α(v) is zero.We infer that the curve α(v) is a plane curve and the plane on which the curve α lies parallels to π .Hence x satisfies either (1) or (2) of Theorem 2. As for the case (ii), similar to the proof in the case 1, we know that this is a contradictory system of equations.
Next, when β is constant, we have β ∧ e = 0. Or else, by proposition 34 in chapter 7 of (Spivak 1979), we have K ≡ −1.A contradiction.We may assume α , β = 0. ( Thus α(v) is a plane curve and (6) becomes We have β ∧ α , e = 0. Or else, α(v) is a plane curve and lies on a plane parallel to the one spanned by β and e.Then x : D → H 3 is totally geodesic plane H 2 and K ≡ −1.This is contradictory to K > −1.So α , β ∧ α = 0.By ( 14) , we know that α(v) is a straight line in R 3 and x : D → H 3 is either a equidistant surface or a horosphere x 3 = constant > 0. We have proved Theorem 2.
Locally, the ruled surfaces (3) and (4) in Theorem 2 can be represented as a graph is a solution of equation (1), where c 1 = 0 and c 2 are constants.

GLOBAL PROPERTIES
The equidistant surfaces and horosphere x 3 = const > 0 are simply connected, complete, totally umbilics and properly embedded surfaces.Generally, we have the following global properties for the ruled surfaces obtained in theorem 2.
THEOREM 3. The following are simply connected, complete and properly embedded surfaces in H 3 , where c > 0, c 1 > 0 and c 2 > 0 are constants. PROOF.
(1) The map x : D 1 → H 3 is one-to-one.For any compact subset S in H 3 , it is easy to know that the relative closed set x −1 (S)(⊂⊂ D 1 ), as the subset of R 2 , is bounded and closed set of R 2 .Hence x −1 (S) is a compact set.So, x : D 1 → H 3 is a proper map and x(D 1 ) is a complete surface in H 3 .On the basis of this, we can infer that x : D 1 → r (D 1 ) is closed.So x maps D 1 homeomorphically onto its image with the induced topology and x : D 1 → H 3 is embedded.
(2) By a parameter transformation u = ū and v Similar to the proof of (1), we can prove ( 2).

SHUGUO SHI
We have obtained the translational surfaces with conformal normal Gauss map in nonparameter form (Shi 2004), satisfying (1).Now, the parameter form of these translational surfaces is locally given by x(u, v) = (a cos u, b cos v, a sin u + b sin v).(2) For 0 < a < b, the image of the map x(u, v) = (a cos u, b cos v, a sin u + b sin v) : D 2 → H 3 is a complete and properly embedded surface diffeomorphic to S 1 × R, where PROOF.
(1) By a parameter transformation u Similar to the proof of (1) in Theorem 3, we can prove (1).
(2) It is easy to know that image x(D 2 ) is diffeomorphic to S 1 × R. For any divergent curve α : [0, 1) → D 2 , when t → 1−, α(t) tends to either the boundary curve, v = − arcsin a b sin u and v = π + arcsin a b sin u or ∞.For the former, if the length of x(α(t)) is finite, then there exist a compact set S in H 3 containing completely the curve x(α(t)).However, when restricting x(D 2 ) on S, there exists a positive constant ε 0 such that a sin u + b sin v ≥ ε 0 .We may assume ε Noting that 2π 0 a| sin u| a sin u+b du > 0, we have L = +∞.In a word, the length of x(α(t)) is always infinite.Hence the image x(D 2 ) is complete.
Next, we prove that the image x(D 2 ) is a proper surface.When restricting x(D 2 ) on any compact set S in H 3 mentioned above, the first fundamental form of x : D 2 → H 3 satisfies ds 2 ≤ C ε 2 0 (du 2 + dv 2 ), where C is constant and depend only on a and b.So, we infer that the restriction of S to x(D 2 ) is bounded closed subset of x(D 2 ) and hence is compact subset of x(D 2 ).We complete the proof.
REMARK.The curves in Theorem 3, corresponding respectively to v = π 2 on x(D 1 ) and to v = ± π 2 on x(D 2 ) and the curves in Theorem 4, corresponding to u = 0,.u = π, v = 0 and v = π on x(D 1 ) and x(D 2 ) are all geodesics of H 3 follow which K = −1 and accordingly the second fundamental forms are degenerate.Furthermore, every such a geodesic is mapped to a point by the normal Gauss map.
The map x(u, v) = (cos u, cos v, sin u + sin v) : D 1 → H 3 is a simply connected, complete and properly embedded surface, where D 1 is a simply connected open domain of R 2 enclosed by four straight lines v = u ± π, v = −u + 2π and v = −u.
does not tend to the boundary of D 2 , which is contradictory that α(t) tends to the boundary of D 2 .So the length of x(α(t)) is infinite.For the latter, as t → 1−, u(t) → ∞.The first fundamental form of x : D 2 → H 3 satisfiesds 2 = a 2 du 2 + 2ab cos u cos vdudv + b 2 dv 2 (a sin u + b sin v) 2 ≥ a 2 sin 2 udu 2 (a sin u + b) 2 .