The Gauss map and second fundamental form of surfaces in R 3

The Gauss map and second fundamental form of surfaces in R 3 J. A. Gálvez A. Martínez Departamento de Geometría y Topoloía, Facultad de Ciencias, Universidad de Granada, 18071 GRANADA. SPAIN. e-mail: jaalvez@oliat.ur.es; amartine@oliat.ur.es Abstract Given a surface S, a map N from S to S and a conformal structure on S, we solve the problem of existence and uniqueness of an immersion x : S R 3 with Gauss map N and such that the conformal structure on S is the induced by the second fundamental form. 1991 Mathematics Subject Classification: 53A05, 53A07, 58E0. Key Words and Phrases: Gauss map, second fundamental form, Weierstrass representation, harmonic maps. 1 Introduction The properties of the Gauss map on a submanifold in R n and the extent to which the Gauss map determines the immersion of the submanifold have been of reat interest in Differential Geometry see [1], [7], [8], [9], [10], [13], [14], [15], [16]. The existence and uniqueness of an immersion from a surface or a hypersurface into R n with a iven metric or conformal structure and a iven Gauss map has been studied by several authors. D. A. Hoffman, R. Osserman and K. Kenmotsu [8], [9], [10] researched the existence and uniqueness problem for surfaces. They proved there exists a conformal non-minimal immersion from a simply-connected surface in R n with a iven Gauss map if and only if a set of differential equations dependin on the conformal structure and the Gauss map is satisfied. The uniqueness problem for hypersurfaces was also studied by K. Abe and J. Erbacher see [1]. Our object in this paper is to study properties of the Gauss map of a surface immersed Research partially supported by DGICYT Grant No. PB97-0785 1

in R 3, particularly those related to the eometry of the immersion and the conformal structure determined by its second fundamental form. The main problem considered is the existence and uniqueness of an immersion x : S R 3 from a surface S with prescribed conformal structure that yields a iven Gauss map and for which the second fundamental form is a conformal metric on S. Amon the results we obtain are the followin: A For a simply-connected surface with non-zero Gauss curvature, the existence of such immersion is equivalent to that the Gauss map is a local diffeomorphism and a third-order differential equation involvin the conformal structure and the Gauss map is satisfied. Moreover, we recover the immersion by a representation similar to the Enneper-Weierstrass formula. Theorems 3 and 5. B If the set of points where the Gauss curvature vanishes has empty interior then x is uniquely determined, up to similarities, by the Gauss map and the conformal structure iven by the second fundamental form. Otherwise, if the hypothesis about the Gauss curvature is not satisfied, the immersion is, in eneral, non-unique. Corollary 4 and Remark 5. C A hypersurface in R n with non-deenerate second fundamental form has constant Gauss-Kronecker curvature if and only if its Gauss map is harmonic from the hypersurface with the metric iven by its second fundamental form. Thus, surfaces with non-zero constant Gauss curvature can be recovered from each harmonic local diffeomorphism into the unit sphere usin the above mentioned formula, Corollaries, 3 and Theorem 6. E. A. Ruh and J. Vilms proved in [15] similar results involvin constant mean curvature and the conformal structure induced by the first fundamental form. Some of these results will be used in [5] in order to estimate the heiht, area, curvature and enclosed volume of a surface with positive constant Gauss curvature in R 3 boundin a planar curve. Surfaces in R 3.1 Surfaces with Positive Gauss Curvature Let S be a smooth surface and x : S R 3 an immersion with positive Gauss curvature. Then some deleted neihborhood n p of any point p on S lies to one side of the tanent plane T p S to S at p. A smooth unit normal vector field N : S S R 3 is obtained by assinin at each point p of S the unit normal vector to the same side of T p S as n p. This orients S and makes the quadratic form σ associated with the second fundamental

form defined by σ p v, w =< dn p v, w >, p S, v, w T p S, into a positive definite metric. Here <, > is the usual inner product in R 3. Throuhout.1, S will be considered as a Riemann surface with the conformal structure induced by σ. Let z = u + iv be a conformal parameter, E =< x u, x u >, F =< x u, x v >, G =< x v, x v >, e = σx u, x u, 0 = σx u, x v, e = σx v, x v, where e > 0 and, for instance, x u = x 154-155 in [] or p. 143 in [17] state that N u = u. e EG F Gx u + F x v, N v = 1 Then the Weinarten equations see pp. e EG F F x u Ex v. Let us denote by : S C { } the composition of the usual stereoraphic projection with N, that is, = N 1 + in /1 N 3, 3 where N = N 1, N, N 3. We will also call the Gauss map of the immersion. Thus we have: Theorem 1 Let x : S R 3 be an immersion with positive Gauss curvature, K, : S C { } its Gauss map and z = u + iv a conformal parameter. Then x z = i1 + z + 1 + K z, 4 x 3 z = z K z, i u v and by bar we will denote the complex conju- where x = x 1, x, x 3, = 1 z ation. Proof : Use 3 to et x 1 z = 1 z 1 K z, z = 1 N1 + in 1 N 3 z z = 1 N1 + in 1 N 3 z N 3 N 1 + in z N 3 N 1 + in z + N 3 z N 1 + in, + N 3 z N 1 + in, 5 3

where / z = / u + i / v = / z. Now use 1,, N = x u x v x u x v, 6 and K = e /EG F to express the second and third terms within brackets in 5 as products of first partials with respect to u or v of the components of x. After simplification, this ives z = 1 1 N 3 z N 1 + in + K z x 1 + ix, z = 1 1 N 3 Since < N, N >= 1, 3 ives Then 7 can be written or equivalently, z N 1 + in K z x 1 + ix N 1 = +, N = i, N 3 = 4 z = 4 z = z z + K x 1 + ix z, K x 1 + ix z,. 7. 8 z + z = K x 1 + ix z, 9 z + z = K x 1 + ix z. 10 The two first equations of 4 are obtained from 9 and the conjuated equation of 10. Now, since 6 ives < x u, N >=< x v, N >= 0, one has and usin 8 x 3u = N 1x 1u + N x u N 3, x 3v = N 1x 1v + N x v N 3, x 3z = + 1 x 1z i 1 x z. 11 The third equation of 4 follows from 9, 10 and 11. 4

A straiht computation ives us Corollary 1 With the above notation, the first and second fundamental forms of the immersion are iven, respectively, by ds = 4 K z z dz + z z + z z dz z z dz, σ = 4 z z K dz. Now, we study the structure equations of the immersion. Theorem If x : S R 3 is an immersion with positive Gauss curvature K, then the Gauss map satisfies: 4K zz z z = K z z + K z z. E Moreover, the Gauss curvature is determined, up to multiplication by positive constants, by the Gauss map. Proof : From Theorem 1, x iz z = x i z z i = 1,, 3 if and only if 8K + 3 z z + 3 z z + Kz 1 z 1 z + K z 1 z 1 z + 4K 1 zz 1 zz = 0, 1 8K + 3 z z + + 3 z z + Kz 1 + z + 1 + z + K z 1 + z + 1 + z 4K 1 + zz + 1 + zz = 0, 13 8K z z + z z + Kz z z + K z z z + 4K zz zz = 0. 14 If we take 1 minus 13 plus times 14, then we obtain E. Moreover, E and its conjuated equation yield lo K z z + lo K z z = 4 zz z z, lo K z z + lo K z z = 4 zz z z. Now, z z > 0 ives 5

lo K z = 4 z z z zz z zz + z z z z. L From L it is clear that K is determined, up to multiplication by positive constants, by. Remark 1: By Lemma 5 in [1], K > 0 is constant on S if and only if K < x z, x z > is holomorphic. Use the fact that ds and σ satisfy the Codazzi-Mainardi equations on p. 35 of [] or p. 144 of [17]. Since Theorem 3 in [1] shows that K < x z, x z > is holomorphic on S if and only if N : S S is harmonic as defined in [3], [4] or [1], it follows from Theorem that N : S S is harmonic if and only if holds wherever on S. zz z z = 0, Remark : If x : S R 3 is an immersion with constant K > 0, then computation based on Theorems 1 and, or formula 9 from [11] ives σ x = N, where σ = 1/e is the Laplacian for σ and is the usual Laplacian in the u,v-plane. Thus, in a similar way to constant mean curvature see, for instance, [6], the existence of a simply-connected surface with constant Gauss curvature K > 0 and boundary a Jordan curve Γ R 3 is equivalent to solve the followin Plateau problem: x : Ω R 3 such that a x = Kx u x v, b detx uu x vv, x u, x v = 0 = detx uv, x u, x v conformality, c x : Ω R 3 is an admissible representation of the Jordan curve Γ. where Ω is the unit disk and det the usual determinant. Remark 3: The equation L is equivalent to E. And E is satisfied if and only if 1, 13 and 14 are satisfied. Theorem 3 Let S be a simply connected Riemann surface and N : S S R 3 a differentiable map. Then, there exists an immersion x : S R 3 with Gauss map N and such that the conformal structure on S is the induced one by the second fundamental form if and only if z z > 0, 15 { } 4 Im z z z z zz z zz + z z z z = 0, 16 6

where is as in 3. Moreover, the immersion is unique, up to a similarity transformation of R 3 and it can be recovered usin the equations { x 1 = Re 1 z 1 } K z dz + c 1, { x = Re i 1 + z + 1 + } K z dz + c, 17 { } z x 3 = Re 4 K z dz + c 3, where { } 8 lo K = Re z z z zz z zz + z z z z dz + λ, c 1, c, c 3, λ are real constants and the interals are taken alon a path from a fixed point to a variable point. Proof : If S is a Riemann surface with conformal structure iven by the second fundamental form of an immersion x : S R 3, then K > 0, so that lo K and lo K/ z z must both be real. The result follows from Theorems 1, and Corollary 1. Conversely, since S is simply connected, there exists ϕ : S R, such that K = e ϕ, satisfyin L if and only if 16 is satisfied. Now, from Remark 3, it is easy to check that L or equivalently, E is the complete interability condition for 4. Moreover, if x 1, x : S R 3 are two immersions as above with Gauss curvature K 1, K, respectively, then lo K 1 z = lo K z and K 1 = rk for some positive constant r. Thus x z = rx 1z and x = rx 1 + c, c R 3. Corollary Let S be a simply connected Riemann surface. Then S can be immersed in R 3 with constant Gauss curvature and the conformal structure on S is iven by its second fundamental form if and only if there exists a harmonic local diffeomorphism from S to S. Proof : If x : S R 3 is an immersion with constant Gauss curvature K, since S is a Riemann surface with conformal structure iven by the second fundamental form, then K must be positive. Consequently, N is a local diffeomorphism and, from Remark 1, N : S S must be harmonic. Conversely, if N : S S is a harmonic local diffeomorphism, then usin N for N, if necessary, to make z > z for the obtained from 3, both 15 and 16 are satisfied, so the immersion can be calculated usin 17. Finally, from Theorem 3 lo K z = 0, so that K must be a positive constant. 7

. Surfaces with Neative Gauss Curvature Let S be an orientable smooth surface and x : S R 3 an immersion with neative Gauss curvature. Then σ is a Lorentz metric and S can be considered as a Lorentz surface see [17], p. 13. Now, we choose a unit normal vector field N on S compatible with proper σ-null coordinates u, v, so that, σx u, x u = 0, f = σx u, x v > 0, σx v, x v = 0. Any other proper σ-null coordinates û, ˆv are related to u, v by û = ûu, ˆv = ˆvv with û u ˆv v > 0. The Weinarten equations now take the form N u = f EG F F x u Ex v, N v = f EG F Gx u + F x v. 18 Proceedin as in.1, use of 1, 3, 6, 18 and K = f /EG F ives x 1 u = Im1 u K x u = Re1 + u K Im x 3 u = 4 u K x 1 v = Im1 v K x v = Re1 + v K Im x 3 v = 4 v K so that E = 4 u u K G = 4 v v K F = u v + v u K u f = i v v K u. Settin x 1 uv = x 1 vu and x uv = x vu, one obtains the followin analo of Theorem. Theorem 4 Let S be an oriented surface, x : S R 3 an immersion with neative Gauss curvature K, then the Gauss map must satisfy: 4K uv u v = K u v + K v u. Moreover, the Gauss curvature is determined, up to positive constants, by the Gauss map. 8

Remark 4: Fact 3 and Lemma 8 from [1] show that K < 0 is constant on S if and only if K < x u x u > and K < x v, x v > depend only on u and v respectively. Here aain, use the fact that ds and σ satisfy the Codazzi-Mainardi equations. Lemma 6 and Theorem 4 from [1] then show that K < 0 is constant on S if and only if N : S S is harmonic. Thus N : S S is harmonic on S if and only if uv u v = 0 holds wherever. Theorem 5 Let S be a simply connected Lorentz surface and N : S S R 3 a differentiable map. Then, there exists an immersion x : S R 3 such that the structure iven by the second fundamental form is the one iven on S and N its Gauss map if and only if i u v u v > 0, and 1 Im v u u v u uv + v Im u = v = 1 Im v u v u u uv + u Im v, v where is as in 3. Moreover, the immersion is unique, up to a similarity transformation of R 3. And it can be calculated as follows: x 1 = Im1 u du + Im1 v dv + c K K 1, x = Re1 + u K du + Re1 + v K dv + c, Im x 3 = 4 u du + 4 Im v dv + c K K 3, where 8 lo K = Im u v u u uv + u Im v du+ v 8 + Im v u v u uv + v Im u dv + λ, v c 1, c, c 3, λ are real constants and the interals are taken alon a path from a fixed point to a variable point. Corollary 3 Let S be a simply connected Lorentz surface. Then S can be immersed in R 3 with constant Gauss curvature and the conformal structure on S is iven by the second fundamental form if and only if there exists a harmonic local diffeomorphism from S to S. Moreover, the immersion can be calculated as in the above theorem. 9

.3 Uniqueness of the immersion. Now we will prove the followin uniqueness result. Corollary 4 Let S be a connected, oriented surface, χ i : S R 3, i=1,, two immersions with the same Gauss map and conformal structure of the second fundamental form. If the set S 0 = {p S/ dn p is not injective} has empty interior then the immersions aree, up to a similarity of R 3. Proof : Let <, > i be the induced metric on S by the immersion χ i, that is, <, > i = χ i <, >, i = 1,. So, since S 0 is a closed set, if q S 0 then from theorems 3 and 5 there exists a simply connected, open neihbourhood of q on which χ 1 = µqχ +bq, where µq 0 and bq R 3 are constants. Since the interior of S 0 is empty and <, > 1 q = µq <, > q for all q S S 0, the above equality is true everywhere. Moreover, µ is a differentiable function such that dµ = 0 on S S 0. Therefore, µ is constant and µ is constant on each connected component of S S 0 and equal to r or r, with r 0. If p S 0 then there is a neihbourhood U of p such that µ is constant on U S S 0. Otherwise, there would exist two sequences of points p m, p n which tend to p and dχ 1 pm = rdχ pm, dχ 1 pn = rdχ pn. So, we obtain, rdχ p = dχ 1 p = rdχ p. Thus we can assume µ = r on S and since dχ 1 rχ = 0 the proof is completed. Remark 5: If the set of points, where dn p is not injective, has not empty interior then the corollary does not remain true. For instance, we can consider the immersions χ 1 u, v = cos u, 1 sin u, v, cos u χ u, v = 1 4 cos u + 4 sin u, sin u 1 4 cos u + 4 sin u, v, with the same Gauss map and σ 1 = 1 3 17 15 cosu 3 σ. But, from the expression of χ 1 and since χ u, v S 1 R, it is clear that there does not exist a similarity ϕ such that χ 1 = ϕ χ. 3 Hypersurfaces of constant Gauss-Kronecker curvature It is known that the Gauss-Kronecker curvature of a hypersurface is zero if and only if the second fundamental form is deenerate everywhere. Now, we prove that if the 10

second fundamental form is non-deenerate, then the Gauss-Kronecker curvature is constant if and only if the Gauss map is harmonic for the metric iven by the second fundamental form. Theorem 6 Let M n be an orientable n-manifold and x : M n R n+1 an immersion with non-deenerate second fundamental form. We consider M n with the metric σ, then the Gauss-Kronecker curvature of the immersion is constant if and only if the Gauss map N : M n S n is harmonic. Moreover, in that case the Laplacian of N for σ is iven by σ N = nhn, where H is the mean curvature of the immersion. Proof : Let E 1,..., E n be an orthonormal movin frame for σ in a neihbourhood of a point p M n, that is, σe i, E j = ε i δ ij, with ε i = ±1 and δ ij the Kronecker delta, such that σ E i E j p = 0. Let us calculate < σ N, E j > at p: < σ N, E j > = i ε i < E i E i N, E j >= i ε i < Ei Ei N, E j > = i = i ε i E i < Ei N, E j > < Ei N, Ei E j > ε i σe i, Ei E j, 19 where is the Levi-Civita connection of R n+1. If we denote by kl =< E k, E l > and lk the inverse matrix of G = kl, then we have < Ei N, E l >= σe i, E l = ε i δ il, 0 and Ei N = l ε i il E l. 1 Since the Lie bracket [E i, E j ]p = 0, usin Koszul formula, we obtain from 19 and 0 < σ N, E j > = il < Ei E j, E l >= 1 il E i < E j, E l > +E j < E i, E l > i,l i,l E l < E i, E j > = 1 il E j il = 1 i,l traceg 1 E j G. The Gauss-Kronecker curvature K satisfies K = 1/detG and E j lodetg = E jdetg detg = traceg 1 E j G, 11

where det denotes the usual determinant in R n. Thus < σ N, E j >= 1 E jlo K. Therefore, N is harmonic if and only if K is constant. Moreover, from 1 < σ N, N > = i ε i < E i E i N, N > = i ε i < Ei Ei N, N > = i ε i E i < Ei N, N > < Ei N, Ei N > = i,l il < E l, Ei N > = i,l il ε i δ il = i ii ε i = nh. References 1. Abe K. and Erbacher J.: Isometric immersions with the same Gauss map, Math. Ann., 15 1975, 197-01.. do Carmo M. P.: Differential Geometry of curves and surfaces, Prentice-Hall, 1976. 3. Eells J. and Lemaire L.: A report on harmonic maps, Bull. London Math. Soc., 10 1978, 1-68. 4. Eells J. and Lemaire L.: Another report on harmonic maps, Bull. London Math. Soc., 0 1988, 385-54. 5. Gálvez J. A. and Martínez A.: Estimates in surfaces with positive constant Gauss curvature, Proc. Am. Math. Soc. to appear. 6. Hildebrandt S.: On the Plateau problem for surfaces of constant mean curvature, Comm. Pure Appl. Math., 31970, 97-114. 7. Hoffman D. A. and Osserman R.: The eometry of the eneralized Gauss map, Mem. Amer. Math. Soc. 36 Providence, R. I., 1980. 8. Hoffman D. A. and Osserman R.: The Gauss map of surfaces in R n, J. Differential Geometry, 18 1983, 733-754. 9. Hoffman D. A. and Osserman R.: The Gauss map of surfaces in R 3 and R 4, Proc. London Math. Soc., 3 50 1985, 7-56. 1

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