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Anais da Academia Brasileira de Ciências

Print version ISSN 0001-3765

An. Acad. Bras. Ciênc. vol.85 no.4 Rio de Janeiro  2013  Epub Nov 10, 2013

http://dx.doi.org/10.1590/0001-3765201376911 

Mathematical Sciences

The Gauss Map of Complete Minimal Surfaces with Finite Total Curvature

PEDRO A. HINOJOSA1 

GILVANEIDE N. SILVA2 

1Universidade Federal da Paraíba-UFPB, CCEN, Departamento de Matemática, 58051-900 João Pessoa, PB, Brasil

2Universidade de Pernambuco-UPE, Rua Amaro Maltez, 201, 55800-000 Nazaré da Mata, PE, Brasil

ABSTRACT

In this paper we are concerned with the image of the normal Gauss map of a minimal surface immersed in ℝ3 with finite total curvature. We give a different proof of the following theorem of R. Osserman:The normal Gauss map of a minimal surface immersed in3 with finite total curvature, which is not a plane, omits at most three points of𝕊2

Moreover, under an additional hypothesis on the type of ends, we prove that this number is exactly 2.

Key words: Gauss map; minimal surfaces; Finite total curvature; Image of the Gauss map

RESUMO

Neste trabalho, estuda-se a imagem pela aplicação normal de Gauss de uma superfície mínima imersa em ℝ3com curvatura total finita. Apresentamos uma prova diferente do seguinte Teorema de Osserman: A aplicação normal de Gauss de uma superfície mínima imersa em3 com curvatura total finita, que não é um plano, omite no maximo três pontos de 𝕊2: A seguir, usando uma hipótese adicional sobre o tipo de fins, provamos que este numero é exatamente dois.

Palavras-Chave: aplicação de Gauss; superfícies mínimas; curvatura total finita; Imagem da aplicação de Gauss

INTRODUCTION

Complete minimal surfaces immersed or embedded in ℝ3constitute a honorable field of research. In particular, many geometers studied with special emphasis the Gauss map properties of such surfaces. Several important results have been shown and many other problems have emerged, among them the following one: let X : M → ℝ3 be a complete non-planar minimal immersion and let N : M → 𝕊2 ≅ ℂ ∪ {∞} be its Gauss map. How many points can be omitted by N?

In 1961 R. Osserman (Osserman 1961) proved that N does not omit more than one set of logarithmic capacity zero. In 1981 F. Xavier (Xavier 1981) proved that N can omit at most six points. In 1988, H. Fujimoto (Fujimoto 1988) refined Xavier's result showing that Nomits at most four points. This is the best possible result since there are complete immersed minimal surfaces in ℝ3 whose Gauss map omits exactly four values. An example of such a surface is the Scherk's surface and other examples can be found in the book of R. Osserman (Osserman 1986). However, these examples have infinite total curvature. This is an expected fact since R. Osserman (Osserman 1964) proved that ifM has finite total curvature, then N omits at most three points. Until now examples of complete minimal surfaces with finite total curvature whose Gauss map omits three points of 𝕊2 are not known. Hence, the maximum number of points that can be omitted by N is either two or three. By trying to answer this question we naturally ask what are the conditions for the existence of such a surface. However little progress has been made in this direction since the question was posed by R. Osserman. We mention that a result of R. Osserman (Osserman 1961) states that if M is a complete, orientable minimal surface in ℝ3; then its total curvature is a multiple of −4π: We recall that even if a minimal surfaceM is not orientable, its oriented double covering is an orientable minimal surface. In that way we can study the Gauss map of . Here we consider only the orientable case. In 1987, Weitsman and Xavier (Weitsman and Xavier 1987) proved that if N omits three values, the total curvature ofM is less than or equal to −16π:

In this paper we give a new proof of the Osserman's theorem in the case of finite total curvature (Osserman 1964). We believe that a simpler proof, using fewer arguments of complex variables, could lead to a better understanding and even to the possibility of improving the result. Our proof uses the Weierstrass representation, the fact that M is a compact surface minus a finite number of points and the Poincaré's theorem about the sum of indices of a differentiable vector field with isolated singularities on a compact surface.

We consider certain particular functions on the surface and we apply the Poincaré's theorem to the gradient vector field of these functions. This permits us to obtain relations between the Euler characteristic of ; the number of ends, the index of each end, the total curvature ofM and the number of points omitted by the spherical image ofM: Finally, under additional hypothesis on the type of ends of the surface, we obtain results that improve the Osserman's theorem.

Let X : M→ ℝ3 be an isometric minimal immersion of a Riemann surface M into ℝ3: Let N be Gauss map and C (M) := ∫M KdM the total curvature of M. IfM has finite total curvature (ie |C(M)| < ∞), then we know the following results (see Osserman 1986).

1.1. The Riemann surface M is diffeomorphic to a compact surface of genus γ minus a finite set points, that is,

1.2. The Gauss mapping N extends continuously to a function : So, is a branched covering of 𝕊2(1);

1.3. If m represents the number of times that covers each point of 𝕊2 then the total curvature of M is given by −4πm:

Let π: 𝕊2→ ℂ ∪ {∞} be any identication of 𝕊2 withℂ ∪ {∞}. Represent by g the compositiong:=π ο N and set .

For each choice of π there is a meromorphic one-formω on satisfying:

1.4. at each point of M where g has a pole of order v the form ω has a zero of order 2v;

1.5. the form has no real periods on M;

1.6. the Weierstrass representation gives the following integral expression X = 2ℜ (∫zα).SinceM is complete with the induced metric, thenα must have poles at each point of . We know that (see, for example, Osserman 1986):

1.7. the poles of α have order greater than or equal to two.

The points of , or punctured neighborhoods of such points, are calledends of M. L. Jorge and W. Meeks (Jorge and Meeks 1983) have shown that:

1.8. for ρ > 0 sufficiently large, the set Mρ := {pM, |X (p)| ≥ρ} is a disjoint union of closed sets E 1, E 2, ..., En , where each Ei is an end of M and Vi := Ei ?{pi } is a disk neighborhood of pi in for i = 1, ..., n;

1.9. if γiρ: 𝕊1 → M parametrizes ∂Vi and , then Ciρ converges, in the C topology, to the great circle of 𝕊2(1) perpendicular to covered I(pi) times;

1.10. the number I(pi) is an integer greater than or equal to one.

If I(pi) = 1 then, for ρ sufficiently large, X (Ei) is graph of a function f : 𝒰i → ℝ, where 𝒰i is some neighborhood of infinite in the subspace perpendicular to :

HEIGHT FUNCTIONS

For each ξ ∈ 𝕊2(1)we define:

Hence, a simple computation yields:
where ξT (p) means the orthogonal projection ofξ into TpM. The singularities of this vector fields are the points in N –1 {ξ, − ξ}.

Proposition 2.1. If p ∈ N–1 {ξ, − ξ} then grad(fξ)(p) = 0 and the index of this singularity is −v(p), where v(p) is the order of p as a zero of g − g(p).

Proof. By a change of coordinates in ℝ3; if necessary, we may suppose that ξ = ±(0, 0, 1). Consider the identication AQUI π : 𝕊2(1) → ℂ ∪ {0}, for which it holds that π(0, 0, 1) = ∞ and π (0, 0, −1)=0. As N(p) = ±ξ we haveg(p) = 0 org(p) = ∞. We may suppose that g(p) = 0. Letz = u + iv be a local complex parameter of M such that pcorresponds to z = 0. Then, in a neighborhood ofp, we have:

It follows from (1.5) and (1.6) that:
where h is a holomorphic function such thath(0) = 0 andh′(0) ≠ 0: Hence, we get:
and
what implies that the index of grad(fξ ) at p is −v(p).

In order to determine the behavior of the vector fieldgrad (fξ ) at the points we consider its restriction to the curves γiρ defined in (1.9).

Proposition 2.2. For ρ sufficiently large, the index of grad(fξ ) along γiρ is

Proof. Choose coordinates on ℝ3 so thatξ = ± (0,0,1) and ĝ (Pi) ≠ ∞. With respect to the local parameter z =u + iv around pi we have:

It follows from (1.7) thatµ(pi ) ≥ 2. As in the proof of the Proposition 2.1 we have:
It follows that one of the following cases can occur:

  1. . In this case, ĝ(0) ≠ 0 and we can write where and h 1(0) ≠ 0.

  2. . Now ĝ (0) = 0 and where and h 2(0) ≠ 0.

These cases correspond respectively to:

  1. (A′) and

  2. (B′) , where n 1 = −µ, n 2 = vµ, the functions Wi , i = 1,2 are holomorphic and W'1 (0) ≠ 0.

As before, consider the gradient vector field

Then, take large enough so that γiρ(𝕊1) is contained in the neighborhood of pi we are considering. It follows that the index ofgrad(fξ ) along γiρ is µ in the case (A) and vµ in the case (B).

The following lemma completes the proof of Proposition 2.2.

Lemma 2.3. If , ĝ(p) ≠ ∞ and ω has a pole of order µ at p, then I(p) =µ − 1.

Proof. We may choose coordinates in 𝕊2(1) in such a way that ĝ(p) = 0. By using (1.6) we have:

where f 1 and f 2 are power series with at most terms in .

Setting z = re and we obtain

Therefore, for ρ sufficiently large,X 1 + iX 2 takes the circle of radius r centered atp into a closed curve that rotates (µ − 1)-times around the origin, that isI(p) =µ − 1.

Corollary 2.4. Let be a complete minimal isometric immersion with finite total curvature. If m is the number of times that covers 𝕊2(1) then:

Choose ξ such that {−ξ, ξ} does not intersect . We have:
If the genus of is γ, then . Hence, we obtain . Since I(pi ) ≥ 1 for all i∈ {1,2, ...,n} it follows that:
Note that the equality holds if and only ifI(pi ) = 1 for all i∈ {1,2, ...,n}. In this case every end of M are embedded. Moreover, (5) is equivalent to:

γ ≤ m - n+1.

SUPPORT TYPE FUNCTIONS

Now we consider, for each ξ ∈ 𝕊2, the support type function

A simple computation shows that the singular points of grad(𝔰ξ) are the points whereξT = 0 or K = 0 (branch points ofg). By a change of coordinates in ℝ3 we may always assume that ξ = (0, 0, 1).

Since:

(see, for example, Barbosa and Colares 1986), we have:

Consider a point p such that . In terms of local coordinates z =u+iv; with pcorresponding to z = 0; we have ; ; when v is a nonzero integer.

If v > 0 then 𝔰ξ(0) = -1 and

The point p is then a local minimum for 𝔰ξ with index + 1.

If v < 0 then 𝔰ξ(0) = 1 and

The point p is then a local maximum and so its index is also +1.

When , then , , where v ≥ 1. It follows that and

Considering z = re one obtains

where . Setting ā0b0 = Re, we have:

From this expression it follows that the difference 𝔰ξ (Z) ― 𝔰ξ (0) changes sign in any neighborhood of z = 0. It follows easily that the index of grad(𝔰ξ) at p is 1 − v(p).

Therefore we have proved:

Proposition 3.1. If p is a critical point of 𝔰ξ bof index 1. If and p is a critical point of 𝔰ξthen g has a branch point of order v at p and the index of grad(𝔰ξ) at p is 1 − v.

THE OSSERMAN'S THEOREM

The following theorem was proved by Osserman (Osserman 1964) using a different method.

Theorem 4.1. Let be a complete minimal isometric immersion with finite total curvature, − 4πm. Suppose that { ξ1, ξ2, ..., ξk } = 𝕊2 (1) \ g(M). If k ≥ 4, then M is flat.

Proof. Set:

and C := {p ∈ M : v(p) > 1 } Assume M is not flat and let ξ ∈ g(M) be a regular value of g. Counting indices of the singularities of grad(𝔰ξ) we obtain:
Observe that and . Hence,
Now using 2.4, we have . Using (9) it follows that:
Thus (4 − k)m > 0. Thereforek < 4.

Corollary 4.2. Let be a complete minimal isometric immersion with finite total curvature −4πm. If N : M → 𝕊2 (1) omits 3 points then . Moreover, if we have:

  1. m = n;

  2. B = C = ϕ;

  3. The ends of M are embedded.

Proof. It follows from equation (9) that ifk = 3 then:

On the other hand, from (5) we have: . Therefore nm ≤ 0, where . Moreover, if then m = n andB = C =ϕ. It is consequence of (5) that the endsM are embedded.

Corollary 4.3. Under the same hypothesis of the previous corollary it holds that ifN : M → 𝕊2omits 3 points then m ≥ 3:

Proof. From the proof of the previous corollary we have . Hence mn: However, for each point q ∈ 𝕊2 \ M (M)there is at least one point such that . Hence, we have n ≥ 3.

RESULTS

Now we study the behavior of M at infinity. For this purpose we consider as before and let (g,ω) be the the corresponding Weierstrass data. After a rotation we may assume, for j fixed, that ĝp(j) = 0 . Let z be a local parameter in an open subset with pj∈ D such that (pj ) = 0. Hence, locally we have:

Moreover, as the immersion is complete it follows that fhas pole at pj whose order is greater than or equal to 2. Hence we have nearpj that:

By Lemma (2.3) we haveI(pj ) = µ − 1: Thus,pj is an embedded end if and only ifµ(pj ) = 2. Suppose that pj is an embedded end. Then α 3:= = gfdzand

Hence, if v > 1 then gf is holomorphic and X3(p) = ℜ(∫P⨍ω) satisfies

Thus, the end Fj is asymptotic to the plane {X 3 = L}. In this case, we say thatFj is a planar end of order v − 1:

If v = 1, thengf(z) = a 0 b 0 z –1+ (holomorphic function). Since: ℜ(∮cgfdz) = 0 for any closed curve C ⊂ M, we have ℜ(2πia0b0) = 0 and then a0b0 ∈ ℝ. Therefore: X3 = a0b0 |z| +R(z), whereR is an harmonic function in a neighborhood of 0 ∈ ℂ and R(0) = 0. The numbera 0 b 0 is the logarithmic growth of the end Fj . In this case the end Fj is of catenoidal type. This means geometrically that Fj is asymptotic to a catenoid in ℝ3.

In the general case we have . Now consider N 0 := vµ. IfN 0 ≥ 0 then fg is holomorphic. Otherwise, ifN 0 < 0 then holomorphic function.

Denition 5.1. Let be a end of M. We say that is a non-degenerate end if, and only if, N 0(p) :=v(p) −µ(p) ≤ 0.

For non-degenerate ends we have:

Theorem 5.2. Let be a complete non-flat minimal immersion with finite total curvature and non-degenerate ends. If the Gauss map omit k points, then k ≤ 2:

Proof. By the Osserman's theorem (4.1) we know thatk ≤ 3: Suppose that k = 3 and letξ 1, ξ 2 and ξ 3 be the omitted points. As in the proof of the Theorem 4.1 consider the sets:

and

Counting the indices of the singularities ofgrad(fξ ) we obtain:

  1. in the case ξ 2 = −ξ 1 :

  2. in the case ξ 2 ≠ −ξ 1 and ξ 3 ≠ −ξ 1:

Moreover, since N 0 ≤ 0 we have:

In both cases, we conclude what contradicts Corollary 4.2. Therefore k= 2.

In the case k = 2 with non-degenerate ends we have:

Theorem 5.3. Let be a complete non-at minimal immersion with finite total curvature and non-degenerate ends. Suppose that the Gauss map omits two points. Then M is topologically a sphere minus two points.

Proof. Let ξ1, ξ2 ∈ 𝕊2 be the two points omitted. Similarly to the proof of Theorem 5.2 we have:

  1. in the case ξ 2 = −ξ 1:

  2. in the case ξ 2 ≠ −ξ 1 :

In both cases, what implies that is homeomorphic to 𝕊2 and therefore: M = 𝕊2 \ {p1; p2, ...., pn}

We claim that n ≤ 3.

By contradiction, suppose that n > 3 and letp 1, p 2, p 3, p 4 four distinct points in . Consider two simple and differentiable curvesα 1 and α 2 joining p 1 to p 2 and p 3 to p 4 respectively. Assume also that the curves α 1 and α 2 do not intersect each other and are contained in Mexcept for its endpoints.

Cutting M along the curves α 1 and α 2 we obtain another surface topologically equivalent to a cylinder. Joining two copies of the later surface along the curves α 1 and α 2 we obtain another compact surface of genus 1 and 2(n − 4) + 4 points removed.

Denote by M′ this new surface and letR : M′ → M the canonical covering map. Now X ο R : M'→ ℝ3 is a minimal immersion with the same image ofX. The Gauss map of X ○ R is given by N' = N ○ R and has the same image as the Gauss map of X. For the ends of M′ we still have N 0 < 0. In fact, the ends that are not endpoints of the curvesαi , i = 1, 2, remain with the same local behavior and hence the same I(p) andv(p). Thus the valueN 0 = vµ =vI −1 remains unaltereted. However when the ends are endpoints of the curves αi , i = 1, 2, some changes occur. For instance, a neighborhood of p 1 now covers twice a old neighborhood. It follows thatI(p 1) is changed for 2I(p 1) and g(z) is changed for . Therefore, v(p 1) becomes 2v(p 1). Since I(p 1) − v(p 1) is changed for 2I(p 1) − 2v(p 1) the inequality I(p 1) − µ(p 1) + 1 > 0. is preserved. Thus X ○ R : M' → ℝ3 is an immersion whose ends are not degenerate. Moreover, this immersion satisfies the hypothesis of the theorem. From the first part of the proof we have . This is a contradiction because . Hence n ≤ 3 as we claimed.

Now we claim that n = 2.

Clearly n ≥ 2 (nk = 2). Assume n = 2 and let p 1, p 2 and p 3 be the ends of M. Consider a curveα joining p 1 to p 2 as before.

The previous construction of M′ now leads to a surface with n = 4 which contradicts n ≤ 3. Therefore, n = 2.

Acknowledgements

Work partially supported by “Programana Nacional de Cooperação Acadêmica (PROCAD-NF)” and Grant 620108-2008-8.

REFERENCES

Barbosa JL and Colares G. 1986. Minimal Surfaces in ℝ3: Lectures Notes in Math. Vol. 1195, Springer-Verlag. [ Links ]

Fujimoto H. 1988. On the number of exceptional values of the Gauss map of minimal surface. J Math Soc Japan 40: 235-247. [ Links ]

Jorge LP and Meeks WH. 1983. The topology of complete minimal surfaces of finite total Gaussian curvature. Topology 22(2): 203-221. [ Links ]

Osserman R. 1961. Minimal Surfaces in the Large. Comm Math Helv 35: 65-76. [ Links ]

Osserman R. 1964. Global properties of minimal surfaces in E3 and En. Ann of Math 80: 340-364. [ Links ]

Osserman R. 1986. A survey of minimal surfaces. 2nd edition, Dover Publications, New York. [ Links ]

Weitsman A and Xavier F. 1987. Some function theoretic properties of the Gauss map for hyperbolic complete minimal surfaces. Michigan Math J 34: 275-283. [ Links ]

Xavier F. 1981. The Gauss map of a complete non-at minimal surface cannot omit 7 points of the sphere. Ann of Math 113: 211-214. [ Links ]

Received: June 29, 2011; Accepted: October 9, 2012

Pedro A. Hinojosa E-mail: hinojosa@mat.ufpb.br

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