The Gauss Map of Complete Minimal Surfaces with Finite Total Curvature

In this paper we are concerned with the image of the normal Gauss map of a minimal surface immersed in R 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 in R with finite total curvature, which is not a plane, omits at most three points of S. Moreover, under an additional hypothesis on the type of ends, we prove that this number is exactly 2.


INTRODUCTION
Complete minimal surfaces immersed or embedded in R 3 constitute 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 → R 3 be a complete non-planar minimal immersion and let N : M → S 2 = ~ 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 N omits at most four points.This is the best possible result since there are complete immersed minimal surfaces in R 3 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 if M 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 S 2 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 R 3 ; then its total curvature is a multiple of −4π: We recall that even if a minimal surface M is not orientable, its oriented double covering M f is an orientable PEDRO A. HINOJOSA and GILVANEIDE N. SILVA minimal surface.In that way we can study the Gauss map of M f .Here we consider only the orientable case.In 1987, Weitsman andXavier (Weitsman andXavier 1987) proved that if N omits three values, the total curvature of M 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 M c 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 M c ; the number of ends, the index of each end, the total curvature of M and the number of points omitted by the spherical image of M: Finally, under additional hypothesis on the type of ends of the surface, we obtain results that improve the Osserman's theorem.
Let X : M → R 3 be an isometric minimal immersion of a Riemann surface M into R 3 : Let N be Gauss map and C(M) := ∫ M KdM the total curvature of M. If M 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 M c of genus ° minus a finite set points, that is, For each choice of π there is a meromorphic one-form ω on M c 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; ¢ has no real periods on M; 1.6.the Weierstrass representation gives the following integral expression Since M is complete with the induced metric, then α must have poles at each point of M c \ M. 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 M c \ M, or punctured neighborhoods of such points, are called ends of M. L. Jorge and W. Meeks (Jorge and Meeks 1983) have shown that: 1.10.the number I(p i ) is an integer greater than or equal to one.
, where U i is some neighborhood of infinite in the subspace perpendicular to N b (p i ):

HEIGHT FUNCTIONS
For each » 2 S 2 (1) we define: (1) Hence, a simple computation yields: where » T (p) means the orthogonal projection of » into T p M. The singularities of this vector fields are the is the order of p as a zero of g -g(p).
Proof.By a change of coordinates in R 3 ; if necessary, we may suppose that » = ±(0, 0, 1).Consider the identication π : S 2 (1) → C [ {0} for which it holds that π (0, 0, 1) = ∞ and π (0, 0, -1)=0.As N(p) = ±» we have g(p) = 0 or g(p) = ∞.We may suppose that g(p) = 0. Let z = u + iv be a local complex parameter of M such that p corresponds to z = 0.Then, in a neighborhood of p, we have: It follows from (1.5) and (1.6) that: where h is a holomorphic function such that h(0) = 0 and h'(0) ≠ 0: Hence, we get: and In order to determine the behavior of the vector field grad (f » ) at the points p i 2 M c \ M we consider its restriction to the curves °i ρ defined in (1.9).Proposition 2.2.For ρ sufficiently large, the index of grad (f » ) along °i ρ is PEDRO A. HINOJOSA and GILVANEIDE N. SILVA Proof.Choose coordinates on R 3 so that » = ± (0,0,1) and g b (p i ) ≠ ∞.With respect to the local parameter z = u + iv around p i we have: , where g ~(0) ≠ 0 and As in the proof of the Proposition 2.1 we have: It follows that one of the following cases can occur: In this case, g b(0) ≠ 0 and we can write @f » @z = ±z -μ h 1 (z) where These cases correspond respectively to: As before, consider the gradient vector field Then, take large enough so that °i ρ (S 1 ) is contained in the neighborhood of p i we are considering.It follows that the index of grad(f » ) along °i ρ is μ in the case (A) and v -μ in the case (B).
The following lemma completes the proof of Proposition 2.2.Lemma 2.3.If p 2 M c \ M, g b(p) ≠ ∞ and ω has a pole of order μ at p, then I(p) = μ -1.
Proof.We may choose coordinates in S 2 (1) in such a way that g b(p) = 0.By using (1.6) we have: where f 1 and f 2 are power series with at most terms in 1 z μ-2 .

GAUSS MAP OF COMPLETE MINIMAL SURFACES
Therefore, for ρ sufficiently large, X 1 + iX 2 takes the circle of radius r centered at p into a closed curve that rotates (μ -1)-times around the origin, that is I(p) = μ -1.

SUPPORT TYPE FUNCTIONS
Now we consider, for each » 2 S 2 , the support type function A simple computation shows that the singular points of grad(s » ) are the points where » T = 0 or K = 0 (branch points of g).By a change of coordinates in R 3 we may always assume that » = (0, 0, 1).
Since: Barbosa and Colares 1986), we have: Consider a point p such that N b (p) = ±».In terms of local coordinates z = u+iv; with p corresponding to z = 0; we have g(z) = z v g ~(z); g ~(0) ≠ 0; when v is a nonzero integer.
If v > 0 then s » (0) = -1 and 1222 PEDRO A. HINOJOSA and GILVANEIDE N. SILVA The point p is then a local minimum for s » with index + 1.If v < 0 then s » (0) = 1 and The point p is then a local maximum and so its index is also +1. When ) .
Considering z = re iθ one obtains where b 0 := g ~(0).Setting a 0 b 0 = Re iα , we have: From this expression it follows that the difference s » (z) -s » (0) changes sign in any neighborhood of z = 0.It follows easily that the index of grad(s » ) at p is 1 -v(p).
Therefore we have proved: and p is a critical point of s » then g has a branch point of order v at p and the index of grad (s » ) at p is 1 -v.

Proof. Set:
Assume M is not flat and let » 2 g(M) be a regular value of g.Counting indices of the singularities of grad(s » ) we obtain: Now using 2.4, we have Â(M c ) n ∑ i=1 I(p i ) + 1) -2m.Using (9) it follows that: (c) The ends of M are embedded.
Proof.It follows from equation ( 9) that if k = 3 then: On the other hand, from (5) we have: It is consequence of (5) that the ends M are embedded.
Corollary 4.3.Under the same hypothesis of the previous corollary it holds that if N : M → S 2 omits 3 points then m ≥ 3: Proof.From the proof of the previous corollary we have 0 ≥ Â(M c ) ≥ 2 (n -m).Hence m ≥ n: However, for each point q 2 S 2 \ N (M) there is at least one point p 2 M c \ M such that N b (p) = q.Hence, we have n ≥ 3.

RESULTS
Now we study the behavior of M at infinity.For this purpose we consider X : M = M c \ {p 1 , p 2 , ..., p n } → R 3 as before and let (g, ω) be the the corresponding Weierstrass data.After a rotation we may assume, for j fixed, that g b(p j ) = 0. Let z be a local parameter in an open subset D ½ M c with p j 2 D such that (p j ) = 0.
Hence, locally we have: Moreover, as the immersion is complete it follows that f has pole at p j whose order is greater than or equal to 2. Hence we have near p j that: PEDRO A. HINOJOSA and GILVANEIDE N. SILVA By Lemma (2.3) we have I(p j ) = μ -1: Thus, p j is an embedded end if and only if μ(p j ) = 2. Suppose that p j is an embedded end.Then α 3 := gω = gfdz and Thus, the end F j is asymptotic to the plane {X 3 = L}.In this case, we say that F j is a planar end of order v -1: If v = 1, then gf(z) = a 0 b 0 z -1 + (holomorphic function).Since: for any closed curve C ½ M, we have <(2πia 0 b 0 ) = 0 and then a 0 b 0 2 R. Therefore: where R is an harmonic function in a neighborhood of 0 2 C and R(0) = 0.The number a 0 b 0 is the logarithmic growth of the end F j .In this case the end F j is of catenoidal type.This means geometrically that F j is asymptotic to a catenoid in R 3 .
In the general case we have fg Denition 5.1.Let p 2 M c \ M be a end of M. We say that p 2 M c \ M 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 X : M = M c \ {p 1 , p 2 , ..., p n } → R 3 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 that k ≤ 3: Suppose that k = 3 and let » 1 , » 2 and » 3 be the omitted points.As in the proof of the Theorem 4.1 consider the sets: Counting the indices of the singularities of grad(f » ) we obtain: GAUSS MAP OF COMPLETE MINIMAL SURFACES (a) in the case » 2 = -» 1 : We claim that n ≤ 3.By contradiction, suppose that n > 3 and let p 1 , p 2 , p 3 , p 4 four distinct points in M c \ M. Consider two simple and differentiable curves α 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 M except for its endpoints.

1. 2 .
The Gauss mapping N extends continuously to a function N b : M c → S 2 : So, N b is a branched covering of S 2 (1); 1.3.If m represents the number of times that N b covers each point of S 2 then the total curvature of M is given by −4πm: Let π : S 2 → C [ {∞} be any identication of S 2 with C [ {∞}.Represent by g the composition g := π ○ N and set g b:= π ○ N b .