UNICITY RESULTS FOR GAUSS MAPS OF MNINIMAL. SURFACES IMMERSED IN R m

Transcription

1 UNICITY RESULTS FOR GAUSS MAPS OF MNINIMAL SURFACES IMMERSED IN R m A Dissertation Presented to the Faculty of the Department of Mathematics University of Houston In Partial Fulllment of the Requirements for the Degree Doctor of Philosophy By Jungim Park, student July 016

2 UNICITY RESULTS FOR GAUSS MAPS OF MNINIMAL SURFACES IMMERSED IN R m Jungim Park, student APPROVED: Min Ru, Ph.D. Chairman, Dept. of Mathematics Shanyu Ji, Ph.D. Dept. of Mathematics Gordon Heier, Ph.D. Dept. of Mathematics Qianmei Feng, Ph.D. Dept. of Industrial Engineering Dean, College of Natural Sciences and Mathematics ii

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4 UNICITY RESULTS FOR GAUSS MAPS OF MNINIMAL SURFACES IMMERSED IN R m An Abstract of a Dissertation Presented to the Faculty of the Department of Mathematics University of Houston In Partial Fulllment of the Requirements for the Degree Doctor of Philosophy By Jungim Park, student July 016 iv

5 Abstract The purpose of this thesis is to discuss the theory of holomorphic curves in order to study value distributions of the (generalized) Gauss map of complete minimal surfaces immersed in R m. The study was initiated by S.S. Chern and R. Osserman [4] in 1960s. Since then, it has been developed by F. Xavier [7], H. Fujimoto [7], M. Ru [], etc. In this thesis, we prove a unicity theorem for two conformally dieomorphic complete minimal surfaces immersed in R m whose generalized Gauss maps f and g agree on the pre-image q f 1 (H j ) for given hyperplanes H j, 1 j q, in P m 1 (C), located in general position, under the assumption that k+1 f 1 (H ij ) =. In the case when k = m 1, the result obtained gives an improvement of the earlier result of Fujimoto [10]. v

6 Contents 1 Introduction 1 Basic Facts about Complete Minimal Surfaces in R m 8.1 Minimal Surfaces in R Minimal Surfaces in R m The Generalized Gauss Map of Minimal Surfaces in R m Completeness of Minimal Surfaces Theory of Holomorphic Curves Holomorphic Curves The Associated Curves Results in Nevanlinna Theory Previous Studies of the Unicity Results Unicity Theorem for meromorpic functions of C Unicity Theorem for Gauss Maps of complete Minimal Surfaces Comparison of the Unicity Theorem for holomorphic curves of C and the Unicity Theorem for Gauss Maps of Complete Minimal Surfaces The Auxiliary Function and the Main Lemma The Auxiliary Function vi

7 5. Pseudo-metric on (R) with Negative Curvature The Main Lemma The Main Theorem Proof of the Main Theorem Conclusion Main Point Future Work Bibliography 87 vii

8 List of Figures 1.1 Catenoid Soap Film Chordal Distance The Classical Gauss Map The Generalized Gauss Map The graph of a doubly periodic Scherk's surface in R Color Wheel Graph of f(z) = e z The relation among M, M, and (R) L a0 and its image Γ a0 when z 0 M The image Γ a0 of L a0 tending to the boundary of M viii

9 List of Tables.1 Non-at complete minimal surfaces immersed in R Comparison of the Recent Results of Unicity Theorems ix

10 Chapter 1 Introduction Minimal surfaces are surfaces which have minimal areas for all small perturbations. Minimal Figure 1.1: Catenoid Soap Film surfaces have been studied not only in mathematics but also in many other elds such as physics, biology, architecture, art, and so on. Physicians have studied the minimal surface using soap lms. The picture on the right is the catenoid created from a soap bubble by a physician. After Alan Schoen discovered a minimal surface called a gyroid, many biologists have found gyroid structures in the membrane of butteries' wings and in certain surfactant or lipid 1

11 mesophases. In recent years, new types of architecture and art using minimal surfaces have been created and developed. In short, interest in the minimal surface has been growing in many elds over the past several decades. Minimal surface theories was studied for the rst time by Joseph-Louis Lagrange who in 176 considered the variational problem of nding the surface of least area stretched across a given closed contour. Since Lagrange's initial investigation, there have been multiple discoveries on minimal surface theories by many great mathematicians. In 1967, the value distribution theory of Gauss map of complete minimal surfaces started to be studied by S.S. chern and R. Osserman [4]. Since then, it have been studied by many mathematicians such as F. Xavier, H. Fujimoto, M. Ru, S.J. Kao, H.P. Hoang, etc. From 199 onward, the unicity theorem for Gauss maps of complete minimal surfaces also have been studied by H. Fujimoto, but there have been no particular improvements. That is, until now. In this thesis, we shall give an improvement of Fujimoto's unicity theorem of Gauss maps of complete minimal surfaces. For a minimal surface x := (x 1,, x m ) : M R m immersed in R m with m 3, its generalized Gauss map G is dened as the map which maps each p M to the point in Q m (C) := {(w 1 : : w m ) w1 + + wm = 0} corresponding to the oriented tangent plane of M at p. We may regard M as a Riemann surface with a conformal metric and G as a holomorphic map of M into P m 1 (C). With this setting, many value-distribution-theoretic properties of holomorphic curves in the complex projective space can be carried into the study of the Gauss maps of complete minimal surfaces in R m. This thesis specically focuses on the corresponding unicity results

12 for two conformally dieomorphic complete minimal surfaces immersed in R m. We begin with by recalling H. Fujimoto's result [10]. Let x : M R m and x : M R m be two oriented non-at complete minimal surfaces immersed in R m and let G : M P m 1 (C) and G : M P m 1 (C) be their generalized Gauss maps. Assume that there is a conformal dieomorphism Φ of M onto M. Then the Gauss map of the minimal surface x Φ : M R m is given by G Φ. Fujimoto obtained the following result. Fujimoto s Theorem (H. Fujimoto, [10]). Under the notations above, consider the holomorphic maps f = G : M P m 1 (C) and g = G Φ : M P m 1 (C). Assume that there exist hyperplanes H 1,, H q in P m 1 (C) located in general position such that (i) f 1 (H j ) = g 1 (H j ) for every j = 1,, q, q (ii) f = g on f 1 (H j ). If q > m + m(m 1)/, then f g. Main Theorem. Assume that both f = G : M P m 1 (C) and g = G Φ : M P m 1 (C) are linearly non-degenerate (i.e. the images of f and g are not contained in any linear subspaces of P m 1 (C)) and that there exist hyperplanes H 1,, H q in P m 1 (C) located in general position and a positive integer k > 0 such that (i) f 1 (H j ) = g 1 (H j ) for every j = 1,, q, k+1 (ii) f 1 (H ij ) = for any {i 1,, i k+1 } {1,, q}, 3

13 q (iii) f = g on f 1 (H j ). If q > (m + m + 4k) + (m + m + 4k) + 16k(m )m(m + 1), (1.1) 4 then f g. When k = 1, the condition (1.1) becomes q > (m + m + 4) + (m + m + 4) + 16(m )m(m + 1). 4 In this case, notice that 3m + m(m 1) > (m + m + 4) + (m + m + 4) + 16(m )m(m + 1) 4 for all m 3, then the Main Theorem gives the following corollary. Corollary 1. Under the notations above, consider the holomorphic maps f = G : M P m 1 (C), g = G Φ : M P m 1 (C). Assume that f and g are linearly non-degenerate, and that there exist hyperplanes H 1,, H q in P m 1 (C) located in general position such that (i) f 1 (H j ) = g 1 (H j ) for every j = 1,, q, (ii) f 1 (H i Hj ) = for i j, q (iii) f = g on f 1 (H j ). If q 3m + m(m 1), then f g. 4

14 When k = m 1, since the condition (ii) in the Main Assumption automatically holds under the assumption that H 1,, H q are in general position, we can omit it. Then, the Main Theorem gives the following Corollary. Corollary. Under the notations above, consider the holomorphic maps f = G : M P m 1 (C), g = G Φ : M P m 1 (C). Assume that f and g are linearly non-degenerate, and that there exist hyperplanes H 1,, H q in P m 1 (C) located in general position such that (i)f 1 (H j ) = g 1 (H j ) for every j = 1,, q, q (ii)f = g on f 1 (H j ). If q > (m +5m 4)+ (m +5m 4) +16(m )(m 1)m(m+1), then f g. 4 Furthermore, if m = 3 and k = m 1 = in the Main Theorem, we obtain q > 6, which matches the result in Fujimoto's paper [8] for the case of the complete minimal surface immersed in R 3 exactly. Now we shall compare the number of hyperplanes, q, in order to show that the result in this thesis gives an improvement of Fujimoto's theorem mentioned before. Let q 1 = m + m(m 1) as in Fujimoto's result and q = m + 5m 4 + (m + 5m 4) + 16(m )(m 1)m(m + 1) 4 5

15 as in our Corollary, we verify that q 1 q for m 3. Indeed, for m 3, (5m 7m + 4) ( 17m 4 m 3 + m 8m + 16) = 8m 4 48m m 48m = 8m(m 1)(m )(m 3) 0. This implies that 5m 7m m 4 m 3 + m 8m Hence, q 1 q = 5m 7m m 4 m 3 + m 8m for m 3. The same argument show that q 1 > q for m > 3. Thus our Corollary gives an improvement of Fujimoto's result mentioned above. The graph below gives how q 1 and q grows as m becomes larger. We outline here the strategies of proving our Main Theorem: (a) Instead of using Fujimoto's approach in [10], we use the method of Ru [19] which gave an explicit construction of the negative curvature on the unit disc; (b) Instead of the auxiliary 6

16 function χ constructed in Fujimoto's paper [10], we use the new auxiliary function constructed in Chen-Yan [14] (see also [15] or [16]) which allows us to improve Fujimoto's result, as well as working for general k (while Fujimoto's case is only for k = m 1). 7

17 Chapter Basic Facts about Complete Minimal Surfaces in R m.1 Minimal Surfaces in R 3 In this section, we dene the classical Gauss map and study the relation between the generalized Gauss map (G = [φ 1 : φ : φ 3 ] : M P (C)) and the classical Gauss map (g : M C) of a surface in R 3. Via the Weierstass-Enneper Representations, the Gauss map of a minimal surface is considered a meromorphic function on the corresponding Riemann surface. From this, we can see a remarkable analogy between the value distribution theory and the minimal surface theory. That analogy applies to their respective unicity theorems as well, so in this section we prove the Enneper-Weierstrass Representation theorem and see some examples. 8

18 In R 3, each oriented plane P G,R m is uniquely determined by the unit vector N such that it is perpendicular to P and the system {X, Y, N} is a positive orthonormal basis of R 3 for arbitrarily chosen positively oriented orthonormal basis {X, Y } of P. For an oriented surface in R 3 the tangent plane is uniquely determined by the positively oriented unit normal vector. On the other hand, the unit sphere S of all unit vectors in R 3 is bijectively mapped onto the extended complex plane C = {C } by the stereographic projection ω. Definition.1.1. For a minimal surface M immersed in R 3 the classical Gauss map g : M C of M is dened as the map which maps each point p M to the point ω(n p ) C, where N p is the positively oriented unit normal vector N p of M at p. We begin by studying the stereographic projection ω. For an arbitrary point (ξ, η, ζ) S set z = x + 1y := ω(ξ, η, ζ) C, which means that, if P (0, 0, 1), then the points N(0, 0, 1), P (ξ, η, ζ), and P (x, y, 0) are collinear and, otherwise, z =. Using γ(t) =N+t(P N) := P, we obtain x = ξ 1 ζ, y = Then by elementary calculation, we see ξ = η 1 ζ, ξ + η + ζ = 1. z + z z + 1, η = 1 z z z + 1, ζ = z 1 z + 1. (.1) For two points (ξ 1, η 1, ζ 1 ) and (ξ, η, ζ ) in S we denote by θ (0 θ π) the angle between two vectors P 1 (ξ 1, η 1, ζ 1 ), P (ξ, η, ζ ) and α = x 1 + 1y 1 := ω(ξ 1, η 1, ζ 1 ), β = x + 1y := ω(ξ, η, ζ ). 9

19 Figure.1: Chordal Distance Dene α, β = sin θ ( 1). Appling Law of Cosine, by denition we have P 1 P = cos θ = sin θ = α, β Thus, geometrically α, β is the chordal distance between P 1 and P. If α and β, by (.1) α, β = 1 P 1P = 1 (ξ1 ξ ) + (η 1 η ) + (ζ 1 ζ ) = 1 (1 ξ1 ξ η 1 η ζ 1 ζ ) = 1 = ( ) ( ) ( ) ( ) α + α β + β α α β β α + 1 β α + 1 β + 1 α β α + 1 β + 1. ( α 1 α + 1 ) ( β ) 1 β

20 If β =, then (ξ, η, ζ ) = (0, 0, 1), so α, β = 1 P 1P = 1 (ξ1 ) + (η 1 ) + (ζ 1 1) = 1 (1 ζ1 ) = 1 α 1 α + 1 = 1 α + 1. We dene the chordal distance between α and β by α, β. By denition, we see 0 α, β 1. Now we take an arbitrary point [w 1 : w : w 3 ] Q 1 (C). Write w i = x i 1y i (1 i 3) with x i, y i R, and set W := (w 1, w, w 3 ), X := (x 1, x, x 3 ), Y := (y 1, y, y 3 ). Since W Q 1 (C), we have w1 + w + w3 = 0 (x 1 1y 1 ) + (x 1y ) + (x 3 1y 3 ) = 0 x 1 + x + x 3 = y1 + y + y3 and x 1 y 1 + x y + x 3 y 3 = 0 X = Y and X, Y = 0. Without loss of generality, we may assume that X = Y = 1. Then, the unit normal vector of the plane which has a positive basis {X, Y } is given by N := X Y = (x y 3 x 3 y, x 3 y 1 x 1 y 3, x 1 y x y 1 ) = Im{(w w 3, w 3 w 1, w 1 w )} 11

21 For the case where w 1 1w, we assign to W the point z = w 3 w 1 1w (.) and, otherwise, the point z =. This correspondence is continuous inclusively at. To see this, we rewrite (.) as z = w 3(w 1 + 1w ) w 1 + w = w 1 + 1w w 3. If W tends to the point with w 1 = 1w, z tends to because w 1 + w + w 3 = 0 ( 1w ) + w + w 3 = 0 w 3 = 0. If w 1 1w, from z = w 1 + 1w w 3 and 1 z = w 1 1w w 3 we have w 1 w 3 = 1 (1 z z) and w w 3 = 1 (1 + z). (.3) z Since w 1 + w + w 3 = x 1 + y 1 + x + y + x 3 + y 3 = X + Y =, we get w 3 = w 1 w 3 + w w = 4 z ( z + 1). (.4) Then, (.3) and (.4) yield that N = Im{(w w 3, w 3 w 1, w 1 w )} {( ( = w 3 w w1 Im, w 3 = w 3 ), w 1 w 3 ( Re z z + 1, Im z z + 1, z 1 z ( w w 3 ). ) )}

22 By (.1) this shows that the point in S corresponding to [w 1 : w : w 3 ] Q 1 (C) is mapped to the point z = w 3 C by the stereographic projection. Figure.: The Classical Gauss Map Now we go back to the study of surfaces in R 3. Let x = (x 1, x, x 3 ) : M R 3 be a non-at surface immersed in R 3. Then its generalized Gauss map G is not a constant, and M may be considered as a Riemann surface (we shall discuss this more in the section.) with a conformal metric ds. For a holomorphic local coordinate z = u + 1v, G is represented as G = [φ 1 : φ : φ 3 ] = [f 1 : f : f 3 ], where φ i = f i dz = x i dz. (.5) z 13

23 Set hdz = φ 1 1φ ( 0) and g := f 3 f 1 1f (.6) By the above discussion, the function g is the classical Gauss map of M. Since the above correspondences are all biholomorphic, we obtain the following proposition which is the special case of Proposition.3.. Refer to Proposition.3. for the proof of the following proposition. Proposition.1.. For a surface M immersed in R 3, M is a minimal surface if and only if the classical Gauss map is meromorphic on M. Now we explain the following Enneper-Weierstrass representaion theorem for minimal surfaces. Theorem.1.3. (H. Fujimoto, [9]) Let x = (x 1, x, x 3 ) : M R 3 be a non-at minimal surface immersed in R 3. Consider the holomorphic forms φ 1, φ, φ 3, hdz and the meromorphic function g which is dened by (.5) and (.6) respectively. Then, (i) it holds that φ 1 = 1 (1 g )hdz, φ = 1 (1 + g )hdz, and φ 3 = ghdz (.7) and (x 1, x, x 3 ) = ( z z z ) Re φ 1 + x 1 (z 0 ), Re φ + x (z 0 ), Re φ 3 + x 3 (z 0 ), z 0 z 0 z 0 (ii) the metric induced from the standard metric on R 3 is given by (.8) ds = (1 + g ) h dz, (.9) 14

24 (iii) the holomorphic form h has a zero of order k when and only when g has a pole of order k. Proof. Consider the function f i and h for a holomorphic local coordinate z. Obviously, ghdz = φ 3. Since f1 + f + f3 = 0, we have ( 1 (1 g )hdz = 1 ( ) ) f 3 1 f 1 (f1 ) 1f dz 1f = f 1 f 1f 1 f f 3 (f 1 1f ) = f 1 dz dz Similarly, = φ 1 1 (1 + g )hdz = 1 f 1 f 1f 1 f + f 3 (f 1 1f ) On the other hand, by (.), for i = 1,, 3, we have obtained dz = f dz = φ dx i = Re(φ i ). This implies x i (z) x i (z 0 ) = Re z z 0 φ i, so the assertion (i) holds. The assertion (ii) is shown by the direct calculations ds = ( f 1 + f + f 3 ) dz = 1 ( 1 g g + 4 g ) h dz = 1 ( (1 g )(1 g ) + (1 + g )(1 + g ) + 4 g ) h dz = ( 1 + g ) h dz If h has a zero at a point p where g is holomorphic, then f 1, f, f 3 have a common zero at p, which contradicts the denition [φ 1 : φ : φ 3 ] = [f 1 : f : f 3 ]. On the other 15

25 hand, if g has a pole of order k at a point p, then h has a zero of exact order k at p. It is because otherwise some f i has a pole or f i 's have a common zero, which contradicts the assumption φ i = f i dz (i = 1,, 3) are holomorphic forms and the denition [φ 1 : φ : φ 3 ] = [f 1 : f : f 3 ]. Thus, the assertion (iii) holds. Furthermore, from (iii), we see that the metric ds in (ii) is continuous. Q.E.D. Theorem.1.4. (H. Fujimoto, [9])Let M be an open Riemann surface, hdz be a nonzero holomorphic form and g a nonconstant meromorphic function on M. Assume that hdz has a zero of order k when and only when g has a pole of order k and that the holomorphic forms φ 1, φ, φ 3 dened by (.7) have no real periods. Then, for the functions x 1, x, x 3 dened by (.8), the surface x = (x 1, x, x 3 ) : M R 3 is a minimal surface immersed in R 3 whose classical Gauss map is the map g and whose induced metric is given by (.9). Proof. We shall prove the general case (R m ) of this theorem in Theorem.3.3. Example.1.5. We regard a helicoid as Riemann surface C, and by (.7) it may be obtained from g = e z and h = 1 e z which give us (φ 1, φ, φ 3 ) = ( 1 sinh z dz, 1 cosh z dz, 1 dz). 16

26 Now using (.8) we compute x = (x 1, x, x 3 ) in the following way. z 1 z 1 x 1 = Re sinh z dz + x 1 (z 0 ) = Re (e z e z )dz + x 1 (z 0 ) z 0 z 0 = 1 Re( 1e z 1e z ) = 1 Re[ 1e u (cos v + 1 sin v) 1e u (cos v 1 sin v)] = 1 (eu e u ) sin v = sinh u sin v z 1 z x = Re z 0 cosh z dz + x 1 (z 0 ) = Re z 0 (ez + e z )dz + x (z 0 ) = 1 Re(e z e z ) = 1 Re[e u (cos v 1 sin v) e u (cos v + 1 sin v)] = 1 (e u e u ) cos v = sinh u cos v z 1 x 3 = Re z 0 dz = Re( 1z) = Re( 1u v) = v Thus, we have obtained the helicoid x(u, v) = (sinh u sin v, sinh u cos v, v). The following picture shows the graph of the helicoid and more examples. 17

27 Table.1: Non-at complete minimal surfaces immersed in R 3 Catenoid Helicoid Enneper's Surface M = C \ { } M = C M = C g = e 1z, h = 1 e 1z g = e z, h = 1 e z g = z, h = 1 φ 1 = 1 sin z dz φ 1 = 1 sinh z dz φ 1 = 1 (1 z )dz φ = 1 cos z dz φ = 1 cosh z dz φ = 1 (1 + z )dz φ 3 = 1 dz φ 3 = 1 dz φ 3 = zdz x 1 = cos u cosh v x 1 = sinh u sin v x 1 = u u3 3 + uv x = sin u cosh v x = sinh u cos v x = v + v3 3 u v x 3 = v x 3 = v x 3 = u v 18

28 . Minimal Surfaces in R m Let M be an oriented real -dimensional dierentiable manifold immersed in R m and x = (x 1,, x m ) : M R m be an immersion. For a point p M, take a local coordinate system (u 1, u ) around p which is positively oriented. The tangent plane of M at p is given by T p (M) = { λ x u 1 p + µ x u p } : λ, µ R and the normal space of M at p is given by N p (M) = { N : N, x p u 1 = N, x p u } = 0 where X, Y denotes the inner product of vectors X and Y. The metric ds on M induced from the standard metric on R m is called the rst fundamental form on M and given by where g ij = ds = dx x u i, x u j = dx, dx x = du 1 + x du, u 1 u = g 11 du 1 + g 1 du 1 du + g du x du 1 + x du u 1 u for 1 i, j, and the second fundamental form of M with respect to a unit normal vector N is given by dσ = b 11 (N)du 1 + b 1 (N)du 1 du + b (N)du 19

29 where b ij (N) = x u i u j, N for 1 i, j. Then the mean curvature of M for the normal direction N at p is dened by H p (N) = g 11b (N) + g b 11 (N) g 1 b 1 (N). (g 11 g g1) Definition..1. A surface M is called a minimal surface in R m if H p (N) = 0 for all p M and N N p (M). Definition... A local coordinate system (u 1, u ) on an open set U in M is called isothermal on U if ds can be represented as ds = λ (du 1 + du ) for a positive smooth function λ on U. This means that λ := g 11 = g and g 1 = 0. Theorem..3. (S. S. Chern, [3]) For every surface M, there is a system of isothermal local coordinates whose domains cover the whole M. Propositon..4. (H. Fujimoto, [9]) For an oriented surface M with a metric ds, if we take two positively oriented isothermal local coordinate (x, y) and (u, v), then w = u + 1v is a holomorphic function in z = x + 1y on the common domain. Proof. By assumption, there exists a positive dierentiable function λ such that du + dv = λ (dx + dy ). Then, we have du + dv = λ (dx + dy ) (u x dx + u y dy) + (v x dx + v y dy) = λ (dx + dy ) (u x + vx)dx + (u x u y + v x v y )dxdy + (u y + vy)dy = λ (dx + dy ) A := u x + vx = u y + vy, u x u y + v x v y = 0 0

30 This means that the Jacobi matrix J := u x u y v x v y satises the identity J t J = u x + vx u x u y + v x v y u x u y + v x v y u y + vy = u x + vx 0 0 u x + vx where I is the unit matrix of degree. We then have J 1 = 1 t J and A J 1 = 1 det(j) v y v x u y u x = 1 A u x v x u y v y. = A I On the other hand, since [det(j)] = det(j t J) = det(a I ) = A and det(j) > 0, we have det(j) = A. These imply that u x = v y and u y = v x. Therefore, the function w = u + 1v is holomorphic in z. Q.E.D. Let x = (x 1,, x m ) : M R m be an oriented surface with a Riemannian metric ds. With each isothermal local coordinate (u, v), we associate the complex function z = u + 1v. Let z = u 1v. Notice that u = z + z and v = z z 1. Then, x(u, v) = (x 1 (u, v),, x m (u, v)) may be written as x(z, z) = (x 1 (z, z), x (z, z),, x m (z, z)). By Proposition..4, we can see that the surface M has a complex structure, and these complex valued functions dene holomorphic local coordinates on M. Thus, 1

31 we may regard M as a Riemann surface. By the use of complex dierentiations x i z = 1 ( xi u 1 x ) i v and x i z = 1 ( xi u + 1 x ) i v we have x i z = 1 x i 4 u 1 x i v [ = 1 ( xi ) + 4 u ( ) ] xi. v From this property, we obtain that x 1 z + + x m z [ = 1 m ( xi ) ( ) ] xi + 4 u v i=1 = 1 [ x 4 u, x x + u v, x ] v = 1 x u, x u (.10) and dz = dz d z = (du + 1dv)(du 1dv) = du + dv. Then, by (.10) we may rewrite ds = λ (du + dv ) x = u, x (du + dv ) u = ( x 1 z + x z + + x m z ) dz. ( Let λ x z = 1 + x + + ) x m z z z. Then M may be considered as a Riemann surface with a metric ds = λ z dz where λ z is a positive C function in terms of a holomorphic local coordinate z.

32 We denote by = u coordinate z = u + 1v. + v, the Laplacian in terms of the holomorphic local Proposition..5. (H. Fujimoto, [9]) For each isothermal local coordinate (u, v), we have the following property: Proof. By the assumption, we have (i) x, X = 0 for each X T p (M) (ii) x, N = λ H(N) for each N N p (M) λ = x u, x x = u v, x, v Deerentiating these identities, we have These imply x u, x x = u u v, x, v x u, x = 0 v x v u, x x + v u, x = 0 v x, x x = u u, x x + u v, x x = u v u, x x + v u, x = 0 v By similar method, we have Since x u x, x = 0 v x and generate the tangent plane, we write T v p(m) = {a x + b x u v For each X = a x + b x T u v p(m) we have x, X = x, a x u + b x v = a x, x + b x, x = 0 u v a, b R}. as desired in (i). On the other hand, for every normal vector N to M it holds that H(N) = b 11(N) + b (N) x, N + x, N u = v x, N = λ λ λ 3

33 because λ = g 11 = g and g 1 = 0. This shows (ii). Q.E.D Theorem..6. (H. Fujimoto, [9]) Let x = (x 1,, x m ) : M R m be a surface immersed in R m, which is considered as a Riemann surface as above. Then, M is minimal if and only if each x i is a harmonic function on M, namely, ( x i = u + v ) x i = 0 for every holomorphic local coordinate z = u + 1v. Proof. ) Assume that M is a minimal surface. Then, by denition H(N) = 0. Now we apply proposition..5 to get x, N = λ H(N) = 0. Since each N N p (M) is not equal to the zero vector, we have x i = 0. Thus, each x i is a harmonic function on M. ) Assume that each x i is a harmonic function on M. Then, by denition, each x i = 0. By Proposition..5, we have 0 = x = λ H(N). Since λ 0, we have H(N) = 0. Therefore, M is a minimal surface. Q.E.D. Corollary..7. (H. Fujimoto, [9]) There exists no compact minimal surface without boundary in R m. Proof. For a minimal surface x = (x 1,, x m ) : M R m immersed in R m, if M is compact, then each x i takes the maximum value at a point in M. By the maximum principle of harmonic functions, x i is a constant. This is impossible because x is an 4

34 immersion. Q.E.D. By the uniformization theorem, a simply connected Riemann surface is conformally equivalent to either the sphere S, the complex plane C, or the unit disk = {z : z < 1}. Because of Corollary..7. the rst case is excluded, and we obtain the following corollary. Corollary..8. For a minimal surface x = (x 1,, x m ) : M R m immersed in R m, if M is simply connected, then M is conformally equivalent to the complex plane or the unit disk..3 The Generalized Gauss Map of Minimal Surfaces in R m First, we consider the set of all oriented -dimensional planes in R m which contain the origin and denote it by G,R m. To clarify the set G,R m, we regard it as a subspace of the (m 1)-dimensional complex projective space P m 1 (C) as following. To each P G,R m, taking a positively oriented basis {X, Y } of P such that X = Y, X, Y = 0, we assign the point Φ(P ) = π(x 1Y ) where π denotes the cannonical projection of C m \ {0} onto P m 1 (C), namely, the map which maps each P = (w 1,, w m ) (0,, 0) to the equivalence class [w 1 : : w m ] := {(cw 1,, cw m ) c C \ {0}}. 5

35 For another positive basis { X, Ỹ } of P satisfying X = Ỹ and X, Ỹ = 0, we can nd a real number θ such that X = r(x cos θ + Y sin θ) Ỹ = r( X sin θ + Y cos θ) where r := X / X = Ỹ / Y. Therefore, we can write X 1Ỹ = reiθ (X 1Y ). This shows that the value Φ(P ) does not depend on the choice of a positive basis of P but only on P. Since X 1Y, X 1Y = X 1 X, Y Y = 0, we have w w m = 0 via Φ. Thus, Φ(P ) is contained in the quadric Q n (C) := {[w 1 : : w m ] w w m = 0} P m 1 (C). Conversely, take an arbitrary point Q Q n (C). If we choose some W C m \ {0} with π(w ) = Q and write W = X 1Y with real vectors X and Y, then X and Y satisfy the conditions X = Y, X, Y = 0. It is easily seen that there is a unique oriented -dimensional plane W such that Φ(W ) = Q. This shows that Φ is bijective. Therefore, in the following sections, there will be no confusion if we identify the set of all oriented -dimensional planes in R m, G,R m, with Q n (C). Now, we consider a surface x = (x 1,, x m ) : M R m immersed in R m. For each point p M, the oriented tangent plane T p (M) is canonically identied with an element of G,R m after the parallel translation which maps p to the origin. 6

36 Definition.3.1. The (generalized) Gauss map of a surface M is identied as the map of M into Q n (C) which maps each point p M to Φ(T p (M)). Figure.3: The Generalized Gauss Map For a positively oriented isothermal local coordinate (u, v), the vectors X = x u, Y = x v give a positive basis of T p (M) satisfying the conditions X = Y, X, Y = 0. Therefore, the Gauss map G is locally given by G = π(x [ x1 1Y ) = z : : x ] m z (.11) where z = u + 1v. Take a reduced representation G = ( x 1,, ) xm z z. Then we have ds = ( x 1 z + x z + + x m z 7 ) dz = G dz

37 We may write G = [φ 1 : : φ m ] with globally dened holomorphic 1-forms φ i = x i dz, 1 i m. z Proposition.3.. (H. Fujimoto, [9]) A surface x = (x 1,, x m ) : M R m is minimal if and only if the Gauss map G : M P m 1 (C) is holomorphic. Proof. ) Assume that M is minimal. Then, from x = ( 1 x 1 x z u v ( 1 x + ) 1 x u v we have ( ) x = 1 [ ( 1 x z z u u 1 x ) + 1 v by Theorem..6. This shows that x z the Gauss map G is holomorphic. ) and x z = ( x 1 v u 1 x )] = 1 v 4 x = 0 satises Cauchy-Riemann equation. Hence, ) Assume that G is holomorphic. For a holomorphic local coordinate z we set f i = x i z (1 i m). After a suitable change of indices, we may assume that f m has no zero. Since f i f m ( xi 1 4 x i = z = f i f m are holomorphic, we have ) = ( ) [ fi f m = z z f m z ] [ ] 1 fm = f i z [ fm z for i = 1,,, m. Write f m 1 f m ( fi f m [ ] fm = h 1 + 1h z with real-valued functions h 1, h. Then we have x i = 4 f 1 [ ] fm i f m z = 4 x i z (h 1+ 1h ) = 4 1 )] f m + f i f m Take the real parts of both sides of the above equation to see ( x x = u h 1 + x ) v h T p (M). 8 [ ] fm z ( x u 1 x ) (h 1 + 1h ) v

38 According to Proposition..5. (i), we obtain x, x = 0 because x T p (M) by the above. Thus, we get x = 0. This implies that M is a minimal surface by Theorem..6. Q.E.D. We say that a holomorphic form φ on a Riemann surface M has no real period if for every closed curve γ in M. Since dx i = x i z dz + x i z d z = 1 we have ( xi u 1 x i v Re γ φ = 0 ) (du + 1dv) + 1 ( xi u + 1 x ) i (du 1dv) v = x i u du + x i v dv [ ( 1 xi = Re u 1 x ) i (du + ] 1dv) v ( ) xi = Re z dz = Re(φ i ), for a piecewise smooth curve γ z z 0 x i (z) = Re γ z z 0 φ i + x i (z 0 ) (.1) in M joining z 0 and z. If φ has no real period, then the quantity x(z) = Re γ z z 0 φ + x(z 0 ) 9

39 depends only on z, and so x is a well-dened function of z on M, which we denote by x(z) = Re z z 0 φ + x(z 0 ) from here on. Related to Proposition.3., we state here the following construction theorem of minimal surfaces. Theorem.3.3. (H. Fujimoto, [9]) Let M be an open Riemann surface and let φ 1, φ,, φ m be holomorphic forms on M such that they have no common zero, no real periods and locally satisfy the identity ( ) x1 + z ( ) x + + z ( ) xm = 0 (.13) z for holomorphic functions x i z with φ i = x i dz. Set z x i = Re z z 0 φ i + x i (z 0 ) for an arbitrarily xed point z 0 of M. Then, the surface x = (x 1,, x m ) : M R m is a minimal surface immersed in R m such that the Gauss map is the map G = [ x1 z : : xm z ] : M Q m (C) and the induced metric is given by ds = ( x 1 z + x z + + x m z ) dz. Proof. By the assumption, the x i are well-dened single-valued functions on M. 30

40 Consider the map x = (x 1,, x m ) : M R m. By (.13) we have x u, x x 1 u u, x x v v, x x = v u 1 x v, x u 1 x v = x z, x z [ ( x1 ) = 4 + z = 0 ( ) x + + z ( ) ] xm z for z = u + 1v. This gives that Moreover, By (.10) (x i, x j ) (u, v) i<j = x u, x x = u v, x, v x u, x = 0 v x u, x x u v, x x v u, x x = v u, x u = 4 ( x 1 z + x z + + x m z ) > 0 Hence, x is an immersion. Then, the induced metric is given by ds = x u, x u (du + dv ) = ( x 1 z + x z + + x m z ) dz and (u, v) gives a system of isothermal coordinates for the induced metric ds. On the other hand, by (.11) the Gauss map G of M is given by G = [ x 1 : : ] xm z z with holomorphic functions x i, and so holomorphic. According to Proposition.3., z the surface M is a minimal surface. Q.E.D. 31

41 Let M be a Riemann surface with a metric ds which is conformal, namely represented as ds = λ z dz with a positive C function λ z in term of a holomorphic local coordinate z. Definition.3.4. For each point p M we dene the Gaussian Curvature of M at p by K ds = ln λ z λ z For a minimal surface M immersed in R m consider the system of holomorphic functions G = (f 1,, f m ) and set G = ( f f m ) 1/ for f i = x i. From the z denition of Gaussian curvature, we get K ds = 1 ln G = G z z G 6 ( ) fi f j f j f i i<j (.14) This implies that the Gaussian curvature of a minimal surface is always nonpositive. Definition.3.5. A surface with a metric ds = λ z dz is called to be at if the Gaussian curvature K ds vanishes identically. If a minimal surface is at, then (.14) implies that (f i /f i0 ) = 0, 1 i m for some i 0 with f i0 0, which means f i /f i0 is constant. Therefore, the Gauss map G is a constant..4 Completeness of Minimal Surfaces We will prove the Main Theorem of this paper using the Completeness of Minimal Surfaces, so we shall explain it in this section. 3

42 Definition.4.1. A divergent curve on a Riemann manifold M is a dierentiable map γ : [0, 1) M such that for every compact subset K M there exist a t 0 (0, 1) with γ(t) K for all t > t 0. That is, γ leaves every compact subset of M. Definition.4.. Riemann manifold M is said to be complete if every divergent curve γ : [0, 1) M has unbounded length. Definition.4.3. A divergent curve on a minimal surface x : M R m is a continuous map Γ : [0, 1) R m of the form Γ = x γ where γ : [0, 1) M is a divergent curve on the Riemann maniod M endowed with the metric of R m via the mapping x. Definition.4.4. A minimal surface x : M R m immersed in R m is said to be complete if every divergent curve Γ : [0, 1) R m on x has unbounded length. Example.4.5. The helicoid in Example.3.4. x(u, v) = (sinh u sin v, sinh u cos v, v) obtained from g = e z and h = 1 e z is a non-at complete minimal surface. Proof. By.3.3. we have ds = (1 + g ) h dz. Then the length of a curve γ is dened as d(γ) = γ ( 1 + g ) h dz. where γ : [0, 1) M, t γ(t) is a dieomorphism. The only way a curve can be divergent here is when it tends to. As the curve γ(t) tends to, ( ) 1 + g h dz = (1 + e z 1 ) γ e z dz = 1 ( ) 1 γ e z + ez dz γ 33

43 tends to. Thus d(γ) is unbounded. Therefore, we conclude the helicoid is complete. As a matter of fact, all helicoid are complete minimal surfaces. Q.E.D. Now we consider a doubly periodic Scherk's surface in R 3. Let M = C \ {1, 1, 1, 1}, g = z, and h = 4. Then z 4 1 ( ( φ 1, φ, φ3 ) = z + 1 dz, 1 z 1 dz, ) 4z z 4 1 dz. But then x = Re φ is not well dened because the 1-form φi, i = 1,, 3 have real periods on M as we see the picture below. To solve this, we dene M as the universal covering of M, and we take g and h as the lifts of g and h to M, respectively. Thus, x = Re φ is well dened. Figure.4: The graph of a doubly periodic Scherk's surface in R 3 As we have seen in example of the doubly periodic Scherk's surface, sometimes we need to deal with surfaces x : M R 3 which is dened on Riemann surfaces M, 34

44 where a simpley connected manifold M of the same dimension as M is regarded as the universal covering of M. Any minimal surfaces x : M R 3 can be lifted from M to M as a minimal surface x : M R 3, and we shall see that x is complete if and only if x is complete. If M is a Riemann surface with conformal structure c, then π 1 induces a conformal sturcture c on M such that the projection map π : (M, c) ( M, c) becomes a holomorphic mapping of the Riemann surface (M, c) onto the Riemann surface ( M, c). Consequently if x : M R 3 is a minimal surface with M as parameter domain, and if π : M M is the universal map, then x := x π : M R 3 denes a minimal surface. We call this map the universal covering of the minimal surface x. Note that x is regular if and only if x is regular, and the images of the Gauss maps G of M and the image of the Gauss maps G of M coincide. Proposition.4.6. (U.Dierkes, S.Hildebrandt, and F.Sauvigny, [17]) A minimal surface x : M R 3 is complete if and only if its universal covering x : M R 3 is complete. Proof. ) Suppose that x is complete. We consider an arbitrary divergent curve Γ on x. Lifting Γ to the covering surface x, we obtain a divergent curve Γ on x which must have innite length as x is complete. Since π : M M is a local isometry, it follows that Γ has innite length, and we conclude that x is complete. ) Now Suppose that x is complete. Consider an arbirary divergent curve Γ on x given by Γ = x γ, γ : [0, 1) M. Let Γ := x γ be the curve on x with γ = π γ on 35

45 M. Then, we have Γ := x γ = x π γ = x γ = Γ. We have to show that the length of Γ is innite. If γ is divergent on M, then by denition Γ is divergent on x, so the completeness of x implies that Γ has innite lengh, and hence Γ has innite length since π is locally an isometry. On the other hand, if γ is not divergent, then there is a compact subset K of M and a sequence of parameter values tn in [0, 1) converging to 1 such that γ(t n ) belongs to K for all n. We may assume that the points γ(t n ) converges to a point p M. Then we choose a chart ϕ : U R around p such that ϕ( p) = 0, and that π 1 (U) is the disjoint union of open sheets V i. Since the branch points are isolated, there is an ε > 0 such that Ω ε := B ε (0) \ B ε/ (0) is contained in ϕ(u) and that the metric of M is positive denite on ϕ 1 (Ω ε ). Since the points γ(t n ) converge to p, almost all of them belong to the compact set ϕ ( 1 B ε/ (0) ). Since γ is divergent on the universal covering M, the points γ(t n ) are distributed over innitely many sheets V i. From this fact we infer that the path ϕ γ has to cross Ω ε an innite number of times. This implys that the length of γ is innite. Hence the length of γ is innite via π, so is Γ. 36

46 Chapter 3 Theory of Holomorphic Curves 3.1 Holomorphic Curves Let f be a holomorphic curve in P n (C) dened on an open Rieman surface M, which means a nonconstant holomorphic map of M into P n (C). For a xed system of homogeneous coordinates [w 0 : : w n ] we set V i := {[w 0 : : w n ] : w i 0}, 1 i q. Then, every z 0 M has a neighborhood U of z 0 such that f(u) V i for some i and f has a representation f = [f 0 : : f i 1 : 1 : f i+1 : : f n ] on U with holomorphic functions f 0,, f i 1, f i+1,, f n. Definition For an open subset U of M we call a representation f = [f 0 : 37

47 : f n ] to be a reduced representation of f on U if f 0,, f n are holomorphic functions on U and have no common zero. We dene the reduced representation as F = (f 0,, f n ) in C n+1. As stated above, every holpomorphic map of M into P n (C) has reduced representation on some neighborhood of each point in M. Let f = [f 0 : : f n ] be a reduced representation of f. Then, for an arbitrary nowhere zero holomorphic function h, f = [f 0 h : : f n h] is also a reduced representatioin of f. Conversely, for every reduced representation f = [g 0 : : g n ] of f, each g i can be written as g i = hf i with a nowhere zero holomorphic function h. Definition For a nonzero meromorphic function h on M, we dene the divisor ν h of h as a map of M into the set of integers such that for z 0 M m if h has a zero of order m at z 0, ν h (z 0 ) = m if h has a pole of order m at z 0, 0 otherwise. We now take n + 1 holomorphic functions f 0,, f n on M at least one of which does not vanish identically. Take a nonzero holomorphic function g such that ν g (z 0 ) = min{ν fi (z 0 ) : f i 0, 0 i n} for z 0 M. Then, f i /g (0 i n) are holomorphic functions without common zeros. We can dene a holomorphic map f with a reduced representation f = [f 0 /g : : f n /g], which we call the holomorphic curve dened by f 0,, f n. Definition Let a 0, a 1,, a n C be scalars not all equal to 0. Then the set 38

48 H consisting of all homogeneous coordinates [w 0 : w 1 : : w n ] in P n (C) such that a 0 w 0 + a 1 w a n w n = 0 is a subspace of one dimension less than the demension of P n (C) is called a hyperplane in P n (C). In other words, H = {[w 0 : w 1 : : w n ] : a 0 w 0 + a 1 w a n w n = 0}. Take a holomorphic map f of M into P n (C) and a hyperplane H in P n (C) not including the image f(m) of f. For each point z M choosing a reduced representation f = [f 0 : : f n ] or F = (f 0,, f n ) on a neighborhood on U of z, we consider the holomorphic function F (H) := a 0 f a n f n on U. Since ν F (H) depends only on f and H, we can dene ν F (H) on M, which we call pull-back of H considered as a divisor. Proposition (H. Fujimoto, [9])Every holomorphic map f of an open Riemann surface M into P n (C) has a reduced representation on M. Proof. Set H i := {w i = 0} (0 i n) for a xed system of homogeneous coordinates [w 0 : : w n ]. Changing indices if necessary, we may assume that f(m) H 0, and so ν F (H0 ) is well-dened. Then there is a nonzero holomorphic function g such that ν g = ν F (H0 ). On the other hand, we can take an open covering {U k : k I} of M such that f has a reduced representation f = [f k0 : : f kn ] on each U k. Then, ν g = ν fk0. Let g ki := g f k0 f ki (0 i n), which is holomorphic on U k. If U kλ := U k U λ φ, then there is a nowhere zero holomorphic function h with f λi = hf ki (0 i n) on U kλ. We have g ki = gf ki f k0 = ghf ki hf k0 = gf λi f λ0 = g λi on U kλ for each i. Therefore, we can dene the function g i on M which equals g ki 39

49 on each U k. For these functions, f = [g 0 : : g n ] is a reduced representation of M. Q.E.D. Now, we consider k arbitrarily given holomorphic functions f 0,, f k on M. For a holomorphic local coordinate z on an open subset U of M, we denote by (f (l) i ) z, or simply by f (l) i, l-th derivative of f i with respect to z, where we set (f (0) i ) z := f i. By denition, the Wronskian of f 0,, f k is given by ( W (f 0,, f k ) W z (f 0,, f k ) := det (f (l) i ) ) z : 0 i, l k. Proposition (H. Fujimoto, [9])For two holomorphic local coordinate z and ζ, it holds that W ζ (f 0,, f k ) W z (f 0,, f k ) Proof. Set F = (f 0,, f k ) and ( F (l)) z = ( (f (l) ( ) k(k+1)/ dz. dζ ). Then, we have 0 ) z,, (f (l) W ζ (f 0,, f k ) = det (t ) (F (0) ) ζ, t (F (1) ) ζ,, t (F (k) ) ζ ( ( ) ) k = det t (F (0) ) z, t (F (1) dz dz ) z dζ,,t (F (k) ) z dζ = det (t ) ( ) ( ) ( ) k (F (0) ) z, t (F (1) ) z,, t (F (k) dz dz dz ) z dζ dζ dζ ( ) k(k+1)/ dz = W z (f 0,, f k ) dζ k ) z where t F (l) denotes the transpose of the vector F (l). Q.E.D The following proposition is a well-known property, so we will not prove it. 40

50 Proposition For holomorphic functions f 0,, f k on M, the following conditions are equivalent: (i) f 0,, f k are linearly dependent over C (ii) W z (f 0,, f k ) 0 for some (or all) holomorphic local coordinate z. Definition A holomorphic map f of M into P n (C) is said to be (linearly) non-degenerate if the image of f is not included in any hyperplane in P n (C). If f = [f 0 : : f n ] is non-degenerate, then f 0,, f n are linearly independent over C. 3. The Associated Curves Let f be a linearly non-degenerate holomorphic map of (R) := {z : z < R}( C) into P n (C) where 0 < R. Take a reduced representation F = (f 0,, f n ) of f with P(F ) = f where P is the canonical projection of C n+1 \{0} onto P n (C). Denote the k-th derivative of F by F (k) and dene F k = F (0) F (k) : (R) k+1 C n+1 for k = 0, 1,, n. Evidently, F n+1 0. Note that F 0 = F. Take a basis {E 0,, E n } of C n+1. Then the set {E i0 E ik : 0 i 0 < < i k n} gives a basis of k+1 C n+1. Then we see that F k = W (f i0,, f ik )E i0 E ik (3.1) 0 i 0 < <i k n and F k = W (f i0,, f ik ) 0 i 0 < <i k n 41

51 where W (f i0,, f ik ) is the Wronskian of f i0,, f ik. If f = [f 0 : : f n ] is nondegenerate, then f 0,, f n are linearly independent over C. Therefore, by (3.1) and Proposition we have F k 0 for 0 k n 1. Let P : k+1 C n+1 \ {0} P(k+1 C n+1 \ {0}) P N k (C) be the canonical projection where N k = n+1 C k+1 1 = (n+1)! (n k)!(k+1)! 1. Definition The curve f k := P(F k ) : (R) P N k (C), k = 0, 1,, n is called the k-th associated curve of f. Note that f 0 = P(F 0 ) = P(F ) = f. Let ω k be the Fubini-Study form on P N k (C), and let Ωk = (f k ) ω k, k = 0, 1,, n, be the pullback via the k-th associated curve. Then Ω k = dd c ln F k = 1 F k 1 F k+1 dz dz 0 π F k 4 for 0 k n, and by convention F 1 1. Note that Ω 0 = dd c ln F, and Ω n 0 since F n+1 0. It follows that ( ) Ric Ω k : = dd c Fk 1 F k+1 ln F k 4 = dd c ln F k 1 + dd c ln F k+1 dd c ln F k = Ω k 1 + Ω k+1 Ω k. Let H = {[z 0 : : z n ] a 0 z a n z n = 0} be a hyperplane in P n (C) with a a n = 1. Take a basis {E 0,, E n } of C n+1. Then the set {E i0 E ik : 0 i 0 < < i k n} 4

52 gives a basis of k+1 C n+1. We dene ( ) F k (H) = a i0 W (f i0,, f ik ) E i1 E ik. 0 i 1 < <i k n i 0 i 1, i n Then F 0 (H) = a 0 f a n f n = F (H). Note that and F k (H) = 0 i 1 < <i k n i 0 i 1, i n a i0 W (f i0,, f ik ). Definition 3... F (H) = a 0 f a n f n. F k (H)(z) F k (z) is said to be the projective distance from the k-th associated curve f k (z) to the hyperplane H. Definition We dene the k-th contact function of f for H by Then φ 0 (H) = φ k (H) := F k(h) F k. F (H) F and φ n (H) = W (f 0,,f n) n i=0 a i F n = W (f 0,,f n) F n = 1. Note that 0 φ k (H) φ k+1 (H) 1 for 0 k n Results in Nevanlinna Theory We shall introduce a few portions of the Nevalinna Theory in order to enforce the proof of the Main Theorem. 43

53 Definition Let f be a meromorphic function on (R), where 0 R and let r < R. Denote the number of poles of f on the closed disc (R) by n f (r, ), counting multiplicity. We then dene the counting function N f (r, ) to be N f (r, ) = r 0 n f (t, ) n f (0, ) dt + n f (0, ) ln r, t here n f (0, ) is the multiplicity if f has a pole at z = 0. For each complex number a, we then dene the counting function N f (r, a) to be N f (r, a) = N 1 (r, ). f a Definition The Nevanlinna's proximity function m f (r, ) is dened by m f (r, ) = 1 π ln + f(re iθ ) dθ π 0 where ln + x = max{0, ln x}. For any complex number a, the poximity function m f (r, a) of f with respect to a is then dened by m f (r, a) = m 1 (r, ). f a Definition The Nevanlinna's characteristic function of f is dened by T f (r) = m f (r, ) + N f (r, ). Here, T f (r) measures the growth of f. Example (K.S. Charak [7]) Consider the rational function f(z) = P (z) Q(z) = a nz n + a n 1 z n a 0 b m z m + b m 1 z m b 0, a n, b m 0. 44

54 Distingquish the following two cases: Case 1. When m n. In this case lim z f(z) is nite, so there is a positive real number r 0 such that n f (r, ) = m for all r r 0. Thus N f (r, ) = r0 0 n f (t, ) n f (0, ) dt + t r n f (t, ) n f (0, ) r 0 t dt + n f (0, ) ln r = (m n f (0, )) (ln r ln r 0 ) + n f (0, ) ln r + O(1) = m ln r m ln r 0 + n f (0, ) ln r 0 + O(1) = m ln r + O(1) Next, note that for polynomial P (z) = a n z n + a n 1 z n a 0 with a n 0, given positive ɛ there is an r 0 > 0 such that for all r = z > r 0 we have (1 ɛ) a n r n P (z) (1 + ɛ) a n r n. Thus, for all r r 0 we can assume that P (z) = a n r n (1 + o(1)) and Q(z) = b m r m (1 + o(1)). This implies that ln + f = O(1), and so m f (r, ) = O(1). Hence in this case T f (r) = m ln r + O(1) = O(ln r). Case. When m < n. by the same arguments used in Case 1 we get T f (r) = T 1 (r) + O(1) = n ln r + O(1) = O(ln r). f Thus for a rational function f we have T f (r) = O(ln r). Also, the converse of this statement holds. That is, if f is a meromorphic function with T f (r) = O(ln r), then f is a rational function. Furthermore, by letting m = 0 and n = 0, we have that T f (r) = O(1) if and only if f is constant. 45

55 Example (K.S. Charak [7]) Consider f(z) = e z. Then for all r we have m f (r, ) = 1 π = 1 π = 1 π = r π. π 0 π 0 π π ln + e reiθ dθ ln + e r cos θ dθ r cos θdθ Since e z is an entire function, for all r N f (r, ) = 0 and so T f (r) = r π. Theorem (Nevanlinna's First Main Theorem) Let f be a non-constant meromorphic function on (R), R. Then, for any r (0 r < R) and a C T f (r) = m f (r, a) + N f (r, a) + O(1) or T f (r) = T 1 (r) + O(1) f a holds. This theorem implies N f (r, a) T f (r). Example Consider the function in Example If a = 0, then by the same arguments in Example we have m f (r, 0) = m 1 (r, ) = r f π and N f (r, 0) = N 1 (r, ) = 0. f for all r. Thus we get T f (r) = r = m π f(r, 0) + N f (r, 0) = T 1 (r). Therefore, the result f in Theorem holds. In this case, letting r the Nevanlinna's First Main Theorem tells us in a sense that f is close to 0 on the left half plane and close to on the 46

56 right half plane. We can see this fact from Figure 3.1: Color Wheel Graph of f(z) = e z the Color Wheel Graph of the exponential function f(z) = e z on the right. The transition from dark to light colors shows that the magnitude of the exponential function is increasing to the right. The periodic horizontal bands indicate that the exponential function is periodic in the imaginary part of its argument. Now we dene N f (r, a) in the same way as en.wikipedia.org/wiki/exponential_function N f (r, a) but without taking multiplicity into account. Theorem (Nevanlinna's Second Main Theorem) Let f be a non-constant meromorphic function on (R), R, and let a 1,, a q be distinct complex numbers in C { }. Then, the inequality (q )T f (r) exc q N f (r, a j ) + S(r, f) holds where S(r, f) is the small error term and exc means that the inequality holds for all r [0, ) outside of a set of nite Lebesgue measure. Following are some of the known estimates of S(r, f). Theorem (W.K. Hayman [8], K.S. Charak [7]) Let f be a non-constant meromorphic function z < R +. Then 47

57 (a) if R = +, S(r, f) = O(ln + T f (r)) + o(ln r) as r through all values if f is of nite order, and as r outside a set E of nite linear measure otherwise. (b) if 0 < R < +, ( ) S(r, f) = O ln + 1 T f (r)) + ln R r as r outside a set E with E dr R r < +. As an immediate deduction from Theorem 3.3.9, we have Theorem (W.K. Hayman [8], K.S. Charak [7]) Let f be a non-constant meromorphic function z < R +. Then S(r, f) T f (r) 0 as r R, (3.) with the following provisos: (a) if R = + and f is of nite order, then (3.) holds without any restriction. (b) if R = + and f is of innite order, then (3.) holds as r outside a set E of nite length. (c) if R = + and lim inf r R T f (r) ln( 1 R r ) = +, then (3.) holds as r R through a suittable sequence of values of r. From Theorem we see that S(r, f) is the small error term with the property that S(r, f) = o(t f (r)) as r. Let ɛ > 0. Then by replacing o(t f (r)) by ɛ T f (r) 48

58 we can restate the inequality (q ɛ)t f (r) exc q N f (r, a j ) as r R. Definition We say that q hyperplanes in P n (C) are in general position if for 1 k n, k hyperplanes of them intersect in an (n k)-dimensional plane, and for k > n, any k hyperplanes of them have empty intersection. In other words, q hyperplanes are in general position if any subset of k normal vectors of the q hyperplanes is linearly independent whenever 1 k n. Theorem (M. Ru [4], Cartan's Second Main Theorem with Truncated Counting Functions) Let H 1,, H q be hyperplanes in P n (C) in general position. Let f : C P n (C) be a linearly non-degnerate holomorphic curve. Then, the inequality (q (n + 1))T f (r) exc q N (n) f (r, H j ) + O(ln + T f (r)) + o(ln r) holds where exc means that the inequality holds for all r [0, ) outside of a set of nite Lebesgue measure. As we see before, S(r, f) = O(ln + T f (r)) + o(ln r) is the small error term with the property that S(r, f) o(t f (r)) as r. Thus by replacing o(t f (r)) by ɛ T f (r) we can restate the inequality (q (n + 1) ɛ)t f (r) exc q N (n) f (r, H j ). 49

59 Chapter 4 Previous Studies of the Unicity Results The uniqueness theory mainly studies conditions under which there exists essentially only one function. Here we shall introduce two types of unicity theorems. 4.1 Unicity Theorem for meromorpic functions of C The Finnish mathematician Rolf Nevanlinna is the person who made the decisive contribution to the development of the theory of value distribution by introducing the charicteristic function T f (r) for the meromorphic function f and proved the rst unicity theorem for meromorpic functions in 196. The theorem is as follows: 50

60 Theorem (R. Nevanlinna [], 196) If two non-constant meromorphic functions f, g : C P 1 (C) have the same inverse images ignoring multiplicities for ve distinct complex values, then f g. Proof. Suppose that f g. Let a 1,, a q be distinct complex numbers in P 1 (C) and dene χ = f g. We apply Nevanlinna's Second Main Theorem to f and g. Then, for ɛ > 0 we have (q ɛ)t f (r) exc q N f (r, a j ) (4.1) (q ɛ)t g (r) exc q N g (r, a j ) (4.) as r. Let T (r) = T f (r) + T g (r). Since f 1 (a j ) = g 1 (a j ) for j = 1,, q, there exists a z such that f(z) = a j = g(z) for each j. From this, we get the equation χ(z) = f(z) g(z) = 0. Now by combining (4.1) and (4.) and applying Nevanlinna's First Main Theorem, we can get the inequality q ( (q ɛ)t (r) exc Nf (r, a j ) + N g (r, a j ) ) exc N χ (r, 0) exc T χ (r) exc T (r) This is equivalent to (q 4 ɛ)t (r) exc 0. From this we obtain q 4 + ɛ, which contradicts the assumption q = 5. Therefore, f g. Q.E.D. 51

61 This theorem, tells that any non-constant meromorphic functions can be uniquely determined by ve values. However, the `ve' in the theorem cannot be reduced to `four'. For example, e z and e z share four values which are 0, 1, 1, and, but they are not identical. Hirotaka Fujimoto, in 1975, extended Nevanlinna's result to non-degenerate holomorphic curves f, g : C P n (C). During the four decades following 1975, this problem has been studied intensively by H. Fujimoto, W. Stoll, L. Smiley, S. Ji, M. Ru, S.D. Quang, Z.H. Chen, Q.M. Yan and other mathematicians. In 1983, Smiley considered holomorphic maps f, g : C P n (C) which share 3n + hyperplanes in P n (C) without counting multiplicity and proved the following theorem. Theorem (L. Smiley [11], 1983) Let f, g : C P n (C) be linearly nondegenerate meromorphic mappings. Assume that there are q hyperplanes H 1,, H q in P n (C) located in general position satisfying (i) f 1 (H j ) = g 1 (H j ) for all 1 j q, (ii) f 1 (H i ) f 1 (H j ) = φ for all 1 i < j q, q (iii) f = g on f 1 (H j ). If q 3n +, then f g. Proof. Suppose that f g. Fix reduced (global) representations F = (f 0,, f n ) with T fi (r) T f (r) and G := (g 0,, g n ) with T gi (r) T g (r) such that f = P(F ) and g = P(G) where P denotes the canonical projection of C n \{0} onto P n (C). Since f g, there exist indices 1 i, j q such that χ = F (H i )G(H j ) G(H i )F (H j ) 0. 5

62 We now apply Cartan's Second Main Theorem to f and g. Then, for ɛ > 0 we have (q (n + 1) ɛ)t f (r) exc q N (n) f (r, H j ) (4.3) (q (n + 1) ɛ)t g (r) exc q N (n) g (r, H j ) (4.4) as r. Let T (r) = T f (r) + T g (r). Then, by combining (4.3) and (4.4), we can get the inequality q ( (q (n + 1) ɛ)t (r) exc If F (H i )(z 0 ) = 0, then z 0 exc n N (n) f ) (r, H j ) + N g (n) (r, H j ) q ( Nf (r, H j ) + N g (r, H j ) ) q f 1 (H i ), so f(z 0 ) = g(z 0 ) by the assumption (iii), and hence χ(z 0 ) = 0. Also, note that F (H j )(z 0 ) 0 for all j i, so q N f (r, H j ) N χ (r, 0). Similarly, q N g (r, H j ) N χ (r, 0). By these, together with by Nevanlinna's First Main Theorem, N χ (r, 0) T χ (r) T (r), implies This is equivalent to (q (n + 1) ɛ)t (r) exc n N χ (r, 0) exc n T χ (r) exc n T (r) (q (3n + 1) ɛ)t (r) exc 0. From this we obtain q 3n ɛ, which contradicts the assumption q 3n +. Therefore, f g. Q.E.D. In 009, Z.H. Chen and Q.M. Yan [14] improved Smiley's result for q 3n + 53

63 to n + 3 with the same assumption. Theorem (Z.H. Chen and Q.M. Yan [14], 009) Let f and g be two linearly non-degenerate holomorphic maps of C into P n (C) over C and let H 1,, H q (q n) be hyperplanes in P n (C) in general position. Assume that (i) f 1 (H j ) = g 1 (H j ) for all 1 j q, (ii) f 1 (H i ) f 1 (H j ) = φ for all 1 i < j q, (iii )f = g on q f 1 (H j ). If q n + 3, then f g. In 01, H. Giang, L. Quynh and S. Quang [15] generalized previous results by changing the condition (ii) to a more general one using the auxiliary function q [ χ = F (Hi )G(H σ(i) ) G(H i )F (H σ(i) ) ] 0 i=1 Their theorem is stated as follows: Theorem (H. Giang, L. Quynh and S. Quang [15], 01) Let f and g be two linearly non-degenerate holomorphic maps of C into P n (C) and let H 1,, H q (q n) be hyperplanes in P n (C) in general position. Assume that for a positive integer k with 1 k n, (i) f 1 (H j ) = g 1 (H j ) for all 1 j q, (ii) f 1 ( k+1 H i j ) = φ for all 1 i 1 < < i k+1 q, q (iii )f = g on f 1 (H j ). If q (n + 1)k + n +, then f g. Also in 01, after [15] was published, using the same auxiliary function q [ χ = F (Hi )G(H σ(i) ) G(H i )F (H σ(i) ) ] 0, i=1 54

64 F. Lü [16] reduced the number of hyperplanes in the theorem of Giang, Quynh and Quang to q 1 + 8k k + n + k +, 4 which is derived from q < q k+kn [q (n+1)]. A multitude of unicity theorems have kn been proved, but only those theorems that are relevant to this thesis were mentioned here. 4. Unicity Theorem for Gauss Maps of complete Minimal Surfaces Hirotaka Fujimoto is the rst person who used the above technique to prove the unicity theorem for Gauss maps of minimal surfaces immersed in R m. He also showed that generalized Gauss maps of complete minimal surfaces in R m are holomorphic curves in P m 1 (C), so many value-distribution-theoretic properties of holomorphic curves in the complex projective space could be used. The following theorem is his rst unicity theorem submitted in 199, which is for Gauss maps of minimal surfaces immersed in R 3. Theorem (H. Fujimoto [8], submitted in 199) Let x := (x 1, x, x 3 ) : M R 3 and x := ( x 1, x, x 3 ) : M R 3 be two non-at minimal surfaces immersed in R 3 and assume that there is a conformal dieomorphism Φ of M onto M. Consider the maps f := π G and g := π G Φ where π is the stereographic projection and G and G are the Gauss maps of M and M respectively. Suppose that there are q distinct points α 1, α,, α q such that f 1 (α j ) = g 1 (α j ) for 1 j q. 55

65 (a) If q 7 and either M or M is complete, then f g. (b) If q 6 and both of M or M are complete and have nite total curvature, then f g. Since it will take several pages to prove this theorem completely, we just give the outline of this proof. The proving process is very similar to the process of proving our Main Theorem. Outline of the proof of (a) in Theorem First, suppose that f g in order to get a contradiction, and assume that α q =. Second, we dene a pseudo-metric dη and show that dη is continuous on M and has strictly negative curvature on the set {dη 0}. dη is dened as follows: Set λ := ( q ( ) ) 1+ɛ ( q ( ) ) 1+ɛ a0 f, α j ln, f, α j λ a0 := g, α j ln g, α j for a 0 > 0 and ɛ with q 4 > qɛ > 0 and dene f dη := f, g g λ λ 1 + f 1 + g dz outside the set E := q f 1 (α j ) and dη = 0 on E where, is the cordal distance between two complex values. Third, we apply Ahlfors-schwartz Lemma for dη in order to get for a constant C. dη 4R C (R z ) dz (4.5) Fourth, we assume that M is complete, and we consider M as open Riemann surfaces with induced metric ds = h (1 + f )(1 + g ) dz from R 3 where h is a nowhere zero holomorphic function. 56

66 Fifth, we take some δ with q 6 > qδ > 0, set τ := q 4 qδ < 1 for q 7, and dene the pseudo-metric dσ by dσ := h 1 τ ( q 1 ( f α j g α j ) 1 δ f g f g q 1 (1 + α j ) 1 δ ) τ 1 τ dz (4.6) which does not depend on a choice of holomorphic local coordinate z and so welldened on M := M \ D where D := {z M : f (z) = 0, g (z) = 0, or f(z) = g(z) = α j for some j}. Note that dσ is at on M. Sixth, take an arbitrary point z in M. Since dσ is at on M, we can take R ( ) such that there is a holomorphic map B : (R) M with B(0) = z which is a local isometry with respect to the standard metric on (R) and the metric dσ on M. Then the pseudo-metric B dη on (R) also has strictly negative curvature. Since there is no metric with strictly negative curvature on C, we have necessarily R <. Seventh, we choose a point a with a = R such that for the line segment L a : w = ta (0 t < 1), the image Γ a of L a by B tends to the boundary of M as t tends to 1. Then choosing the suitable δ in the denition τ, we can actually show that Γ a tends to the boundary of M as t 1. Eighth, Since B is a local isometry, we may take the coordinate w as a holomorphic local coordinate on M and we may write dσ = dw. Then, from (4.6) we obtain ( f g f g q 1 h = (1 + α ) j ) 1 δ τ q 1 ( f α. j g α j ) 1 δ 57

67 Ninth, by (4.5) with ɛ = δ/ we have the inequality ds = h (1 + f )(1 + g ) dz ( ) τ C f, g f g λ λ dw (1 + f )(1 + g ) ( ) τ R C dw R w Finally, we calculate the length of Γ a R ( ) τ ds = B R ds C dw < since τ < 1, Γ a L a 0 R w which contradicts the assumption of completeness of M. Therefore, we have necessarily f = g. Q.E.D. Fujimoto, in his paper [8], gave an example which shows that the number seven in (a) is the best possible number. Here is an example: Take a number α with α 0, 1, 1 and consider the homomorphic functions h(z) := 1 z(z α)(αz 1), g(z) = z and the universal covering surface M of C \ {0, α, 1/α}. Then by the Enneper- Weierstrass representations we can construct a minimal surface x = (x 1, x, x 3 ) using the following formulas: x 1 := Re z (1 g )hdz, x := Re z z 1(1 + g )hdz, x 3 := Re ghdz As we have seen in chapter, the map g is the classical Gauss map of M. It is easily seen that M is complete (the proof is similar to Example.4.5.). We can also construct another minimal surface x = ( x 1, x, x 3 ) in a similar manner from h(z) := 1 z(z α)(αz 1), g(z) = 1 z 58

68 Since M is isometric with M, the identity map Φ : z M z M is a conformal dieomorphism. For the classical Gauss maps g and g, we have g 1 (α j ) = g 1 (α j ) for six values α 1 := 0, α :=, α 3 := α, α 4 := 1 α, α 5 := 1, α 6 := 1. However, g g. Hence, the number seven in Theorem cannot be reduced to six. In 1993, Fujimoto [10] generalized Theorem 4..1 into R m. The generalized theorem is stated as follows. Theorem 4... (H. Fujimoto [10], 1993) Let x := (x 1,, x m ) : M R m and x := ( x 1,, x m ) : M R m be two oriented non-at complete minimal surfaces immersed in R m and let G : M P m 1 (C) and G : M P m 1 (C) be their generalized Gauss maps. Assume that there is a conformal dieomorphism Φ of M onto M. Then the Gauss map of the minimal surface x Φ : M R m is given by G Φ. Consider the holomorphic maps f = G : M P m 1 (C), g = G Φ : M P m 1 (C). Assume that there exist hyperplanes H 1,, H q in P m 1 (C) located in general position such that (i) f 1 (H j ) = g 1 (H j ) for every j = 1,, q, q (ii) f = g on f 1 (H j ). If q > m + m(m 1)/, then f g. The unicity theorem for meromorphic functions in the value distribution theory has been improved for several decades by many mathematicians, but there has been no improvement of the unicity theorem for Gauss maps in the minimal surface theory 59

69 since Theorem 4.. was published. Thus we are going to improve upon the theorem through this thesis. 4.3 Comparison of the Unicity Theorem for holomorphic curves of C and the Unicity Theorem for Gauss Maps of Complete Minimal Surfaces In order to get some idea to make an improvement of Theorem 4.., we now compare the recent unicity theorem for holomorphic maps on C to the recent unicity theorem for Gauss maps on complete minimal Surface M. See the table below. Table 4.1: Comparison of the Recent Results of Unicity Theorems Holomorphic Curves f, g : C P n (C) Gauss maps f, g : M P n (C) with f 1 (H i H j ) φ without with f 1 (H i H j ) φ without q > n + q > (n + 1) (No result q > (n + 1) + n(n+1) (Chen and Yan [14], (Giang, Quynh, and in the previous studies) (Fujimoto [10], 1993) 009) Quang [15], 01) with f 1 ( k+1 H i j ) φ q > (n + 1) + with f 1 ( k+1 H i j ) φ knq q > (n + 1) + knq + n(n+1) q k+kn q k+kn (F. Lü [16], 01) (The result in this thesis, 016) 60

70 As we see the table above, the unicity theorem for holomorphic curves and the unicity theorm for Gauss maps are very related and the term n(n + 1)/ is the key. 61

71 Chapter 5 The Auxiliary Function and the Main Lemma 5.1 The Auxiliary Function In this section, we construct a (new) auxiliary function, similar to the auxiliary function used in Chen-Yan [14] (see also [15] or [16]), which will be used later. Let M be a simply connected Riemann surface and f, g : M P n (C) be two linearly non-degenerate holomorphic maps. Fix a reduced (global) representations F = (f 0,, f n ) and G := (g 0,, g n ), i.e. f = P(F ), g = P(G) where P denotes the canonical projection of C n \ {0} onto P n (C) and f 0,, f n (resp. g 0,, g n ) are holomorphic functions on M without common zeros such that T fi (r) T f (r) and T gi (r) T g (r). For a hyperplane H = {[z 0 : : z n ] a 0 z a n z n = 0} in P n (C), we dene 6

72 F (H) := a 0 f a n f n and G(H) := a 0 g a n g n. Assume that f g on M, and let H 1,, H q be hyperplanes in general position in P n (C). We dene an equivalence relation on L := {1,, q} as i j if and only if F (H i) G(H i ) F (H j) G(H j ) 0. Set {L 1,, L s } = L/. Since f g and H 1,, H q are in general position, we have that #(L k ) n for all k = {1,, s} where #(L k ) is the number of elements in L k. Without loss of generality, we assume that L k := {i k 1 + 1, i k 1 +,, i k } for 1 k s, where 1 < i 1 < < i s = q, i.e. F (H 1 ) G(H 1 ) F (H ) G(H ) F (H i 1 ) F (H i 1 +1) G(H i1 ) G(H }{{} i1 +1) F (H i 1 +) G(H i1 +) F (H i ) G(H i ) }{{} L 1 group L group F (H i +1) G(H i +1) F (H i +) G(H i +) = F (H i 3 ) F (H i s 1 +1) G(H i3 ) G(H }{{} is 1 +1) F (H i s ). G(H is ) }{{} L 3 group L s group Dene the map σ : {1,, q} {1,, q} by i + n if i + n q, σ(i) = i + n q if i + n > q. Then obviously σ is bijective and σ(i) i n assuming q n. This implies that i and σ(i) belong two distinct sets of {L 1,, L s }, so F (H i ) G(H i ) F (H σ(i)) G(H σ(i) ) 0. Let χ i := F (H i )G(H σ(i) ) G(H i )F (H σ(i) ). Then χ i 0. Dene χ := q χ i = i=1 q [F (H i )G(H σ(i) ) G(H i )F (H σ(i) )] 0. (5.1) i=1 63

73 For a nonzero meromorphic function h on M, we dene the divisor ν h of h as a map of M into the set of integers such that for z 0 M m if h has a zero of order m at z 0, ν h (z 0 ) = m if h has a pole of order m at z 0, 0 otherwise. We note that ν f(h) (z) is the intersection multiplicity of the images of f and the hyperplane H at f(z). Lemma (F. Lü [16], see also [14] and [15]) Let f, g : M P n (C) be two linearly nondegenerate holomorphic mappings. Let H 1,, H q be hyperplanes in P n (C) located in general position with q n such that (i) f 1 (H j ) = g 1 (H j ) for every j = 1,, q, (ii) k+1 f 1 (H ij ) = for any {i 1,, i k+1 } {1,, q}, q (iii) f = g on f 1 (H j ). Then the following holds on the domain of each holomorphic local coordinate z of M: ( ) q q k + kn ( ) ν χ (z) νf n (H kn j )(z) + νg(h n j )(z) where ν n F (H j ) (z) = min{n, ν F (H j )(z)}. Proof. If z q f 1 (H j ), then ν F (Hj )(z) = 0, so this lemma is obviously true. Thus, we only need to consider the case when z q f 1 (H j ). Let J = {j : F (H j )(z) = 0, 1 j q} and denote by #(J) the number of elements of J. Then, #(J) k by the assumption (ii). If j J, then z is a zero of F (H j ), and hence z 64

74 is a zero of χ j with multiplicity at least min{ν F (Hj )(z), ν G(Hj )(z)}. We now dene σ 1 (J) = {j : σ(j) J}. If l {1,, q} \ (J σ 1 (J)), then z is a zero of χ l with multiplicity at least 1 by the assumption (iii), and so ν χl 1. Therefore, ν χ (z) min{ν F (Hj )(z), ν G(Hj )(z)} + j J, j σ 1 (J) min{ν F (Hj )(z), ν G(Hj )(z)} + l {1,,q}\(J σ 1 (J)) j J, j σ 1 (J) l {1,,q}\(J σ 1 (J)) min{ν F (Hj )(z), ν G(Hj )(z)} + q #(J σ 1 (J)) j J min{ν F (Hj )(z), ν G(Hj )(z)} + q k j J ν χl 1 j J [min{n, ν F (Hj )(z)} + min{n, ν G(Hj )(z)} n] + q k. = [νf n (H j )(z) + νg(h n j )(z) n νf 1 (H j )(z)] + q k k k j J q [νf n (H j )(z) + νg(h n j )(z) n νf 1 (H j )(z)] + q k k q q [νf 1 (H j )(z) + νg(h 1 j )(z)] [ν n F (H j )(z) + ν n G(H j )(z)] + [ q k k n] q [νf 1 (H j )(z) + νg(h 1 j )(z)] q [νf n (H j )(z) + νg(h n j )(z)] + [ q k q kn 1] [νf n (H j )(z) + νg(h n j )(z)] ( ) q q k + kn ( ) = νf n (H kn j )(z) + νg(h n j )(z), where, in above, we use the facts that min{a, b} min{a, n} + min{b, n} n for all positive integers a and b, νf 1 (H j )(z) = 1 for j J, as well as k #(J) q min{1, ν F (H j )(z)} = q ν1 F (H j )(z). This nishes the proof. Q.E.D. 65

75 Lemma (M. Ru [1]) Let H 1,, H q be hyperplanes P n (C) in general position and let F = (f 0,, f n ) be the reduced representation of a linearly non-degenerate holomorphic map f : M P n (C). Then the following holds on the domain of each holomorphic local coordinate z of M: q ν F (Hj )(z) ν Fn (z) q νf n (H j )(z) where F n = W (f 0,, f n ) which is the Wronskian of f 0,, f n. Proof. If z 0 q f 1 (H j ), then ν F (Hj )(z 0 ) = 0, so this lemma is obviously true. Thus, we only need to consider the case when z 0 q f 1 (H j ). Then there are integers h j 0 and nowhere vanishing holomorphic function g j in a neighborhood U of z 0 such that F (H j ) = (z z 0 ) h j g j for j = 1,, q Let J = {j : F (H j )(z 0 ) = 0, 1 j q}. Then #(J) n because H 1,, H q are in general position. We re-order H 1,, H q so that the multiplicity h j, 1 j q are in descending order. Without loss of generality, we assume that h j h j+1 for 1 j j 1 and h j = 0 for j j q where j 1 n. Assume that h 1 h h j0 n for 1 j 0 j 1. Since each F (H j ), 1 j q is a linear combination of f 0,, f n, by the property of the wronskian, F n = W (f 0,, f n ) = cw (F (H 1 ),, F (H n+1 )) = j 0 (z z 0 ) h j n ξ(z) where ξ(z) is a holomorphic function dened on U and c is a constant. Therefore, ν Fn (z 0 ) j 0 (h j n). 66

76 Thus, q n ν F (Hj )(z 0 ) ν Fn (z 0 ) = ν F (Hj )(z 0 ) ν Fn (z 0 ) = = = n ν F (Hj )(z 0 ) n j=j 0 +1 n j=j 0 +1 Proposition The function ν F (Hj )(z 0 ) + ν F (Hj )(z 0 ) + n νf n (H j )(z 0 ) = j 0 kn (h j n) j 0 j 0 χ q k+kn Fn G n q F (H j)g(h j ) [ν F (Hj )(z 0 ) (h j n)] n q νf n (H j )(z 0 ). Q.E.D. is continuous on the domain of each holomorphic local coordinate z of M, where χ is dened in (5.1). Proof. Denote by P := χ kn q k+kn Fn G n q F (H j)g(h j ) and let E := q f 1 (H j ). Then P is obviously continuous at z 0 for all z 0 E. Now assume that z 0 E. By Lemma and Lemma 5.1.., we get kn q q ν P (z 0 ) q k + kn ν χ(z 0 ) + ν Fn (z 0 ) + ν Gn (z 0 ) ν F (Hj )(z 0 ) ν G(Hj )(z 0 ) q ( ) νf n (H j )(z 0 ) + νg(h n j )(z 0 ) q νf n (H j )(z 0 ) q νg(h n j )(z 0 ) = 0. Therefore, P is continuous. Q.E.D. 67

77 5. Pseudo-metric on (R) with Negative Curvature Let f : (R) P n (C) be a linearly non-degenerate holomorphic map where (R) := {z : z < R} with 0 < R <. Let H 1,, H q be hyperplanes located in general position in P n (C). Min Ru (see M. Ru [19]) constructed a pseudo-metric Γ = 1 π h(z)dz dz on (R) \ q {φ 0(H j ) = 0} whose Gauss curvature is less than or equal to 1, i.e. Ric Γ Γ (where Ric Γ := dd c ln h), for q n + as follows: Lemma 5..1 (M. Ru [19]). Assume q n + and q < q (n+1) N n(n+) q ) [ βn q n 1 h(z) := c ( 1 φ 0 (H j ) where h p = F p 1 F p+1 F p 4, β n = 1 n 1 p=0 λ 1 p p=0 h βn/λp p (N ln φ p (H j )) βn and dene ] 1, λ p = n p+(n p) q, and c > 0 is some positive N constant. Let Γ = 1 π h(z)dz dz. Then Ric Γ Γ on (R) \ {φ 0(H j ) = 0}. (In our case, since hyperplanes H 1,, H q are located in general position, θ = 1 and q ω(j) = q where θ is the Nochka constant and ω is the Nochka weight.) Note that, Min Ru (see at the end of the proof of the Main Lemma in [19]) indeed proved the following slightly stronger result: Ric Γ β n [ ( q (n + 1) (n + n) q N ) n Ω 0 + p=1 ] q N Ω p + q N Ω n 1 + Γ. (5.) It is obvious that the right-hand side of (5.) is bigger than Γ because q (n + 1) (n + n) q N Noticing that > 0 by the assumption. We will use (5.) in our proof. n 1 h p=0 1 λp p = n 1 ( Fp 1 F p+1 p=0 F p 4 ) n p+(n p) q N = F 0 (n+1) (n +n 1) q N F1 8q N Fn 1 8q 4q + N Fn N, 68

78 F 0 = F and φ 0 (H j ) = F (H j) F, we get [ ] F S ( F 1 F n 1 ) 4q q βn 1+ N F n N h(z) := c q F (H j) n 1 q p=0 (N ln φ (5.3) p(h j )) where S = q (n + 1) (n + n 1) q N. Let f, g : (R) := {z : z < R}( C) P n (C) be two linearly nondegenerate holomorphic maps with reduced representations F = (f 0,, f n ) and G = (g 0,, g n ) respectively, where 0 < R <. Let H 1,, H q be hyperplanes in P n (C) in general position. Denote by and, according to (5.3), we let φ F k (H) = F k(h), φ G F k k (H) = G k(h) G k [ ] F S ( F 1 F n 1 ) 4q q βn 1+ N F n N h 1 := c q F (H j) n 1 q p=0 (N ln, (5.4) φf p (H j )) [ ] G S ( G 1 G n 1 ) 4q q βn 1+ N G n N h := c q G(H j) n 1 q p=0 (N ln. (5.5) φg p (H j )) Lemma 5... Assume q n + 1, q (n + 1) (n + n 1) q knq > 0 N q k+kn and take N such that q < q (n+1) N n(n+). Let Θ = 1 η(z)dz dz with π ( η(z) := c χ(z) F (z) q G(z) q ) knβn q k+kn h1 (z)h (z), (5.6) where χ is the auxiliary function given in (5.1), h 1, h are given in (5.4) and (5.5) respectively. Then η(z) is continuous on (R) and Ric Θ CΘ for some positive 69

79 constant C. Proof. Plugging h 1 (z) and h (z) into η(z), we have ( F G ) S knq kn q k+kn χ q k+kn n 1 η = c p=1 ( F p G p ) 4q q 1+ N ( F n G n ) N q ( F (H j)g(h j ) ) n 1 q [ p=0 (N ln φ F p (H j ))(N ln φ G p (H j )) ] βn. By Proposition , η(z) is continuous on (R). Now we calculate Ric Θ. Set Γ 1 = 1 π h 1(z)dz dz and Γ = 1 π h (z)dz dz where Γ 1, Γ are given in (5.4) and (5.5). Using (5.3) (which holds for h 1, h as well), the assumption q (n + 1) (n + n 1) q N knq q k + kn > 0, and the fact that χ is holomorphic, we have Ric Θ = dd c ln η(z) = 1 ddc ln h 1 (z) + 1 knβ n [ ddc ln h (z) + dd c ln χ q dd c (ln F + ln G ) ] q k + kn 1 Ric Γ Ric Γ knqβ n ( dd c ln F + dd c ln G ) q k + kn 1 ( β n q (n + 1) (n + n) q ) n Ω f0 N + q N Ωf p + q N Ωf n Γ 1 p=1 + 1 ( β n q (n + 1) (n + n) q ) n Ω g0 N + q N Ωg p + q N Ωg n Γ 1 β n knqβ n q k + kn Ωg 0 knqβ n q k + kn Ωf 0 [( q (n + 1) (n + n) q N [( q (n + 1) (n + n) q + 1 β n N 1 1 (Γ 1 + Γ ) h1 (z)h (z)dz dz π p=1 knq q k + kn ) knq q k + kn Ω f 0 ] + 1 Γ 1 ) Ω g 0 ] + 1 Γ 70

80 ( 1 π C = CΘ. χ(z) F (z) q G(z) q ) knβn q k+kn h1 (z)h (z)dz dz In the last step, we used the fact that χ(z) C F (z) q G(z) q for some positive constant C. Q.E.D. 5.3 The Main Lemma The following Ahlfors-Schwarz lemma is well-known (see M. Ru [0]): Ahlfors-Schwarz Lemma. Let Γ = 1 h(z)dz d z be a continuous pseudo-metric π on (R) whose Gaussian curvature is bounded above by 1 (i.e., RicΓ Γ). Then, for all z (R), h(z) ( ) R. R z Now we are ready to prove our Main Lemma. Let ψj,p F = a j,l W (f l, f i1,, f ip ) and ψj,p G = l i 1,,i p l i 1,,i p a j,l W (g l, g i1,, g ip ). By the assumption that f and g are linearly non-degenerate, ψj,p F and ψj,p G do not vanish identically, and thus have only isolated zeros since they are both holomorphic. Note that, from denition, ψ F j,p < F p (H j ) and ψ G j,p < G p (H j ). (5.7) 71

81 knq Main Lemma. Assume q n+1, S > 0 q k+kn with S = q (n+1) (n +n 1) q q, and take N such that N N C 0 such that, for all z (R), knq S F (z)g(z) q k+kn χ(z) < q (n+1). Then there exists some positive constant n(n+1) kn q k+kn Fn (z)g n (z) q F (H j)(z)g(h j )(z) ( R C 0 R z q 1+ n 1 q N p=0 ) n(n+1) [ + n 1 q p=0 (n p) N ]. ( ψ F j,p (z)ψ G j,p(z) ) 4 N Proof. From Lemma 4, we know that η(z) is a continuous map and Ric Θ CΘ. Applying the Schwarz lemma for Θ, we get ( ) R η(z) C 0. (5.8) R z for some constant C 0 > 0. On the other hand, from (5.4), (5.5), the denition of η, 1 and the fact that β n = n 1 p=0 λ 1 p η [ n(n+1) + n 1 q p=0 (n p) N ] = 1 n 1 p=0(n p+(n p) q N ) = 1 n(n+1) + n 1 p=0 (n p) q N, we get = knq F G S q k+kn χ kn q F (H j)g(h j ) n 1 p=0 q q k+kn ( F1 F n 1 G 1 G n 1 ) 4q q 1+ N F n G n N [ (N ln φ F p (H j ))(N ln φ G p (H j )) ]. For a given /N > 0, it holds that lim x 0 x /N (N ln x) <, so we can set K := sup 0<x 1 x /N (N ln x). Since 0 < φ F p (H j ) < 1 for all p and j, we have, by using (5.7), 1 N ln φ F p (H j ) 1 K φf p (H j ) /N = 1 F p (H j ) 4/N 1 ψj,p F 4/N K F p 4/N K F p. 4/N Similarly, 1 N ln φ G p (H j ) 1 ψj,p G 4/N K G p. 4/N 7

82 Hence, η [ n(n+1) + n 1 q p=0 (n p) N ] C knq F G S q k+kn χ kn q 1+ n 1 q N p=0 q k+kn Fn G n q F (H j) q G(H j) ( ψ F j,p 4/N ψ G j,p 4/N) for some constant C > 0. This, together with (5.8), proves the lemma. Q.E.D. 73

83 Chapter 6 The Main Theorem 6.1 Proof of the Main Theorem Main Theorem. Assume that both f = G : M P m 1 (C) and g = G Φ : M P m 1 (C) are linearly non-degenerate (i.e. the images of f and g are not contained in any linear subspaces of P m 1 (C)) and that there exist hyperplanes H 1,, H q in P m 1 (C) located in general position and a positive integer k > 0 such that (i) f 1 (H j ) = g 1 (H j ) for every j = 1,, q, k+1 (ii) f 1 (H ij ) = for any {i 1,, i k+1 } {1,, q}, (iii) f = g on q f 1 (H j ). If q > (m + m + 4k) + (m + m + 4k) + 16k(m )m(m + 1), then f g. 4 74

84 Proof. First, we let n = m 1 for simplicity. Then we obtain the equivalent inequility from the assumption for our q > 0 as the following: q > (m + m + 4k) + (m + m + 4k) + 16k(m )m(m + 1) 4 q > (n + 3n + + 4k) + (n + 3n + + 4k) + 16k(n 1)(n + 1)(n + ) 4 n(n + 1) knq q (n + 1) q k + kn > 0. (6.1) Let x : M R m and x : M R m be two oriented non-at minimal surfaces immersed in R m and let G = [ x 1 z : : x m z ] : M P m 1 (C) and G = [ x 1 z : : xm z ] : M P m 1 (C) be their generalized Gauss maps of M and M respectively. Assume that there is a conformal dieomorphism Φ of M onto M. Then the induced metric on M and on M from the standard metric on R m are given ( by ds x = ) ( x m z z dz and d s x = ) x m z z dz respectively. Assume that M and M are complete with respect to the induced metrics. Note that we may regard M and M as Riemann surfaces and Φ as a conformal diffeomorphism between M and M. Consider the linearly non-degenerate holomorpic maps f := G : M P n (C) and g := G Φ : M P n (C) with n = m 1 and take their reduced (global) representations F := (f 0,, f n ) and G := (g 0,, g n ) respectively with f = P(F ), g = P(G) where P is the canonical projection of C n+1 \ {0} onto P n (C). By Proposition... f 0,, f n (resp. g 0,, g n ) are holomorphic functions on M without common zeros. Set 75

85 F = x x m z z and G = x x m z z. Then, since G = G Φ, we can rewrite the (induced) metric by ds = F dz = F P 1 (f) dz = F P 1 (G) dz = F P 1 ( G Φ) dz = F P 1 (g) dz = F G dz which is, by the assumption, complete. In order to prove the unicity theorem, we will use ds = F G dz because it has both F and G. We are now ready to prove the Main Theorem. By taking the universal cover of M and lifting the maps, if necessary, we can assume that M is simply connected. Then, by the Unifomization theorem, the Riemann surface M is conformally equivalent to either C or the unit-disc := {z : z < 1}. In the case when M = C, it is known that f g from [8]. We enclose a proof here for the sake of completeness. Assume that f g. We will use Cartan's Second Main Theorem to derive a contradiction. To do so, we use some standard notations in Nevanlinna theory. From Lemma , we get ( ) q q k + kn ( N χ (r, 0) kn N (n) f ) (r, H j ) + N g (n) (r, H j ), and, by the First Main Theorem, N χi (r, 0) T f (r) + T g (r) (by choosing and xing the reduced representation F = (f 0,, f n ) and G = (g 0,, g n ) of f and g respectively). On the other hand, from H. Cartan's Second Main Theorem, we have that for ɛ > 0 and all r (0 r < R, R ) except for a set E with nite Lebesgue 76

86 measure, the inequality (q (n + 1) ɛ)t f (r) q N (n) f (r, H j ) + o(t f (r)) hols as r. It is also holds for g. Hence, we get q (T f (r) + T g (r)) q N χj (r, 0) After simplifying, we have ( q k + kn q kn N χ (r, 0) ( ) q q k + kn ( kn ( q k + kn kn N (n) f ) (r, H j ) + N g (n) (r, H j ) ) (q (n + 1) ɛ)(t f (r) + T g (r)) ) (q (n + 1)) q (n + 1) which gives a contradiction under the assumption (6.1) for our q. knq q k + kn 0 So we only need to consider the case M = := {z : z < 1}. We now have two linearly non-degenerate holomorphic maps f, g : P n (C). Assume that f g. We will derive a contradiction. We will use the notations as being introduced in the previous sections. Since q (n + 1) n(n+1) knq > 0 and q k+kn n + n 1 + n 1 p=0 (n p) = n + n 1 + n(n+1) + n 1 p=0 p(p + 1), we can thus choose N > 0 such that [ q n + n 1 + ] n 1 p=0 (n p) q (n + 1) n(n+1) < N < knq q k+kn [ + q n + n 1 + n(n+1) + n 1 q (n + 1) n(n+1) knq q k+kn p=0 p(p + 1) ]. 77

87 Let ρ = S [ n(n+1) + knq + n(n + 1) q + n 1 p(p + 1) q ] where S = q (n + q k+kn N N 1) (n + n 1) q N. Then p=0 ρ > 0 and ρn > 1. (6.) Moreover, from we get N > [ q n + n 1 + ] n 1 p=0 (n p) q (n + 1) n(n+1) knq q k+kn [ ] n(n + 1) knq N[q (n + 1)] qn(n + ) > N + + q q k + kn and n (n p) > 0 q (n + 1) (n + n 1) q N knq n(n + 1) > + q n 1 (n p) > 0, q k + kn N and thus p=0 p=0 q N < q (n + 1) n(n + ) and q (n + 1) (n + n 1) q N knq q k + kn > 0. Hence we can apply the Main Lemma. To do so, let ρ = 1 ρ and let ( q u = F (H j)g(h j ) kn χ q k+kn Fn G n 1+q/N n 1 q ( p=0 ψ F j,p ψj,p G ) 4/N ) ρ (6.3) which, by Proposition 5..3., is a strictly positive continuous function (i.e. it has no zeros) on M = \D where D := {F n = 0} {G n = 0} j,p {z : ψ F j,p(z) = 0 or ψ G j,p(z) = 0}. 78

88 Figure 6.1: The relation among M, M, and (R) Let π : M M be the universal covering map of M. Then ln(u π) is a harmonic function since F (H j ), G(H j ), F n, G n, ψj,p, F ψj,p G and χ are all holomorphic functions and u has no zeros. So we can take a holomorphic function β on M such that β = u π. Without loss of generality, we may assume that M contains the origin o of C. Let õ M denote the point corresponding to o M. For each point p of M we take a continuous curve γ p : [0, 1] M with γ p (0) = o and γ p (1) = π( p), which corresponds to the homotopy class of p. Set w = X( p) = β(z)dz γ p where z denotes the holomorphic coordinate on M induced from the holomorphic global coordinate on M by π. Then X is a single-valued holomorphic function on M satisfying the condition X(õ) = 0 and dx( p) 0 for every p M. Furthermore, for 79

89 the metric β dz on M ln β we have the Gaussian curvature K = β which means that = ln u u = 0, M is at. Thus, we can choose the largest R( ) such that X maps an open neighborhood U of õ biholomorphically onto an open disc (R) in C, and consider the map B = π (X U ) 1 : (R) M. By Liouville's theorem, R = is impossible. Thus, we conclude that R <. Also, by denition of w = γ p β(z)dz, we have dw = β(z) dz = u(z) dz. (6.4) For each a (R) consider the curve L a : w = ta, 0 t < 1 and the image Γ a of L a by B. We shall show that there exists a point a 0 (R) such that Γ a0 tends to the boundary of M. To this end, we assume the contrary. Then, for each a (R), there is a sequence {t ν : ν = 1,,... } such that lim ν t ν = 1 and z 0 = lim ν B(t ν a) is in M. Suppose that z 0 M. Then either F n (z 0 ) = 0, Figure 6.: L a0 and its image Γ a0 when z 0 M G n (z 0 ) = 0, ψ F j,p(z 0 ) = 0, or ψ G j,r(z 0 ) = 0. In any one of these cases, there is a 80

90 constant C such that in a neighborhood of z 0. Thus, u C z z 0 ρ N dw R = dw = L a Γ a dz dz = 1 u(z) dz C dz = Γ a Γ a z z 0 ρ N since ρ N = ρn > 1. This is a contradiction because R <. Therefore, we have z 0 M. Take a simply connected neighborhood V of z 0 which is relatively compact in M. Set C = min z V u(z) > 0. Then B(ta) V (t 0 < t < 1) for some t 0. In fact, if not, Γ a goes and returns innitely often from V to a suciently small neighborhood of z 0 and so we get the absurd conclusion R = dw = L a u(z) dz C Γ a dz =. Γ a Thus we see that lim t 1 B(ta) = z 0. Since π maps each connected component of π 1 (V ) bioholomorphically onto V, there exists the limit p 0 = lim t 1 (X U ) 1 (ta) M. Thus (X U ) 1 has a biholomorphic extension to a neighborhood of a. Since a is arbitrarily chosen, X maps an open neighborhood Ū biholomorphically onto an open neighborhood of (R). This contradicts the property of R, i.e.,r is the largest radius such that X maps an open neighborhood U of õ biholomorphically onto an open disc (R) in C. Therefore, there exists a point a 0 (R) such that Γ a0 tends to the boundary of M. 81

91 Figure 6.3: The image Γ a0 of L a0 tending to the boundary of M Our goal is to show that Γ a0 has nite length, contradicting the completeness of the given minimal surface M. From (6.3) and (6.4), we have dw = u(z) dz = ( q F (H j)(z)g(h j )(z) kn χ(z) q k+kn Fn (z)g n (z) q 1+ n 1 N p=0 q ψf j,p (z)ψg j,p (z) 4 N ) ρ dz. Let Z F (w) = F (B(w)), Z G (w) = G(B(w)), and let ψ Z F j,p, ψz G j,p be dened from Z F, Z G in the same way as ψ F j,p, ψ G j,p from F, G respectively. Then, because ( Z F Z F Z (n) dz F = (F F F (n) ) dw ( Z G Z G Z (n) dz G = (G G G (n) ) dw ( ) p(p+1) ψ Z F dz j,p = ψf j,p dw ) n(n+1) ) n(n+1) ( ) p(p+1) and ψ Z G dz j,p = ψg j,p, dw it is easy to see that, if we let h = n(n+1) + n(n + 1) q + n 1 q N p=0 p(p + 1), then N ρ q Z F (H j )Z G (H j ) [ dw dz = kn χ q k+kn (ZF ) n (Z G ) n 1+q/N n 1 dz q p=0 ψz F j,p ψz G j,p 4/N dw,, h ] ρ. 8