Complete bounded curves

Complete bounded holomorphic curves immersed in C2with arbitrary genus

Francisco Martin Departamento de Geometría y Topología, Universidad de Granada, 18071 Granada, Spain. Masaaki Umehara Department of Mathematics, Graduate School of Science, Osaka University, Toyonaka, Osaka 560-0043, Japan  and  Kotaro Yamada Faculty of Mathematics, Kyushu University, Fukuoka 812-8581, Japan
October 29, 2008
Abstract.

In [MUY], a complete holomorphic immersion of the unit disk into whose image is bounded was constructed. In this paper, we shall prove existence of complete holomorphic null immersions of Riemann surfaces with arbitrary genus and finite topology, whose image is bounded in . To construct such immersions, we apply the method in [L] to perturb the genus zero example in [MUY] changing its genus.

As an analogue the above construction, we also give a new method to construct complete bounded minimal immersions (resp. weakly complete maximal surface) with arbitrary genus and finite topology in Euclidean 3-space (resp. Lorentz-Minkowski 3-spacetime).

1. Introduction

In [MUY], the authors constructed a complete bounded minimal immersion of the unit disk into whose conjugate is also bounded. As applications of our results, we show in this paper that the technique developed by F. J. López in [L] can be suitably modified to give the following three things:

• The first examples of complete bounded complex submanifolds with arbitrary genus immersed in ;

• A new and simple method to construct complete, bounded minimal surfaces with arbitrary genus in the Euclidean 3-space;

• A method to construct weakly complete, bounded maximal surfaces with arbitrary genus in the Lorentz-Minkowski 3-spacetime.

Actually, a complete and bounded conformal minimal immersion of into whose conjugate is also bounded is the real part of a complete and bounded null holomorphic immersion

 X0:D⟶C3.

As it was shown in [MUY], if we consider the projection

 (1.1) π:C3∋(z1,z2,z3)⟼(z1,z2)∈C2,

the map gives a complete bounded complex submanifold immersed in . By a perturbation of , considering a similar deformation like in [L], we shall prove the following

Theorem A.

Let be a closed Riemann surface. For an arbitrary positive integer , there exist points , , …, on , neighborhoods of which are homeomorphic to the unit disc , and a holomorphic immersion

 X:M∖e⋃l=0¯¯¯¯¯Ul⟶C2

which is complete and bounded.

If we do not specify the choice of with the fixed arbitrary genus, we can choose so that the number of ends is at most (see [L, Lemma 3]).

Among other things, López in [L] gave a method to construct complete minimal surfaces between two parallel planes of arbitrary genus. He constructed these examples as a perturbation of the minimal disk given by Jorge and Xavier in [JX]. However, the global boundedness was not treated in [L], because López’ technique is not of use when you apply it directly to Nadirashvili’s minimal disk [N]. So, the aim of this paper is to introduce some simple but appropriate changes in López’ machinery in order to obtain complete bounded minimal surfaces with nontrivial topology. Also the deformation parameter given in [L] is not sufficient to kill the periods of the holomorphic immersions in , and so we must consider another deformation space and calculate the new periods.

On the other hand, the proof of the above theorem allow us to conclude that

Corollary B.

Let be a closed Riemann surface. For an arbitrary positive integer , there exist points ,…, on , neighborhoods of which are homeomorphic to the unit disc , and a complete conformal minimal immersion into the Euclidean -space (resp. a weakly complete maximal surface in the sense of [UY3], which may have singular points, whose first fundamental form is conformal on the regular set, in the Lorentz-Minkowski -spacetime )

 x:M∖e⋃l=0¯¯¯¯¯Ul⟶R3(resp.\ L3)

whose image is bounded.

Though general methods to produce bounded complete minimal immersions into with higher genus are already known (cf. [AFM], [FMM], [LMM1] and [LMM2] ), this corollary gives a new and short way to construct such examples. On the other hand, for a maximal surface in , this corollary produces the first examples of weakly complete bounded maximal surfaces with arbitrary genus. The weak completeness of maximal surfaces, which may admit certain kind of singularities was defined in [UY3]. In [A] a weakly complete disk satisfying a certain kind of boundedness was given. After that, a bounded example of genus zero with one end was shown in [MUY].

Finally, we remark that the technique in this paper does not produce bounded null holomorphic curves immersed in with arbitrary genus because of the difficulty in obtaining the required deformation parameters. If we do succeed to find an example, the above three examples could be all realized as projections of it. So it is still an open problem to show the existence of complete bounded null holomorphic immersion into with arbitrary genus.

2. Preliminaries

Let be a compact Riemann surface of genus , and fix a point . Let

 (2.1) 1=d1

be the Weierstrass gap series at , that is, there exists a meromorphic function on which is holomorphic on such that is the pole of order if and only if . Then there exists a meromorphic function on which is holomorphic on and is the pole of order (). Let be the set of branch points of . Then the divisor of is written as

 (2.2) (df)=∏el=1QmllQm0+10(m0>dκ),

where () are positive integers, and the divisor is expressed by the multiplication of these branch points.

Remark 2.1.

In the latter construction of surfaces in Theorem A and Corollary B, the ends of surfaces in or correspond to the points , …, . We set

 fj+1:=(fj−cj)2,f0=f(j=1,2,3,…)

inductively, where the ’s are complex numbers. Then, by replacing by , , ,…, we can make an example such that is greater than the given number .

We write

 (2.3) M0:=M∖{Q0,…,Qe}.

Denoting by and the (first) holomorphic de Rham cohomology group of and (as the vector space of holomorphic differentials on factored by the subspace of exact holomorphic differentials), respectively, then

 dimH1hol(M)=κ,dimH1hol(M0)=n

hold, where we set, for the sake of simplicity,

 (2.4) n:=2κ+e.

Take a basis of .

Lemma 2.2 (cf. III.5.13 in [F]).

One can take a basis of

 {ζ1,…,ζκ;ξ1,…,ξκ;η1,…,ηe},

where is a meromorphic -form on which is holomorphic on , and is a pole of order , and is a meromorphic -form on which is holomorphic on , and and are poles of order .

For simplicity, we set

 (2.5) ζκ+j :=ξj (j=1,…,κ), ζ2κ+l :=ηl (l=1,…,e).

Then () is a basis of .

Lemma 2.3 ([L, Lemma 1]).

There exists a meromorphic function on with the following properties:

1. is holomorphic on ,

2. is a pole of whose order is greater than or equal to , and

3. is a pole of whose order is greater than .

Proof.

For each , let be a meromorphic function on which is holomorphic on , and so that is a pole of order . On the other hand, take a meromorphic function on which is holomorphic on and is a pole of order . Then is a desired one. ∎

Using as in (2.2) and as in Lemma 2.3, we define

 (2.6) GΛ:=λ0v+1dfn∑j=1λjζj:M⟶C∪{∞},

where

 Λ:=(λ0,λ1,…,λn)∈Cn+1.
Lemma 2.4 ([L, Section 3]).

The function as in (2.6) is a meromorphic function on such that

1. is holomorphic on ,

2. if , is nonconstant on , and

3. if , has poles at .

Proof.

(1) and (2) are obvious. (3) follows from the fact that has a pole of higher order than , …, at for each . ∎

We write

 |Λ|=√|λ0|2+|λ1|2+⋯+|λn|2,

and consider the unit sphere in the space of :

 S1:={Λ∈Cn+1;|Λ|=1}.

The following assertion is a modification of [L, Lemma 2], which is much easier to prove. For our purpose, this weaker assertion is sufficient.

Proposition 2.5.

Let be a point in satisfying . Then there exist and a neighborhood of in such that if and , the set is conformally equivalent to a compact surface of genus minus pairwise disjoint discs with analytic regular boundaries. In particular, there are no branch points of on the boundary .

Proof.

Since the poles of are exactly , …, and , for sufficiently small , the inverse image of the unit disk by is homeomorphic to a compact surface of genus minus pairwise disjoint discs with piecewise analytic boundaries. Moreover, since the set of branch points of does not have any accumulation points, has no branch points for sufficiently small , and consists of real analytic regular curves in .

Furthermore, since , are holomorphic near , …, for any which is sufficiently close to . Thus has the same properties as . ∎

Under the situation in Proposition 2.5, let be a family of loops on which is a homology basis of . On the other hand, take a loop on for each , surrounding a neighborhood of (as in Proposition 2.5).

We define the period matrix as

 (2.7) P=(pkj),pkj:=∫γkζj,

which is a nondegenerate matrix.

3. Proof of the main theorem

In this section, we give a proof of Theorem A in the introduction.

The initial immersion

Let be a complete holomorphic null immersion whose image is bounded in (as in Theorem A in [MUY]), where is the unit disk. We write

 (3.1)

where is a canonical coordinate on and () are holomorphic functions on . Since is null, it holds that

 (3.2) (φ1)2+(φ2)2+(φ3)2=0.

Let be the Weierstrass data of , that is,

 (3.3) φ1=(1−g2)ω0,φ2=i(1+g2)ω0,φ3=2gω0,

where .

Lemma 3.1.

Let be a null holomorphic immersion as in (3.1) whose image is not contained in any plane. Then there exists a point and a complex orthogonal transformation in ( is the transpose of ) such that, up to replacing by , the following properties hold:

1. ,

2. and , where ,

3. and .

Moreover, if is complete and bounded, then so is .

We shall now assume our initial satisfies the three properties above, and set by a coordinate change of .

Proof of Lemma 3.1.

Since the image of is not contained in any plane, at least one of , and , say , is not constant. Then we can take such that and . Moreover, if or vanishes identically, this contradicts that the image of is not contained in any plane. So we may also assume that

 φ1(z0)≠±iφ2(z0).

When , we replace by , where is the linear map associated with a complex orthogonal matrix

 ⎛⎜⎝−c(1+c2)−1/2−c(1+c2)−1/20\par−c(1+c2)−1/2−c(1+c2)−1/20001⎞⎟⎠(c=φ2(z0)/φ1(z0)).

Since is orthogonal, is also a null holomorphic immersion. So we get the property (1). Then (3.2) implies that . Replacing by if necessary, we may assume . Differentiating (3.2), we have . Then

 −φ3(z0)φ′3(z0)=φ1(z0)φ′1(z0)+φ2(z0)φ′2(z0)=iφ3(z0)φ′2(z0)

holds. In particular, it holds that .

Next we prove the boundedness and completeness of under the assumption that is complete and bounded. Since is bounded and is continuous, and are both contained in the closed ball in of a certain radius centered at the origin. We denote by the canonical metric on , and consider the pull-back metric by . Now we apply the following Lemma 3.1 in [MUY] for in . Then there exist positive numbers and () such that on . Now, we consider the pull back of the metric and by the immersion . Since

 (X0)∗h1=(X0)∗(T∗h0)=(T∘X0)∗h0=(X1)∗h0,

we have that

 a(X0)∗h0<(X1)∗h0

on . Since is complete, is a complete Riemannian metric on . Then the above relation implies that is also a complete Riemannian metric on , that is, is also a complete immersion. ∎

A family of holomorphic immersions

Let

 Λ=(λ0,…,λn)∈Cn+1,Δ=(δ1,…,δn)∈Cn,

and take a meromorphic function as in (2.6). Define a meromorphic function as

 (3.4) FΔ=1dfn∑j=1δjζj:M⟶C∪{∞}.

and Weierstrass data on as

 (3.5) ^g=^g(Λ,Δ):=hΔ⋅(g∘GΛ),^ω=^ω(Λ,Δ):=ω0∘GΛdfhΔ(hΔ:=expFΔ),

where is the meromorphic function as in (2.2), and define holomorphic -forms on as

 (3.6) Ψ1=(1−^g2)^ω,Ψ2=i(1+^g2)^ω,Ψ3=2^g^ω.

The following lemma is a modified version of [L, Theorem 3]. (In fact, our data (3.4) and (3.5) for the surfaces are somewhat different from those in [L].)

Lemma 3.2.

If as in (3.1) is a complete immersion, the metric

 d^s2:=(1+|^g|2)2|^ω|2

determined by as in (3.5) is a complete Riemann metric on for a sufficiently small .

Proof.

As in the equations (15) and (17) in [L], there exists a positive constant () such that

 a<|hΔ|<1aand∣∣∣dfdGΛ∣∣∣>aon G−1Λ(D).

Then, (setting )

 (1+|^g|2)|^ω| =(1+|ghΔ|2)∣∣∣ω0hΔ∣∣∣∣∣∣dfdGΛ∣∣∣|dz| ≥(a2+|ag|2)|aω0|(a|dz|)=a4(1+|g|2)|ω|.

Thus we have the conclusion. ∎

Thus, for each (sufficiently small) , there exists a complete null immersion

 (3.7)

where denotes the universal cover of . In fact, the line integral () from a base point depends on the choice of the path, but can be considered as a single-valued function on .

Then we get the following assertion, which can be proven exactly in the same way as Corollary B in [MUY]:

Proposition 3.3.

Let be the projection as in (1.1). Then is a complete immersion of into .

The period map

Under the situations above, we define the period map

 (3.8) Per1:C2n+1∋(Λ,Δ)⟼t(∫γ1Ψ1,…,∫γnΨ1,∫γ1Ψ2,…,∫γnΨ2)∈C2n,

where (see (2.4)), “” is the transposing operation for matrices, ’s are loops as in (2.7), and and are as in (3.6). The following assertion is an analogue of [L, Theorem 2]:

Proposition 3.4.

Suppose that satisfies the three conditions as in Lemma 3.1. Then the matrix

 (3.9) J1:=(∂Per1∂λ1,…,∂Per1∂λn,∂Per1∂δ1,…,∂Per1∂δn)∣∣∣(Λ,Δ)=(0,0)

is nondegenerate.

Proof.

Note that

 (3.10) GΛ|Λ=0=FΔ|Δ=0=0,hΔ|Δ=0=1.

By the definitions, we have

 ∂GΛ∂λj∣∣∣Λ=0=ζjdf,∂hΔ∂δj∣∣∣Δ=0=∂expFΔ∂δj∣∣∣Δ=0=ζjdf

for . Then

 ∂^g∂λj∣∣∣(Λ,Δ)=(0,0) =(dgdz∣∣∣z=0)(∂GΛ∂λj∣∣∣Λ=0)=g′(0)ζjdf, ∂^ω∂λj∣∣∣(Λ,Δ)=(0,0) =ω′0(0)ζj, ∂^g∂δj∣∣∣(Λ,Δ)=(0,0) =g(0)ζjdf,∂^ω∂δj∣∣∣(Λ,Δ)=(0,0)=−ω0(0)ζj,

where . Hence we have

 ∂Ψ1∂λj∣∣∣(Λ,Δ)=(0,0) =(−2^g∂^g∂λj^ω+(1−^g2)∂^ω∂λj)∣∣ ∣∣(Λ,Δ)=(0,0) =(−2g(0)g′(0)ω0(0)+(1−g(0)2)ω′0(0))ζj =((1−g2)ω)′∣∣z=0ζj=φ′1(0)ζj, ∂Ψ1∂δj∣∣∣(Λ,Δ)=(0,0) =−(1+{g(0)}2)ω0(0)=iφ2(0)ζj,

where ’s are holomorphic functions on as in (3.3). Similarly, we have

 ∂Ψ2∂λj=φ′2(0)ζj,∂Ψ3∂λj=φ′3(0)ζj,∂Ψ2∂δj=−iφ1(0)ζj,∂Ψ3∂δj=0

at . Thus, we have that

 (3.11) ∂∂λj∫γkΨ1 =φ′1(0)∫γkζj, ∂∂δj∫γkΨ1 =iφ2(0)∫γkζj, ∂∂λj∫γkΨ2 =φ′2(0)∫γkζj, ∂∂δj∫γkΨ2 =−iφ1(0)∫γkζj, ∂∂λj∫γkΨ3 =φ′3(0)∫γkζj, ∂∂δj∫γkΨ2 =0

hold at , for . Hence the matrix in (3.9) is written as

because of Lemma 3.1, where is the nondegenerate period matrix as in (2.7). Hence is nondegenerate. ∎

The period-killing problem

Since , Proposition 3.4 yields that there exists a holomorphic map such that

 Per1(c,λ1(c),…,λn(c),δ1(c),…,δn(c))=0

for sufficient small . We set

 Gc=GΛ(c),whereΛ(c):=(c,λ1(c),…,λn(c)).

Since , there exists an analytic function such that near . Now we can apply Proposition 2.5 to . Then, for sufficiently small , is conformally equivalent to minus pairwise disjoint discs with analytic regular boundaries, and the map

 (3.12) Xc:=π∘X(Λ(c),Δ(c)),Λ(c)=(c,λ1(c),…,λn(c)),Δ(c)=(δ1(c),…,δn(c))

is well-defined on . Moreover, by Corollary 3.3, is a complete immersion for any sufficient small .

Boundedness of Xc

By Proposition 2.5, does not vanish on for sufficiently small . Then if we choose a real number sufficiently close to , we have

 dGc≠0onG−1c(¯¯¯¯D∖Dr),

where . Moreover, is exactly a union of closed annular domains surrounding the points in . To show the boundedness of , it is sufficient to show that the image of each annular domain by is bounded. For the sake of simplicity, we show the boundedness of at . We denote by the closed annular domain surrounding the point . Then

 G:=Gc|¯¯¯Ω:¯¯¯¯Ω⟶¯¯¯¯D∖Dr

gives a holomorphic finite covering.

Since is compact and is holomorphic, there exists a positive constant such that

 (3.13) |Xc|≤K0onG−1c(¯¯¯¯Dr).

We denote by the set of interior points of , and fix arbitrarily. Let () be a line segment such that and , that is,

 σ(t):=(1−t)G(q)+trG(q)|G(q)|.

Since is a covering map, there exists a unique smooth curve such that and . Moreover, there exists a neighborhood of the line segment and a holomorphic map which gives the (local) inverse of . By definition, we have . We set

 q1:=~σ(1)∈∂Ω∖G−1c(∂D),

that is, lies on the connected component of which is further from , see Figure 1.

By (3.13), it is sufficient to show that

 (3.14) ∣∣Xc(q1)−Xc(q)∣∣=∣∣∣∫~σ(Ψ1,Ψ2)∣∣∣

is bounded from above by a constant which does not depend on . Here we have

 (3.15) ∣∣∣∫~σΨ1∣∣∣=∣∣∣∫~σ(1−(hΔ)2(g∘Gc)2)ω0∘GchΔdf∣∣∣=∣∣ ∣∣∫10⎛⎝(1−(g∘σ(t))2(h(t))2)ω0∘σ(t)h(t)d(f∘H(z))dz∣∣ ∣∣z=σ(t)dσdt⎞⎠dt∣∣ ∣∣=|I1+I2|,

where and

 I1 =12∫10((φ1(z)−iφ2(z))∣∣z=σ(t)dσdt)(1hΔ∘H(z)df∘H(z)dz∣∣∣z=σ(t))dt, I2 =−12∫10((φ1(z)+iφ2(z))∣∣z=σ(t)dσdt)(hΔ∘H(z)df∘H(z)dz∣∣∣z=σ(t))dt.

Here we used the relations and . To estimate , we shall apply Lemma A.1 in the appendix for

 a(t)=(φ1(z)−iφ2(z))∣∣z=σ(t)dσdtandb(t)=1hΔ∘H(z)df∘H(z)dz∣∣∣z=σ(t).

Let us check the hypotheses:

 (A(s):=)∫s0a(t)dt=∫σ([0,s])(φ1(z)−iφ2(z))dz=(X1(z)−iX2(z))∣∣z=σ(s)z=σ(0),

where as in (3.1). Since is bounded, is bounded for all . On the other hand,

 b(t)=~b(σ(t)),where~b(z)=1hΔ∘H(z)d(f∘H(z))dz.

Since