CUBO A Mathematical Journal

Vol.10, N
o
¯ 03, (83–91). October 2008

Many-Ended Complete Minimal Surfaces

Between Two Parallel Planes in R
3

Francisco Brito∗

Universidade Federal de Pernambuco – UFPE,

CCEN-Departamento de Matemática,

50.740.540, Recife-PE, Brazil

email: brito@dmat.ufpe.br

ABSTRACT

We use some special convergent Hadamard gap series to provide examples of complete

minimal surfaces of many different conformal types between two parallel planes in three

dimensional Euclidean space.

RESUMEN

Nosotros usamos algunas series convergentes especiales de Hadamard para dar ejemplo

de superficies mı́nimas completas de varios diferentes tipos conforme entre dos planos

paralelos en espacios euclideanos de dimensión tres.

Key words and phrases: Complete minimal surfaces, Hadamard gap series.

Math. Subj. Class.: 53A10.

∗I am grateful to Profs. Marcus V. M. Wanderley and Maria Elisa Oliveira for helpful hints and discussions.



84 Francisco Brito CUBO
10, 3 (2008)

1 Introduction

In [7] F. Xavier and L. P. M.Jorge established the existence of complete non-planar minimal surfaces

between two parallel planes in R3. Their technique consisted of an artful use of Runge’s Theorem

to prove the existence of holomorphic functions on the unit disc D with the right properties they

needed. Later their method was adapted by others to produce new surfaces as above with new

features, like having cylindrical type as in [6] or being non-orientable as in [3].

Another way for rendering complete minimal surfaces between two parallel planes in R3 was

developed in [1]. This method consisted mainly in proving the existence of bounded holomorphic

functions h in D, given by lacunary power series, and such that

∫

γ

|h′(z)|2|dz| = ∞ for all divergent

paths γ in the unit disc.

In this paper we intend to show the flexibility of the second method by producing examples

of complete minimal surfaces between two parallel planes in R3 of the following conformal types:

1. A disc with finitely many points removed.

2. Any annulus, 0 < r < |z| < R.

3. Any annulus as above with finitely many points removed.

This work is organized as follows: In §2 we give some definitions and prove the lemmas that

will be needed in the other sections. In the three remaining sections we describe the examples of

the types above.

Remark 1.1 This work was written about fourteen years ago and circulated as a preprint for some

time. Meanwhile N.Nadirashvili proved in [4] the existence of bounded complete minimal surfaces

in R3.

2 Some definitions and lemmas

Lacunary power series were defined in [1] with the restriction that they would have radius of

convergence 1. This is just a mild technical point. Here we use any positive real number R as

radius of convergence and make the necessary changes for having an analogue of Theorem 2 of [1].

Definition 2.1 A convergent power series

∞
∑

k=0

akz
nk is lacunary if there exists a real number q > 1

such that
nk+1

nk
≥ q for all k = 0, 1, . . . .

Lemma 2.2 Let h(z) =

∞
∑

k=0

akz
nk be a lacunary power series of radius of convergence R > 0, and

suppose that the following three conditions hold:



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Many-Ended Complete Minimal Surfaces ... 85

a)

∞
∑

k=0

|ak|R
nk converges.

b) lim
k→∞

R
nk |ak|min

{

nk+1

nk
,

nk

nk−1

}

= ∞.

c)

∞
∑

k=0

R2nk |ak|
2
nk diverges.

Then h is bounded in DR = {z ∈ C; |z| < R}, and for all divergent paths γ in DR, one has

that

∫

γ

|h′(z)|2|dz| = ∞.

Proof. The change of variable z = Rw together with (a) show that h is bounded. The same

change of variable and Theorem 2 of [1] finish the proof.

Lemma 2.3 If h satisfies the conditions of the above lemma in DR, H(z) = h(z
−1) is a bounded

holomorphic function in AR = {z ∈ C; |z| > R
−1}, and for all divergent paths γ in AR tending to

a point of |z| = R−1 one has that

∫

γ

|H
′

(z)|2|dz| = ∞.

Proof. The change of variable z = w−1 and Lemma 2.2 prove this assertion.

Now, given r,R ∈ R, 0 < r < R, let ΩR,r be the annulus r < |z| < R.

Lemma 2.4 Suppose that h1(z) =

∞
∑

k=0

akz
nk and h2(z) =

∞
∑

k=0

bkz
mk are lacunary power series

that satisfy the conditions of Lemma 2.2, and have radii of convergence R and r−1 respectively,

with 0 < r < R. Suppose further that h
′

1
(z) and h

′

2
(z−1) do not vanish in |z| = r and |z| = R

respectively. Then, for all divergent paths γ in ΩR,r one has that

∫

γ

|h
′

1
(z)|2|(h2(z

−1))′|2|dz| = ∞.

Proof. A divergent path in ΩR,r either approach |z| = r or |z| = R. Suppose that γ is a divergent

path that approaches |z| = r. Since h
′

1
is holomorphic in a neighborhood of that circle and does

not vanish at any point of it, it follows that there is a perhaps smaller neighborhood U of |z| = r

having compact closure U , and such that inf
z∈γ∩U

{|h
′

1
(z)|2} = A > 0. Consequently, if γ̃ denotes the

portion of γ inside U one has that

∫

γ

|h
′

1
(z)|2|(h2(z

−1))
′

|2|dz| ≥ A2
∫

γ̃

|(h2(z
−1))

′

|2|dz| = ∞

by Lemma 2.3. The rest of the proof follows in a similar way.

In the next sections we will use mainly the Weierstrass representation for minimal surfaces in

R
3 (see [5]) and the three lemmas above.



86 Francisco Brito CUBO
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3 Minimal immersions of a disc with finitely many points

removed

Let Ω = DR − {a1, a2, . . . , an}, where DR is the open disc of radius R centered at the origin and

a1,a2,. . . ,an are distinct points of DR.

Theorem 3.1 There exist complete minimal immersions M of Ω between two parallel planes of

R
3. Furthermore the ends of M corresponding to the points a1,a2,. . . ,an are all planar and have

index one.

Proof. To avoid notational inconveniences we will prove separately the cases n = 1 and n > 1. In

the first case we take Ω = DR − {a}, for some a ∈ DR. Using the Weierstrass representation, set

f (z) = (z − a)−2 and g(z) = (z − a)2h′(z)

where h is any function as in Lemma 2.2. Clearly the data above defines a minimal immersion M

of Ω in R3. Moreover, M is also complete for the metric is given by

λ(z)|dz| = 1
2
{|z − a|−2 + |z − a|2|h′(z)|2}|dz|,

and because x3(z) = Re(h(z)), it follows from the properties of h that the third coordinate of M

is bounded.

The end corresponding to a is of course planar because f g is holomorphic at that point and

have index one because f has a pole of order two at a and f g2 vanishes at that same point. For

information on the behavior of ends of complete minimal surfaces see [2].

Now we consider the case n > 1. In the Weierstrass representation for M set

f (z) = F (z)−1 exp







n
∑

j=1

Aj Fj (z)







and g(z) = h′(z)F (z),

where h is a function as in Lemma 1.1, F (z) =

n
∏

j=1

(z − aj )
2 the functions Fj satisfy (z−aj)

2F
′

j (z) =

F (z), and the Aj are constants to be chosen so that

∫

σ

f (z)dz = 0 for all closed curves σ in Ω.

Since f g and f g2 have holomorphic extensions to all of |z| < R, it follows that this will be

enough to exclude the possibility of real periods appearing in the Weierstrass representation of

M. An easy computation shows that the choice Aj = F
′′

j (aj )(F
′

j (aj ))
−2, j = 1, . . . , n solves the

problem.



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Many-Ended Complete Minimal Surfaces ... 87

Observe that the metric λ(z)|dz| on M is given by

2λ(z)|dz| = {|F (z)|−1 + |F (z)||h
′

(z)|2}

∣

∣

∣

∣

∣

exp

{

n
∑

k=1

AkFk(z)

}
∣

∣

∣

∣

∣

|dz|,

and since there is a positive real C such that

∣

∣

∣

∣

∣

exp

{

n
∑

k=1

AkFk(z)

}
∣

∣

∣

∣

∣

≥ C, z ∈ Ω, it follows

from the properties of h and that f has poles of order 2 at the aj that M is complete. Also,

x3(z) = Re

∫

h
′

(z) exp{

n
∑

j=1

Aj Fj (z)}dz is bounded in Ω. This can be seen in the following way:

exp

{

n
∑

k=1

AkFk(z)

}

and its derivatives as well as h are all bounded holomorphic functions in

Ω. As a matter of fact they are bounded in DR, so, by integration by parts, it follows that
∫

h
′

(z) exp{
n

∑

j=1

Aj Fj (z)}dz is bounded in DR, so x3 is also bounded.

By an argument similar to the one done in the case n = 1 we conclude that the ends corre-

sponding to the points aj are all planar and have index one.

4 Complete minimal annuli between two parallel planes in R
3

All the examples of complete minimal annuli in [6] have the conformal type of an annulus of the

form R−1 < |z| < R. Here we give examples of all possible annuli 0 < r < |z| < R < ∞.

Theorem 4.1 Given any annulus ΩR,r there is a complete minimal immersion of it in R
3 with

one coordinate bounded.

Proof. Take any two functions h1 and h2 as in Lemma 1.3, say h1(z) =

∞
∑

k=0

akz
nk and h2(z) =

∞
∑

l=0

alz
ml , with radii of convergence R and r−1 respectively, and such that all the nk and ml are

simultaneously either even or odd. Then in the Weierstrass representation we set in DR,r,

f (z) = 1 and g(z) = h
′

1
(z)H

′

2
(z),

where H2(z) = h2(z
−1). Because f is constant, and g is an even function where defined, it follows

that for all closed curves γ in ΩR,r one has that

∫

γ

f dz =

∫

γ

f gdz =

∫

γ

f g2dz = 0.



88 Francisco Brito CUBO
10, 3 (2008)

Thus, the minimal surface so obtained is in fact well defined. Furthermore, since the metric λ(z)|dz|

is given by

2λ(z)|dz| =
(

1 + |h
′

1
(z)|2|H

′

2
(z)|2

)

|dz|

it follows from Lemma 1.3 that it is complete. It remains to prove only that one of the coordinates

of that immersion is bounded.

Since x3(z) = Re

∫

gdz = Re

∫

h
′

1
(z)H

′

2
(z)dz, it is enough to prove that

∫

gdz is a bounded

holomorphic function in ΩR,r. Consider ρ ∈ R such that r < ρ < R and define the sets

A1 = ΩR,r ∩ {z ∈ C such that |z| ≤ ρ} and A2 = ΩR,r ∩ {z ∈ C such that |z| ≥ ρ}.

We then observe that h
′

1
(z), its derivatives and H2 are bounded in A1 and the same happens to H

′

2
,

its derivatives and h1 in A2. So, by integration by parts we can conclude that

∫

gdz is bounded

in both A1 and A2, hence in ΩR,r.

5 The case of a annulus with finitely many points removed

Let A = {a1, . . . , an} be a set of distinct points of ΩR,r such that A
⋂

(−A) = ∅, and set Ω =

ΩR,r − A.

Theorem 5.1 There is a complete minimal immersion of Ω between two parallel planes of R3.

Furthermore, the ends corresponding to a1, . . . , an are planar.

Proof. First suppose n = 1 and take Ω = ΩR,r − {a} with a ∈ ΩR,r. Consider holomorphic func-

tions h1 and h2 as in Theorem 3.1, and define the surface M using the Weierstrass representation

by taking

f (z) = (z − a)−2 and g(z) = (z − a)2h
′

1
(z)H

′

2
(z),

keeping the notation of Theorem 3.1. Then f is holomorphic in a neighborhood of |z| ≤ r, and has

residue zero at a, so

∫

σ

f dz = 0, for all closed curves σ inside Ω.

Because f g = h
′

1
H

′

2
is an even holomorphic function in ΩR,r it follows that

∫

σ

f gdz = 0 for

all closed curves σ in Ω too.

Now we must make one more choice in order to have

∫

σ

f g2dz = 0 for all closed curves σ in Ω.

The idea is to start with h1 and h2 having no low powers in their power series expansions. Since

f (z)g2(z) = (z − a)2(h
′

1
(z))2(H

′

2
(z))2 is holomorphic in Ω, by expanding (z − a)2, it is clear that

the only term that may cause problems is −2az(h
′

1
(z))2(H

′

2
(z))2 because the other two terms are

even functions. Hence if the functions h1 and h2 are chosen to satisfy Theorem 3.1 and have the



CUBO
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Many-Ended Complete Minimal Surfaces ... 89

form

h1(z) =

∞
∑

k=1

akz
1+2

2
k

and h2(z) =

∞
∑

k=1

bkz
1+2

2
k

there is no term in z−1 in the Laurent expansion of z(h
′

1
(z))2(H

′

2
(z))2. Hence, for all closed curves

σ in Ω,

∫

σ

f g2dz = 0 as wanted. Besides, as in Theorem 2.1, the end corresponding to the point a

is planar and have index one.

In order to study the case n > 1 define the following holomorphic functions in Ω: F (z) =
n

∏

j=1

(z − aj )
2, Fk(z) = (z − ak)

−2F (z) and G
′

k
(z) = zFk(z)Fk(−z) for k = 1, . . . , n, and finally

H(z) =

n
∑

j=1

Aj Gj (z), where the constants Aj are to be determined so that, if the immersion M of

Ω is defined in terms of the Weierstrass representation by setting

f (z) = (F (z))−1 exp H(z) and g(z) = F (z) exp

{

−
1

2
H(z)

}

h
′

1
(z)H

′

2
(z),

then f has residue zero at all the points aj . As before, the functions h1 and h2 are chosen

satisfying the conditions of Lemma 1.3 and the exponents are chosen so that

∫

|z|=ρ

f (z)g(z)2dz = 0,

for r < ρ < R.

It must be pointed out that once these constants Aj are determined the rest is done quite

easily as follows: First we observe that H is an even holomorphic function in ΩR,r, and the same

happens to h
′

1
and H

′

2
, thus f g is an even holomorphic function in ΩR,r and so

∫

σ

f gdz = 0 for all

closed curves σ in ΩR,r as wanted.

Furthermore,

2λ(z)|dz| = |F (z)|−1| exp H(z)| + |F (z)||h
′

1
(z)|2|H

′

2
(z)|2,

hence, repeating the reasoning in the proof of Theorem 2.1 we conclude that λ(z)|dz| is complete

and x3 is bounded.

Now we determine the constants Aj . First, we observe that all the poles of f have order two

and that for j = 1, . . . , n, (z − aj )
2f (z) =

exp H(z)

Fj (z)
, thus

d

dz
{(z − aj )

2f (z)} =
exp H(z)

F 2
j
(z)

[

H
′

(z)Fj (z) − F
′

j (z)
]

=
exp H(z)

F 2
j
(z)

[

Fj (z)

{

n
∑

k=1

AkzFk(z)Fk(−z)

}

− F
′

j (z)

]

.



90 Francisco Brito CUBO
10, 3 (2008)

So, the residue of f at aj is zero if and only if

Fj (aj )

{

n
∑

k=1

Akaj Fk(aj )Fk(−aj )

}

− F
′

j (aj ) = 0.

Since Fk(aj ) = 0 for k 6= j, Fj (aj ), Fj (−aj ) 6= 0 and aj 6= −ak, for 1 ≤ j, k ≤ n, it follows

that aj F
2

j (aj )Fj (−aj )Aj − F
′

j (aj ) = 0, for each j, 1 ≤ j ≤ n, hence

Aj =
F

′

j (aj )

aj F
2

j
(aj )Fj (−aj )

, for j = 1, . . . , n.

To finish the proof it is enough to show that we can choose h1 and h2 in such a way that
∫

|z|=ρ

f (z)g(z)2dz = 0, for r < ρ < R.

Since f (z)g2(z) = F (z)[h
′

1
(z)H

′

2
(z)]2, and F has degree 2n, if we define

h1(z) =
∞
∑

k=1

akz
1+2n+2

2
k

and h2(z) =
∞
∑

k=1

bkz
1+2n+2

2
k

we are done. Also the observations about the ends in the other cases are valid here without change.

The assumption that (−A) ∩ A = ∅ is not really needed. It is just a technical difficulty that

can be easily overcome as follows:

Corollary 5.2 If A is any finite subset of ΩR,r there is a complete minimal immersion of Ω =

ΩR,r − A between two parallel planes of R
3. Furthermore, the ends corresponding to the points of

A are planar.

Proof. Induction on the number of elements of A shows that there exists a transformation

π : Ω −→ Ω
′

,

where π(z) = z2
p

for some positive integer p and Ω
′

satisfies the condition of Theorem 5.1.

It is clear that (π , Ω) is an unramified covering of Ω
′

, and by Theorem 5.1, there is a complete

minimal immersion X of Ω
′

between two parallel planes of R3. So X◦π is also a complete minimal

immersion of Ω in R3 with the same properties as before.

Received: February 2008. Revised: June 2008.



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Many-Ended Complete Minimal Surfaces ... 91

References

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Duke Math. Journal 68 (1992), 297–300.

[2] L.P.M. Jorge and W. Meeks, The topology of complete minimal surfaces opf finite Gaussian
curvature, Topology 22 (1983), 203–221.

[3] F. Lopez, A non orientable complete minimal surface in R3 between two parallel planes,
Proceedings of the Amer. Math. Soc 103 (1988), 913–917.

[4] N. Nadirashvili, Hadamard and Calabi-Yau’s conjectures on negatively curved and minimal
surfaces, Invent. Math. 126 (1996), 457–465.

[5] R. Osserman, A survey of minimal surfaces, Van Nostrand, New York, 1969.

[6] H. Rosenberg and E. Toubiana, A cylindrical type minimal surface in a slab in R3, Bull.
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[7] F. Xavier and L.P.M. Jorge, A complete minimal surface in R3 between two parallel planes,
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