HOW TO CONSTRUCT COMPLETE BOUNDED MINIMAL SURFACES

HOW TO CONSTRUCT COMPLETE BOUNDED MINIMAL
SURFACES
F. MART´
IN
Departamento de Geometr´ıa y Topolog´ıa. Universidad de Granada
18071 Granada Spain
E-mail: fmartin@ugr.es
S. MORALES
Departamento de Geometr´ıa y Topolog´ıa. Universidad de Granada
18071 Granada Spain
E-mail: santimo@ugr.es
In this paper, we have used the Weierstrass representation and Runge’s theorem
in order to produce complete bounded minimal surfaces with certain topology in
the 3-dimensional Euclidean space.
1
Introduction
The Weierstrass representation of a minimal surface has provided a profitable
relationship between the Complex Analysis and the Minimal Surface Theory.
The Weierstrass representation is the most useful tool to construct minimal
surfaces. We will see a simply version of the Weierstrass representation in
section 3.
The Weierstrass representation has encouraged the use of Complex Anal-
ysis to solve problems in Minimal Surface Theory. For instance, Runge’s the-
orem has been used to solve the problem of the existence of complete bounded
minimal surfaces immersed in Euclidean spaces:
In the 1960’s, Calabi raised the question of the existence of complete, non
planar, minimal surfaces immersed in a half-space of R3. The followings two
problems have arisen from Calabi’s question:
Are there any complete bounded minimal immersions in R3?
And, if the answer is affirmative:
What is the geometry of these surfaces?
Initially, a partial solution to these problems was given by Jorge and
Xavier4 in 1980. They used the Weierstrass representation and Runge’s the-
orem to construct complete minimal disks, non flat, immersed into a slab of
R3, that is, between two parallel planes.
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During the following years, more examples of complete minimal surfaces
in a slab appeared. The principal tool was again the Weierstrass represen-
tation and Runge‘s theorem, but used in a more elaborate way, in order to
produce examples with non trivial topology. So, Rosenberg and Toubiana10
constructed annuli, and L´
opez5 constructed Moebius strips.
Apart from this, Fortes Brito found out an innovative way to construct
complete, minimal disks and annuli immersed in a slab1,2. His new method
was based on the use of power series with Hadamard gaps, and it does not
use Runge’s technique. Furthermore, Costa and Sim˜
oes3, as well as L´
opez6
constructed minimal surfaces with arbitrary genus immersed in a slab.
A more definitive solution to the problem was given by Nadirashvili8. He
constructed complete minimal disks immersed in a ball of R3. This construc-
tion was based on the Weierstrass representation and Runge’s theorem, like
Jorge and Xavier’s proof, and also used the L´
opez-Ros transformation.
As before, we can discover new examples of complete bounded minimal
surfaces with topologies more complicated. This is what this paper tries to
do. It generalizes the techniques used by Nadirashvili to obtain complete
bounded minimal annuli immersed into a ball7. It also intend to illustrate
the used of the Weierstrass representation and Runge’s theorem to obtain
complete bounded minimal surfaces.
2
Notation
We denote Dr = {z ∈ C : |z| < r}, Sr = {z ∈ C : |z| = r}, D∗ = D1 \ {0},
Br = {p ∈ R3 : p < r} and Int(α) as the interior domain bounded by a
Jordan curve α.
Let α be a curve in D∗, by long(α, X) we mean the length of α with
the metric associated to immersion X. For T ⊂ D∗ we define the following
distance: If a, b ∈ T let dist(X,T)(a,b) = inf{long(α,X)|α : [0,1] → T, α(0) =
a, α(1) = b}. If A ⊂ T, dist(X,T)(z,A) means the distance between point z
and set A.
3
The Weierstrass representation of a minimal surface
A version of the Weierstrass representation for minimal immersions defined
over an open subset of the complex plane, is the following:
Let Ω an open subset of the complex plane, and X : Ω → R3 a conformal
minimal immersion. Then, fixed a orthogonal coordinates, there exist three
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holomorphic functions φ1, φ2, φ3 : Ω → C3,
∂X
∂X
φ
j
j
j =
,
j = 1, 2, 3,
(z = u + iv),
(1)
∂u − i ∂v
with real residues at 0, verifying
3
φ 2(z) = 0 and
3
j=1
j
j=1 |φj(z)|2 = 0,
∀z ∈ Ω. So, the immersion has the form:
z
X(z) = Re
(φ1(w), φ2(w), φ3(w)) dw + c,
z0 ∈ Ω, c ∈ R3,
(2)
z0
If we define
φ
f = φ
3
1 − iφ2, g =
,
(3)
φ1 − iφ2
then g is a meromorphic function on Ω that coincides with the stereographic
projection of the Gauss map. The behavior of f is determined by the rule
that f is holomorphic on Ω, with zeroes precisely at the poles of g, but with
twice order
Conversely, if f, g are a holomorphic and meromorphic functions, respec-
tively, on Ω such that
f
f
φ1 =
(1
(1 + g2),
φ
2
− g2), φ2 = i2
3 = f g,
(4)
are holomorphic functions on Ω, and they have no real periods in zero, (when
Ω is a simply connected set, this property trivially holds), then X : Ω → R3,
defined as (2), is a conformal minimal immersion. It is usual to label φ =
(φ1, φ2, φ3) as the Weierstrass representation of the minimal immersion X.
We can write the conformal metric associated to the immersion X,
λ2X (z) < ·,· >, in terms of the Weierstrass representation as follows:
1
φ(z)
λX (z) =
.
(5)
2 |f (z)|(1 + |g(z)|2) =
√2
We remark that the Weierstrass representation of a minimal immersion
depends on the choosen orthogonal coordinates.
For more details minimal surfaces we refer to 9.
3.1
L´
opez-Ros transformation
If we have a Weierstrass representation (f , g) of a minimal immersion X : Ω →
R3, we can define new parameters (f h, g/h) where h : Ω → C is a holomorphic
nonnull function. Let X from (2) and (4) where f = f h and g = g/h. If X
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is well defined, then X is a new minimal immersion such that has the same
third coordinate as the immersion X.
This transformation can be used with functions h, where h by using
Runge’s theorem. For instance, let X a minimal immersion over the set Ω
with Weierstrass representation (f, g), K be a compact set of Ω and K a open
neighbourdhood of K. From Runge’s theorem exists h holomorphic such that
|h−1| < in Ω\K and |h−τ| < in K. If (fh,g/h) becomes a well defined
minimal immersion Y , then:
• for > 0 small enough, Y and X are similar in Ω \ K ,
• for τ large enough, the intrinsic metric of Y grows with respect to the
intrinsic metric of X in K: λY >> λX , in K,
• Y and X have the same third coordinate.
3.2
z2-type Weierstrass representation
In this paper, we deal with Weierstrass representation defined over domains
that have the conformal type of an annulus, so the immersion may not be well
defined. For this reason, we use a special kind of Weierstrass representation,
called z2-type.
We say φ is a z2-type Weierstrass representation when φ has the form
φj(z) = φj(z2), j = 1, 2, 3, where φj are holomorphic functions.
Therefore, if φ is z2-type defined over D∗, the immersion X, from (2),
will be well defined.
A z2-type Weierstrass representation gives rise to certain symmetries on
the immersion: so X(z) + X(−z) is constant. For this, we use certain sym-
metrical domains in C: domains bounded by polygonal pairs.
4
Polygonal pairs
A polygonal pair, (P, Q) is a pair of closed, simple curves in R2 formed by a
finite number of straight segments verifying:
• D1/3 ⊂ Int(Q) ⊂ Int(Q) ⊂ D2/3,
• D2/3 ⊂ Int(P) ⊂ Int(P) ⊂ D1,
• −z ∈ P, ∀z ∈ P and −z ∈ Q ∀z ∈ Q.
We denote T (P, Q) = Int(P ) \ Int(Q), (see the figure 1).
For ξ > 0 small enough, a polygonal pair parallel to (P, Q) is a new
polygonal pair (P ξ, Qξ) such that:
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Figure 1. Polygonal pairs.
• P and Pξ are parallel, and ξ apart,
• Q and Qξ are parallel, and ξ apart,
• Tξ = T(Pξ,Qξ) ⊂ T.
5
Construction of a complete bounded minimal annulus
Theorem There exists a complete minimal immersion X : A → R3 such
that X is bounded and A ⊂ C is biholomorphic to an open annulus.
The immersion X is defined as the limit of a sequence of conformal mini-
mal immersions Xn. The limit is taken in the harmonic functions spaces with
the usual topology of uniform convergence over compact sets. The immer-
sions Xn are defined over a sequence of domains Tn = T (Pn, Qn) bounded by
polygonal pairs.
The sequence of immersions Xn and domains Tn must verify the followings
properties:
(A ) (1
1/i < dist
1/i,
n
− kn) ni=1
(Xn,Tn)(z, S2/3) <
n
i=1
n
∀z ∈ Pn ∪ Qn,
where kn
1/i < 1/(n + 1),
i=1
(B ) (1
1/i < dist
(z, S
1/i,
n
− kn−1) n−1
i=1
(X
2/3) <
n−1
n
∀z ∈
1 ,T ξn
)
i=1
−
n
1
−
P ξn
n−1 ∪ Qξn
n−1,
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(C ) X
, where r
,
n
n(Tn) ⊂ Brn
1 > 1, and rn =
r2n−1 + (2/n)2 + 1
n2
(D ) X
n
n(z) + Xn(−z) = 0, ∀z ∈ D∗,
z
(E ) X
φn(w)dw
+ c
n
n(z) = Re
n, where cn ∈ R3 and φn is z2−type,
2/3
(F )
X
in T ξn ,
n
n − Xn−1 < 1n2
n−1
(G ) λ
in T ξn , where 0 < α
α
n
Xn ≥ αnλXn 1
i < 1 and
i
−
n−1
{ ni=1 }n → 1/2,
◦
◦
◦
(H ) T
T
1
T ξn , T ξn
T
n
n ⊂ n−1, T ξn−
n−2 ⊂ n−1
n−1 ⊂ n .
From (Gn), Xn is a Cauchy sequence and so, Harnack’s theorem im-
plies that Xn converges. (An) ensure that the limit immersion is complete.
From (Cn) the limit immersion is bounded. From (Hn), we can ensure that
◦
A =
T
n∈N
n is nonempty and biholomorphic to an annulus.
So, the sequence Xn combines together two properties: the intrinsic dis-
tance of Xn(Tn) is divergent, but the extrinsic distance of Xn(Tn) in R3 is
bounded. The theorem is a consequence of the existence of a sequence of
immersions Xn verifying these properties.
The sequence Xn is constructed by induction. The most difficult step in
the proof of the theorem is it constructs the term Xn starting from the term
Xn−1. To construct the sequence Xn, we will need the following lemma:
Lemma Let X : D∗ → R3 be a conformal minimal immersion. Consider
(P, Q) polygonal pair, and ρ, r > 0, and 1 > k > 0, satisfying:
(1) (1 − k)ρ < dist(X,T)(z,S2/3) < ρ, ∀z ∈ P ∪ Q,
(2) X(T ) ⊂ Br,
z
(3) X(z) = Re
φ(w)dw
+ c, where c ∈ R3 and φ is z2−type,
2/3
(4) X(z) + X(−z) = 0, ∀z ∈ D∗.
Then, for any ε > 0, and for any s, ξ, k > 0 verifying:
(1 − k)ρ < dist(X,Tξ)(z,S2/3) < ρ,
∀z ∈ Pξ ∪ Qξ,
ρk < s , ρ < (1 − k )(ρ + s),
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Figure 2. Proof of the lemma.
there exist a polygonal pair (P , Q) and a conformal minimal immersion Y :
D∗ → R3, such that:
(1) (1 − k )(ρ + s) < dist
(z, S
(Y, eT) 2/3) < ρ+s, ∀z ∈ P ∪Q,
(2) Y (T ) ⊂ BR, R = r2 + (2s)2 + ε,
z
(3) Y (z) = Re
ψ(w)dw
+ c , where c ∈ R3 and ψ is z2−type,
2/3
(4) Y (z) + Y (−z) = 0, ∀z ∈ D∗,
(5)
Y − X < ε in Tξ,
◦
◦
(6) T ξ ⊂T and T ⊂T.
The proof of the lemma is quite technical. We give an sketch of the proof
and see how to use Weierstrass representation, L´
opez-Ros transformation, and
Runge theorem to obtain the proof of the lemma. I focus on properties (1)
and (2) of Y . The other properties are more easy.
Let X : T → Br ⊂ R3 the minimal immersion of the lemma.
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Roughly speaking, we modify the Weierstrass representation of X, only
around the boundary of T to obtain a Weierstrass representation of a new
minimal immersion Y . This immersion will proof the lemma.
For a finite collection of small sets, ω1, . . . , ωn around all the boundary
of T , we can successively repeat the following process in order to modify the
immersion X.
ω
ω
ω
X = X
1
2
n
0 → X1 → . . . → Xn = Y
The process of obtaining Xi from Xi−1 is the following: firstly, we consider
X(ωi). Let pi a vector in the direction of X(ω1), pi = 1, (see figure 2).
Using Runge’s Theorem and the L´
opez-Ros transformation, for a suitable
orthogonal coordinates (with third coordinate equal to pi), we can obtain a
Weierstrass representation of a new minimal immersion Xi, such that, for a
> 0 and τ , (see section 3.1):
(i) Xi−1 is close to Xi in T \ ωi:
Xi−1
Xi in T \ ωi,
(ii) The intrinsic metric of Xi grows with respect to the intrinsic metric of
Xi−1 in ωi:
λX >> λ
in ω
i
Xi 1
i,
−
(iii) Xi and Xi−1 are identical in the direction of pi:
Xi, pi = Xi−1, pi , in T.
We define Y as the immersion obtained in last step: Xn. We check that
Xn verifying the properties (1) and (2) of Y :
Property (1) From the growth of intrinsic metric around the boundary of T ,
(property (ii)), we can take a domain T , inside T , bounded by a polygonal
pair (P , Q) such that the property (1) of Y holds. Then the distance to
the boundary of T , with the metric associated to the immersion Y , grows
with respect to the distance with the metric associated to X.
Property (2) We are going to check that Y (z) < R, ∀z ∈ T.
• If z ∈ T \(ω1∪...∪ωn), it checks easily, since Y (z) is close to X(z),
(property (i)), and X(z) < r < R.
• Instead, if z ∈ ωi, i = 1,... ,n:
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– From the properties (1) of the immersion X and Y , we can say,
roughly speaking, X(z) − Y (z) < s, ∀z ∈ T.
– Y and Xi are close in ωi then,
Y, pi
Xi, pi =,
and Xi and Xi−1 are equal in the direction of pi then,
= Xi−1, pi
X, pi
The scalar product of X and pi is less than r in ωi, then
Y, pi < r in ωi.
So, we have this situation in the figure 2: the vector X(z) is close to
pi and Y (z)−X(z) is almost orthogonal to X(z). Then Pythagoras’
theorem says Y (z) < √r2 + s2 < R.
References
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minimal surfaces, Duke Math. J. 68 (1992), 297-300.
2. F.F. de Brito, Many-ended complete minimal surfaces between two paral-
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3. C. Costa and P.A.Q. Simoes, Complete minimal surfaces of arbitrary
genus in a slab of R3, Ann. Inst. Fourier (Grenoble) 46 (1996), 535-546.
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parallel planes, Ann. of Math. (2) 112 (1980), 203-206.
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opez, A nonorientable complete minimal surface in R3 between two
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opez, Hyperbolic complete minimal surfaces with arbitrary topology,
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curved and minimal surfaces, Invent. math. 126 (1996), 457-465.
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