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IntroductionWeierstrass representation
Weierstrass representation for minimal surfaces
Arnold Kowalski
GRADUATE STUDENT'S WORKSHOP
on
ALGEBRA, LOGIC and ANALYSIS
Department of Mathematics and Physics
University of Szczecin
Szczecin
June 3, 2019
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Consider the soap lm created by dipping a cube frame into soap
solution. The soap lm creates the minimum surface area for a
surface with a cube as its boundary.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Why?
Water molecules exert a force on each other. Near the surface of
the water there is a greater force pulling the molecules toward the
center of the water, creating surface tension that tends to minimize
the surface area of the shape.
Soap solution has a lower surface tension than water and this
permits the formation of soap lms that also tend to minimize
geometric properties such as length and area.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Other examples of soap lms
By dipping into soap solution a wire frame of a slinky (or helix)
with a straw connecting the ends of the slinky, we can create part
of the minimal surface known as the helicoid.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Other examples of soap lms
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Other examples of soap lms
By dipping a 3-dimensional version of the wire frame (a box frame
missing two parallel edges on the top and two parallel edges on the
bottom) into soap solution we can create part of the minimal
surface known as Scherk's doubly periodic surface.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Other examples of soap lms
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Plateau's Problem
Physical interpretation of
minimal surfaces is related to
J. Plateau work and is called
Plateau's Problem.
J.A.F. Plateau, 18011883
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Plateau's Problem
Physical interpretation of
minimal surfaces is related to
J. Plateau work and is called
Plateau's Problem.
J.A.F. Plateau, 18011883
Plateau described results and
experiments on surface tension in
1873 in Statique expérimentale et
théorétique des liquides soumis
aux seules forces moleculaires.
However Plateau's Problem was
raised by Joseph-Louis Lagrange
in 1760.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Plateau's Problem
Plateau's Problem
For any closed curve γ ∈ R3 of the nite length there is a surface S
of the lowest area spanned on the curve γ.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Plateau's Problem solution
Plateau's Problem was solved independently in years 1960-1961 by
Jesse Douglas and Tibor Radó. J.Douglas got Field's medal in 1936
for this result!
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Minimal surfaces
In mathematics the minimal surface is a surface, such that the
main curvature is equal to zero on every point on this surface.
Let us explain what does it mean:
consider the surface S ∈ R3 and any point p ∈ S ,
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Minimal surfaces
In mathematics the minimal surface is a surface, such that the
main curvature is equal to zero on every point on this surface.
Let us explain what does it mean:
consider the surface S ∈ R3 and any point p ∈ S ,
determine normal vector −→n to the surface S in p (,i.e.
normalized vector perpendicular to S in p),
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Minimal surfaces
In mathematics the minimal surface is a surface, such that the
main curvature is equal to zero on every point on this surface.
Let us explain what does it mean:
consider the surface S ∈ R3 and any point p ∈ S ,
determine normal vector −→n to the surface S in p (,i.e.
normalized vector perpendicular to S in p),
consider all planes P including −→n ,
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Minimal surfaces
In mathematics the minimal surface is a surface, such that the
main curvature is equal to zero on every point on this surface.
Let us explain what does it mean:
consider the surface S ∈ R3 and any point p ∈ S ,
determine normal vector −→n to the surface S in p (,i.e.
normalized vector perpendicular to S in p),
consider all planes P including −→n ,
get the curve α as an intersection of plane P and surface S ,
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Minimal surfaces
In mathematics the minimal surface is a surface, such that the
main curvature is equal to zero on every point on this surface.
Let us explain what does it mean:
consider the surface S ∈ R3 and any point p ∈ S ,
determine normal vector −→n to the surface S in p (,i.e.
normalized vector perpendicular to S in p),
consider all planes P including −→n ,
get the curve α as an intersection of plane P and surface S ,
determine curvature of α (,i.e. score |α′′|),
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Minimal surfaces
In mathematics the minimal surface is a surface, such that the
main curvature is equal to zero on every point on this surface.
Let us explain what does it mean:
consider the surface S ∈ R3 and any point p ∈ S ,
determine normal vector −→n to the surface S in p (,i.e.
normalized vector perpendicular to S in p),
consider all planes P including −→n ,
get the curve α as an intersection of plane P and surface S ,
determine curvature of α (,i.e. score |α′′|),determine normal curvature in the xed direction w , i.e.
k(w) = α′′ · n
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Minimal surfaces
The normal curvature measures how much the surface bends
towards −→n as you travel in the direction of the tangent vector w
starting p. As we rotate the plane about −→n , we get a set of curves
on the surface each of which has a value for its curvature. Let k1and k2 be the maximum and minimum curvature values at p.
The directions in which the normal curvature attains its absolute
maximum and absolute minimum values are known as the principal
directions.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Minimal surfaces
Denition
The mean curvature (i.e., average curvature) of a surface S at p is
H :=k1 + k2
2
Denition
A minimal surface is a surface S with mean curvature H = 0 at
all points p ∈ S .
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Minimal surfaces
Denition
Let us consider the surface S ⊂ R3. Let D ⊂ R2 be na open set.
Then S can be presented by mapping x : D → R3, where
x(u, v) = (x1(u, v), x2(u, v), x3(u, v)),
,i.e. S is a image of the set x(D) by map x. We demand that
x ∈ C 2(D). The mapping above is called a parametrization of the
surface S .
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Minimal surfaces
Theorem
Let S be the surface with parametrization
x(u, v) = (x1(u, v), x2(u, v), x3(u, v))
and p be any point on this surface. Let us denote:
E : =3∑
j=1
(∂xj∂u
)2
, F : =3∑
j=1
∂xj∂u
∂xj∂v
, G : =3∑
j=1
(∂xj∂v
)2
,
and for vector −→n = [n1, n2, n3] normal to S in the point p:
e : =3∑
j=1
∂2xj∂u2
· nj , f : =3∑
j=1
∂2xj∂u∂v
· nj , g : =3∑
j=1
∂2xj∂v2
· nj
Then
H =Eg + Ge − 2Ff
2(EG − F 2).
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Minimal surfaces
Corollary
The surface S is minimal if
Eg + Ge − 2Ff
2(EG − F 2)= 0.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Examples
Enneper's surface (Alfred Enneper, 1864)
x(u, v) =
(u − 1
3u3 + uv2, v − 1
3v3 + u2v , u2 − v2
),
where (u, v) ∈ (x , y) ∈ R2 : x2 + y2 = 1.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Examples
Catenoid (Leonhard Euler, 1744 )
x(u, v) = (a cosh(v) cos(u), a cosh(v) sin(u), av) ,
where (u, v) ∈ (x , y) ∈ R2 : 0 ≤ u ≤ 2π, −π ≤ v ≤ π.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Examples
Helikoid (Leonhard Euler in 1774 and Jean Baptiste Meusnier in
1776)
x(u, v) = (a sinh(v) cos(u), a sinh(v) sin(u), au) ,
where (u, v) ∈ (x , y) ∈ R2 : −π ≤ u ≤ π, −23π ≤ v ≤ 2
3π.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Examples
Scherk's doubly periodic surface:(Heinrich Scherk, 1834)
x(u, v) =
(u, v , ln
(cos(u)
cos(v)
)),
where (u, v) ∈ (x , y) ∈ R2 : −π2≤ u ≤ π
2, −π
2≤ v ≤ π
2.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Examples
Pieces of Scherk's doubly periodic surface can be put together in
the xy -plane in a checkerboard fashion. They repeat (or are
periodic) in two directions, x and y.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Examples
Scherk's singly periodic surface (Heinrich Scherk, 1834)
x(u, v) = (arcsinh(u), arcsinh(v), arcsin(uv)),
where (u, v) ∈ (x , y) ∈ R2 : −π2≤ u ≤ π
2, −π
2≤ v ≤ π
2.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Examples
Individual pieces of Scherk's singly periodic surface can t together
creating a tower in the z direction. You can visualize adding two
pieces together by taking one piece of Scherk's singly periodic
surface and adding it to another piece that has been reected
across the xy -plane and shifted up in the z direction.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Examples
Hennenberg's surface (Ernst Lebrecht Henneberg, 1875)
x(u, v) = (−1 + cosh(2u) cos(2v),− sinh(u) sin(v)
− 1
3sinh(3u) sin(3v),− sinh(u) cos(v)
+1
3sinh(3u) cos(3v)),
where (u, v) ∈ (x , y) ∈ R2 : 0 ≤ u ≤ 2π, −π ≤ v ≤ π.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Examples
Catalan's surface (Eugene Charles Catalan, 1843)
x(u, v) = (1− cos(u) cosh(v), 4 sin(u
2) sinh(
v
2), u − sin(u) cosh(v)),
where (u, v) ∈ (x , y) ∈ R2 : 0 ≤ u ≤ 2π, −π ≤ v ≤ π.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Examples
There is an extensive list of minimal surfaces, but we have no way
of listing all of them.
So, we often focus on trying to classify minimal surfaces. This
means, we try to nd results that include all possibilities for minimal
surfaces with specic properties. The simplest example of this is:
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Examples
There is an extensive list of minimal surfaces, but we have no way
of listing all of them.
So, we often focus on trying to classify minimal surfaces. This
means, we try to nd results that include all possibilities for minimal
surfaces with specic properties. The simplest example of this is:
A nonplanar minimal surface in R3 that is also a surface of
revolution is contained in a catenoid.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Isothermal parametrization
Denition
A parametrization x is isothermal if E = xu · xu = xv · xv = G and
F = xu · xy = 0.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Isothermal parametrization
Denition
A parametrization x is isothermal if E = xu · xu = xv · xv = G and
F = xu · xy = 0.
Because E ,F , and G describe how lengths on a surface are
distorted as compared to their usual measurements in R3, if
F = xu · xy = 0 then xu and xv are orthogonal and if E = G then
the amount of distortion is the same in the orthogonal directions.
Thus, we can think of an isothermal parametrization as mapping a
small square in the domain to a small square on the surface.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Sometimes an isothermal parametrization is called a conformal
parametrization, because the angle between a pair of curves in the
domain is equal to the angle between the corresponding pair of
curves on the surface.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Theorem
Every minimal surface in R3 has an isothermal parametrization.
Theorem
If the parametrization x is isothermal, then
xuu + xvv = 2EH−→n , (1)
where E is a coecient of the rst fundamental form and H is the
mean curvature.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Corollary I
A surface S with an isothermal parametrization
x(u, v) = (x1(u, v), x2(u, v), x3(u, v)) is minimal if and only if x1,
x2, and x3 are harmonic.
Proof:
(⇒)If S is minimal, then H = 0 and so by (1) we have xuu + xvv = 0,
and hence the coordinate functions are harmonic.
(⇐)Suppose x1, x2, and x3 are harmonic. Then xuu + xvv = 0. So by
(1) we have 2(xu · xu)H−→n = 0. But −→n 6= 0 and E = xu · xu 6= 0.
Hence H = 0 and S is minimal.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Weierstrass representation
Suppose S is a minimal surface with an isothermal parametrization
x(u, v). Let z = u + iv be a point in the complex plane, so
z = u − iv . Solving for u, v in terms of z , z we get
u =z + z
2v =
z − z
2i
The parametrization of the minimal surface S can be written in
terms of the complex variables z and z as
x(z , z) = (x1(z , z), x2(z , z), x3(z , z)).
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Weierstrass representation
Lemma I
Let f (u, v) = x(u, v) + iy(u, v) be a complex function. Using
u = z+z2
and v = z−z2i
, we have
i.∂f
∂z=
1
2
(∂x
∂u+∂y
∂v
)+
i
2
(∂y
∂u− ∂x
∂v
),
∂f
∂z=
1
2
(∂x
∂u− ∂y
∂v
)+
i
2
(∂y
∂u+∂x
∂v
).
ii. f is analytic ⇔ ∂f∂z = 0.
iii. 4(∂∂z
(∂f∂z
))= fuu + fvv .
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Weierstrass representation
Theorem I
Let S be a surface with parametrization x = (x1, x2, x3) and let
φ = (φ1, φ2, φ3), where
φk =∂xk∂z
.
Let φ2 denote (φ1)2 + (φ2)2 + (φ3)2.Then x is isothermal ⇔ φ2 ≡ 0.
If x is isothermal then S is minimal ⇔ each φk is analytic.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Weierstrass representation
Proof of theorem I:
Applying the complex dierential operator ∂f∂z from lemma I and
squaring the terms, we have
(φk)2 =
(∂xk∂z
)2
=
[1
2
(∂xk∂u− i
∂xk∂v
)]2=
1
4
[(∂xk∂u
)2
−(∂xk∂v
)2
− 2i∂xk∂u
∂xk∂v
]
Also,
xu · xu =
(∂x1∂u
)2
+
(∂x2∂u
)2
+
(∂x3∂u
)2
=3∑
k=1
(∂xk∂u
)2
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Weierstrass representation
Similarly we have xv · xv =3∑
k=1
(∂xk∂v
)2. Hence,
φ2 = (φ1)2 + (φ2)2 + (φ3)2
=1
4
[3∑
k=1
(∂xk∂u
)2
−3∑
k=1
(∂xk∂v
)2
−3∑
k=1
∂xk∂u
∂xk∂v
]
=1
4(xu · xu − xv · xv − 2i(xu · xv ))
=1
4(E − G − 2iF )
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Weierstrass representation
Thus x is isothermal ⇔ E = G , and F = 0 ⇔ φ2 ≡ 0.
Suppose that x is isothermal. By corollary I, it it suces to show
that for each k , xk is harmonic ⇔ φk is analytic. By lemma I this
follows because
∂2xk∂u∂u
+∂2xk∂v∂v
= 4
(∂
∂z
(∂f
∂z
))= 4
(∂
∂z(φk)
)= 0.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Weierstrass representation
Let's apply this theorem. Suppose we have analytic functions φk
and we want to nd the functions xk . If x is isothermal, then
|φ2| =
∣∣∣∣∂x1∂z∣∣∣∣2 +
∣∣∣∣∂x2∂z∣∣∣∣2 +
∣∣∣∣∂x3∂z∣∣∣∣2 =
1
4
(3∑
k=1
(∂xk∂u
)2
+3∑
k=1
(∂xk∂u
)2)
=1
4(xu · xu + xv · xv ) =
1
4(E + G ) =
E
2.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Weierstrass representation
Let's apply this theorem. Suppose we have analytic functions φk
and we want to nd the functions xk . If x is isothermal, then
|φ2| =
∣∣∣∣∂x1∂z∣∣∣∣2 +
∣∣∣∣∂x2∂z∣∣∣∣2 +
∣∣∣∣∂x3∂z∣∣∣∣2 =
1
4
(3∑
k=1
(∂xk∂u
)2
+3∑
k=1
(∂xk∂u
)2)
=1
4(xu · xu + xv · xv ) =
1
4(E + G ) =
E
2.
So if |φ2| = 0 , then the coecients of the rst fundamental form
are zero and S is a point.
We need to solve φk = ∂xk∂z for xk since the parametrization of the
surface is given as x = (x1, x2, x3).
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Weierstrass representation
Since xk is a function of the two variables u and v , we can write
dxk =∂xk∂u
du +∂xk∂v
dv . (2)
Also dz = du + idv so by lemma I we have
φkdz =∂xk∂z
dz =1
2
(∂xk∂u− i
∂xk∂v
)(du + idv)
=1
2
[∂xk∂u
du +∂xk∂v
dv + i
(∂xk∂u
dv − ∂xk∂v
du
)],
φkdz =∂xk∂z
dz =1
2
(∂xk∂u
+ i∂xk∂v
)(du − idv)
=1
2
[∂xk∂u
du +∂xk∂v
dv − i
(∂xk∂u
dv − ∂xk∂v
du
)].
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Weierstrass representation
Adding we get
∂xk∂u
du +∂xk∂v
dv = φ+ kdz + φkdz = 2Reφkdz (3)
Combining (2) and (3) we have
dxk = 2Reφkdz
Therefore xk = 2Re∫φkdz + ck .
Since adding ck translates the image by a constant amount and
multiplying a coordinate function by 2 scales the surface, the
constants do not aect the geometric shape of the surface. Hence,
we do not need them and we will let our coordinate function be:
xk = Re
∫φkdz .
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Weierstrass representation
Corollary
If we have analytic functions φk (k = 1, 2, 3) such that φ2 ≡ 0 and
|φ2| 6= 0 and is nite, then the parametrization
x =
(Re
∫φ1(z)dz ,Re
∫φ2(z)dz ,Re
∫φ3(z)dz
)(4)
denes a minimal surface.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Weierstrass representation
For example, consider the functions p(z) and q(z) such that
φ1 = p(1 + q2)
φ2 = −ip(1− q2)
φ3 = −2ipq.
Then
φ2 = [p(1 + q2)]2 + [−ip(1− q2)]2 + [−2ipq]2
= [p2 + 2p2q2 + p2q4]− [p2 − 2p2q2 + p2q4]− [4p2q2]
= 0,
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Weierstrass representation
and
|φ|2 = |p(1 + q2)|2 + | − ip(1− q2)|2 + | − 2ipq|2
= |p|2[(1 + q2)(1 + q2) + (1− q2)(1− q2) + 4qq]
= |p|2[2(1 + 2qq + q2q2)],
= 4|p|2(1 + |q|2)2 6= 0.
For φk to be analytic, p, pq2, and pq have to be analytic. If p is
analytic with a zero of order 2m at z0, then q can have a pole of
order no larger than m at z0. A function that is analytic in a
domain D except possibly at poles is a meromorphic function in D.
This leads to the following result.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Examples and applications
Weierstrass representation
Weierstrass Representation
Every regular minimal surface has a local isothermal parametric
representation x = (x1, x2, x3) of the form
x1(z) = Re
z∫a
p(1 + q2)dz
,
x2(z) = Re
z∫a
−ip(1− q2)dz
,
x3(z)) = Re
z∫a
−2ipq dz
,
where p is an analytic function and q is a meromorphic function in
a domain Ω ∈ C, having the property that where q has a pole of
order m, p has a zero of order at least 2m, and a ∈ Ω is a constant.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Examples and applications
Example
For p(z) = 1 and q(z) = iz we get:
x(z)
=
Re
z∫
0
(1− z2)dz
,Re
z∫
0
−i(1 + z2)dz
,Re
z∫
a
2zpq dz
=
(Re
z − 1
3z3,Re
−i(z +
1
3z3)
,Rez2)
.
Letting z = u + iv , this yields
x(z) =
(u − 1
3u3 + uv2, v − 1
3v3 + u2v , u2 − v2
),
which gives Enneper's surface.
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Examples and applications
References
Brilleslyper M.A., Dor M.J., McDougall J.M., Explorations in Complex
Analysis, Mathematical Association of America, Hardcover 2012.
Vekua I.N., The Basics of Tensor Analysis and Theory of Covariants,Nauka, Moscow (1978, Russian).
Arnold Kowalski Weierstrass representation
IntroductionWeierstrass representation
Examples and applications
Thank you for patience and listening.
Arnold Kowalski Weierstrass representation