Page 327 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes The normal N is easily calculated:

  cos(v)  sin(v) sinh(u) 
N(u,v)   , , 
 cosh(u) cosh(u) cosh(u) 

If  (t) is a meridian, then  (t) = N(t, v) is its spherical image under the Gauss map, and
v
v
differentiating with respect to t, we get the principal curvature associated with meridians:
k(u, v) = –1/cosh(u). Similarly, the principal curvature associated with parallels is: 1/cosh(u).
Thus, we conclude that
1
H  0, K   .
cosh(u) 2
Definition 3. A parametric surface X is minimal if it has vanishing mean curvature H = 0.
For example, the catenoid is a minimal surface. The justification for the terminology will be
given in the next section.
Proposition 3. Let X be a minimal surface. Then X has non-positive Gauss curvature K  0, and
K(u) = 0 if and only if u is a planar point.
We will set out to construct a large class of minimal surfaces. We will use the Weierstrass
Representation.

Definition 4. A parametric surface X is conformal if the first fundamental form satisfies g = g 22
11
and g = 0. A parametric surface X is harmonic if X = X + X = 0.
12
22
11
Proposition 4. Let X : U   be a parametric surface which is both conformal and harmonic.
3
Then X is a minimal surface.
 
g
Proof. We can write the first fundamental form   , its inverse g , and the second fundamental
ij
ij
 
form k ij as:
11
12
g
      0   , g ij   1 0  , k ij   X  N X  N   .
  
   
1 
ij

 0    0    X  N X  N 
12
22
Thus, the mean curvature vanishes:
ij
H  g k    1 X  X 22  N  0.
ij
11
In order to construct parametric surfaces which are both conformal and harmonic, we will use
h
f
complex analysis in the domain U. Let  = u+iv where i denotes  1, and let    and    be
two complex analytic functions on U. Define
2
F = f – h , F   i f  h 2  , F = 2fh.
2
2
2
3
1
We have:
2
2
2
F
F
F

        0.
3
1
2
If we write F =  + in , then this can be written as:
j
j
j
3 2 2 2 3
n
      j   2i   j n  0.


j
j
j 1     j 1


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