Page 309 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes with A,B  linearly independent. Then X is a plane. After reparametrization, we may assume
3
that A and B are orthonormal. In that case, the first fundamental form is:

ds = du + d .
2
2
2
Furthermore, |A × B| = 1, and N = A × B is constant, hence k = 0. In particular, all the points of
X are planar, and we have for the mean and Gauss curvatures: H = K = 0.
It is of interest to note that if all the points of a parametric surface are planar, then X(U) is
contained in a plane. We will later prove a stronger result: X has a reparametrization which is
linear.

3
Proposition 5. Let X : U   be a parametric surface, and suppose that its second fundamental
form k = 0. Then, there is a fixed vector A and a constant b such that X A = b, i.e., X is contained
in a plane.
Proof. Let A be the unit normal N of X. Let 1  i  2, and note that N is tangential. Indeed, N  N
i
= 1, and differentiating along u , we get N N = 0. However, since k = 0 it follows from (2.6) that
i
i
N  X = –k = 0. Thus, N = 0 for i = 1, 2, and we conclude that N is constant. Consequently, (X N) i
j
i
ij
i
= X N = 0, and X N is also constant, which proves the proposition.
i
2
24.4.2. Spheres. Let U (0, ) (0,2 )      2 , and let X : U   be given by:
X(u, v) = (sin u cos v, sin u sin v, cos u).
The surface X is a parametric representation of the unit sphere. A straightforward calculation
shows that the first fundamental form is:
ds = du + sin u dv ,
2
2
2
2
and the unit normal is N = X. Thus, N = X , and consequently k = –N  X = –X  X = –g , i.e., k =
i
i
i
ij
ij
j
j
i
–g. In particular, the principal curvatures are both equal to –1 and all the points are umbilical.
We have for the mean and Gauss curvatures:
H = –1, K = 1
Proposition 6. Let X : U   be a parametric surface and suppose that all the points of X are
3
umbilical. Then, X(U) is either contained in a plane or a sphere.
Proof. By hypothesis, we have
N = X . i ...(12)
i
We first show that is a constant. Differentiating, we get N = X + X . Interchanging i and j,
ij
ij
i
j
subtracting these two equations, and taking into account N – N = X – X = 0, we obtain  X – X i
ij
j
j
i
ji
ij
ji
= 0, e.g.,
 X –  X = 0.
1
2
2
1
Since X and X are linearly independent, we conclude that  =  = 0 and it follows that  is
2
2
1
1
constant. Now, if  = 0 then all points are planar, and by Proposition 6, X is contained in a plane.
Otherwise, let A = X –  N, then A is constant:
-1
A = X –  N = 0,
-1
i
i
i
and X  A    1 is also constant, hence X is contained in a sphere.
24.4.3. Ruled Surfaces. A ruled surface is a parametric surface of the form:
X(u,v)   (u) vY(u)

302 LOVELY PROFESSIONAL UNIVERSITY