Page 324 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 26: Lines of Curvature

Proposition 1. Let  be a curve on a parametric surface X with unit normal N, and let   N  be Notes

its spherical image under the Gauss map. Then  is a line of curvature if and only if

    0. ...(1)

Proof. Suppose that (1) holds, then we have:

 N    0.
 
Let  be the linear transformation on T X associated with k. Then, we have for every Y T X :

u
u

 
 ( ),Y 
g    k( ,Y)     N Y   g(  ,Y).
,
Thus,       and   is a principal direction. The proof of the converse is similar.


It is clear from the proof that  in (1) is the associated principal curvature. The coordinate curves
of a parametric surface X are the two family of curves  (t) = X(t, c) and  (t) = X(c, t). A surface is
c
c
parametrized by lines of curvature if the coordinate curves of X are lines of curvature. We will
now show that any non-umbilical point has a neighborhood in which the surface can be
reparametrized by lines of curvature. We first prove the following lemma which is also of
independent interest.
Lemma 1. Let X : U   be a parametric surface, and let Y and Y be linearly independent
3
1 2
vector fields. The following statements are equivalent:
1. Any point u  U has a neighborhood U and a reparametrization : V  U such that if

0
0
0
0


X  X  then X  Y  .


i
i

2. Y ,Y  0.
1
2
 

Proof. Suppose that (1) holds. Then Equation shows that X ,X   0. However, since the
2 

1

commutator is invariant under reparametrization, it follows that Y ,Y  0.
2
1
j
j
j

b
Conversely, suppose that Y ,Y  0. Express X  a Y j and Y  b X , j note that   is the
i
i
1
i
i
2
i
a
j
inverse of   . We now calculate:
i
0    X ,X 
i
j 
l
k
   a Y ,a Y 
j 
l
k
i
k
l
l

  a Ya l k j  a Ya k i  Y  a a l j Y ,Y l 
k
j
i
k
l
i
l
m
l
k
m
  a b  m a  a b  m a k i  Y k
l
i
j
j
l
  a i k j   j a k i  Y .
k
Since Y and Y are linearly independent, we conclude that:
2
1
k
 i a   j i k 0. ...(2)
a 
j
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