Page 277 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes for all s  L. Now we obtain, by (15), (20), (21) and (22),

det[t( (s)),n ( (s),n ( (s)),n (j(s))] det[t(s),n (s),n (s),n (s)] 1





3
2
1
1
2
3
for all s  L. And we obtain
t( (s),n ( (s))  0, n ( (s)),n ( (s))   ij




i
j
i
for all s  L and i, j = 1, 2, 3. Thus the frame {t ,n ,n ,n } along C is of orthonormal and of
3
1
2
positive. And we obtain
dn (s)
k ( (s)) = ds s  (s) ,n ( (s))
2


3
3
2
=   1k (s)k (s)
1
3
2
 '(s) ( k (s) k (s))  (k (s)) 2


3
2
1
> 0
for all s  L. Thus curve C is a C -special Frenet curve in E . And it is trivial that the Frenet
4

(1, 3)-normal plane at each point c(s) of C coincides with the Frenet (1, 3)-normal plane at
4

corresponding point c(s)  c( (s)) of C. Therefore C is a (1, 3)-Bertrand curve in E .
Thus (i) and (ii) complete the proof of theorem B.
22.4 An Example of (1, 3)-Bertrand Curve
Let a and b be positive numbers, and let r be an integer greater than 1. We consider a C -curve

C in E defined by c : L  E ;
4
4
  r  
 acos  2 2 2 s  
  r a  b 
  r  
 asin  s  
  r a  b 2  
2
2
c(s)   
  r  
 bcos  2 2 2 s 
  r a  b 
  r  
 bsin  s 

2
   r a  b 2  
2
for all s  L. The curve C is a regular curve and s is the arc-length parameter of C. Then C is a
special Frenet curve in E and its curvature functions are as follows:
4
4
2
r a  b 2
k (s) = r a  b 2 ,
2
2
1
2
k (s) = r(r  1)ab ,
2
2
2
2
2
4
(r a  b ) r a  b 2
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