Page 257 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes Proof. Let  = (s ) be M Bishop spherical image of a regular curve  = (s). Then, the equations
1

(4) and (6) hold. Using (4) in (6), we have
d 
k 1 ds  k   ds 
2
ds
k
T     1  . (8)
2

k  k 2 2
1
Substituting (5) to (8) and integrating both sides, we have (7) as desired.
In the light of the Propositions 2 and 3, we state the following theorems without proofs:
Theorem 3. Let  = (s ) be M Bishop spherical image of a regular curve  = (s). If  is a general
1

helix, then, Bishop curvatures of  satisfy
'

k 2 k  
2
1 
 k 1  constant.
3 
2
2 2
(k  k )
1
2
Theorem 4. Let  = (s ) be the M Bishop spherical image of a regular curve  = (s). If  is a slant
1

helix, then, the Bishop curvatures of  satisfy
 2 k  '  '

 k 1  2  
2
2 4
  k 1   (k  k ) constant.
2
1
 3  3 
2
 (k  k )    k  '2 2 
2 2
1
2
2 3
3
2
    k  2   (k  k ) 
1 
2
1
 k 1   

We know that  is a spherical curve, so, by the Proposition 3 one can prove.
Theorem 5. Let  be the M Bishop spherical image of a regular curve  = (s). The Bishop
1
curvatures of the regular curve  = (s) satisfy the following differential equation
'

k 2 k   '
2
1 
 k 1   k k 2  constant.
1
3   2 2  
2
2 
(k  k )   k  k 
2 2
1
2
1
s 
Remark 2. Considering  = T ds and using the transformation matrix, one can obtain the
 
 
0
Bishop trihedra {T , M , M } of the curve  = (s ).
 1 2 
21.4 M Bishop Spherical Images of a Regular Curve
2
Definition 11. Let  = (s) be a regular curve in E . If we translate of the third vector field of
3
Bishop frame to the center O of the unit sphere S , we obtain a spherical image of  = (s ). This
2

curve is called the M Bishop spherical image or the indicatrix of the curve  = (s).
2
Let  = (s ) be M spherical image of the regular curve  = (s): We can write
 2
d ds 
'     k T.
ds  ds 2
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