Page 275 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes 1 dt(s)
= .
k ( (s)) ds s  (s)

1
1
=
2
2





 ( k (s) k (s))  (k (s)) . {( k (s) k (s) . n (s) k (s) . n (s)}
1
1
3
2
2
3
3
1
for all s  L. Thus, we can put
n ( (s)) = (cos (s)) . n (s) + (sin (s)) . n (s), (17)

1
1
3
where

cos (s) =  k (s) k (s) (18.1)
2
1
2


 ( k (s) k (s))  (k (s)) 2
3
1
2
k (s)
sin (s) = 3  0 (18.2)
2
 ( k (s) k (s))  (k (s)) 2


3
1
2
for all s  L. Here,  is a C -function on L. Differentiating (17) with respect to s and using the

Frenet equations, we obtain
dn (s) dcos (s)


 '(s) . 1   k (s)(cos (s)) . t(s)  . n (s)
ds s  (s) 1 ds 1
dsin (s)


 {k (s)(cos (s)) k (s)(sin (s))} . n (s)  ds . n (s)


3
3
2
2
Differentiating (c) with respect to s, we obtain
(gk (s) k (s)k (s) ( k (s) k (s))k (s) 0. (19)
'
'
'





2
3
3
1
1
2
Differentiating (18.1) and (18.2) with respect to s and using (4.19), we obtain
dcos (s)  0, dsin (s)  0,


ds ds
that is,  is a constant function on L with value  . Thus, we obtain
0
 k (s) k (s) = cos  , (18.1)’

1
2
2
 ( k (s) k (s))  (k (s)) 2 0


1
2
3
k (s)
3
0
2


 ( k (s) k (s))  (k (s)) 2 = sin  0. (18.2)’
1
2
3
From (17), it holds
n ( (s)) = (cos  ) . n (s) + (sin  ) . n (s). (20)

1
0
1
3
0
Thus we obtain, by (15) and (16),
2


( k (s) k (s))  (k (s)) 2



k ( (s)) . t( (s))  1 2 3 . ( . t(s)  n (s)),
2
1
2
2


 '(s)(  1) ( k (s) k (s))  (k (s)) 2
3
1
2
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