Page 271 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes we obtain
’(s)  0, ’(s)  0,

that is,  and  are constant functions on L with values a and b, respectively. Therefore, for all
s  L, (1) is rewritten as


c(s)  c( (s)) = c(s) +  . n (s) +  . n (s), (1)’
1 3
and we obtain

 '(s) . t( (s)) = (1 – k (s)) . t(s) + (k (s) – k (s)) . n (s). (3)
1 2 3 2
Here, we notice that
(’(s)) = (1 – k (s)) + (k (s) – k (s))  0 (4)
2
2
2
2
3
1
for all s  L. Thus, we can set
t( (s)) = (cos T(s)) . t(s) + (sin T(s) . n (s) (5)

2
cos T(s) = (1 – k (s))/(’(s))
1
sin T(s) = (k (s) – k (s))/(’(s))
2
3
where T is a C -function on L. Differentiating (5) with respect to s and using the Frenet equations,

we obtain
dcos(T(s))



 '(s)k ( (s)) . n ( (s)) = ds .t(s) {k (s)cos(T(s)) k (s)sin(T(s))}.n (s)

1
2
1
1
1
dsin(T(s))

 .n (s) k (s)sin(T(s)).n (s).
ds 2 3 3
Since n ( (s)) is expressed by linear combination of n (s) and n (s), it holds that

1
1
3
dcosT(s)  0, dsin T(s))  0,
ds ds
that is, T is a constant function on L with value T . Thus, we obtain
0
t( (s)) = (cosT ) . t(s) (sinT ) . n (s) 0 2 (5)’

0
’(s) cos T = 1 – k (s) (6.1)
0
1
’(s) sin T = k (s) – k (s) (6.2)
2
0
3
for all s  L. Therefore, we obtain
(1 – k (s)) sin T = (k (s) – k (s)) cos T 0 (7)
2
0
1
3
for all s  L.
If sin T = 0, then it holds cos T = ±1. Thus, (5)’ implies that t( (s)) = ±t(s). Differentiating this

0
0
equality, we obtain
 '(s)k ( (s)) . n ( (s))   k (s) . n (s),


1
1
1
1
that is,

n ( (s))   n (s),
1
1
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