Page 232 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 19: Serret-Frenet Formulae

for all i, with 1  i  n, where R is the linear isometry associated with h (in fact, a rotation). Notes
Consider the curve f = h  f. The hypotheses of the lemma and Lemma 4, imply that


 ij(t) =   ij (t) =  ij (t), f'(t) = f'(t)  f'(t) ,
and, by construction,

(e 1(t ), . . . , e (t )) = (e (t ), . . . , e (t )) and f(t ) = f(t ).



n
0
n
0
0
1
0
0
0
Let
n
(t) =  (e i(t) e (t) (e (t) e (t)).  i  i   i
i 1

Then, we have
n '
’(t) = 2  (e i(t) e (t) (e  i  i (t) e (t))  '
i 1 i

n
'
'
= 2  (e (t) e (t) e (t) e (t)).
 


i
i
i 1 i i

Using the Frenet equations, we get
n n n n
’(t) = 2   ij i   j 2   ij e e  i

e e 

j
i 1 j 1 i 1 j 1




n n n n
= 2   ij i   j 2   ij i  j

e e 
e e 
i 1 j 1 j 1 i 1




n n n n
= 2   ij i   j 2   ij i  j
e e 
e e 

i 1 j 1 j 1 i 1




= 0,
since  is skew symmetric. Thus, (t) is constant, and since the Frenet frames at t agree, we get
0
(t) = 0.
Then, e (t) = e (t) for all i, and since f'(t)   f'(t) , we

i
i
have

f '(t)  f '(t) e (t)   f'(t) e (t)  f'(t),

1
1



so that f(t) f(t)  is constant. However, f(t ) = f(t ), and so, f(t) = f(t), and f = f = h f.
0
0
Lemma 8. Let  , . . .,  be functions defined on some open ]a, b[ containing 0 with
1
n–1
 C n–i–1 continuous for i = 1, . . ., n – 1, and with  (t) > 0 for i = 1, . . ., n – 2 and all t  ]a, b[ . Then,
i
i
there is curve f : ]a, b[  E of class C , with p  n, satisfying the non-degeneracy conditions of
n
p
Lemma 2, so that f'(t) = 1 and f has the n – 1 curvatures  (t), . . .,  (t).
1
n–1
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