Page 229 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes Lemma 2. Let f: ]a, b[  E (or f: [a, b]  E ) be a curve of class C , with p  n, so that the
p
n
n
derivatives f (t), . . ., f (n–1) (t) of f(t) are linearly independent for all t  ]a, b[. Then, there is a
(1)
unique Frenet n-frame (e (t), . . ., e (t)) satisfying the following conditions:
1
n
(1) The k-frames (f (t), . . ., f (t)) and (e (t), . . ., e (t)) have the same orientation for all k, with
(1)
(k)
k
1
1  k  n – 1.
(2) The frame (e (t), . . ., e (t)) has positive orientation.
n
1
Proof. Since (f (t), . . ., f (n–1) (t)) is linearly independent, we can use the Gram-Schmidt
(1)
orthonormalization procedure to construct (e (t), . . ., e (t)) from (f (t), . . ., f (n–1) (t)). We use the
(1)
1
n–1
generalized cross-product to define e , where
n
e = e × · · · × e .
n–1
1
n
From the Gram-Schmidt procedure, it is easy to check that ek(t) is C for 1  k  n – 1, and since
n–k
the components of e are certain determinants involving the components of (e , . . ., e ), it is also
n
n–1
1
clear that en is C . 1
The Frenet n-frame given by Lemma 2 is called the distinguished Frenet n-frame. We can now
prove a generalization of the Frenet-Serret formula that gives an expression of the derivatives
of a moving frame in terms of the moving frame itself.
Lemma 3. Let f: ]a, b[  E (or f: [a, b]  E ) be a curve of class C , with p  n, so that the
n
n
p
derivatives f (t), . . ., f (n–1) (t) of f(t) are linearly independent for all t  ]a, b[ . Then, for any moving
(1)
'
frame (e (t), . . ., e (t)), if we write  (t) = e (t) e (t), j we have
n
ij
i
1
n
'
e (t)    ij (t)e (t),
i
j
j 1

with
 ij (t)   ij (t),
and there are some functions  (t) so that
i
n
f'(t)    i (t)e (t).
i
i 1

Furthermore, if (e (t), . . ., e (t)) is the distinguished Frenet n-frame associated with f, then we
n
1
also have
 1 (t) = f'(t) , (t) = 0 for i 2,


i
and
 (t) = 0 for j > i + 1.
ij
'
Proof. Since (e (t), . . ., e (t)) is a moving frame, it is an orthonormal basis, and thus, f’(t) and e (t)
t
n
1
are linear combinations of (e (t), . . ., n(t)). Also, we know that
1
e
n
'
'

e (t)   (e (t) e (t))e (t),
j
i
i
j
j 1

'
and since e (t) · e(t) =  , by differentiating, if we write  (t) = e (t) e (t), j we get
i
ij
j
i
ij
 (t) = – (t).
ij
ji
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