Page 213 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes Moreover,
E × E = E , E = E × E , and E = E × E .
2
3
1
1
2
1
3
3
2
In the remainder of this section, we will always assume that small Latin letters run from 1 to 3.
Note that given two different frames, x,E ,E ,E and x,E ,E ,E , there is exactly one affine motion
3
2
1
3
1
2
of Euclidean space taking x to x and taking E to E . When x(t),E (t),E (t),E (t) is a family of frames
i
i
1
2
3
depending on a parameter t, we say we have a moving frame along the curve.
Proposition 4. A family of frames x(t), E1(t), E (t), E (t) satisfies a system of differential equations:
3
2
x’(t) = p (t)E (t)
i
i
E (t) = q (t)E(t)
'
t
ij
j
where p (t) = x’(t) . E (t) and q j(t) = E (t) . E(t).
'
t
i
j
i
i
Since E (t) . E(t) = 0 for i  j, it follows that
j
i
q (t) + q (t) = E (t) . E(t) + E (t) . E (t) = 0
'
'
t
j
ij
i
j
ji
i.e. the coefficients q (t) are anti-symmetric in i and j. This can be expressed by saying that the
ij
matrix ((q j(t))) is an anti-symmetric matrix, with 0 on the diagonal.
i
In a very real sense, the function p (t) and q (t) completely determine the family of moving
i
ij
frames.
Specifically we have:
Proposition 5. If x(t), E (t), E (t), E (t) and x(t), E (t), E (t), E (t) are two families of moving frames
2
3
3
2
1
1
such that p (t) = p (t) and q (t) = q (t) for all t, then there is a single affine motion that takes x(t),
ij
i
ij
i
E1(t), E (t), E (t) to x(t),E (t),E (t),E (t)) for all t.
3
2
2
3
1
Proof. Recall that for a specific value t , there is an affine motion taking x(t ), E (t ), E (t ), E (t )
0
0
0
1
0
3
0
2
to x(t ), E (t ), E (t ), E (t ). We will show that this same motion takes x(t), E (t), E (t), E (t) to x(t),
0
0
1
0
2
0
3
1
2
3
E (t), E (t), E (t) for all t. Assume that the motion has been carried out so that the frames x(t0),
1
3
2
E (t ), E (t ), E (t ) and x(t ), E (t ), E (t ), E (t ) coincide.
1
0
0
3
0
0
2
1
0
2
0
0
3
Now consider
'

(E (t) . E (t))’ = E (t) E (t) ' j  i   Ei(t) E (t)
i
i
i
= q (t)E(t) . E (t) + E (t) . q (t)E(t)
j
ij
j
i
i
ij
= q (t)E(t) . E (t) + q (t)E (t) . E(t)
i
j
i
ij
j
ij
= q (t)E(t) . E (t) + q (t)E(t) . Ei(t)
ji
i
j
ij
j
= 0 .
It follows that
E (t) . E (t) = E (t ) . E (t ) = E (t ) . E (t ) = 3
0
i
i
i
0
i
i
0
i
0
for all t. But since |E (t) . E (t)|  1 for any pair of unit vectors, we must have E (t) . E (t) = 1 for
i
i
i
i
all t. Therefore, E (t) = E (t) for all t.
i
i
Next, consider
(x(t) – x(t))’ = p (t)E (t) – p (t)E (t) = p (t)E (t) – p (t)E (t) = 0 .
i
i
i
i
i
i
i
i
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