Page 239 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes Proposition 1 (Frenet frame equations). The Frenet frame X = (T, N, B) of a curve in  satisfies:
3
X’ = X. ...(2)

The Frenet frame equations, Equation (2), form a system of nine linear ordinary differential
equations.
Definition 6. A rigid motion of  is a function of the form R(x) = x + Qx where Q is orthonormal
3
0
with det Q = 1.
Note that if X is the Frenet frame of  and R(x) = x + Qx is a rigid motion of  , then QX is the
3
0
Frenet frame of R   This follows easily from the fact that Q is preserves the inner product and
.
orientation of  .
3
Theorem 1 (Fundamental Theorem). Let  > 0 and  be smooth scalar functions on the interval [0,
L]. Then there is a regular curve  parametrized by arclength, unique up to a rigid motion of  ,
3
whose curvature is  and torsion is .
Proof. Let  be given by (1). The initial value problem
X’ = X,
X(0) = I
can be solved uniquely on [0,L]. The solution X is an orthogonal matrix with det X = 1 on [0, L].
Indeed, since  is anti-symmetric, the matrix A = XX is constant. Indeed,
t
A’ = X X + X X = X( +  )X = 0,
t
t
t
t
t
and since A(0) = I, we conclude that A  I, and X is orthogonal. Furthermore, detX is continuous,
and detX(0) = 1, so detX = 1 on [0,L]. Let (T,N,B) be the columns of X, and let    T, then (T, N,
B) is orthonormal and positively oriented on [0, L]. Thus,  is parametrized by arclength, '  T,
and N = k T’ is the principal normal of . Similarly, B is the binormal, and consequently,  is the
-1
curvature of  and  its torsion.
Now suppose that  is another curve with curvature  and torsion , and let X be its Frenet



frame. Then there is a rigid motion R(x) = Qx + x of  such that R (0)    (0), and QX(0)  X(0).
3
0
By the remark preceding the theorem, QX is the Frenet frame of the curve R   and thus, both
,
QX and X satisfy the initial value problem:

Y’ = Y,
Y(0) = QX(0).
X.
By the uniqueness of solutions of the initial value problem, it follows that QX   In particular,
)'
',

(R     and since R   (0)    (0) we conclude R      .
Assuming (0)  0, the Taylor expansion of  of order 3 at s = 0 is:
1 2 1 3 4
 (s)   '(0)s   "(0)s   "'(0)s  O(s ).
2 6
Denote T = T(0), N = N(0), B = B(0),  = (0), and  = (0). We have '(0)  T , "(0)   0 N , and


0
0
0
0
0
0
0
 "'(0)   '(0)N     k T 0 0   0 B 0 . Substituting these into the equation above, decomposing
0
into T, N, and B components, and retaining only the leading order terms, we get:
4 
3 
3
3
  B
2
 (s)   s O s   T      s  O s      6 s  O s  
  N  
2
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