Page 269 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes
points c(s) and c(s)  c( (s)) under . Here s and s arc-length parameters of C and C respectively.

In this case, C is called a Bertrand mate of C. The following results are well-known:
Theorem (the case of n = 2). Every C -plane curve is a Bertrand curve.

Theorem (the case of n = 3). A C -special Frenet curve in E with 1-curvature function k and
3

1
2-curvature function k is a Bertrand curve if and only if there exists a linear relation
2
ak (s) + bk (s) = 1
2
1
for all s  L, where a and b are nonzero constant real numbers.
The typical example of Bertrand curves in E is a circular helix. A circular helix has infinitely
3
many Bertrand mates.
We consider the case of n  4. Then we obtain Theorem A.
Proof of Theorem A. Let C be a Bertrand curve in En (n  4) and C a Bertrand mate of C. C is

distinct from C. Let the pair of c(s) and c(s)  c( (s)) be of corresponding points of C and C .

Then the curve C is given by

c(s)  c( (s))  c(s)   (s) . n (s) (1)
1
where  is a C -function on L. Differentiating (1) with respect to s, we obtain

dc(s)
'
j'(s) .  c'(s)   '(s) . n (s)   (s). n (s).
ds s (s) 1 1
Here and hereafter, the prime denotes the derivative with respect to s. By the Frenet equations,
it holds that


 '(s) . t( (s)) (1   (s)k (s)) . t(s)   '(s) . n (s)   (s)k (s) . n (s).

1
1
2
2
Since t( (s)),n ( (s)) 1   0 and n ( (s))   n (s), we obtain, for all s  L,
1
1
’(s) = 0,
that is,  is a constant function on L with value a (we can use the same letter without confusion).
Thus, (1) are rewritten as

c(s)  c( (s)) c(s)   . n (s), (1)’

1
and we obtain
’(s) . t( (s)) = (1 – k (s)) . t(s) + k (s) . n (s) (2)

1
2
2
for all s  L. By (2), we can set
t( (s)) = (cos (s)) . t(s) + (sin (s)) . n (s), (3)

2
where  is a C -function on L and

cos (s) = (1 – k (s))/’(s) (4.1)
1
sin (s) = k (s)/’(s). (4.2)
2

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