Page 220 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 18: Theory of Space Curves

We have, however, the following lemma: Notes

Lemma 2. There exists a continuous function (s)  such that (s)   (s) mod 2.
Proof. We suppose C to be a closed curve of total length L. The continuous mapping g is
uniformly continuous. There exists therefore a number  > 0 such that for |s – s | < , T(s ) and
2
1
1
T(s ) lie in the same open half-plane. Let s (= O) < s < · · · < s (= L) satisfy |s – s | <  for i = 1,
2
i–1
0
1
i
i
. . ., m. We put (s )  0 = (s ). For s  s  s , we define (s)  to be (s )  0 plus the angle of rotation
1
0
0
from g(s ) to g(s) remaining in the same half-plane. Carrying out this process in successive
0
intervals, we define a continuous function (s)  satisfying the condition in the lemma. The
 
difference (L)    (O) is an integral multiple of 2. Thus, (L)    (O) = 2. We assert that the
 
integer  is independent of the choice of the function .  In fact let '(s)  be a function satisfying
the same conditions. Then we have '(s)  – (s)  = n(s)2p where n(s) is an integer. Since n(s) is
continuous in s, it must be constant. It follows that '(L)– (O)      t(L)    (O), which proves the
independence of  from the choice of .  We call the rotation index of C. In performing integration
over C we should replace t(s) by   in (8). Then we have
1 1 .

2   k ds  2   d   ...(10)
We consider the mapping h which sends an ordered pair of points x(s ), x(s ), O  s  s  L, of C
2
1
2
1
into the end-point of the unit vector through O parallel to the secant joining x(s ) to x(s ). These
2
1
ordered pairs of points can be represented as a triangle  in the (s , s )-plane defined by O  s 1
2
1
 s  L. The mapping h of  into  is continuous. Moreover, its restriction to the side s = s is the
2
1
2
tangential mapping g in (9).
To a point p   let (p) be the angle of inclination of Oh(p) to the x -axis, satisfying O  (p) <
1
2. Again this function need not be continuous. We shall, however, prove that there exists a
,

 
continuous function (p),p such that (p)    (p) mod 2 . In fact, let m be an interior point
of . We cover  by the radii through m. By the argument used in the proof of the above lemma
,
we can define a function (p),p such that (p)   (p) mod 2, and such that it is continuous
 
 
on every radius through m. It remains to prove that it is continuous in .
For this purpose let p  4. Since h is continuous, it follows from the compactness of the segment
0
mp that there exists a number  = (p ) > 0, such that for q  mp and for any point q   for
0
0
0
0
which the distance d(q, q ) <  the points h(q) and h(q0) are never antipodal. The latter condition
0
can be analytically expressed by

  (q)    (q ) mod . ...(11)
0

Now let R / > 0, R / < be given. We choose a neighborhood U of p such that U is contained
2 0
in the -neighborhood of p and such that, for p  U, the angle between Oh(p ) and Oh(p) is
0
0
< R. This is possible, because the mapping h is continuous. The last condition can be expressed in
the form
 (p) – t(p ) = R' + 2k(p) , ...(12)

0
LOVELY PROFESSIONAL UNIVERSITY 213