Page 242 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 20: Curves

.
defines  everywhere in . It remains to show that  is continuous. Let y  Since x y is Notes
0
0
0
compact, it is possible to choose  small enough that the following holds: y' x y and y  y'  

0
0
implies e(y) e(y')  2 or equivalently e(y) and e(y’) are not antipodal. Let 0 <  < . Then there
exists a neighborhood U  B (y ) of y such that y  U implies (y)   (y )  2 k(y)  '(y)


0
0
0

where  '(y)   and k(y) is integer-valued. It remain to prove that k  0. Let y  U and consider
the continuous function:


 (s)   x  s(y x 0  )   x0 s(y 0  x 0  ) , 0  s 1.
0
Since
x  s(y x 0  )  (x  s(y  x ))  s(y  y )   ,

0
0
0
0
0

it follows from our choice of  that  e x  s(y x 0  ) and e x  s(y  x 0  ) are not antipodal.

0
0
0
.
s
Thus, (s)   for all 0   1, and since (0) = 0 we conclude that    In particular,
2 k(y)   '(y)   (y)   (y )   (1)   ,

0
and it follows that


2 k(y)  2 k(y)  '(y)   '(y)  2 .

Since k(y) is integer-valued this implies k(y) = 0.
Proof of the Rotation Theorem. Pick a line which intersects the curve  and pick a last point p on
this line, i.e., a point with the property that one ray of the line from p has no other intersection
points with . Let h be the unit vector pointing in the direction of that ray. We assume without
loss of generality that  is parametrized by arclength,  (0)   (L)  0.  Now, let
1
  (t ,t )  2 : 0  t  t  L  , and note that  is star-shaped. Define the  -valued function:
2
1
2
1

 '(t ) t ;
1
   1 if t  2

e(t ,t )   '(0) if (t ,t ) (0,L);

2
1
1
2
  (t )   (t )
 2 1 otherwise.
   (t )   (t )
1
2
It is straightforward to check that e is continuous on . By the Lemma, there is a continuous
function :    such that e = (cos , sin ). We claim that (L,L) – (0, 0) = ±2 which proves the
theorem, since (t, t) is a continuous function satisfying (3) in Proposition 6, and thus can be used
on the right-hand side of (4) to compute the rotation number.
To prove this claim, note that, for any 0 < t < L, the unit vector
 (t)   (0)
e(0,t) 
 (t)   (0)
is never equal to h. Hence, there is some value  such that (0, t) – (0, 0)  + 2k for any integer

k. Thus, (0,t)   (0,0)  2 , and since e(0,L) = –e(0, 0) it follows that (0,L) – (0, 0) = ±.
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