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Complex Analysis and Differential Geometry
Notes Proof. Let a = t < t < · · · < t = b be a partition of [a, b] so that the diameter of e ([t , t ]) is less
0
1
1
i
i-1
n
than 2, i.e., e restricted to each subinterval maps into a semi-circle. Such a partition exists since
1
e is uniformly continuous on [a, b]. Choose (a) so that (3) holds at a, and proceed by induction
1
on i: if is defined at t then there is a unique continuous extension so that (3) holds. If is any
i
)
other continuous function satisfying (3), then k (1/2 )( is a continuous integer-valued
function, hence is constant. Finally, e = (sin , cos ) hence
2
e ke ( sin ,cos ),
1
2
and we obtain = k.
20.3.2 Global Theory
n
n
Definition 7. A curve :[a,b] is closed if (k) (a) (k) (b). A closed curve :[a,b] is
simple if (a,b) is one-to-one. The rotation number of a smooth closed curve is:
1
n (a) (b) , ...(4)
2
where is the function defined in Proposition 2.
We note that the rotation number is always an integer. For reference, we also note that the
rotation number of a curve is the winding number of the map e . Finally, in view of the last
1
statement in Proposition 2, we have:
1
n k ds.
2 [0,L]
2
Theorem 5 (Rotation Theorem). Let :[0,L] be a smooth, regular, simple, closed curve.
Then n In particular,
1.
1 1.
2 [0,L] k ds
For the proof, we will need the following technical lemma. We say that a set is star-
n
shaped with respect to x if for every y the line segment x y lies in .
0
0
1
,
Lemma 1. Let Rn be star-shaped with respect to x and let e : be a continuous
n
0
function. Then there exists a continuous function : such that:
e = (cos , sin ). ...(5)
2
Moreover, if is another continuous function satisfying (5), then k where k is a
constant integer.
In fact, it is sufficient to assume that is simply connected, but we will not prove this more
general result here.
Proof. Define (x ) so that (5) holds at x . For each x , define continuously along the line
0
0
segment x x as in the proof of Proposition 2. Since is star-shaped with respect to x , this
0
0
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