Page 246 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 20: Curves

n-1
Proof. Suppose, by contradiction, that there is   such that     0. Notes
Then
L
0 
0     L      0    ds  0.
If     0, then the same inequality shows that     0, hence,  lies in the plane perpendicular


to  through (0).


Lemma 3. Let n  3, and let  :[0,L]  n 1 be a regular closed curve on the unit sphere
parametrized by arclength.
1. If the arclength of  is less than 2 then  is contained in an open hemisphere.
2. If the arclength of  is equal to 2then  is contained in a closed hemisphere.
Proof
1. First observe that no piecewise smooth curve of arclength less than 2 contains two antipodal
points. Otherwise the two segments of the curve between p and q would each have length
at least , and hence, the length of the curve would have to be at least 2. Now pick a point
p on  and let q on  be chosen so that the two segments  and  from p to q along  have
2
1
equal length. Note that p and q cannot be antipodal. Let  be the midpoint along the
shorter of the two segments of the great circle between p and q. Suppose that  intersects
1


the equator, the great circle  x 0. Let  be the reflection of  with respect to , then the
1
length of    is the same as the length of  hence is less than 2. But    contains


1
1
1
1
two antipodal points, a contradiction. Thus,  cannot intersect the equator. Similarly,  2
1
cannot intersect the equator, and we conclude  stays in the open hemisphere  x 0.

2. If the arclength of  is 2, we refine the above argument. If p and q are antipodal, then both
 and  are great semi-circle, thus,  stays in a closed hemisphere. So we can assume that
1
1
2
p and q are not antipodal and proceed as before, defining  to be the midpoint on the
shorter arc of the great circle between p and q. Now, if  crosses the equator, then     1
1
1
contains two antipodal points on the equator, and the two segments joining these points
enter both hemispheres. Thus, these segments are not semi-circle, and consequently both
have arclength strictly greater than . Thus the arclength of    is strictly larger than

1
1
2 a contradiction. Similarly,  does not cross the equator, and we conclude that  stays in
2

the closed hemisphere  x 0.
Proof of Fenchel’s Theorem. Note that the total curvature is simply the arclength of the spherical
image of . By Lemma 2  is not contained in an open hemisphere, so by Lemma 3
 

K  I  ds 2 .

If the arclength of  is 2, then by Lemma 3,  is contained in a closed hemisphere, and by
Lemma 2,  maps into an equator. If n > 3, we may proceed by induction until we obtain that is
1.
planar. Once we have that  is planar, the Rotation Theorem gives n   Without loss of

generality, we may assume that n  1. Hence,
2

1 In fact, since  is smooth,  and  are contained in the same great circle, and hence  is itself a great
1
1
circle.
2 Reversing the orientation of  if necessary
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