4
$\begingroup$

Let $S\subset R^3$ be a regular surface homeomorphic to a sphere. Let $\alpha\subset S $ be a simple closed geodesic in S,let A and B be a regions of S which have $\alpha$ as a common boundary. Let N:$S->S^2$ be the Gauss map of S. prove that N(A) and N(B) have the same area.

My thoughts: I think I need to use Gauss-Bonnet theorem,since the curve $\alpha$ can be mapped into a sphere,and the mapped the curve should be smooth(But I am not sure whether it is still geodesic),so use Gauss-Bonnet Theorem, we have $$\iint_{N(A)}Kds+\int_{0}^lk_{g}ds=2\pi X(s)$$,where X(s) is the Euler-poincare characteristic of a regular surface. if the geodesic curvature $k_{g}$=0,I can know that the area of N(A) should be $2\pi$,since the Gaussian curvature k=1 in the unit sphere,but I have no idea whether the geodesic curve in the original surface is still be geodesic in the sphere. Any help?

$\endgroup$
1
  • $\begingroup$ the mapped geodesic curve should be still closed in the sphere,so it can divide the surface of the sphere into two parts,but not sure whether it is still geodesic? $\endgroup$
    – user144600
    Apr 24, 2015 at 21:15

1 Answer 1

3
$\begingroup$

I suppose you need to assume that the Gauss curvature of $S$ is positive. In this case, $N$ is a diffeo from $S$ onto the sphere. Otherwise, you should define what area of $N(A)$ means.

In any case, if you use the Gauss-Bonnet theorem on $A$, then you obtain that $$ \int_A K=2\pi, $$ because the geodesic curvature of $\alpha$ vanishes, and the Euler characteristic of the disk $A$ is one.

On the other hand, $\int_A K$ is the spherical area of $A$ (that is, the area of $N(A)$ counting multiplicities). This can easily be seen using that $$ N_u\times N_v=K X_u\times X_v $$ for a local parameterization $X(u,v)$ of the surface. And integrating, $$ area(N(A))=\int_{N(A)} 1=\int_A |K|=2\pi. $$

Analogously, $area(N(B))=2\pi$.

Observe that, in general $N(\alpha)$ is not a geodesic.

$\endgroup$
1
  • $\begingroup$ Thanks, it is really helpful $\endgroup$
    – user144600
    Apr 24, 2015 at 23:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.