This is Problem 11 of this year's Miklos Schweitzer.

(a) Consider an ellipse in the plane. Prove that there exists a Riemannian metric which is defined on the whole plane, and with respect to which the ellipse is a geodesic. Prove that the Gaussian curvature of any such Riemannian metric takes a positive value.

(b) Consider two nonintersecting, simple closed smooth curves in the plane. Prove that if there is a Riemmanian metric defined on the whole plane and the two curves are geodesics of that metric, then the Gaussian curvature of the metric vanishes somewhere.

I do not know how this can be approached, and my knowledge of Riemmanian geometry is limited. I posted this question because I'm curious what do we need to do in order to find the solution.


(b) [Reference : do Carmo's books, Riemannian geometry, Differential geometry of curves and surface]

Consider two curves $c_i$.

Then there exists a minimizing geodesic $c$ between them : $$ c(0)\in c_1,\ c(l)\in c_2 $$ There exists a parallel vector field $V$ along $c$ s.t. $$ V(0)\in T_{c(0)} c_1,\ V(l)\in T_{c(l)} c_2 $$

Then we have a variation $$ f(s,t) = {\rm exp}_{c(t)}\ sV(t)$$

And for energy second variation formula implies $$ E(s)=\frac{1}{2} \int_0^l |f_t|^2\ dt,\ E''(0) = \int_0^l - {\rm Rm}(f_s,f_t,f_t,f_s) \ dt \geq 0 $$

Hence Gaussian curvature $K$ is nonpositive.

Since $c_1$ is a simple, so it bound a disk $D$. From Gauss-Bonnet Theorem, $$ \int\int_D K = 2\pi $$ so that there exists a point whose Gaussian curvarture is positive.

So we complete the proof.


For part (a), take a round semiinfinite cylinder of unit radius and glue it to round unit hemisphere. This surface is just $C^1$ but you can smooth it by replacing hemisphere by a suitable surface of revolution. Your closed geodesic will be the boundary circle of the cylinder.


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