# Determining if there can be smooth closed geodesic given its curvature

In my class on differential geometry I have been given the following question on which I am stuck:

Let S be a regular orientable surface in $R^3$ with Gaussian curvature $K$ (not necessarily constant) and we are to check if there can be a smooth closed geodesic on S in the following case:

$K>0$

$K=0$

$K<0$

and we are to give an example when it is possible that the closed geodesic bounds a simply connected region

I know for starters about the first one with positive curvature I may take the sphere and a great circle which is a geodesic and obviously bounds a simply connected subset of the sphere. But with the zero and negative curvatures I have no idea how to tell if a closed geodesic exists

I think the Gauss Bonnet theorem has something to do with this but I cannot really proceed from there and I am posting here in the hope of getting help

******* Progress: I can work out what happens with Gauss Bonnet if I have the constraint that the geodesic bounds a simply connected region as this is simple but the fact of the matter is where I am stuck is the general existence or non existence if no constraint is given on the region bounded

• For starters (regarding existence), do you know any examples? For the second part (bounding a simply-connected region), have you written down what Gauss-Bonnet guarantees if a closed geodesic bounds a simply-connected region? – Andrew D. Hwang Feb 13 '16 at 15:39
• @AndrewD.Hwang : for the second part I can do this I don't know anything on general existence without topological properties of bounded region – kroner Feb 13 '16 at 18:53
• You might have a look at surfaces of rotation. :) – Andrew D. Hwang Feb 13 '16 at 21:03

First i want to state something that is much more general.

Every closed (compact with no boundary) Riemannian manifold carry a closed smooth geodesic!

The above result is due to Lyusternik and Fet. In the case where the manifold is not simply connected the idea of the existence is relatively easy. You just go to each non trivial free homotopy class of loops and you minimize the length (equivalently the energy) producing that way a closed geodesic. If you are interested for the simply connected case take a look here: (L. A. Lyusternik and A. I. Fet. Variational problems on closed manifolds. Doklady Akad. Nauk SSSR (N.S.), 81:17–18, 1951)

Therefore, for closed surfaces you have always the existence of a closed smooth geodesic! On the other hand if you have a complete non-compact surface, in each one of your three categories on the Gauss curvature, then this doesn't imply in general the existence of a closed smooth geodesic. I think the following (counter)examples will do the job:

1. $K>0$ Paraboloid.
2. $K=0$ Plane.
3. $K<0$ Helicoid.

Now, on your second question. In the case where the Gauss curvature is either everywhere negative or zero, then if you have a smooth closed geodesic $c$ then $c$ cannot bound a simply connected region. To prove this, let's suppose it did. Then if you apply the Gauss-Bonnet theorem you obtain $$\int_R K=2\pi\chi(R),\ \ \ (\dagger)$$ where $R$ is the region bounded by $c$. Since $R$ is simply connected we have $\chi(R)=1$. Thus, we reach a contradiction due to $(\dagger)$. Therefore this can only happen in the case of positive Gauss curvature. An immediate example where that happens is the sphere. You know that the geodesics are the great circles. Just pick your favourite and you have that it bounds a simply connected region.