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Can we characterize surfaces $S$, for which cut locus $C_p(S)$ with respect to a point $p$ on $S$, is itself a geodesic between the points it passes through? This holds for example for a cylinder and therefore surfaces isometric to it which do have a cut locus. I am looking for more examples.

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Do you assume that the cut locus is connected? – wspin Feb 24 '14 at 18:27
@wspin: For closed (connected) manifolds, the cut locus will be connected automatically. The idea is that for any $p\in M$, $M\setminus \{p\}$ deformation retracts onto the cut locus. Since $M\setminus\{p\}$ is connected (unless $M = \mathbb{R}$), this implies the cut locus is as well. When $M = \mathbb{R}$, every metric is isometric to the standard one, so there is no cut locus. I don't know what happens for non-closed manifolds. – Jason DeVito Feb 25 '14 at 19:16

For the real projective plane with a round metric the cut locus of a point is a real projective line hence a geodesic.

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If the manifold is compact a topological classification can be obtained as follows:

First observe that the cut locus of a closed connected manifold is connected, as noted in Jason De Vitos comment. In fact for $p \in S$ and $q \in S \setminus \{p\}$ consider a minimal geodesic $\gamma$ from $p$ to $q$. Mapping $q$ to $\gamma(t)$, where $t$ is the first time such that $\gamma$ can not be extended to a minimal geodesic from $p$ to $\gamma(t + \epsilon)$ for small $\epsilon > 0$ defines a map from $M \setminus \{p\}$ to $Cut(p)$. It is well known that this map is continous (but not trivial). Hence $Cut(p)$ is connected if $M$ is.

Let us assume that $S$ is compact and connected. Then $Cut(p)$ is compact as well. From your assumption and the above it follows that $Cut(p)$ is either homeomorphic to $\mathbb S^1$ or to a point. Via the map in the above argument one can moreover construct a homotopy equivalence $S\setminus\{p\} \cong Cut(p)$. Thus $S\setminus\{p\}$ is either homotopy equivalent to a point or a circle. In the first case it follows that $S$ is homeomorphic to $\mathbb S^2$. In the second case it follows that $S$ is homeomorphic to $\mathbb RP^2$, which is a bit harder to see, but follows from the calssification of surfaces.

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