Compact open topology on the space of geodesics

I'm new in the field, so I'm sorry in advance if my question is too naive.

Let's consider $S$ a surface of genus $g\ge 2$ with an hyperbolic metric $g$. Let's call $\mathcal{S}(S)$ the set of closed geodesics on $S$ with respect to $g$.

Usually on $\mathcal{S}(S)$ there is the topology induced by the hausdorff metric (indentifying a geodesic with its image, we can view $\mathcal{S}(S)$ as a subset of $\mathcal{C}(S)$, the set of closed sets of $S$).

Lately I've stumbled upon the "$\textit{compact-open topology}$" on $\mathcal{S}(S)$: this topology has as a subbase the sets $V(K,U)$ of isometries $h:K\rightarrow U$, where $K$ is a compact subset of $\mathbb{R}$ and $U$ is an open set in $S$.

My questions are:

1) Why one should use the compact-open topology instead of the topology induced by the hausdorff metric on $\mathcal{S}(S)$? Maybe it has something to do with conververgence of sequences of geodesics?

2) When is a subset $V\subset\mathcal{S}(S)$ closed for the compact open topology?

3) Can you point me out some references to study properties of the compact open topology on $\mathcal{S}(S)$?

Thank you

• To be precise, V(K,U) consists of maps R-->S which send K to U. – ThiKu Nov 20 '15 at 19:41
• Probably you are aware of this, but just to mention: if S is compact, then the Arzela-Ascoli theorem implies compactness of the space of geodesics (with the K-O-topology), which is often useful. – ThiKu Nov 20 '15 at 19:48
• It is important to note that these are topologies on different sets: The c/o topology is on the space of parameterized geodesics while the Hausdorff topology is on the space of unparameterized geodesics. Of course, there is a map from one to the other, but the sets are different. One also frequently uses the Hausdorff topology on the set of all closed subsets of $S$ so that the limit of a sequence of unparameterized geodesics is not a geodesic but a different object, say, a geodesic lamination in the context of simple geodesics. – Misha Nov 20 '15 at 19:59
• yes, of course you're right. So I guess my question #1 becomes "which is the advantage of considering parametrized geodesics with compact open topology instead of unparametrized geodesics with hausdorff topology?" – gustav hertric Nov 20 '15 at 20:31
• @gustavhertric isn't it tautological? A parametrized geodesic is a function with a domain. – Ryan Budney Nov 21 '15 at 21:42

Consider the set $\tilde{\mathcal G}(S)$ of all unit speed parametrised geodesics, closed or not. Then regard the space of closed parametrised geodesics $\tilde{\mathcal S}(S)$ as a subspace with the respective subspace topology. Denote the quotients by $\mathbb R$ as $\bar{\mathcal G}(S)$ and $\bar{\mathcal S}(S)$, so these are sets of directed geodesics. Quotienting out $\mathbb Z/2$ gives the spaces $\mathcal G(S)$ and $\mathcal S(S)$ of unparametrised, undirected geodesics. The $C^0$ topology on $\tilde{\mathcal S}(S)$ is closest in spirit to the Hausdorff topology on $\mathcal S(S)$.
In hyperbolic space $\mathbb H^2$, any two parametrised geodesics that are not reparametrisations of each other will separate. So $\tilde{\mathcal G}(\mathbb H^2)$ in the $C^0$ topology is the topological union of infinitely many copies of $\mathbb R$, one for each unparametrised, directed geodesic in $\bar{\mathcal G}(\mathbb H^2)$. You get the same behaviour for $\tilde{\mathcal G}(S)$. Similarly, the Hausdorff topology on $\mathcal G(\mathbb H^2)$ and $\mathcal S(S)$ will be discrete.
On the other hand, there is a bijection between $\tilde{\mathcal G}(S)$ and the unit tangent bundle of $S$, were a unit speed geodesic $c$ is mapped to $\dot c(0)$. With respect to the compact-open topology, this is a homeomorphism.