One way to distinguish points of a hyperbolic surface $S$ is to show that the local geometry of the Voronoi graph of a point $p$ is changed when $p$ is perturbed. The Voronoi graph is a connected 1-complex consisting of the locus of points on $S$ for which there are two or more shortest paths to $p$; its complement is an open 2-cell which is the locus of points for which there is a unique shortest path to $p$. Equivalently, it is the projection to $S$ of the Voronoi diagram for the full pre-image of $p$ under a locally isometric universal covering map $\mathbb{H}^2 \mapsto S$. You need a nontrivial fundamental group to get a nonempty Voronoi graph, and a noncyclic fundamental group to get a Voronoi graph more complicated than a bi-infinite geodesic and therefore possessing at least one 0-cell (point of valence three or more).

Suppose that $x$ is a certain 0-cell of the Voronoi graph, a point at which there are three or more shortest paths to $p$. If $x$ has valence three, which is the "generic" situation, then $p$ can be perturbed so that the three angles of the 1-cells incident to $x$ are changed; this can be verified by some simple hyperbolic trigonometry calculations. If $x$ has valence four or more, then $p$ can be perturbed to change the local topology of the Voronoi graph near $x$, breaking $x$ into two or more vertices of smaller valence.

By the way, with regard to the two topological types of closed Euclidean surfaces, while the torus is always homogeneous in any Euclidean metric, the Klein bottle is never homogeneous, because an orientation reversing simple closed geodesic is unique in its homotopy class.

Further remarks: I addressed Igor's first question in my comment below. To address his second question, the answer I am offering to the OP's question is that points $p \in S$ can be locally distinguished by the local geometry of their Voronoi graph $V(p)$. By this I mean two things: (1) for each $p,q \in S$, there exists an isometry of $S$ taking $p$ to $q$ if and only if $V(p)$ is locally isometric to $V(q)$; (2) for $q$ near $p$ there does not exist a local isometry between $V(p)$ and $V(q)$.

To fill this in somewhat: the edges of $V(p)$ are geodesic arcs, and the local geometry of $V(p)$ is determined by the topological type of its embedding into its regular neighborhood $Nhd(V(p))$, by the lengths of its edges, and by the angles around its vertices. To prove the ``if'' direction of (1), once the local isometry type of $V(p)$ is determined, there is a unique way to fill in $Nhd(V(p)) - V(p)$ by an open hyperbolic disc. The cheapest proof of (2) is to use (1) together with the fact that the isometry group of $S$ acts discretely on $S$, which of course requires already knowing the latter fact, using a proof that depends for example on facts about geodesics explained in the other answers.

As alluded to, there is a much more expensive proof of (2) which uses hyperbolic trigonometry, which under special assumptions goes like this. Assume first that vertices of $V(p)$ have valence 3. Cut $S$ open along $V(p)$ to get a convex hyperbolic polygon containing $p$. Assume further that there is a perpendicular geodesic from $p$ to each side of this polygon (not true in general). Use those perpendiculars, and the geodesics from $p$ to the vertices, to cut the polygon into hyperbolic right triangles. Write down equations relating the shapes of these right triangles to the local geometry of the Voronoi graph. Use these equations to prove (with much agony) that as $p$ is perturbed the local geometry of the Voronoi graph changes.

notlook the same locally! $\endgroup$ – Qfwfq Aug 6 '12 at 13:43compactgroup $G$ acts transitively on a smooth manifold, then by averaging one can produce a Riemann metric on that manifold which is $G$-invariant. $\endgroup$ – Liviu Nicolaescu Aug 6 '12 at 13:536more comments