# How do you see that higher genus surfaces are not homogeneous?

I am trying to get some intuition about why the torus and the sphere are the only surfaces which can be realised as homogeneous spaces. On the one hand, I know this is true because there is the result that homogeneous spaces must have non-negative Euler characteristic:

A Structure Theorem for Homogeneous Spaces, Mostow, G, Geometriae Dedicata, (114) 2005, 87-102

However, on the other hand, a higher genus surface can be realised as a quotient of hyperbolic space by a group of isometries. The latter would seem (in my head) to give rise to a hyperbolic surface where the points look the same; all of the points have the same curvature, for instance.

My question is then: How do you distinguish the points of such a hyperbolic surface?

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After reading two answers I'm led to ask: in the question, is "homogeneus space" intended to include a Riemannian metric? There is also the purely differentiable notion of homogeneus space (as, e.g., per en.wikipedia.org/wiki/Homogeneous_space ). Maybe there is also a reasonably easy proof just in the differentiable setting... – Qfwfq Aug 6 '12 at 13:21
@Claudio: "I think your intuition is right when you say that "the points look the same" in the sense that the surface is not (globally) homogeneous but it is certainly locally homogeneous as a Riemannian manifold." As I am sure you realize, that is true for every Riemannian manifold. Perhaps the OP should make his question more precise: obviously "homogeneous" in the sense of admitting a transitive action of a (finite dimensional) real Lie group is very different from "the points look the same" in the sense that sufficiently small charts are isomorphic. So what does the OP want? – Jason Starr Aug 6 '12 at 13:22
@Jason: Riemannian manifolds in general do not look the same locally! – Qfwfq Aug 6 '12 at 13:43
The group of diffeomorphisms of a smooth connected manifold acts transitively on that manifold. In that sense, any smooth connected manifold is homogeneous. On the other hand, if a compact group $G$ acts transitively on a smooth manifold, then by averaging one can produce a Riemann metric on that manifold which is $G$-invariant. – Liviu Nicolaescu Aug 6 '12 at 13:53

There is a classical theorem of Hurwitz which states that the group of biholomorphisms of a Riemann surface of genus $g>1$ contains at most $84(g-1)$ elements. Such a Riemann surface cannot be a homogeneous space because the (orientation preserving) isometries are conformal maps and thus they are also biholomorphic maps.

The proof is of Hurwitz' theorem is based on the concept of Weierstrass point. There are at least $2g+2$ such points on any Riemann surface of genus $g>1$, but their number is bounded from above by a universal function of $g$. The set of such points is a biholomorphism invariant. The clincher is that a biholomorphic map of a Riemann surface cannot have too many fixed points. For more details see Eric Reyssat's Quelques Aspects des Surfaces de Riemann, Birkhauser or Rick Miranda's Algebraic Curves and Riemann Surfaces, Amer. Math. Soc.

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Another way to see it is this: there is a unique geodesic in each nontrivial homotopy class because the surface is hyperbolic so that energy is a convex function on loops. If the surface were homogeneous (i.e. admitted a transitive Lie group action (EDIT: by isometries)) then you would be able to move geodesics around to fill out the space. You get distinguished points this way as intersections between geodesics. The universal cover is homogeneous, of course, but there is no nontrivial homotopy class.

EDIT: As the OP asked about hyperbolic surfaces, I interpreted homogeneous space to mean Riemannian homogeneous space/Klein geometry. I interpreted the question to be "Given that the geometry is locally homogeneous, how can we distinguish special points?". You have to use a global tool.

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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.

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Why does this argument not work for the torus? – Igor Rivin Aug 6 '12 at 14:54
Because the deck transformation group of the torus is a subgroup of the translation group. So when you perturb $p$, and lift to the total orbit of $p$, all of the points in that orbit are perturbed by adding the same little vector, and so the entire Voronoi diagram is perturbed by adding the same little vector, and the angles are unaffected. – Lee Mosher Aug 6 '12 at 15:24
I guess this is really just walking through the proof that a torus is homogeneous. – Lee Mosher Aug 6 '12 at 15:29
I think we all agree that the torus is homogeneous, my question was: what about your argument depends on strictly negative curvature? – Igor Rivin Aug 6 '12 at 15:46
@Lee: Your suggestion sounds way too painful. Even proving that generically Voronoi tiling has 3-valent graph is not at all easy, it was done in 2007 by Diaz and Ushijima (their proof could be simplified though). An easier argument is to consider not just Voronoi tiling of the hyperbolic plane associated with the given orbit, but the entire collection of bisectors of points in the orbit. Then you can see how angles change already in the case of cyclic group (angle/distance between bisectors depends on distance from the point to the axis). @Igor: This is where I would use hyperbolicity. – Misha Aug 7 '12 at 21:26