I try to understand some of the topology of the space of pointed non-compact hyperbolic surfaces (with the pointed Gromov-Hausdorff topology). It is known that the fundamental group of a non-compact surface is a free group, so I am interested in free Fuchsian groups (discrete, free groups of direct isometries of the hyperbolic plane, not necessarily finitely generated).

Call ``ideal polygon'' any domain of the hyperbolic plane that is an intersection of half-planes limited by geodesics that are pairwise disjoint.

Is it true that any free fuchsian group has a fundamental domain that is an ideal polygon?

I think that I can manage do do it by hand for the simplest examples (e.g. covering groups of an hyperbolic punctured tori, or a hyperbolic trice punctured sphere) and the result seems plausible, but I feel that either true or false it shall be well-known. Any reference on this, or more generally on uniformization of non-compact, possibly infinite genus hyperbolic surfaces would be welcome.


1 Answer 1


Yes, this is true. Topologically, one may find a locally finite collection of properly embedded arcs in a connected surface whose complement is homeomorphic to $R^2$. Then make each of these arcs geodesic in the hyperbolic metric. The complement will be the fundamental domain of the type you want.

Addendum: I'll add some comments on one way to obtain these properly embedded arcs in the infinite topology case. Ian Richards gave a classification of connected surfaces. In Theorem 3 of that paper, he explains how to construct all surfaces. A planar surface $\Sigma\cong S^2-X$ is obtained by removing a totally disconnected compact set $X\subset S^2$ from $S^2$. As explained in Prop. 5 of the paper, one may consider the totally disconnected set $X\subset S^2$ to be a subset of the Cantor set, and therefore a subset of the interval (including the endpoints) $X\subset I\subset S^2$. Then the properly embedded arcs $I\cap (S^2-X)$ give a decomposition of $\Sigma$ into $R^2$.

If the surface is non-planar, then one removes from $S^2-X$ a properly embedded countable collection of disks $D_1,D_2,\ldots$, and makes identifications of their boundaries. We may assume after an isotopy that these disks are all centered on $I$, and that the identifications either identifies antipodal points, or identifies two disks which are adjacent along a component of $I-X$ with a $\pi$ twist. The complement $U=S^2-(I\cup_i D_i)$ is again homemorphic to $R^2$. If we identify antipodal points of $D_j$, then this identifies two arcs in the boundary of $U$ to obtain an open Mobius strip. We add two arcs connecting antipodal points of $D_j$ to a point $x\in X$ at the end of the interval of $I-X$ which intersects $D_j$, which forms a single arc after identification of antipodal points of $D_j$, and cuts the Mobius strip back up into $R^2$. If adjacent disks $D_i, D_{i+1}$ are identified, then the complement $U$ gives a punctured torus. We add 4 arcs connecting these points to $x$ (again, $x\in X$ is at the end of the interval of $I-X$ containing $D_i$), cutting the surface into $R^2$ again. Continuing in this fashion inductively, we get a locally finite collection of arcs cutting the surface up into $R^2$.

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    $\begingroup$ Agol's proof is correct, but it might be worth saying more about how the properly embedded topological can be replaced by geodesics. To do this, take (closed disk \ limit set) mod the fuchsian group, choose one point on each boundary component, and make all the geodesics go to that point (which comes frorm the circle at $\infty$). The corresponding statement is false in one higher dimension, for free groups in hypebolic 3-space. E.g., There are upper bounds < 2 to the Hausdorff dimension of limit sets of groups with planar fundamental domain boundaries, but the Hausdorff limit sets --> 2. $\endgroup$ Sep 10, 2010 at 20:19
  • 1
    $\begingroup$ There is a canonical way to choose the geodesics. For the case of a surface of finite topology, there is a minimal convex core with totally geodesic boundary. For each proper arc, one may isotope it to be perpendicular to the boundary of the end it exits, or if the end is a cusp, one sends it to the parabolic fixed point. In the non-finite topology case, one may find a locally finite decomposition of the surface into pairs of pants. If an arc crosses infinitely many closed curves of this decomposition, one may show that there is a canonical point at infinity for the endpoint of the geodesic. $\endgroup$
    – Ian Agol
    Sep 11, 2010 at 18:04
  • $\begingroup$ Thanks for the answer and comments; unfortunately I found a flaw in my plan so I need to rethink all of it. References on infinite type Riemann surface are still welcome. $\endgroup$ Sep 12, 2010 at 17:03
  • $\begingroup$ This is stated and proved on p.15 of the following paper : zbmath.org/0265.30023, which I found through this paper : zbmath.org/1095.37008. $\endgroup$
    – Sary
    Apr 22 at 14:14

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