fundamental domains for free fuchsian group. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T22:26:42Z http://mathoverflow.net/feeds/question/38323 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38323/fundamental-domains-for-free-fuchsian-group fundamental domains for free fuchsian group. Benoît Kloeckner 2010-09-10T15:22:13Z 2010-09-11T05:51:28Z <p>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).</p> <p>Call ideal polygon'' any domain of the hyperbolic plane that is an intersection of half-planes limited by geodesics that are pairwise disjoint. </p> <p>Is it true that any free fuchsian group has a fundamental domain that is an ideal polygon?</p> <p>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.</p> http://mathoverflow.net/questions/38323/fundamental-domains-for-free-fuchsian-group/38338#38338 Answer by Ian Agol for fundamental domains for free fuchsian group. Ian Agol 2010-09-10T18:05:34Z 2010-09-11T05:51:28Z <p>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. </p> <p>Addendum: I'll add some comments on one way to obtain these properly embedded arcs in the infinite topology case. Ian Richards gave a <a href="http://www.jstor.org/stable/1993768" rel="nofollow">classification of connected surfaces</a>. 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 $R^2$. </p> <p>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 $R^2$. </p>