Let G be a Fuchsian Schottky group defined by a possibly infinite set of disjoint halfplanes {C_i}_i. Let F be the fundamental domain obtained by intersecting the complements of the C_i's. If H i a subgroup of G then: Can a fundamental domain for H be found in terms of the domain F? Is H of schottky type? Do the above answers change if we require 1) H to be normal in G ? 2) G to be finitely generated ? 3) H to be finitely generated?

It seems quite easy to find a fundamental region for H in terms of F. Yet I run into trouble trying to construct it in such a way that it is actually a domain.

I will be greatful for some good references.


1 Answer 1


Here is a concrete construction of a fundamental domain for $H$ that works regardless of normality and finite generation. The translates of the fundamental domain $F$ by the action of $G$ form a tiling of $\mathbb{H}^2$. Let $T$ denote the dual tree of this tiling, with one vertex for each translate of $F$, and with two vertices connected by an edge if the corresponding translates of $F$ intersect along a line. Note that each oriented edge $E$ of $T$ corresponds to a transversely oriented line $L_E$ of the tiling which subdivides $\mathbb{H}^2$ into two halfplanes, and the transverse orientation on $L_E$ points into one of those half-planes which I'll denote $C_E$.

The action of $G$ on $\mathbb{H}^2$ induces a properly discontinuous action of $G$ on $T$ by simplicial isomorphisms. Consider the restricted action of $H$ on $T$, also properly discontinuous. The quotient graph $T/H$ has fundamental group identified with $H$. Let $\tau$ be a maximal tree in $T/H$. Let $\tilde\tau \subset T$ be a homeomorphic lift of $\tau$. As $E \subset T$ varies over all oriented edges not contained in $\tilde\tau$ but with initial endpoint in $\tau$, the collection of half-planes $C_E$ demonstrates that $H$ is a Schottky group, and the complement of their union is a fundamental domain for $H$.

  • $\begingroup$ Can this construction be found somewhere in a book or a paper? $\endgroup$
    – Anna Z
    Mar 6, 2013 at 18:02
  • $\begingroup$ I don't know of any reference for it, unfortunately. $\endgroup$
    – Lee Mosher
    Mar 6, 2013 at 18:14
  • $\begingroup$ I have found a much more "hands on" proof which #I can include in my project without introducing a lot of additional Background material :) But thank you for your neat answer! $\endgroup$
    – Anna Z
    Mar 7, 2013 at 14:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .