Kleinian Groups Which Are Limits Of Geometrically Finite Groups
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Kleinian Groups which Are Limits of Geometrically Finite Groups
Author: Ken'ichi Ōshika
language: en
Publisher: American Mathematical Soc.
Release Date: 2005
Ahlfors conjectured in 1964 that the limit set of every finitely generated Kleinian group either has Lebesgue measure $0$ or is the entire $S^2$. This title intends to prove that this conjecture is true for purely loxodromic Kleinian groups which are algebraic limits of geometrically finite groups.
Kleinian Groups which are Limits of Geometrically Finite Groups
Author: Kenʼichi Ōshika
language: en
Publisher: American Mathematical Soc.
Release Date: 2005
Ahlfors conjectured in 1964 that the limit set of every finitely generated Kleinian group either has Lebesgue measure $0$ or is the entire $S^2$. We prove that this conjecture is true for purely loxodromic Kleinian groups which are algebraic limits of geometrically finite groups. What we directly prove is that if a purely loxodromic Kleinian group $\Gamma$ is an algebraic limit of geometrically finite groups and the limit set $\Lambda_\Gamma$ is not the entire $S^2_\infty$, then $\Gamma$ is topologically (and geometrically) tame, that is, there is a compact 3-manifold whose interior is homeomorphic to ${\mathbf H}^3[LAMBDA]Gamma$. The proof uses techniques of hyperbolic geometry considerably and is based on works of Maskit, Thurston, Bonahon, Otal, and Canary.
Kleinian Groups which are Limits of Geometrically Finite Groups
Author: Ken_ichi _shika
language: en
Publisher: American Mathematical Soc.
Release Date: 2005-08-03
Ahlfors conjectured in 1964 that the limit set of every finitely generated Kleinian group either has Lebesgue measure $0$ or is the entire $S^2$. We prove that this conjecture is true for purely loxodromic Kleinian groups which are algebraic limits of geometrically finite groups. What we directly prove is that if a purely loxodromic Kleinian group $\Gamma$ is an algebraic limit of geometrically finite groups and the limit set $\Lambda_\Gamma$ is not the entire $S^2_\infty$, then $\Gamma$ is topologically (and geometrically) tame, that is, there is a compact 3-manifold whose interior is homeomorphic to ${\mathbf H}^3/\Gamma$. The proof uses techniques of hyperbolic geometry considerably and is based on works of Maskit, Thurston, Bonahon, Otal, and Canary.