Tema 11: Limit set
For the following, we need the extra structure of a topology on X∪X(∞) (see [Ballmann '95]).
Lemma 1. In this topology, (xn)n⊂X converges to ξ∈X(∞) if and only if d(x,xn)→∞ for some (and hence any) x∈X and the geodesic segments σx,xn converge to σx,ξ.
Given x∈X, denote by Λ(Γ,x) the set of accumulation points of the orbit Γx={γx:γ∈Γ} in X∪X(∞).
Lemma 2. Λ(Γ,x)=¯Γx∩X(∞).
Proof. Since the action of Γ is properly discontinuous, for every K⊂X compact, the set {γ∈Γ:γ(K)∩K≠∅} is finite. This implies the claim.
Lemma 3. Λ(Γ,x) is independent of x∈X.
Proof. Let ξ∈Λ(Γ,x). Let us show that ξ∈Λ(Γ,y) for any y∈X. As ξ∈Λ(Γ,x), there exists (γn)n⊂Γ such that γnx→ξ as n→∞. By the above, d(x,γnx)→∞ and σx,γnx→σx,ξ. Recall that Γ is a group of isometries acting on X. Hence, given y∈X, for every n∈\bN it holds d(γny,γnx)=d(x,y). Hence, d(x,γnx)≤d(x,y)+d(y,γny)+d(γny,γnx)=2d(x,y)+d(y,γny)→∞. It follows that σx,γnx and σy,γny are asymptotic and hence σy,γny→σy,ξ. Hence ξ∈Λ(Γ,y), proving the lemma.
Denote Λ(Γ)=Λ(Γ,x), x∈X arbitrary.
Lemma 4. (See [Knieper '97]) If X/Γ is compact, then Λ(Γ)=X(∞).