My claim is as follows:



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> Let $\Gamma$ be a discrete subgroup of $\operatorname{Isom}(\Bbb{H}^{n})$, the isometries of hyperbolic $n$-space. If $\Gamma$ is a lattice in $\operatorname{Isom}(\Bbb{H}^n)$ then the limit set of $\Gamma$ is $\partial \Bbb{H}^{n}$.
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I'm inclined to believe that the statement above is true purely on intuition gleaned from $\operatorname{PSL}(2,\Bbb{Z})$ in $\operatorname{PSL}(2,\Bbb{R})$ and that it (or probably a discussion of a more general statement) should be in the literature somewhere; however, I have yet to find it through Google or in my go-to print resources (Ratcliffe, Bearon, and Margulis).


Is anyone able to provide insight into the validity of this statement and/or point to potential resources? It would be greatly appreciated. Thank you.



What is the limit set of a hyperbolic lattice?
This question is probably not really research level given that the answer exists in many textbooks. Nevertheless, it may be useful to have different viewpoints.


An elementary geometric argument follows from Chapter 8 of [Thurston's notes](https://dl.dropboxusercontent.com/u/8592391/ThurstonNotes.pdf). In fact, the argument shows that if $\mathbb{H}^n/\Gamma$ does not contain a half-space, the limit set $L\_\Gamma=S^{n-1}\_\infty$. 


Thurston notes in Prop. 8.2.3 that the action of $\Gamma$ on $D\_\Gamma=S\_\infty-L\_\Gamma$ (which is an open subset) is properly discontinuous. The proof is by taking the convex hull of $L\_\Gamma$, and noting that there is a retract from $\mathbb{H}^n\cup S^{n-1}\_\infty$ to this set, which send $D\_\Gamma$ to the boundary of the convex hull. But the action of $\Gamma$ on the boundary of the convex hull is discrete (being in the interior of $\mathbb{H}^n$), hence the action on $D\_\Gamma$ is. 


Now observe that there is a ball in $D\_\Gamma$ which is mapped disjointly from itself by all elements of $\Gamma$, from the proper discontinuity of the action (this requires a bit of argument, but follows from the existence of a free orbit of the action of $\Gamma$ on $\mathbb{H}^n$ and hence on $D\_\Gamma$). Hence there is an embedded half-space in $\mathbb{H}^n/\Gamma$, so it is infinite volume. 


