# What is the limit set of a hyperbolic lattice?

My claim is as follows:

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 $\partial \Bbb{H}^{n}$.

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.

• I am little confused because if you consider $n=2$ and take the fundamental group of any hyperbolic surface of finite area then the limit set is either whole of $\partial\mathbb{H}^2$ or a Cantor set in the boundary. An example of the second is pairs of pants. So why it has has to be the whole of $\partial\mathbb{H}^2$"
– Cusp
Mar 3, 2017 at 6:36
• I don't think the fundamental group of pair of pants is actually a lattice in $\operatorname{Isom}(\Bbb{H}^2)$. Once you embed $\pi_1(\text{Pants})$ into $\operatorname{SL}(2,\Bbb{R})$, the Dirichlet domain for this group would appear to have adjacent sides that are ultraparallel. Mar 3, 2017 at 15:56

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. 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.

• Thank you, Ian. This is the type of argument (and reference) I was hoping to find. I knew it wasn't a research-level question in its own right, but my advisor and I were just stuck on sourcing/justifying the claim for my thesis. Mar 3, 2017 at 18:21

The first (but most likely, not the shortest) argument that comes to my mind is the following one, which uses the geodesic flow on the unit tangent bundle of the quotient manifold (essentially, the Hopf dichotomy).

The limit set is precisely the part of the boundary sphere where the boundary action is minimal; on its complement the action is properly discontinuous, hence dissipative with respect to any quasi-invariant measure. On the other hand, the ergodic components of the geodesic flow on the unit tangent bundle of the quotient manifold with respect to the Liouville measure are in one-to one correspondence with the ergodic components of the boundary action with respect to the Lebesgue measure, and this correspondence preserves the type (conservative or dissipative) of components. Since $\Gamma$ is a lattice, the Liouville measure is finite, so that the geodesic flow has no dissipative components, whence the same is true for the boundary action, whence the limit set is the whole boundary sphere.

• Thank you for this sketch of argument and the prompt response, R W. I have ultimately accepted another answer as it's more aligned with the content of my thesis, but this is definitely an interesting strategy and it's nice to see that there are various proofs out there. Mar 3, 2017 at 18:23

Let $G=\text{Isom}(\mathbb{H}^n)$, $\Gamma<G$ be a lattice and endow $X=G/\Gamma$ with its $G$-invariant probability measure.

For a point $x\in X$ let $G_x$ be its stabilizer in $G$ (which is a lattice conjugated to $\Gamma$), let $\Lambda_x\subset \partial \mathbb{H}^n$ be the limit set of $G_x$ and let $C_x$ by the convex hull of $\Lambda_x$ in $\mathbb{H}^n$. Clearly the maps $x\mapsto G_x,\Lambda_x,C_x$ are all $G$-equivariant.

Fix a base point $o\in\mathbb{H}^n$ and for every $p\in\mathbb{H}^n$ set $$f(p)=\int_X \left( d(p,C_x)-d(o,C_x)\right)dx.$$ Note that the integrand is bounded by $d(p,o)$ thus the integral is defined. For every $p$, $$f(gp)=\int_X \left( d(gp,C_x)-d(o,C_x)\right)dx= \int_X \left( d(p,C_{g^{-1}x})-d(g^{-1}o,C_{g^{-1}x})\right)dx= \int_X \left( d(p,C_x)-d(g^{-1}o,C_x)\right)dx.$$ Specializing for $p=o$, $$f(go)=\int_X \left( d(o,C_x)-d(g^{-1}o,C_x)\right)dx,$$ and we get $f(gp)-f(go)=f(p)$. Setting $\phi(g)=f(go)$ and subtituting $p=ho$ we get $$\phi(gh)=f(gho)=f(ho)+f(go)=\phi(g)+\phi(h),$$ thus $\phi:G\to \mathbb{R}$ is a homomorphism. Since $G$ has no non-trivial homomorphism to $\mathbb{R}$, we conclude that for every $g\in G$, $\phi(g)=0$. Equivalently, for every $p\in \mathbb{H}^n$, $f(p)=0$. Since for every $x\in X$, $p\mapsto d(p,C_x)-d(o,C_x)$ is a convex function on $\mathbb{H}^n$ we conclude that for every $x$, $d(p,C_x)-d(o,C_x)$ is constant. Substituting $p=o$ we get that this constant is $0$. Taking $p\in C_x$ we get that $o\in C_x$ and conclude that for every $p$, $p\in C_x$. That is $C_x=\mathbb{H}^n$. It follows that $\Lambda_x=\partial\mathbb{H}^n$. Applying to the point $x$ with $G_x=\Gamma$ we get that the limit set of $\Gamma$ is $\partial\mathbb{H}^n$.

• Thank you, Uri, for the response and very clear proof. I have ultimately accepted another answer as it's more aligned with the content of my thesis, but it's nice to see other viewpoints in the approach. Mar 3, 2017 at 18:28