# Cusps in hyperbolic manifolds and fundamental group

I am reading the book "The Arithmetic of Hyperbolic Three Manifolds" by Maclachlan and Reid and I am having some problems in understanding something about cusps.

The definition they give of a cusp is the following (1.2.7):

A point $\zeta \in \overline{\mathbb{C}}$, the sphere at infinity, is a cusp of the Kleinian group $\Gamma$ if the stabiliser $\Gamma_\zeta$ contains a free abelian group of rank 2.

This is ok for me. Anyway, I know another definition of cusp (or, better, of "cusp neighborhood"), the one coming from the thick-thin decomposition. Obviously, one cannot expect these to be the same thing, but my claim would be the following:

Let $M= \frac{\mathbb{H}^3}{\Gamma}$ be a hyperbolic (finite volume, complete, connected, orientable) 3-manifold. There is a bijection between the cusps under the action af the group $\Gamma$ (that is, $\zeta \cong \zeta'$ if exists $A \in \Gamma$ s.t. $A\zeta = \zeta'$) and the cusp neighborhoods obtained in the thick-thin decomposition.

Is this true? Can you give me some reference about this?

I have another problem related with this is much more specific, and I think it should be easier, but I find some difficulties. From some consideration made in the rest of the book, i would believe this, but I cannot prove it.

In the same settings as above, let $T_i \times [0, \infty)$ be a cusp neighborhood. Let $\zeta_i$ be the corresponding cusp in $\overline{\mathbb{C}}$. Then $\pi_1(T_i \times \{0 \})$ embeds in $\Gamma$ through $i^*$ (the inclusion) and its image is (after composing, if necessary, with conjugation) $\Gamma_{\zeta_i}$.

Is this true? Can you give me some reference?

I thank you in advance.

• Yes, this is true, and it is a straightforward application of the Margulis Lemma, which is what one uses to derive the thick-thin decomposition. Look for the Margulis Lemma in more general books on hyperbolic manifolds, such as Ratcliffe's textbook. Commented Sep 1, 2018 at 14:33
• @LeeMosher thank you very much. I am going to search for it! Commented Sep 2, 2018 at 9:19

## 1 Answer

This is true for sufficiently small Margulis constant (depending on the manifold). Just make it smaller than the translation length in the smallest Margulis tube.

For the second, there will be a horoball invariant under the action of each cusp. Shrink the horoball until the minimal translation length of a parabolic fixing it is at least one in the horosphere boundary. Then the horoball will embed. See Adam's papers:

Adams, Colin C., Waist size for cusps in hyperbolic 3-manifolds, Topology 41, No. 2, 257-270 (2002). ZBL0985.57012.