Margulis lemma

Remarks

The proof combines a Zassenhaus neighbourhood in a suitable Lie group with a compactness argument. The curvature bounds give uniform control of the isometries that move a point only a little, which is why $\varepsilon_n$ depends only on $n$.

Example: in the upper half-plane model of $\mathbb{H}^2$, let $\Gamma$ be generated by the parabolic isometry $\gamma(z) = z + 1$. At $x = i y$, the distance $d(x, \gamma x) = 2 \operatorname{arsinh}(1/(2y))$ tends to $0$ as $y \to \infty$. So for large $y$ the group $\Gamma_x$ is all of $\Gamma \cong \Z$, which is abelian, as the lemma allows.

The lower curvature bound matters. Multiplying the metric of $\mathbb{H}^2$ by $1/c$ gives curvature $-c$ and divides all displacements by $\sqrt{c}$. For $c$ large, the generators of the fundamental group of a closed hyperbolic surface then move a point less than $\varepsilon_2$. This group is not virtually nilpotent, so the conclusion fails without the bound $-1$.

Going deeper

For a student

The main application is the thick-thin decomposition of a hyperbolic manifold $M = \mathbb{H}^n/\Gamma$. The thin part, where the injectivity radius is less than $\varepsilon_n / 2$, consists of tubes around short closed geodesics and of cusps. This structure underlies the study of volumes of hyperbolic manifolds and of their convergence. A Lie group form, with lattices in place of $\Gamma$, is closely related to the Kazhdan–Margulis theorem.