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On the unit three-ball $B^3$, the distance in the Poincare ball model is the infimum of the lengthsover piecewise smooth paths joining the two points. Equivalently the hyperbolic metric as a metric-space distance isIt has curvature-minus-one normalization; in particular $d(0,x)=2\mathrm{artanh}|x|$.

For a nonempty finite set $p_1,\dots ,p_m$, define $R(x)={\max }_{j}d(x,p_j)$. The triangle inequality gives $|R(x)-R(y)|\le d(x,y)$, so $R$ is continuous. Take $R_0=R(p_1)$. Any improving center satisfies $d(x,p_1)\le R(x)\le R_0$, and thus lies in the closed hyperbolic ball $K=\overline{B}(p_1,R_0)$.

This ball is compact: an isometry sending $p_1$ to the origin identifies it with the Euclidean closed ball of radius $\tanh (R_0/2)<1$, and the Euclidean and hyperbolic topologies agree inside $B^3$. The continuous function $R$ attains a minimum on $K$. Centers outside $K$ have $R>R_0$, so this is also the global minimum. Its associated closed ball proves a minimum enclosing ball in hyperbolic space exists. For the empty finite set, any center and radius zero suffice.

Assume the level set is regular and non-null. The latter hypothesis is necessary: a null hypersurface has a degenerate induced metric and hence no ordinary intrinsic Levi-Civita connection determined by that metric. Normalize the normal vector and define the non-null hypersurface projection byHere $h$ denotes the induced metric restricted to tangent vectors, while its ambient extension annihilates $n$. The intrinsic covariant derivative is $D_{X}Y=P{\nabla }_{X}Y$ for tangent vector fields. Projection of the ambient torsion-free connection and metric compatibility shows that $D$ is the Levi-Civita connection of $h$. Define the extrinsic curvature by $K(X,Y)=g({\nabla }_{X}n,Y)$, or $K_{ab}=P^c{}_a{P}^d{}_b{\nabla }_c{n}_d$. It is symmetric: differentiating the defining function twice and projecting removes the derivative of its normalization, leaving a multiple of its symmetric Hessian. This sign convention makes the scalar second fundamental form $g(n,{\nabla }_{X}Y)=-K(X,Y)$.

Let $S$ be the tangent linear map characterized by $h(SX,Y)=K(X,Y)$. Differentiating $g(n,n)=\epsilon$ and $g(n,Y)=0$ gives the Gauss formulaFix the Riemann curvature tensor conventionApplying the Gauss formula twice givesSubtracting the expression with $X,Y$ interchanged, and then subtracting the bracket derivative, provesThus the Gauss–Codazzi equations for a non-null hypersurface give the intrinsic Riemann curvature tensor asAll free indices here are tangent to the hypersurface. The normal component of the same calculation isAfter antisymmetrizing and using $D_{X}Y-D_{Y}X=[X,Y]$, this yields the curved-ambient Codazzi equationThe extrinsic curvature measures how tangent planes turn in the ambient manifold; the Gauss equation says that this turning contributes to intrinsic curvature even when the ambient Riemann curvature tensor is zero.

For the quadric in Minkowski spacetime, write its position vector as $X^a$. On the hypersurface, $g(X,X)=1$ and differentiating this identity in a tangent direction gives $g(X,Y)=0$. Hence $n=X$ is a unit normal and a spacelike vector, $\epsilon =1$, and ${\nabla }_{Y}n=Y$ in flat Cartesian coordinates. Therefore $K=h$. The ambient Riemann curvature tensor vanishes, soThis is unit-radius de Sitter spacetime. Reversing the normal changes $K$ to $-K$ but does not change this curvature. An opposite overall convention for the Riemann curvature tensor would reverse both curvature signs below.

To retain a four-dimensional Lorentzian metric while reversing the curvature, take a flat ambient metric with two negative directions and the quadricNow the position vector is a unit normal with negative squared norm, so $\epsilon =-1$ while $K=h$ remains true. The Gauss equation givesThis is unit-radius Anti-de Sitter spacetime. If a positive-definite induced metric is acceptable, an even simpler alternative is to keep the original one-time ambient Minkowski metric and change the quadric level to $g(X,X)=-1$. Each sheet then gives four-dimensional hyperbolic space with the same negative curvature, but its induced signature differs from that of de Sitter spacetime.

A model of three-dimensional hyperbolic space is ${\mathbb{H}}^3=\{(z,t):z\in \mathbb{C},t>0\}$ with metric $d{s}^2=(|dz{|}^2+d{t}^2)/t^2$, the complete simply connected metric of curvature $-1$. The translation length of an isometry $g$ is $\ell (g)={\inf }_{P}d(P,gP)$. A hyperbolic isometry has positive translation length attained on its invariant complete geodesic, the axis of a hyperbolic-space isometry. It joins the two fixed ideal points and is translated by distance $\ell (g)$; a three-dimensional axial isometry can also rotate around it.

For the boundary map $z\mapsto kz$, the hyperbolic extension of a Möbius transformation is $(z,t)\mapsto (kz,|k|t)$. Every path from $P$ to $gP$ has length at least the absolute change of $\log t$, namely $|\log |k||$, because the metric length dominates $|dt|/t$. On the vertical line $z=0$ a vertical segment attains that bound. HenceIf $|k|\ne 1$ the vertical line is the axis. If $|k|=1$ the map is a rotation with zero translation length, including the identity when $k=1$; it is not a positive-length hyperbolic isometry.

Identify the Möbius group with ${\mathrm{PSL}}_2(\mathbb{C})$, the orientation-preserving isometry group of three-dimensional hyperbolic space. Here a properly discontinuous group action permits finite stabilizers: for each compact set $K$, only finitely many $g$ satisfy $gK\cap K\ne \varnothing$.

Fix $o\in {\mathbb{H}}^3$. The set of isometries with $d(o,go)\le R$ is compact. To see this, an orientation-preserving isometry is uniquely determined by the image of an oriented orthonormal frame at $o$. The image point lies in a compact closed hyperbolic ball and its frame lies in a compact rotation-group fiber. These choices parametrize the indicated set of isometries continuously.

A discrete subgroup of a topological group is closed: otherwise distinct subgroup elements accumulating at the same point would have quotients tending to the identity, contradicting isolation of the identity. Its intersection with a compact set is therefore finite. If $K\subseteq B(o,R)$ and $gK\cap K\ne \varnothing$, choosing $x,gx\in K$ gives $d(o,go)\le d(o,gx)+d(gx,go)\le 2R$. Hence discreteness implies a properly discontinuous group action. Conversely, if the subgroup is not discrete, distinct $g_n$ tend to the identity. For a compact ball $K$ with $o$ in its interior, eventually $g_{n}o\in K$, so infinitely many $g_{n}K$ meet $K$, contradicting proper discontinuity. This proves the equivalence.

Matrices of determinant one over the Gaussian integers are closed under multiplication and inversion, with inverse $\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$. Quotienting by $\{\pm I\}$ gives the specified group of Möbius transformations. The Gaussian integers form a discrete set in the complex plane, so these matrices form a discrete subgroup of ${\mathrm{SL}}_2(\mathbb{C})$, and the finite quotient remains discrete. Thus this group acts properly discontinuously on ${\mathbb{H}}^3$.

Explicit representatives of all four types areThe first is $z\mapsto -1/z$ and has order two; the second is the nonidentity translation $z\mapsto z+1$. The third has positive real reciprocal eigenvalues $(3\pm \sqrt{5})/2$, giving translation along an axis without rotation. For the last, a nonreal trace excludes both unit-modulus and real eigenvalues. Its multiplier therefore has modulus different from one and a nonzero rotation angle: it is a loxodromic Möbius transformation.

A Kleinian group is a discrete subgroup of ${\mathrm{PSL}}_2(\mathbb{C})$, acting by Möbius transformations on the Riemann sphere, equivalently by orientation-preserving isometries of hyperbolic space ${\mathbb{H}}^3$.

Consider the classes ofThey generate a discrete subgroup because their matrix representatives have integer entries: a sequence of such matrices converging to $I$ or $-I$ is eventually constant. On the real projective line set $X=\{x:|x|>1\}\cup \{\infty \}$ and $Y=\{x:|x|<1\}$. For every nonzero integer $m$,The last inclusion follows from $|2mx+1|\ge 2|x|-1>|x|$ when $|x|>1$; also $B^m(\infty )=1/(2m)\in Y$. The ping-pong lemma therefore says that the subgroup is the free product $⟨A⟩\ast ⟨B⟩$. Both generators have infinite order, soa free group on two generators. The ping-pong inclusions ensure that no nonempty reduced word can act as the identity, which is the essential freeness argument.

The unit ball in Euclidean space with the displayed Riemannian metric models hyperbolic space of curvature minus one. Its ideal boundary is the unit sphere, and its geodesics are diameters or circle arcs orthogonal to that sphere. The metric is invariant under the conformal ball automorphisms.