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A subgroup $G$ of a topological group is discrete when the identity has a neighbourhood containing no other element of $G$. Translation then isolates every group element. In a Hausdorff topological group, such a subgroup is closed: any accumulation at a point would make differences of distinct nearby group elements approach the identity. Consequently its intersection with a compact set is finite. For isometries of hyperbolic space, this converts compactness of the set of isometries moving one point a bounded distance into a properly discontinuous group action.

For example, $(-1/2,1/2)\cap \mathbb{Z}=\{0\}$ isolates the identity element of the additive discrete subgroup $\mathbb{Z}\subset \mathbb{R}$. Its induced subspace topology is the discrete topology, while the surrounding real numbers retain their usual topology; see closed subgroup for why this subgroup is closed but not open.

A surface with a complete Riemannian metric of constant Gaussian curvature $-1$. Its local model is the Poincaré half-plane model; the orientable case is a quotient by a free properly discontinuous group action of orientation-preserving isometries.

If a group has a properly discontinuous group action by isometries that is also a cocompact group action on a nonempty proper metric space that is a geodesic metric space, then the group is finitely generated and every orbit map from the word metric for any finite generating set of a group is a quasi-isometry. Finite stabilizers are permitted.

Choose a finite symmetric generating set $S$ for $\Gamma$, a base point $p$, and put $L={\max }_{s\in S}d(p,sp)$. If the word length of $\gamma$ is at most $m$, the triangle inequality and invariance of Riemannian distance under isometries give $d(p,\gamma p)\le Lm$.

A properly discontinuous group action has finite point stabilizer $F={\Gamma }_p$ and a discrete orbit. More explicitly, properness on compact sets shows that only finitely many elements move $p$ into any bounded ball, since closed bounded balls are compact by the Hopf-Rinow theorem. Thus there is $\rho >0$ such that distinct orbit points have distance more than $2\rho$: choose it using the finitely many orbit points within a fixed small ball of $p$, and translate the resulting separation by isometries. If the whole orbit is one point, the group itself is finite and the conclusion is immediate.

Let $N_m$ be the number of distinct orbit points represented by the word metric ball of radius $m$. Their radius-$\rho$ balls are disjoint, have equal positive volume $v_0=\mathrm{Vol}B(p,\rho )$, and lie inside $B(p,Lm+\rho )$. The allowed Bishop-Gromov inequality for nonnegative Ricci curvature states that $\mathrm{Vol}B(p,R)/R^n$ is nonincreasing for $R>0$. ThereforeEach orbit point has at most $|F|$ preimages inside the word ball, since its full fiber is a coset of $F$. For the growth function of a finitely generated group, this givesHence $\Gamma$ has polynomial growth of a group of degree at most $n$. The finite stabilizer factor makes the proof valid even when the properly discontinuous action is not free. This is polynomial growth of properly discontinuous isometry groups.

Choose a finite symmetric generating set $S$ for $\Gamma$, let $B_S(m)$ be its word metric ball, and fix $p\in M$. Set $D={\max }_{s\in S}d(p,sp)$. The triangle inequality and invariance of distance implyThe properly discontinuous group action has a finite point stabilizer $H={\Gamma }_p$. It also gives uniform separation of distinct orbit points. To see this, only finitely many group elements can send $p$ into a fixed compact ball about $p$, by proper discontinuity and the Hopf-Rinow theorem. The positive distances among this finite list, together with the radius of the ball for all other elements, have a positive lower bound $\delta$. Thus $d(\gamma p,{\gamma }^{\prime }p)\ge \delta$ whenever the centers differ.

Take $0<\epsilon <\delta /2$ and write $v=\mathrm{Vol}B(p,\epsilon )>0$. Balls of radius $\epsilon$ about distinct orbit points are disjoint and all have volume $v$, because the action is by isometries. Those corresponding to words of length at most $m$ lie in $B(p,Dm+\epsilon )$. Each orbit point arises from at most $|H|$ group elements in $B_S(m)$, soThe final inequality is the allowed Bishop-Gromov inequality for nonnegative Ricci curvature. If the entire group fixes $p$, proper discontinuity already makes it finite and the conclusion is immediate. Otherwise the preceding separation argument applies. ThereforeThis is polynomial growth of a group of degree at most $n$, proved by polynomial group growth from orbit packing. Neither freeness of the action nor compactness of the quotient was assumed.

The Lie group $PSL(2,\mathbb{C})$ is a second-countable space. For every point of its discrete subgroup, choose the first basic neighbourhood isolating that point; different elements require different such neighbourhoods. This injects $G$ into a countable set, proving countability.

A discrete subgroup is closed and meets compact sets in finitely many points. Indeed, distinct subgroup elements accumulating anywhere would give quotients converging to identity, contradicting an identity neighbourhood containing no other subgroup element. To apply this, use the positive Hermitian determinant-one model of hyperbolic space. At the base point $I$, the image under a determinant-one representative $g$ is $g{g}^{\ast }$ andThis distance formula follows by unitary diagonalization: eigenvalues are $e^t,e^{-t}$ and the axial hyperbolic displacement is $|t|$. Thus bounded displacement bounds all matrix entries. The determinant-one condition is closed, so the resulting set is compact; the quotient by $\{I,-I\}$ is compact too. Conjugation gives the same conclusion at any base point. These are the bounded displacement sets of hyperbolic isometries are compact facts.

Hence, for each $R$, only finitely many $g\in G$ satisfy $d(p,gp)\le R$. Any enumeration of an infinite group without repetitions therefore has $d(p,g_{n}p)\to \infty$. A finite discrete group is a harmless exception to the wording: it has no infinite enumeration of distinct elements, and proper discontinuity is immediate.

For a compact set $C$, put $R={\max }_{x\in C}d(p,x)$. If $gC\cap C\ne \varnothing$, choose $x,gx\in C$. The triangle inequality gives $d(p,gp)\le d(p,gx)+d(x,p)\le 2R$. Only finitely many such $g$ exist, proving properly discontinuous group action. Conversely, if the subgroup is not discrete, there are distinct $g_n\to e$. A closed ball about $p$ then meets its images under infinitely many $g_n$, violating this compact-set definition. Thus discreteness is equivalent to proper discontinuity. Here proper discontinuity allows finite point stabilizers; a stronger convention requiring freeness would exclude discrete elliptic subgroups and is not the equivalence intended.

A representative of an element of $G$ is a determinant-one matrix $\left(\begin{array}{cc}a& b\\ \overline{b}& \overline{a}\end{array}\right)$ over the Gaussian integers, considered up to its common sign. It acts by a Möbius transformation of the disk. Since $T(0)=b/\overline{a}$ and $|a{|}^2-|b{|}^2=1$, we haveOnly finitely many Gaussian-integer pairs $(a,b)$ can have $T(0)$ in a fixed compact subdisk. More generally, if $T$ carries some point of a compact set $K$ into $K$, and $K$ lies in the hyperbolic ball of radius $R$ about zero, the triangle inequality gives $\rho (0,T0)\le 2R$. Thus there are only finitely many such $T$. This proves the properly discontinuous group action, and in particular discreteness. The specified $A,B$ belong to $G$, so their subgroup $H$ is discrete as well.

The fixed-point equations reduce to $-i(z-1{)}^2=0$ for $A$ and $i(z+1{)}^2=0$ for $B$. Thus their unique fixed points are respectively $1$ and $-1$ on the ideal boundary. Neither transformation is the identity; their displayed determinant-one representatives have trace two. They are therefore parabolic Möbius transformations.

To find the Dirichlet domains, use the Möbius transformationIt takes the disk to the upper half-plane, and direct substitution givesFor $w=x+iy$ and $v$ in the upper half-plane, the hyperbolic distance obeys $\cosh \rho (w,v)=1+|w-v{|}^2/(2y\mathrm{Im}v)$. Since $C(A^{k}0)=i-2k$, the distance inequalities are equivalent toThese reduce to $kx+k^2\ge 0$ for all integers $k$. The cases $k=1,-1$ imply $-1\le x\le 1$, which also suffices for every other $k$. Therefore the cyclic Dirichlet domain becomes the strip $|\mathrm{Re}w|\le 1$.

The two vertical hyperbolic geodesics pull back to arcs joining $1$ to $i$ and $1$ to $-i$. In disk coordinates they are the circles $|z-(1+i)|=1$ and $|z-(1-i)|=1$, restricted to the disk. ThusConjugating by $z\mapsto -z$ gives $B$, so its two hyperbolic geodesics join $-1$ to $i$ and $-1$ to $-i$, andThe requested drawing and their intersection are shown below.

For any $z_0$, the orbit points satisfying $\rho (0,T{z}_0)\le \rho (0,z_0)$ form a nonempty finite set, by proper discontinuity and compactness of closed hyperbolic balls. Choose a closest one, $q=T{z}_0$. For every integer $k$, its minimality givesand the same inequality with $B$. Hence $q$ lies in the ideal hyperbolic quadrilateralIts ideal vertices are $1,i,-1,-i$.

For the quotient identification, it remains to justify that this covering region is a fundamental polygon. The four open corner caps excluded by $P$ are pairwise disjoint inside the disk. The transformation $A$ carries the exterior of the lower-right cap into the upper-right cap; $A^{-1}$ reverses the pairing. Similarly $B$ carries the exterior of the upper-left cap into the lower-left cap, and $B^{-1}$ reverses this. These assertions follow either from the upper-half-plane formulas or by mapping their boundary circles and testing zero: $A0=(1+i)/2$ and $B0=(-1-i)/2$.

Apply a reduced word in $A^{\pm 1},B^{\pm 1}$ from right to left to a point in the interior of $P$. The first letter puts it into its target cap. Every subsequent letter can do the same, since its inverse cap is distinct from the preceding target cap unless the word cancels. Thus the final point lies in the target cap of the leftmost letter, never in the interior of $P$. This ping-pong lemma argument proves freeness and disjointness of distinct translated interiors. Together with orbit coverage it makes $P$ a fundamental polygon. The corresponding argument on closed caps shows that boundary identifications are generated by the stated side pairings.

The generator $A$ pairs the side from $1$ to $-i$ with the side from $1$ to $i$, fixing the ideal endpoint $1$ and sending $-i$ to $i$. The generator $B$ pairs the side from $-1$ to $i$ with the side from $-1$ to $-i$, fixing $-1$ and sending $i$ to $-i$. There are three ideal-vertex classes: $\{1\}$, $\{-1\}$ and $\{i,-i\}$. The first two give parabolic ends from $A,B$; the third does too, since $AB$ is a nonidentity parabolic transformation fixing $i$. Adding one point at each end produces an oriented closed surface with one face, two edges and three vertices. Its Euler characteristic is $1-2+3=2$, so it is a sphere. Removing the three added points gives a thrice-punctured sphere, proving

A subgroup of the Möbius group ${\mathrm{PSL}}_2(\mathbb{C})$ is a discrete subgroup when the identity is isolated in its induced topology, equivalently every group element is isolated. We prove the compact-set criterion for a properly discontinuous group action: for every compact $K\subset {\mathbb{H}}^3$, only finitely many $g\in G$ satisfy $gK\cap K\ne \varnothing$.

Fix an origin $o$ and take $K\subseteq \overline{B}(o,R)$. If $x\in K$ and $gx\in K$, the triangle inequality and invariance of hyperbolic distance giveThe set of Möbius isometries with $d(o,go)\le 2R$ is compact. To see this explicitly, represent an isometry by a determinant-one complex matrix and use singular-value decomposition:The compact factors fix $o$, and the diagonal factor moves it hyperbolic distance $s$, as seen from the vertical dilation by $e^s$ in upper half-space. Thus the bounded-displacement set is the continuous image of $\mathrm{SU}(2)\times [0,2R]\times \mathrm{SU}(2)$, modulo central signs.

By the allowed closedness fact, its intersection with $G$ is a closed discrete subset of a compact space, hence finite. Otherwise an infinite sequence would accumulate in that intersection, contradicting isolation of its limit point. This proves every discrete Möbius subgroup acts properly discontinuously on hyperbolic three-space. Finite stabilizers are allowed; discreteness alone does not imply a free action.

Represent a transformation byMultiplication preserves this form and determinant, and the inverse has the same form with parameters $(\overline{u},-v)$. To check the congruence condition, complex conjugation is the identity modulo $2\mathbb{Z}[i]$. Thus $u+v\equiv 1$ says exactly that $M$ fixes the column $(1,1{)}^T$ modulo two. Products and inverses preserve this property. The identity belongs to the set, and changing the common sign of $(u,v)$ does not change either the transformation or its congruence. Hence these transformations form a group.

Because $|u|>|v|$, the denominator has no zero in the disc. Direct calculation givesThe inverse also maps the disc into itself, so $g$ is a disc automorphism, and these identities show that it preserves the hyperbolic metric $2|dz|/(1-|z{|}^2)$.

The action is properly discontinuous. If $C\subset \mathbb{D}$ is compact, put $R={\max }_{z\in C}\rho (0,z)$. If $gC\cap C\ne \varnothing$, choose $z,w\in C$ with $w=gz$. ThenSince $g0=v/\overline{u}$ and $|u{|}^2-|v{|}^2=1$, its distance bound gives $|v/u|\le \tanh R<1$ and therefore $|u{|}^2=1/(1-|v/u{|}^2)\le {\cosh }^{2}R$. There are only finitely many Gaussian integers $u,v$ with these bounds, so only finitely many $g$ satisfy $gC\cap C\ne \varnothing$. This proves the required properly discontinuous group action directly.

If $g0=0$, then $v=0$ and $u$ is one of the Gaussian units $1,-1,i,-i$. The congruence $u-1\in 2\mathbb{Z}[i]$ leaves only $u=\pm 1$. Both give the identity transformation. Thus the point stabilizer of zero is trivial.

The closed Dirichlet region centered at zero is the set of $z$ with $\rho (z,0)\le \rho (z,g0)$ for every $g$. For a nonidentity element, $v\ne 0$. WriteThe distance inequality is equivalent, by pseudohyperbolic distance, to $|z|\le |(z-q)/(1-\overline{q}z)|$. Squaring and subtracting yieldsSince $|z|<1$ and $|q{|}^2=N/(N+1)$, the bisector half-plane is exactlyFor $u=1+i$ or $1-i$ and $v=i$ or $-i$, all the group conditions hold and the products $uv$ are the four numbers $\pm 1\pm i$, with $N=1$. Their four inequalities forceConversely, for any group element $m^2+n^2=|uv{|}^2=N(N+1)<(N+1{)}^2$. As $m,n$ are integers, $|m|,|n|\le N$. Thus every point satisfying those four inequalities satisfies every other one:This proves, with no missing bisectors, the Gaussian-integer disc group with an ideal-square Dirichlet domain formulaEquivalently it is the part of the unit disc outside all four open circles of radius one centered at $1+i,1-i,-1+i,-1-i$. Its sides are the inward arcs of those circles. They meet the unit circle orthogonally, and their ideal vertices are $\boxed{1,i,-1,-i}$; these vertices themselves lie outside the open disc.

The origin has trivial point stabilizer and the group orbit is locally finite by the discontinuity proof, so this closed bisector intersection has translates covering the disc with disjoint interiors, as in the definition of a Dirichlet domain. Side identifications are allowed on its boundary.

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.

Interpret the printed congruence entrywise in the integer lattice:It then defines the usual level-two principal congruence subgroup of ${\mathrm{SL}}_2(\mathbb{Z})$, despite the PDF's ambient ${\mathrm{SL}}_2(\mathbb{R})$. A literal ideal congruence inside $\mathbb{R}$ would be vacuous because $2\mathbb{R}=\mathbb{R}$, and would make the asserted conclusion false. The integer-lattice interpretation is essential.

Pass to $\overline{\Gamma (2)}=\Gamma (2)/\{\pm I\}$, which has exactly the same action. Reduction modulo two maps the modular group onto ${\mathrm{SL}}_2({\mathbb{F}}_2)$, a group of order six: the reductions of $S$ and $T$ generate it. Its kernel is $\overline{\Gamma (2)}$, so the index of a subgroup is six. The standard fundamental domain of the modular group has hyperbolic area $\pi /3$, and hence the quotient has hyperbolic area $2\pi$.

There are no nonidentity elliptic Möbius transformations in $\overline{\Gamma (2)}$. An integral matrix representing an elliptic Möbius transformation has trace $0$ or $\pm 1$. Here the trace is even, excluding $\pm 1$; trace zero would give $d=-a$ and $1=-a^2-bc\equiv -1(\mathrm{mod}4)$, impossible. Thus the effective action is a free properly discontinuous group action, and the quotient is a Riemann surface.

A cusp of a modular group is represented by a rational boundary point. Their orbits correspond towhich has $6/2=3$ elements. They are represented by $\infty ,0,1$, or by the three nonzero parity vectors of a primitive numerator-denominator pair. Each width of a cusp is two. A union of six copies of the standard fundamental domain of the modular group gives a fundamental region for this subgroup. Removing small horocycle neighbourhoods of its cusps leaves a compact core. Adding one point at each cusp of a modular group, using the local parameter $e^{\pi iz}$ after moving that cusp to infinity, gives a compact Riemann surface $\overline{X}$.

For a finite-area hyperbolic surface of genus $g$ with $r$ cusps, the Gauss-Bonnet theorem gives area $2\pi (2g-2+r)$. Thus $2\pi =2\pi (2g-2+3)$ and $g=0$. A compact genus-zero Riemann surface is the Riemann sphere, by the uniformization theorem. A Möbius transformation sends the three added points to $0,1,\infty$. Restricting it gives

The Milnor–Švarc lemma applies to an isometric properly discontinuous group action that is also a cocompact group action on a nonempty proper metric space that is a geodesic metric space. It concludes that the group is finitely generated and, for a basepoint $x$ and any finite generating set of a group $S$, the orbit mapHere $d_S$ is the word metric, and a quasi-isometry means a map satisfying two-sided coarse distance bounds with a uniformly coarsely dense image. A free action is not required: finite stabilizers are compatible with proper discontinuity. The properness and cocompactness hypotheses ensure that only finitely many group elements move a chosen basepoint by the bounded amount used to generate the group. Replacing a finite generating set by another does not change the conclusion, by equivalence of finite word metrics.

The bounded sequence $(x_n)$ lies in a compact set because $X$ is a proper metric space. After taking a subsequence, $x_n\to x$. Since $d(x_n,{\varphi }_n{x}_n)\to 0$, both $x_n$ and ${\varphi }_n{x}_n$ eventually lie in one compact neighborhood $L$ of $x$. A properly discontinuous group action has only finitely many $g\in \Gamma$ with $gL\cap L\ne \varnothing$, so some conjugate $\psi$ occurs as ${\varphi }_n$ along an infinite subsequence. Continuity then givesThus $\psi$ fixes $x$. Since $\psi =\gamma \varphi {\gamma }^{-1}$ for some $\gamma$, the original element $\varphi$ fixes ${\gamma }^{-1}x$.

A properly discontinuous group action by isometries of a finitely generated group on a complete $n$-dimensional Riemannian manifold of nonnegative Ricci curvature yields the displayed group growth bound. Distinct orbit points are uniformly separated, while a word of length $m$ moves a basepoint by at most $Dm$. Pack disjoint equal-volume small balls about the orbit points into $B(p,Dm+\epsilon )$ and apply the Bishop-Gromov inequality. A finite point stabilizer contributes its order to the bound; freeness is unnecessary.