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An isometry of hyperbolic space with positive attained translation length has an invariant complete geodesic joining its two ideal fixed points. This is its axis. It translates along the axis by its translation length; in three dimensions it may additionally rotate around it. In upper half-space, $(z,t)\mapsto (kz,|k|t)$ with $|k|\ne 1$ has the vertical axis through zero and length $|\log |k||$. Every connecting path has length at least the change in $\log t$, and a vertical segment attains the bound.

In the positive Hermitian determinant-one model of three-dimensional hyperbolic space, the base point is $I$ and a matrix $g\in SL(2,\mathbb{C})$ sends it to $g{g}^{\ast }$. The distance obeys $2\cosh d(I,g{g}^{\ast })=\mathrm{tr}(g{g}^{\ast })$. Bounded displacement therefore bounds every matrix entry, and the determinant-one condition gives a closed bounded subset of a finite-dimensional real vector space. Passing to $PSL(2,\mathbb{C})$ preserves compactness. Conjugating moves any chosen base point to $I$. A discrete subgroup meets each such compact set in finitely many elements, proving escape of distinct elements and proper discontinuity, including finite stabilizers.

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 nontrivial finite-order elliptic Möbius transformation fixes an axis in hyperbolic space. The quotient metric has cone angle $2\pi /m$ along that axis for a stabilizer of order $m$, so it is locally an orbifold metric rather than a smooth hyperbolic-manifold metric. The underlying topological quotient may nevertheless be a manifold.

A Möbius transformation of the ideal two-sphere extends uniquely to an orientation-preserving isometry of three-dimensional hyperbolic space. In upper half-space, translations, complex similarities, and $z\mapsto -1/z$ extend respectively to $(z,t)\mapsto (z+b,t)$, $(\lambda z,|\lambda |t)$, and $(-\overline{z},t)/(|z{|}^2+t^2)$. Each preserves $(|dz{|}^2+d{t}^2)/t^2$; composing these extensions constructs the general case. Ball-origin stabilizers are ordinary rotations, represented on the boundary by $SU(2)$ modulo its central signs.

In static time times round three-dimensional hyperbolic space of curvature radius $R$, a geodesic sphere has area $4\pi R^2{\sinh }^2(\xi /R)$. The Gauss law therefore gives the displayed radial field. It approaches the flat-space inverse-square field near the charge and falls exponentially at large geodesic distance. The curvature changes the area of a sphere, not the conserved flux.

A complete hyperbolic three-manifold is a complete Riemannian manifold of dimension three with constant sectional curvature $-1$. Its universal cover is hyperbolic space ${\mathbb{H}}^3$ and its deck group acts freely and discretely by Riemannian isometries. In the orientable case it is a torsion-free Kleinian group.

A discrete subgroup of ${\mathrm{PSL}}_2(\mathbb{C})$, acting by Möbius transformations on the Riemann sphere or orientation-preserving isometries of hyperbolic space ${\mathbb{H}}^3$. A group may be Kleinian even when some elements fix points; freeness and absence of torsion are additional conditions.

For a discrete isometry group of hyperbolic space, its limit set is the boundary accumulation set of one interior orbit. All interior base points give the same set: corresponding orbit points stay a fixed hyperbolic distance apart, and bounded-distance sequences approaching the boundary have the same endpoint. This definition does not require finite generation.

A Möbius transformation with two distinct fixed points is conjugate to $z\mapsto az$. It is loxodromic when $|a|\ne 1$; on three-dimensional hyperbolic space it translates along the axis joining its fixed points and rotates around that axis by $\arg a$. The positive-real multiplier case has no rotation and is also called hyperbolic. When listing elliptic, parabolic, hyperbolic and loxodromic as separate types, loxodromic refers to the remaining case with nonzero rotation. A determinant-one matrix with nonreal trace has reciprocal eigenvalues that are neither real nor of modulus one, so it supplies this latter case.

For a nonempty finite set $\{p_1,\dots ,p_m\}$ in finite-dimensional hyperbolic space, the continuous function $R(x)={\max }_{j}d(x,p_j)$ attains its minimum. Indeed, a bounded candidate radius confines all possible improving centers to a closed bounded ball about $p_1$. Such a ball is compact in the Poincare ball model, since its points remain away from the ideal boundary. Minimizing $R$ on it supplies a closed ball of smallest enclosing radius. More generally this argument works for any nonempty bounded set in a proper metric space when the supremum-distance function is finite.

For complete finite-volume hyperbolic space manifolds of dimension at least three, an isomorphism of fundamental groups is induced, up to the usual basepoint-path choice, by a unique hyperbolic isometry. The finite-volume noncompact case includes the Prasad extension. Thus complete hyperbolic metrics on a fixed finite-volume three-manifold have no deformation moduli, unlike metrics on closed hyperbolic surfaces.

For a connected smooth Riemannian manifold $(M,g)$ without boundary, let $I(M)$ be the group of all global diffeomorphisms $f$ with $f^{\ast }g=g$, endowed with its compact-open topology. Equivalently these maps preserve the intrinsic distance; the Myers-Steenrod theorem supplies smoothness of distance-preserving bijections and makes $I(M)$ a finite-dimensional Lie group acting smoothly. The group is metric-dependent: a topological manifold can carry a very symmetric metric or one with no continuous symmetries. Connectedness matters here. With infinitely many identical disconnected components, permutations of those components can produce a group which is not a finite-dimensional Lie group.

An isometry is determined by its value and derivative at one point. To prove this, suppose $f(p)=p$ and $d{f}_p=I$. Preservation of the Levi-Civita connection giveswhere the exponential map is defined locally. Thus $f$ is the identity in a normal neighborhood of $p$. The set where its value and derivative agree with the identity is both closed and, by the same argument, open. Connectedness makes it all of $M$. No completeness is needed for this determination argument.

The stabilizer $I(M{)}_p$ consequently acts faithfully and orthogonally on $T_{p}M$, so embeds in $O(n)$. More generally an isometry takes a chosen orthonormal frame at $p$ to one at its image point, providing an injective map into the orthonormal frame bundle. It is a smooth embedding, and the full isometry action is proper. The orbit-stabilizer dimension formula therefore givesIsotropy is compact. If $M$ is compact then $I(M)$ itself is compact, by the same frame description or Arzela-Ascoli applied to isometries and their inverses. Properness also means that the quotient by isometries has well-controlled local orbit structure, even though orbit dimensions may vary.

The infinitesimal equation is the Killing equation. Differentiating $f_t^{\ast }g=g$ at zero givesConversely a vector field satisfying this equation has local flows preserving the metric. A global one-parameter subgroup requires the field to be complete. On a geodesically complete manifold every Killing field is complete: its length is constant along each of its own trajectories, so a finite-time trajectory has bounded speed and stays in a closed bounded ball. The Hopf-Rinow theorem makes that ball compact, and the ordinary differential equation can then be continued. The caveat is real: on the Euclidean interval $(0,1)$, the field ${\partial }_x$ is Killing but its translations are not globally defined for all time. This illustrates Killing fields on incomplete manifolds need not generate global isometries.

The Killing transport identities determine a Killing field from $X(p)$ and the skew-adjoint map $(\nabla X{)}_p$. Along a geodesic the field is a Jacobi field, so uniqueness for its second-order differential equation propagates these initial data. They have $n+n(n-1)/2$ components, reproducing the dimension bound infinitesimally. The important distinction on an incomplete manifold is between all local Killing solutions and the Lie algebra of complete global isometry flows.

In dimension three the bound is six, and rotational isotropy gives stronger information. The Lie subalgebras of $\mathfrak{so}(3)$ have dimensions zero, one or three: identifying the bracket with the cross product, a two-plane is not closed under brackets. Suppose $\dim I(M)=5$. Since an orbit has dimension at most three, its stabilizer would have dimension at least two, and hence exactly three. Its connected isotropy acts as all of $SO(3)$ on the tangent space. But the orbit tangent space is invariant under that action and would have dimension $5-3=2$, impossible because the standard rotational representation is irreducible. Therefore dimension five cannot occur.

If the dimension is six, each orbit has dimension three and each stabilizer dimension three. All orbits are open, so connectedness makes the action transitive. At each point the isotropy is transitive on tangent two-planes, forcing sectional curvature to be independent of the plane; transitivity makes its value independent of the point. Thus a maximally symmetric connected three-manifold has constant sectional curvature. Homogeneous Riemannian manifolds are complete, so its simply connected cover is the round sphere, Euclidean space or hyperbolic space after normalization. Conversely these three simply connected metrics attain dimension six:Here $O^{+}(3,1)$ preserves the chosen sheet of the hyperboloid; it includes orientation-reversing hyperbolic isometries. Constant curvature alone does not imply that a quotient retains six-dimensional global symmetry.

Dimension four likewise forces transitivity: stabilizer dimension three would give a one-dimensional invariant orbit tangent space, again impossible for full rotational isotropy. The stabilizer thus has dimension one and the orbits are three-dimensional. Typical four-dimensional isometry groups occur for $S^2\times \mathbb{R}$ and ${\mathbb{H}}^2\times \mathbb{R}$, where the factor groups have dimensions three and one. The usual Nil metric has three translation symmetries and one rotational isotropy symmetry. In centred Heisenberg coordinates it iswhich exhibits the horizontal rotations. Its distinct vertical and horizontal Ricci eigenvalues bound continuous isotropy by that rotation group, so its full isometry dimension is four. Standard metrics on $\widetilde{S{L}_2(\mathbb{R})}$ and nonround Berger spheres also have dimension four. The standard Sol geometry has a three-dimensional translation group and only finite isotropy, giving dimension three. These examples show how a preferred vertical direction reduces rotational symmetry.

The isometry dimensions in dimension three are precisely $0,1,2,3,4,6$, all realizable. A compact hyperbolic three-manifold gives zero; the product of a circle and a closed hyperbolic surface gives one. For dimension two take $T^2\times (0,1)$ withIts scalar curvature is the constant $-14$, but its Ricci eigenvalues are the distinct constants $-3,-6,-5$ in the two horizontal directions and the vertical direction. Hence connected isotropy is trivial and possible isometries preserve these line distributions. A local isometry preserving them has $t^{\prime }=t+c$ and would require constant dilations $e^{-c}$ and $e^{-2c}$ of the two circle coordinates. A global circle diffeomorphism with such a constant dilation has degree $\pm 1$, forcing $c=0$. The identity component is therefore exactly the two circle translations. A flat three-torus gives dimension three; $S^2\times \mathbb{R}$ gives four; a simply connected constant-curvature model gives six. This construction distinguishes local homogeneity from the smaller global isometry group.

For quotients, let a complete simply connected Riemannian manifold $X$ cover $M=X/\Gamma$. Every isometry of $M$ lifts to an isometry of $X$ which normalizes the deck transformation group. Conversely, every normalizing isometry descends. Two lifts differ by a deck transformation, giving the isometry group of a Riemannian quotient formulaThus one should not replace the isometry group of a quotient by that of its model geometry. For a flat torus, for example,The second factor is finite, whereas translations give the three-dimensional identity component.

There is a useful curvature test for compact manifolds. Contracting and differentiating the Killing equation gives ${\nabla }^{\ast }\nabla X=\mathrm{Ric}(X)$, with the nonnegative rough-Laplacian convention. Integration yields the integrated Bochner identity for Killing fieldsIf Ricci curvature is negative definite, no nonzero Killing field exists. The compact isometry group then has dimension zero and is finite. This explains why closed hyperbolic manifolds have finite global symmetry, despite the six-dimensional isometry group of their universal cover. Compactness is essential to this conclusion.

For a compact flat Riemannian manifold, the same identity makes every Killing field parallel. Parallel fields are exactly the vectors fixed by linear holonomy. ThereforeApplied to the six flat types of Question 1, this gives dimensions three for the torus, one for each of the four nontrivial cyclic types, and zero for the Hantzsche-Wendt manifold. These are identity-component dimensions; disconnected finite symmetries can still be present. Local curvature symmetry, global deck-group constraints, and completeness thus play different roles in determining the isometry group.

Let $\xi$ denote geodesic distance from the charge; $R$ is the fixed curvature radius, not the radial variable. In round three-dimensional hyperbolic space, a sphere has areaUsing the convention that electric flux is $Q/{ϵ}_0$, Gauss's law gives the radial orthonormal electric fieldIts magnitude is $|Q|/A{ϵ}_0$, and a negative charge reverses the direction. In rationalized natural units set ${ϵ}_0=1$; Gaussian conventions instead use flux $4\pi Q$.

For $\xi \ll R$ this approaches $Q/(4\pi {ϵ}_0{\xi }^2)$. For $\xi \gg R$, $E\sim Q{e}^{-2\xi /R}/(\pi {ϵ}_0{R}^2)$. This hyperbolic-space Coulomb field falls rapidly because the sphere area grows exponentially; conserved electric flux is unchanged. A potential vanishing at infinity is $V=Q[\coth (\xi /R)-1]/(4\pi {ϵ}_{0}R)$.

The original PDF actually prints $\sin \theta$ rather than ${\sin }^2\theta$ in the angular metric tensor. The round angular factor used above is required by its description as hyperbolic space. If the printed angular metric tensor is taken literally, the sphere-area coefficient is insteadand a radial Gauss-law solution has $E=Q/[{ϵ}_0{\Omega }_{\ast }{R}^2{\sinh }^2(\xi /R)]$. That angular geometry is not a smooth round sphere, so the literal metric tensor does not describe the asserted homogeneous hyperbolic space. Both readings retain the same radial area-growth factor; the geometric repair is explicit.

We use $[a,b]=a^{-1}{b}^{-1}ab$. To obtain a Wirtinger presentation, choose an oriented regular knot diagram, cut its strands at undercrossings, and give each resulting arc a positively oriented meridian of a knot. The complement of the arcs above the projection plane has a free fundamental group on these meridians. At a crossing, pushing an understrand meridian past the overstrand conjugates it by the overstrand meridian. Thus, if $a$ is the overstrand generator and $b,c$ are the incoming and outgoing understrand generators, the relation is $c=a^{ϵ}b{a}^{-ϵ}$, where $ϵ$ is determined by the crossing orientation. Applying the Seifert-van Kampen theorem while adjoining the crossing neighborhoods gives one such relation per crossing. The remaining three-dimensional region adds no fundamental-group relation. For a connected diagram one crossing relation is redundant. This derives a group presentation from any tame knot diagram, rather than just specifying a recipe without its topological justification.

For the Borromean rings, choose three meridians $x,y,z$ and call the other three arc meridians $x^{\prime },y^{\prime },z^{\prime }$. Base paths and orientations can be chosen so the six crossing relations readEliminate the primed generators with the first row. The second row says respectively that $x$ commutes with $(z^{\prime }{)}^{-1}z=[y^{-1},z]$, that $y$ commutes with $(x^{\prime }{)}^{-1}x=[z^{-1},x]$, and that $z$ commutes with $(y^{\prime }{)}^{-1}y=[x^{-1},y]$. The last relation is redundant, just as one crossing relation is redundant in the Wirtinger presentation. Alternatively the Hall-Witt identity, after conjugating its three factors, shows that any two of these cyclic relations imply the third. Hence the Borromean link group isBoth relators have zero exponent sum in each generator. Passing from the free group to its abelianization therefore imposes no relations beyond commutation, andEquivalently the three linking-number maps to $\mathbb{Z}$ independently detect the three meridians.

Here is an explicit representation, followed by a faithfulness argument:Each determinant is one. Direct multiplication verifies the two relators. For example $[Y^{-1},Z]=\left(\begin{array}{cc}-1& 2i\\ 0& -1\end{array}\right)$, which commutes with $X$. Merely checking those relators would not prove faithfulness, so we identify the image's complete group presentation.

Write $B=PS{L}_2(\mathbb{Z}[i])$, the Gaussian Bianchi group, and use its four generators $T,U,L,A$ from the allowed presentation. In $B$ the three proposed generators areThe last equality is in $PS{L}_2$, so an overall minus sign in matrix multiplication is immaterial. Reduce matrices modulo $4$ in $\mathbb{Z}[i]$, and then quotient by $\{\pm I\}$. The finite image $Q$ has order $1536$: the residue field is ${\mathbb{F}}_2$, $S{L}_2({\mathbb{F}}_2)$ has order six, and each of the three subsequent local-ring layers has a kernel of order $2^3$, before the final quotient by $\{\pm I\}$. Elementary elimination over this local ring shows that the images of $T,U,A$ generate it. The subgroup $\bar{H}=⟨\bar{X},\bar{Y},\bar{Z}⟩$ has order $64$, as multiplication in this finite ring verifies. Let $K$ be its inverse image in $B$; then $[B:K]=24$.

For completeness the following gives the coset computation needed to check the subgroup presentation explicitly. Cosets are right cosets $Kr$; columns record right multiplication by $T,U,L,A$. Representatives are words in the allowed generators.Apply the Reidemeister–Schreier theorem: for a transition $j{⟶}^{g}k$, use generator $r_{j}g{r}_k^{-1}$, omitting the 23 tree transitions. Rewrite each of the eight allowed ambient relators starting at each of the 24 rows. There are 73 remaining Schreier generators. Adjoin the names $x=T$, $y=A{U}^{-2}{A}^{-1}$, $z=A{T}^{-2}{A}^{-1}L{U}^{-1}$ and eliminate the Schreier generators using relations in which they occur once. The remaining relations, with an uppercase letter denoting the inverse of the corresponding lowercase one, are the wordsThis table and the rewriting rule determine every relation without an appeal to a guessed presentation. In this notation cyclically reduced versions of the two desired relators are $r=ZYzXZyzYxy$ and $q$. Elementary substitutions of relator segments, with cyclic reduction at each step, giveusing $r,q$; conversely $r⟶p⟶1$ using $p,q$. These substitutions show equality of the two normal closures. For example the first substitution replaces $xyX$ in the cyclic rotation of $p$ by $zxZyzXZ$, as allowed by $q$, and then the initial $ZYz$ is replaced by $YXyZYzx$ using $r$. Thus $K$ has exactly the presentation of $G$, with $x,y,z$ represented by $X,Y,Z$. In particular these three matrices generate $K$, and

To identify the actual link complement as a hyperbolic three-manifold, rather than only on a manifold with an isomorphic group, cut along the projection-plane regions, making the usual small crossing detours above and below. The octahedral projection graph has eight triangular regions. The upper and lower pieces are ideal octahedra; corresponding faces are matched with one-third-turn corner permutations, with opposite directions on adjacent checkerboard faces. Reversing these cuts reconstructs the link complement. Give both pieces the regular ideal hyperbolic space octahedron metric. All triangular face maps are isometries and each edge cycle has four dihedral angles $\pi /2$, so the metric has no edge singularity. Equal horospherical truncations match in Euclidean squares and assemble into three flat tori, giving complete torus cusps. The compact truncated remainder and these cusps show completeness and finite volume. Thus the Borromean complement is a complete finite-volume hyperbolic three-manifold. This also supplies a geometric discrete faithful holonomy, independently of the subgroup calculation.

Extend the Möbius transformations to isometries of three-dimensional hyperbolic space. For any interior point $o$, the limit set of a Kleinian group is the set of boundary accumulation points of $Go$. Its definition is independent of $o$: two interior orbits stay at a bounded hyperbolic distance, so have the same boundary limits. The Kleinian limit set is closed and $G$-invariant. Iterating a loxodromic Möbius transformation in the positive and negative directions shows that its two fixed points belong to the Kleinian limit set.

The two fixed-point sets in this problem cannot have exactly one common point. Otherwise conjugate that point to infinity and the other fixed point of the first transformation to zero. The transformations then have the forms $f(z)=\alpha z$, with $|\alpha |>1$, and $g(z)=\beta z+b$, with $b\ne 0$. But $f^{-n}g{f}^n(z)=\beta z+b{\alpha }^{-n}$ is a sequence of distinct group elements converging in the Möbius group. Taking quotients of consecutive elements gives nonidentity elements converging to the identity, contrary to discrete subgroup. Consequently the Kleinian limit set contains at least three points.

Let $\xi$ be any point of the Kleinian limit set, and choose distinct $g_n$ with $g_{n}o\to \xi$. The rank-one convergence of divergent Möbius transformations supplies a subsequence converging to $\xi$ uniformly away from one exceptional boundary point $\eta$. To see the lemma directly, normalize determinant-one representing matrices by their norms. The norms tend to infinity, and a subsequential limit has rank one: its kernel is $\eta$ and its image is the attracting point. In the positive-Hermitian-matrix model of hyperbolic space, $g_{n}o$ has this same image limit, so the attracting point is indeed $\xi$.

Choose distinct ${\zeta }_1,{\zeta }_2$ in the Kleinian limit set other than $\eta$. Then $g_n{\zeta }_1$ and $g_n{\zeta }_2$ both tend to $\xi$, belong to the Kleinian limit set, and remain distinct because $g_n$ is injective. If $\xi$ were isolated, both would eventually equal $\xi$, an impossibility. The limit set has no isolated points.

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.