Complete curvature minus one three-manifolds are Kleinian quotients

Choose $p\in M$ and an oriented orthonormal identification $T_{p}M\cong {\mathbb{R}}^3$. By [F1], the pulled-back metric on this whole tangent space is complete and covers $M$. In polar coordinates $(r,\omega )$, let $T$ be the unit radial velocity and extend $v\perp \omega$ to a parallel field $E$ along that radial geodesic by [F2]. Parallel transport preserves inner products, so $E\perp T$ and $\mathrm{\mid }E\mathrm{\mid }=\mathrm{\mid }v\mathrm{\mid }$. The explicit field $J(r)=\sinh rE(r)$ has $J(0)=0$, $D_{r}J(0)=v$, and $D_r^{2}J=\sinh rE$ while $R(J,T)T=-J$; hence $D_r^{2}J+R(J,T)T=0$. Jacobi initial-data uniqueness and the exponential differential formula in [F2] identify this field with the angular variation. Its length is therefore $\sinh r\mathrm{\mid }v\mathrm{\mid }$. Its inner products are ${\sinh }^{2}r$ times the sphere metric because parallel transport preserves inner products. Radial velocity has length one and is orthogonal to these normal fields. Thus the pulled-back metric is exactly $d{r}^2+{\sinh }^{2}rd{\omega }^2$. The map $(r,\omega )\mapsto (\cosh r,\sinh r\omega )$ identifies it with the future unit hyperboloid of Minkowski space; differentiation gives that same metric, and it extends smoothly and bijectively at the origin. The hyperboloid is the upper half-space model via $y=1\mathrm{/}(x_0-x_3)$ and $z=(x_1+i{x}_2)\mathrm{/}(x_0-x_3)$, whose inverse is obtained by solving these equations and $x_0^2-\mathrm{\mid }{x}^{\mathrm{\prime }}{\mathrm{\mid }}^2=1$; direct differentiation gives the upper half-space metric. Thus [F1]'s source is isometric to ${\mathbb{H}}^3$.

We verify the full orientation-preserving isometry group. Identify the hyperboloid with the positive Hermitian determinant-one matrices by $x\mapsto x_{0}I+x_1{\sigma }_1+x_2{\sigma }_2+x_3{\sigma }_3$, where the Pauli matrices satisfy ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{\sum }_k{\epsilon }_{ijk}{\sigma }_k$. The determinant is $x_0^2-\mathrm{\mid }{x}^{\mathrm{\prime }}{\mathrm{\mid }}^2$. The action $X\mapsto AX{A}^{\ast }$ of $A\in {\mathrm{S}\mathrm{L}}_2(\mathbb{C})$ preserves the future hyperboloid and its metric; its kernel is $\{I,-I\}$. It is transitive, because every positive Hermitian determinant-one matrix has a positive Hermitian square root of determinant one, obtained by diagonalizing a two-by-two Hermitian matrix. The stabilizer of $I$ is $\mathrm{S}\mathrm{U}(2)$. On the tangent space of traceless Hermitian matrices it realizes every rotation. Indeed, for a real unit vector $n$, the matrix $U=\cos (\theta \mathrm{/}2)I+i\sin (\theta \mathrm{/}2)n\cdot \sigma$ is unitary of determinant one; the displayed Pauli identity shows its conjugation gives the rotation about $n$ through angle $-\theta$. Every three-dimensional proper orthogonal matrix has an axis and is a planar rotation on its orthogonal complement, so these realize all of $\mathrm{S}\mathrm{O}(3)$. An isometry fixing a point and its tangent frame fixes every geodesic from that point, hence every point by the polar description in step 1.1. Therefore every orientation-preserving model isometry comes from this projective matrix group. Its boundary action on rank-one Hermitian matrices $v{v}^{\ast }$ is $[v]\mapsto [Av]$, the usual fractional linear action on ${\mathbb{C}\mathbb{P}}^1$. This is the Möbius action in [F3].

The deck group of [F1]'s universal cover acts by isometries of the pulled-back metric and preserves its pulled-back orientation. It is therefore a subgroup of the group in step 2.1. It acts freely, since a deck transformation fixing one point fixes every lifted path from that point, hence the connected cover. It is discrete and properly discontinuous: an evenly covered neighborhood has disjoint lifted neighborhoods and a deck transformation intersecting one with itself must be the identity. Such transformations form an isolated neighborhood of the identity in the matrix topology by continuity of evaluation at a point. The group is torsion-free: a finite-order model isometry has a fixed point. In the hyperboloid model, sum the finitely many orbit vectors of any point and normalize their future timelike sum; this is a fixed point. Finally, universal covering path lifting identifies the deck group with ${\pi }_1(M)$ and its orbit quotient with $M$. Under this identification the metric and orientation descend exactly to the given ones, proving the assertion.