$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}$Let $ \Sigma_g $ be a compact orientable surface of genus $ g $. Let the subgroup $ \pi_1(\Sigma) $ of $ \SL_2(\mathbb{R}) $ be a Fuchsian representation of the fundamental group. Then $ \Sigma_g $ admits a hyperbolic structure
$$
\pi_1(\Sigma_g)\backslash \SL_2(\mathbb{R})/\SO_2(\mathbb{R})
$$
Moreover, the 3-dimensional coset space
$$
\pi_1(\Sigma_g)\backslash SL_2(\mathbb{R})
$$
is a locally Riemannian homogeneous space admitting $ \widetilde{\SL_2} $ geometry and it is isometric to the unit tangent bundle of $ \Sigma_g $.
Consider a similar situation for 3-manifolds. Let $ M $ be a hyperbolic 3-manifold
$$
M \cong \pi_1(M) \backslash \SL_2(\mathbb{C})/\SU_2
$$
where $ \pi_1(M) $ a Kleinian representation of the fundamental group of $ M $.
Then what can we say about the six-dimensional manifold
$$
\pi_1(M) \backslash \SL_2(\mathbb{C})
$$
My first thought was that maybe that we could identify $ \pi_1(M) \backslash \SL_2(\mathbb{C}) $ with the space given by compactifying each fiber of the tangent space of $ M $ into a 3-sphere. Upon further reflection, though, that doesn't seem promising. $ M $ is isometrically covered by hyperbolic 3 space, which is contractible. And the tangent bundle over a contractible space is always trivial. So by covering the tangent bundle of $ M $ is also trivial. So I'm worried that taking one point compactifications would just give me a trivial bundle (cartesian product of $ M $ with the three sphere) which doesn't seem quite right. On the other hand the tangent bundle to $ \Sigma_g $ is trivial but the unit tangent bundle isn't trivial. So maybe stuff made from trivial bundles isn't always trivial? Anyway, not sure if I'm missing something obvious but that's my question.
Basically, For a hyperbolic 3-fold $ M $ can we say something interesting about the relationship between $ \pi_1(M) \backslash \SL_2(\mathbb{C}) $ and the tangent bundle of $ M $? (As in the two dimensional case where for a hyperbolic 2-fold $ \Sigma_g $ we have that $ \pi_1(\Sigma_g) \backslash \SL_2(\mathbb{R}) $ is isometric to the unit tangent bundle of $ \Sigma_g $.)
Best Answer
The (orientation-preserving) isometry group $G=PSL(2,{\bf C})$ acts on the bundle of (positively oriented) orthonormal frames on ${\bf H}^3$.
An ad hoc argument using the Iwasawa decomposition $G=KAN$ shows that this action is transitive. Indeed, considering the upper half space model and fixing the standard frame in $(0,0,1)$ you can first use the action of $K=PSU(2)$ to move that frame to any other frame at $(0,0,1)$. Then you can use the action of $A$ (the diagonal matrices) to move it to any frame at any $(0,0,z)$. Finally you can use the action of $N$ (the upper triangular matrices acting as translations) to move that frame to any other frame at any $(x,y,z)$.
This shows transitivity of the action and hence that $G$ bijects to the bundle of positively oriented, orthonormal frames on ${\bf H}^3$. (Injectivity is a standard argument.)
Dividing out $\Gamma=\pi_1M$ you get that $\Gamma\backslash G$ bijects to the bundle of positively oriented, orthonormal frames on $M$.