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$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}$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 $.)

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    $\begingroup$ The space $\Gamma \backslash \mathrm{PSL}_2(\mathbb{C}$ is the oriented orthonormal frame bundle of $\Gamma \backslash \mathbb{H}^3$. The oriented orthonormal frame bundle fibers over the unit tangent bundle by picking the first vector: it is the space $\Gamma \backslash \mathrm{PSL}_2(\mathbb{C}) / M$, where $M \cong \mathrm{U}(1)$ is the compact part of the centralizer of the maximal torus in $\mathrm{PSL}_2(\mathbb{C})$. (1/2) $\endgroup$
    – Toffee
    Jan 1, 2022 at 13:39
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    $\begingroup$ Actually, the same interpretation is true for $\Gamma \backslash \mathrm{PSL}_2(\mathbb{R})$, since the unit tangent bundle is equal to the oriented frame bundle (picking a unit tangent vector automatically determines an oriented frame). (2/2) $\endgroup$
    – Toffee
    Jan 1, 2022 at 13:41
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    $\begingroup$ The tangent bundle of $\Sigma_g$ is not trivial, only that of the universal covering space is. The Euler class of the tangent bundle is the Euler characteristics 2-2g, which is not zero. Actually, the tangent bundle is not even flat (which is harder to prove, it follows from the Milnor-Wood inequality). $\endgroup$
    – ThiKu
    Jan 1, 2022 at 15:49

1 Answer 1

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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 $M$.

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  • $\begingroup$ Wow this is a great answer! ThiKu you and @Toffee seem to have some really great knowledge about bundles and geometry. Any chance either of you might know anything about my question on whether the complex points of quotient of compact groups is just the tangent bundle? Rephrasing in a way that might be more interesting: this says that the tangent bundle of a compact Riemannian homogeneous space always has a transitive action by the complexification of the isometry group. mathoverflow.net/questions/412741/… $\endgroup$ Jan 1, 2022 at 22:45

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