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Let $\mathbb{H}$ be hyperbolic plane, $\Gamma$ is a discrete subgroup of $PSL_2(\mathbb{R}$) so that $\Gamma \backslash \mathbb{H}$ is a compact hyperbolic surface. Maybe it will be very simple to you but I am very confused when I try to construct a homeomorphism $\phi: \Gamma \backslash T_1 \mathbb{H} \longrightarrow T_1(\Gamma \backslash \mathbb{H})$

Suppose $\Gamma (z,v)$ is an element of $\Gamma \backslash T_1 \mathbb{H}$, so what is $\phi (\Gamma (z,v))$? Actually I read in some book and they always accept this fact obviously, but for me it's not obvious.

I'm so sorry if it is not a question of researching.

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  • $\begingroup$ $T_1 \mathbb{H}$ is unit tangent bundle of hyperbolic plane. An element of $PSL(2,\mathbb{R})$ is a orientation-preserving isometry of $\mathbb{H}$, we can express it as a map $z\longmapsto \dfrac{az+b}{cz+d}$, where $ad-bc=1$ $\endgroup$ – Skid Row Jun 14 '17 at 18:34
  • $\begingroup$ Aren't the elements of $\Gamma \setminus T_1 \mathbb{H}$ of the form $\{ (\gamma(p), v \gamma'(p) ) \ \big\vert \ \gamma \in \Gamma\}$ where $p \in \mathbb{H}, v \in \mathbb{C}$ and $v \gamma'(p) = \lim_{h \to 0} \frac{\gamma(p+vh)-\gamma(p)}{h}$ ? And $T_1 (\Gamma \setminus \mathbb{H})$ would be the same $\endgroup$ – reuns Jun 14 '17 at 18:54
  • $\begingroup$ @reuns: Yes, it has that form. But I don't think $T_1(\Gamma \backslash \mathbb{H})$ would be the same, because it is not a quotient space. $\endgroup$ – Skid Row Jun 14 '17 at 18:55
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    $\begingroup$ Why not ? Using fundamental domains, it seems $T_1 (\Gamma \setminus \mathbb{H})$ has the same points/topology/metric $\endgroup$ – reuns Jun 14 '17 at 18:59
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    $\begingroup$ Take an open $U \subset \mathbb{H}$ sufficiently small such that $\varphi : U\to \Gamma(U)$, $\varphi(p) = \Gamma(p)$ is a local homeomorphism (a chart). It extends to a local homeomorphism of tangent bundles. This is how you can define $T_1 (\Gamma \setminus \mathbb{H})$ and see it is the same as $\Gamma \setminus T_1\mathbb{H}$ $\endgroup$ – reuns Jun 14 '17 at 19:35
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I'll define the map and leave the proof that it is a diffeo to you. Assume $M$ is a Riemannian manifold and $\Gamma$ is a torsion free group that acts properly discontinuously by isometries on $M$. Get the quotient map $M\to \Gamma\backslash M$. Check that it is differentiable map between Riemannian manifolds which derivative at every point is an isometry and get an associated map $T^1M\to T^1(\Gamma\backslash M)$. Note that this map is $\Gamma$-invariant, thus factors through the orbit space $\Gamma\backslash (T^1M)$. Call the resulting map $\phi:\Gamma\backslash (T^1M)\to T^1(\Gamma\backslash M)$.

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  • $\begingroup$ Thank you so much. I think we don't need the compactness of surface. We only need $\Gamma$ is torsion free and discrete, then $\Gamma \backslash \mathbb{H}$ will become a hyperbolic surface. Do you agree with that? $\endgroup$ – Skid Row Jun 15 '17 at 6:23
  • $\begingroup$ yes, this is correct. $\endgroup$ – Uri Bader Jun 15 '17 at 8:33
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You only need to think how do you define the structure of Riemannian manifold in $\Gamma \backslash \mathbb{H}$. For this you need an action in the tangent bundle of $\mathbb{H}$ which is same as the one you define to construct $\Gamma \backslash T_1\mathbb{H}$, you can do this because $\Gamma$ is an isometry. So the isomorphism is just the identity.

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  • $\begingroup$ Could you write it clearer, please? Is it the same to use fundamental domain? How to use the compactness of surface? $\endgroup$ – Skid Row Jun 14 '17 at 19:19

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