Fundamental group of the unit tangent bundle on the genus 2 torus? I'm interested in the 3-dimensional model geometries; specifically $\widetilde{SL}(2,\mathbb{R})$. I'm looking for a good (see, easily visualizable) example of a compact manifold formed as a quotient of a group acting on $\widetilde{SL}(2,\mathbb{R})$.
I'm trying to show that the unit tangent bundle (i.e., the unit vectors on the tangent bundle, so a circle bundle) on the closed 2-dimensional genus 2 surface is such a manifold. I call this manifold $UF_2$. In order to do that, I think I need to show that $\pi_1(UF_2)=\left<a,b,c,d,s\big|[a,b][c,d]=s^k\right>$ where $[a,b]=aba^{-1}b^{-1}$ and $k\in\mathbb{Z}$ and $k\not=0$.
I am using Van Kampen's theorem to calculate $\pi_1(UF_2)$ and I finally got a result. I think that $\pi_1(UF_2)=\left<a,b,c,d,s\big|[a,b][c,d]=s^2\right>$, but I'm not that confident in the result. 
My application of Van Kampen's theorem involves splitting $UF_2$ into a unit tangent bundle on 2 different punctured tori. But really, I did a lot of hand waving. I couldn't even write down how I arrived at the result $k=2$. So I'm asking: 
1 Can anyone confirm, using any method, that $\pi_1(UF_2)=\left<a,b,c,d,s\big|[a,b][c,d]=s^2\right>$?
2 Can anyone confirm that $UF_2$ is a quotient of $\widetilde{SL}(2,\mathbb{R})$ by a isometry group?
3 Can anyone write down an explicit method for calculating $\pi_1(UF_2)$?
Thank you all
 A: Any closed surface $\Sigma$ of genus $g \ge 2$ admits a hyperbolic metric, and hence is a quotient $\mathbb{H}/\Gamma$ of the upper half plane $\mathbb{H}$ by a discrete group $\Gamma$ of isometries acting on $\mathbb{H}$ which can be identified, as an abstract group, with the fundamental group $\pi_1(\Sigma)$. 
The isometry group of $\mathbb{H}$ is $PSL_2(\mathbb{R})$, so $\Gamma$ is naturally a discrete subgroup of $PSL_2(\mathbb{R})$. This full isometry group acts freely and transitively on the unit tangent bundle of $\mathbb{H}$, and so the unit tangent bundle of $\mathbb{H}$ can itself be identified with $PSL_2(\mathbb{R})$. Moreover, this identification identifies the unit tangent bundle of $\Sigma$ with the quotient $PSL_2(\mathbb{R})/\Gamma$, which of course can also be written as a quotient of $\widetilde{SL}_2(\mathbb{R})$. 
Using this description, although other approaches are also possible, it follows that the fundamental group of the unit tangent bundle is a certain extension
$$1 \to \mathbb{Z} \to \pi_1(UT(\Sigma)) \to \pi_1(\Sigma)) \to 1$$
of $\pi_1(\Sigma) \cong \Gamma$ by $\mathbb{Z}$, coming from the action of $\Gamma$ on the fundamental groupoid of $PSL_2(\mathbb{R})$. 
A: This is to supplement Qiaochu Yuan's answer which is a bit incomplete.
Proposition. Suppose that $\xi=(p: E\to \Sigma)$ is the oriented circle bundle over oriented closed connected surface $\Sigma$ of genus $g$, such that The Euler number $e(\xi)$ of $\xi$ equals $e\in {\mathbb Z}$. Then $\pi_1(E)$ has the presentation 
$$
\langle a_1, b_1,...,a_g, b_g, t| [a_1,b_1]...[a_g,b_g]t^{-e}, [a_i,t], [b_i,t], 1\le i\le g\rangle.  
$$
In particular, if $\xi$ is the unit tangent bundle of $\Sigma$ then $e(\xi)=2-2g$ and, therefore, the presentation is
$$
\langle a_1, b_1,...,a_g, b_g, t| [a_1,b_1]...[a_g,b_g]t^{2g-2}, [a_i,t], [b_i,t], 1\le i\le g\rangle.  
$$
In particular, if $g=2$ (genus 2 surface) then we get
$$
\langle a_1, b_1,a_2, b_2, t| [a_1,b_1][a_2,b_2]t^{2}, [a_i,t], [b_i,t], 1\le i\le 2\rangle.  
$$
In other words, the presentation written in the question is almost correct, you are missing the commutator relators, i.e. that $t$ is a central element in $\pi_1(E)$. 
A proof is not difficult and boils down to understanding the definition of the Euler number of a circle bundle. Remove an open disk $D$ from $\Sigma$. Since the bundle is oriented, it trivializes over $\Sigma'=\Sigma-D$, 
$$
E'=\Sigma' \times S^1,
$$
hence, 
$$
\pi_1(p^{-1}(\Sigma'))=\pi_1(\Sigma')\times \langle t\rangle,
$$ 
where $t$ represents the fiber of $\xi$. Under this trivialization, the boundary circle of $\Sigma'$, i.e. 
$$
\partial \Sigma' \times \{q\}, q\in S^1,
$$
is a simple loop $s'$ on the torus $T^2=p^{-1}(\partial \Sigma')$. On the level of the fundamental group, $s$ corresponds to the product of commutators
$$
[a_1,b_1]...[a_g,b_g]\in \pi_1(\Sigma')< \pi_1(E')
$$
The manifold $E$ is obtained from $E'$ by attaching to it the solid torus $E''=p^{-1}(cl(D))$. The bundle $\xi$ also trivializes over $cl(D)$, $E''= cl(D)\times S^1$. Accordingly, $\pi_1(T^2)$ has two generators coming from this trivialization of $E''\to D^2$: The "horizontal" generator $s''$ (corresponding to $\partial D \times S^1$) and the "vertical generator" $t$, corresponding to the fiber.  On the boundary circle $\partial D$ the two trivializations do not match, which is algebraically reflected in the equation
$$
s'= s'' + e t
$$
(in $\pi_1(T^2)$), where $e=e(\xi)$ is the Euler number of $\xi$. (I am sloppy here with orientations, in the end, it does not matter.) Now, apply Seifert-Van Kampen Theorem  to the union 
$$
E= E'\cup_{T^2} E''
$$
to obtain the above presentation of $\pi_1(E)$. This is all worked out in great detail in 
 John Hempel,  3-manifolds, Providence, RI: AMS Chelsea Publishing (ISBN 0-8218-3695-1/hbk). xii, 195 p. (2004). ZBL1058.57001.
