Question about the Structure group of a circle bundle over a Riemann surface The last two pages of Appendix C in Milnor's Characteristic Classes gives an example of a flat bundle with nonzero Euler class. I have a question about the structure group of this bundle. 
The example starts with a Riemann surface of genus larger than 2. Its fundamental group, $G$, is represented as a subgroup of $\text{PSL}(2,\Bbb R)$ and acts on the upper half-plane $H$ as the group of covering transformations of the surface.
This action preserves the extended real axis and gives a circle bundle over the surface, $E = (H \times \Bbb{RP}^1)/G \to H/G = S$.
Proof is given that it is isomorphic to the bundle of tangent directions to the surface. 
This bundle has a 2-fold cover corresponding to the cohomology class in $H^1(E;\Bbb Z_2)$ that maps into the Thom class of the associated 2-plane bundle under the connecting homomorphism. This class exists because the second Stiefel-Whitney class of the bundle is zero (the surface has even Euler characteristic).
So the 2-fold cover is another circle bundle and each fiber circle in the original is covered by a fiber circle of the two fold cover.
Now it is claimed that the structure group of this two fold cover comes from a representation of $G$ in $\text{SL}(2,\Bbb R)$. I think this means that it is the quotient $(H\times \Bbb{RP}^1)/G$ where $G$ now acts on the projective line as a subgroup of $\text{SL}(2,\Bbb R)$ rather than $\text{PSL}(2,\Bbb R)$. This I do not understand.
 A: This all comes down to the lift of the action of $PSL(2,R)$ on $RP^1$ to the 2-fold cover:
The unique connected 2-fold cover of $PSL(2,R)$ is $SL(2,R)$, it acts naturally on the unit circle $S^1$, which is understood as the set of oriented lines in $R^2$. The circle $S^1$ is the unique 2-fold cover of $RP^1$ (defined by forgetting the orientation of the lines). Thus, the group $PSL(2,R)$ lifts to $SL(2,R)$ under the covering map $S^1\to RP^1$. The kernel consists of all $SL(2,R)$ matrices which send each line to itself, i.e. equals $\{\pm I\}$. In particular, the diagram  (where the horizontal arrow are group actions and vertical arrow consist of the covering maps and their products)
$$
\begin{array}{ccc}
SL(2,R)\times S^1 & \longrightarrow & S^1\\
\downarrow               &~&           \downarrow\\
PSL(2,R)\times RP^1 & \longrightarrow &RP^1
\end{array}
$$
is commutative. 
The rest is done by applying this consideration to each fiber. Hence, the bundle which is the 2-fold cover Milnor is talking about, is
$$
(H\times S^1)/G
$$ 
where $G$ acts on $S^1$ via its isomorphic lift from $PSL(2,R)$ to $SL(2,R)$. 
The confusion stems from the fact that $S^1$ is homeomorphic to $RP^1$, but in this situation they should be regarded as different entities. 
