Let M be the connected sum of 3 projective planes. Then is the fundamental group of M a Fuchsian group?
This group is an extension of the two element group by the surface group for a genus 2 surface (which is a Fuchsian group). So it seems very close to Fuchsian.
Recall that the connected component of the isometries of the sphere, plane and hyperbolic plane are $ SO_3, SE_2, PSL_2=SO_{2,1} $ respectively.
The nonorientable surface $ N_0 $ (which is just the projective plane) is given by
$$
N_0 \cong SO_3/O_2
$$
where $ SO_2 $ is the connected group of rotations around z axis
$$
SO_2= \left \{ \
\begin{bmatrix}
a & b & 0 \\
-b & a & 0 \\
0 & 0 & 1
\end{bmatrix} : a^2+b^2=1 \right \}
$$
and
$$
r:=\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1
\end{bmatrix}
$$
so that $ <r,SO_2> \cong O_2 $. The orientable double cover is the sphere $ \Sigma_0 $
$$
\Sigma_0 \cong S^2 \cong SO_3/SO_2
$$
Similarly for the the non orientable surface $ N_1 $ (the connected sum of two projective planes, in other words the Klein bottle) we have
$$
SE_2= \left \{ \
\begin{bmatrix}
a & b & x \\
-b & a & y \\
0 & 0 & 1
\end{bmatrix} : a^2+b^2=1 \right \}
$$
there is a connected group $ V $ of translations up each vertical line
$$
V= \left \{ \
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & y \\
0 & 0 & 1
\end{bmatrix} : y \in \mathbb{R} \right \}
$$
There is also an element that shifts horizontally by one unit
$$
b:=\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
$$
Now if we include the rotation by 180 degrees
$$
\tau:=\begin{bmatrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1
\end{bmatrix}
$$
then
$$
N_1 \cong SE_2/<V,b, \tau>
$$
And the orientable double cover is the torus $ \Sigma_1 $
$$
\Sigma_1 \cong T^2 \cong SE_2/<V,b>
$$
Is something similar true for $ N_2 $ the connected sum of 3 projective planes? The orientable double cover here is a genus 2 surface $ \Sigma_2 $. Then
$$
\Sigma_2 \cong \Gamma \backslash SL_2 /SO_2
$$
where $ \Gamma $ is a Fuchsian surface group. Is there another Fuchsian group $ \Gamma' $ containing $ \Gamma $ as a index two subgroup such that
$$
N_2 \cong \Gamma' \backslash SL_2 /SO_2
$$
Best Answer
Yes, you are misinterpreting what is said in the link.
The proof is by contradiction. Suppose that $\pi=\pi_1(M)$ is isomorphic to a Fuchsian subgroup $\Gamma< G= PSL(2,R)$. The group $G$ acts (and properly) conformally on the upper half-plane $U$. Thus, $\Gamma$ acts properly discontinuously and conformally on $U$ as well. Hence, the quotient $S=U/\Gamma$ is a Riemann surface (a 1-dimensional complex manifold) satisfying $\pi_1(S)\cong \Gamma\cong \pi = \pi_1(M)$. Hence, due to asphericity of $M$ and contractibility of $U$, the surfaces $S$ and $M$ are homotopy-equivalent. Since $M$ is compact, you get $H_2(M, Z_2)=Z_2$. Hence, $H_2(S, Z_2)=Z_2$. From this, you conclude that $S$ is also compact. But $S$ is orientable, hence, $H_2(S)\cong Z$, while $H_2(M)=0$.