The aim of this question is to construct a exmaple of quotient manifold.
First, I set notations and difnitions.
Let $\mathbb{R} , \mathbb{C}$ be real and complex numbers with usual topology.
We put the product space $X= \mathbb{C} \times \mathbb{R}$ and define a group $G$ as follows.
$t_1 : X \rightarrow X , (z,x) \mapsto (z+1,x)$
$t_2 : X \rightarrow X , (z,x) \mapsto (z+\xi,x)$
$\alpha : X \rightarrow X , (z,x) \mapsto (\omega z,x+1)$
where $\ $ $i= \sqrt{-1}\ $ , $\ \omega= \frac{-1+\sqrt{3}i}{2}\ $ , $\ \xi= \frac{1+\sqrt{3}i}{2}$
We define $G$ as a group generated by $t_1 , t_2 , \alpha$ and then $G$ acts on $X$.
Next, we consider quotient space $M=X/G$ by the group action of $G$ as a example of quotient manifold. Then I have a next question.
My Question
Does $M$ have a structure $3$-dim compact $C^{\infty}$ manifold $???$
Yes. It will be easier to understand this construction in two stages.
In the first stage you only quotient by $t_1$ and $t_2$. This gives $T^2 \times \mathbb{R}$ where $T^2$ is the $2$-torus, specifically $\mathbb{C}/\Gamma$ where $\Gamma$ is the lattice in $\mathbb{C}$ generated by $1$ and $\xi$.
$\alpha$ descends to a map on $T^2 \times \mathbb{R}$ given by the same formula (which makes sense because multiplication by $\omega = \xi^2$ preserves the lattice $\Gamma$). Quotienting gives the mapping torus of multiplication by $\omega$ acting on $T^2$; more specifically it's a torus bundle (over $S^1$). This is a closed $3$-manifold whose Thurston geometry is Euclidean geometry, as you can see from the fact that $t_1, t_2, \alpha$ are all isometries on $\mathbb{R}^3$.