Let $\Gamma=\langle (1,0),(0,1),(i,i\alpha)\rangle_{\mathbb Z}\subset \mathbb C^2$ where $\alpha\in \mathbb R$. Let $V_\Gamma:=span_\mathbb R(\Gamma)$ and $W_\Gamma:=V_\Gamma\cap iV_\Gamma$.
Now, since $V_\Gamma$ is isomorphic as a vector space to $\mathbb R^3$ consider the map $\sigma:\mathbb C \rightarrow \mathbb R^3$ by $w\mapsto w\cdot(1,\alpha)$
I want to visualize and understand this situation as a subspace of $\mathbb R^3$ like in the image I attached.
My questions are: How can we see that $V_\Gamma/\Gamma=S^1\times S^1\times S^1$ and when $\alpha \in \mathbb R\setminus\mathbb Q$ then $\sigma(\mathbb C)$ is dense in $V_\Gamma/\Gamma$? What can we say about $W_\Gamma$ and its action?
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}$Let $B = (\Basis_{j})_{j=1}^{3}$ be an ordered basis in some real vector space $V$, and let $\Gamma$ be the integer lattice spanned by $B$. The parallelepiped $$ K = \{t_{1} \Basis_{1} + t_{2} \Basis_{2} + t_{3} \Basis_{3} : 0 \leq t_{j} \leq 1\} $$ is a fundamental domain for $\Gamma$ acting by translation (i.e., by addition) in $V$. The quotient $V/\Gamma$ is obtained from $K$ by "gluing opposite faces of $K$", namely by identifying \begin{gather*} t_{2} \Basis_{2} + t_{3} \Basis_{3} \sim \Basis_{1} + t_{2} \Basis_{2} + t_{3} \Basis_{3}, \\ t_{1} \Basis_{1} + t_{3} \Basis_{3} \sim \Basis_{2} + t_{1} \Basis_{1} + t_{3} \Basis_{3}, \\ t_{1} \Basis_{1} + t_{2} \Basis_{2} \sim \Basis_{3} + t_{1} \Basis_{1} + t_{2} \Basis_{2} \end{gather*} for all $t_{1}$, $t_{2}$, $t_{3}$ in $[0, 1]$. This is the direct three-dimensional analogue of the construction of a torus from a parallelogram by gluing opposite edges. (A bit more formally, let $(t_{1}, t_{2}, t_{3})$ be Cartesian coordinates in $\Reals^{3}$, and observe that $\Gamma \leftrightarrow \mathbf{Z}^{3}$, so $V/\Gamma \simeq \Reals^{3}/\mathbf{Z}^{3} \simeq S^{1} \times S^{1} \times S^{1}$.)
To phrase your second question in real terms, fix an irrational number $\alpha$ and consider the map $\sigma:\Reals^{2} \to V/\Gamma$ defined by $$ \sigma(s, t) = s(\Basis_{1} + \alpha \Basis_{2}) + t \Basis_{3}. $$ To see the image of $\sigma$ is dense, note that the curve $C$ where $t = 0$ is an irrational winding, hence is dense in the $2$-torus $T = \{t_{3} = 0\} \subset V/\Gamma$. The image of $\sigma$ is the "cylinder" $C \times S^{1} \subset T \times S^{1} = V/\Gamma$ over $C$.