In this subsection, we will define $3$-dimensional Heisenberg nilmanifolds. Recall the $3$-dimensional Heisenberg group is

(2.5)

$$H(1,\mathbb{R})\equiv\left\{\begin{bmatrix}1&x&t\\ 0&1&y\\ 0&0&1\end{bmatrix}:\ x,y,t\in\mathbb{R}\right\}.$$

To define a Heisenberg nilmanifold, let us define a co-compact group action on $H(1,\mathbb{R})$. First, we define a lattice $\Lambda\equiv\epsilon\mathbb{Z}\langle 1,\tau\rangle\subset\mathbb{R}^{2}_{x,y}=\mathbb{C}$ by choosing

(2.6)

$$\tau_{1}=Re(\tau),\ \tau_{2}=Im(\tau),$$

which is generated by

(2.7)

$\displaystyle\begin{bmatrix}1&\epsilon&0\\ 0&1&0\\ 0&0&1\\ \end{bmatrix},\ \begin{bmatrix}1&\epsilon\tau_{1}&0\\ 0&1&\epsilon\tau_{2}\\ 0&0&1\\ \end{bmatrix}\in H(1,\mathbb{R}).$

Then immediately $A=\Area(\mathbb{R}_{x,y}^{2}/\Lambda)=\epsilon^{2}\tau_{2}$. For $b\in\mathbb{Z}_{+}$, the Heisenberg nilmanifold $\Nil^{3}_{b}(\epsilon,\tau)$ of degree $b$ is the quotient of $H(1,\mathbb{R})$ by the left action generated by

(2.8)

$\displaystyle\begin{bmatrix}1&\epsilon&0\\ 0&1&0\\ 0&0&1\\ \end{bmatrix},\begin{bmatrix}1&\epsilon\tau_{1}&0\\ 0&1&\epsilon\tau_{2}\\ 0&0&1\\ \end{bmatrix},\begin{bmatrix}1&0&\frac{A}{b}\\ 0&1&0\\ 0&0&1\\ \end{bmatrix}.$

Note that these transformations are

(2.9)		$\displaystyle(x,y,t)$	$\displaystyle\mapsto(x+\epsilon,y,t+\epsilon y),$
(2.10)		$\displaystyle(x,y,t)$	$\displaystyle\mapsto(x+\epsilon\tau_{1},y+\epsilon\tau_{2},t+\epsilon\tau_{1}y),$
(2.11)		$\displaystyle(x,y,t)$	$\displaystyle\mapsto\Big(x,y,t+\frac{A}{b}\Big).$

The forms

(2.12)

$\displaystyle dx,dy,\theta_{b}\equiv\frac{2\pi b}{A}(dt-xdy)$

are a basis of left-invariant $1$-forms.

It is clear that $\Nil^{3}_{b}$ is the total space of a degree $b$ circle fibration

(2.13)

$\displaystyle S^{1}\longrightarrow\Nil^{3}_{b}\xrightarrow{\ \pi\ }\mathbb{T}^{2}.$

The following result will be needed later in Proposition 6.6 to determine the Betti numbers of $\mathcal{M}$.

For $b\in\mathbb{Z}_{+}$, we define the Heisenberg nilmanifold $\Nil^{3}_{-b}$ to be the quotient of $H(1,\mathbb{R})$ by the action generated by

(2.15)		$\displaystyle(x,y,t)$	$\displaystyle\mapsto(x+\epsilon,y,t-\epsilon y),$
(2.16)		$\displaystyle(x,y,t)$	$\displaystyle\mapsto(x+\epsilon\tau_{1},y+\epsilon\tau_{2},t-\epsilon\tau_{1}y),$
(2.17)		$\displaystyle(x,y,t)$	$\displaystyle\mapsto\Big(x,y,t-\frac{A}{b}\Big).$

Note that the generated action is conjugate to the previous action by the mapping $(x,y,t)\mapsto(-x,-y,-t)$. The forms

(2.18)

$\displaystyle dx,dy,\theta_{-b}\equiv\frac{2\pi b}{A}(dt+xdy)$

are a basis of left-invariant $1$-forms.