In our interested context, $Y^{3}$ is a Heisenberg nilpotent manifold. We consider a simple example that $Y^{3}\equiv H(1,\mathbb{Z})\setminus H(1,\mathbb{R})$ with

(4.32)

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

and

(4.33)

$$H(1,\mathbb{Z})\equiv\left\{\begin{bmatrix}1&m&p\\ 0&1&n\\ 0&0&1\end{bmatrix}:\ m,n,p\in\mathbb{Z}\right\}.$$

In this case, $Y^{3}$ is a Heisenberg manifold of degree $1$. As a $\mathbb{T}^{2}$ bundle over $S^{1}$, its monodromy is given by $(\begin{smallmatrix}1&1\\ 0&1\end{smallmatrix})\in\SL(2,\mathbb{Z})$. So it is standard that the spectrum consists of two classes of eigenvalues

(4.34)

$$\mathfrak{T}\equiv\Big\{4\pi^{2}(k^{2}+\ell^{2})\Big|k,\ell\in\mathbb{Z}\Big\}\ and\ \mathfrak{S}\equiv\Big\{2\pi|m|(2h+1+2\pi|m|)\Big|m\in\mathbb{Z}\setminus\{0\},h\in\mathbb{N}\Big\}.$$

Detailed discussions can be found in [DS84] and [GW86]. So we can see that the above eigenvalues coincide with the form (4.25).