twist) $$\begin{aligned}
n_{\mu\nu}&=&\left(\begin{array}{cccc}
 0 &  L &  L & L\\
-L &  0 &  L & L\\
-L & -L &  0 & L\\
-L & -L & -L & 0
\end{array}\right),\qquad N=L^2.
\label{EQ:minimal_sym_form}\end{aligned}$$ This twist represents $L^4$ lattice. In order to construct the classical solution for this twist, it is convenient to use the $SL(4, \mathbb{Z})$ transformation for the coordinates on $\mathbb{T}^4$ [@vanBaal:1985na]. Using the $SL(4, \mathbb{Z})$ transformation we can always rewrite the $n_{\mu\nu}$ in the skew-diagonal form $$\begin{aligned}
n_{\mu\nu}&\longrightarrow&n_{\mu\nu}'=V^T n_{\mu\nu}V=
\left(\begin{array}{cc|cc}
 0 &  L &  0 & 0\\
-L &  0 &  0 & 0\\
\hline
 0 &  0 &  0 & L\\
 0 &  0 & -L & 0
\end{array}\right),
\label{EQ:minimal_skew_form}\end{aligned}$$ where $V$ is a $SL(4, \mathbb{Z})$ transformation matrix. This form makes the construction of the twist-eater easy. Here we define $L\times L$ “shift” matrix $\hat{S}_L$ and “clock” matrix $\hat{C}_L$ by $$\begin{aligned}
\hat{S}_L=\left(
\begin{array}{ccccc}
0      & ~1     & 0      & ~\cdots & ~0     \\
\vdots & ~0     & 1      & ~\ddots & \vdots \\
\vdots &        & \ddots & ~\ddots & ~0     \\
0      &        &        & ~\ddots & ~1     \\
1      & ~0     & \cdots & ~\cdots & ~0
\end{array}
\right), \qquad
\hat{C}_L=\left(
\begin{array}{ccccc}
1 &              