& 0\end{array}\right] ,|i|=3.\end{aligned}$$

Since we have $\left[ P_{0},Q_{4}^{2}\right] =0$, the invariant subspace $P_{0}$ contains the even and odd $\mathbb{Z}_{2}$ parity sectors. In addition, $A^{[i]}$ must not contain an invariant subspace whose projector commutes with the operator $Q_{4}$. Similar to the $\mathbb{Z}_{3}$ case, it is necessary that the charge-zerosubalgebras of matrix algebras spanned by $a_{0}^{[i]}$ and $a_{1}^{[i]}$ are simple, respectively. Then the local matrices $A^{[i]}$ span a simple algebra in the $\mathbb{Z}_{4}$ graded sense, but it is semisimple in the $\mathbb{Z}_{2}$ graded sense[@Fidkowski2011]. Because the semisimple algebra can split into irreducible ones, we can reduce the local matrices and break the $\mathbb{Z}_{4}$ symmetry down to $\mathbb{Z}_{2}$ symmetry. To make it explicit, another gauge transformation $$G^{\prime }=\frac{1}{\sqrt{2}}\left[ 
\begin{array}{cccc}
0 & \mathbbm{1} & 0 & -\mathbbm{1} \\ 
i\mathbbm{1} & 0 & -i\mathbbm{1} & 0 \\ 
0 & \mathbbm{1} & 0 & \mathbbm{1} \\ 
\mathbbm{1} & 0 & \mathbbm{1} & 0\end{array}\right] ,$$changes the local matrices into the canonical form: $$\begin{aligned}
A^{[i]} &=&\omega _{4}^{|i|}d^{[i]}\oplus d^{[i]},  \notag \\
d^{[i]} &=&\left[ 
\begin{array}{cc}
a_{1}^{[i]} & 0 \\ 
0 & a_{0}^{[i]}\end{array}\right] ,|i|=0,2;  \notag \\
d^{[i]} &=&\left[ 
\begin{array}{cc}
0 & a_{1}^{[i]} \\ 
a_{0}^{[i]} & 0\end{array}\right] ,|i|=1,3.\end{aligned}$$The block diagonal form of $A^{[i]}$ represents a spontaneous symmetry breaking phase[@Chen-Gu-Wen-2011]. One may question 