why it is not a topological phase? The answer is that it is impossible to transform $A^{[i]}$ into $Y_{2}^{|i|}\otimes d^{[i]}$, which is required in the topological phase. The reducing process actually mixes the charge-0 and charge-2 sectors, as well as the charge-1 and charge-3 sectors. In this sense, the $\mathbb{Z}_{4}$ symmetry breaks down to the $\mathbb{Z}_{2}$ symmetry. After reducing, the sub-block matrices $d^{[i]}$ span an even type simple $\mathbb{Z}_{2}$ graded algebra, shown in Eq. (\[evenZ2\]). So there is no Majorana zero edge modes. The two degenerate ground states are purely resulted from the spontaneous symmetry breaking, and can be transformed into each other via the representation of the $\mathbb{Z}_{4}$ charge symmetry generator $U_{4}=\text{diag}(\mathbbm{1},\omega \mathbbm{1},\omega ^{2}\mathbbm{1},\omega ^{3}\mathbbm{1})$. Therefore, the two parts in the direct sum are connected by $\sum_{i}U_{ij}d^{[j]}=\omega ^{|i|}d^{[i]}$. Moreover, the parafermions $\chi _{i}$ can form bosons $\chi _{i}^{2}$ and its anti-particles $\chi _{i}^{3}$. But the Majorana fermions can not be obtained from $\mathbb{Z}_{4}$ parafermions, so there is no non-trivial topological phase with Majorana edge zero modes in this classification.

3\. $\left[ P_{0},Q_{4}\right] \neq 0$ and $\left[ P,Q_{4}^{2}\right] \neq 0$

Following the general procedure of deriving the irreducible MPS, the standard form of the local matrices can be written as 