one of the two variables, say $\tilde{\sigma}_p$, is ordered (recall that if $N$ is prime, ${\mathbb{Z}}_N$ models can have only a single symmetry broken phase), and that $\mu$ is disordered. Notably, since $\tilde{\mu}_{p'} = \mu \sigma^{p'}$ this also implies that $\tilde{\mu}_{p'}$ is disordered for all $p' \neq p$. We then claim that under these conditions the model is in a “topological phase of type $p$”. Three questions need to be answered to justify this statement: (i) why is this a phase (ii) why do different $p$’s correspond to distinct phases and (iii) why are they topological.

Considering the first point note that the hidden order of the $\sigma$ variables manifested by order in $\tilde{\sigma}_p$ is a robust property. Indeed as argued in the previous case of an Ising symmetry, any local symmetric and defect-free perturbation in original model would map to a local term in the $\tilde{\sigma}_p$ and $\tilde{\mu}_{p'}$ degrees of freedom. Thus robustness of the topological phase is implied by the usual robustness of broken symmetry states. Turning to the second point, and the role of $p$, we can simply note that two different values of $p$ correspond to two different order parameters and thus two different phases. Indeed 