There is a construction of symmetry protected topological (SPT) states which roughly goes as follows. We start with a $d$-dimensional system with symmetry $\mathbb{Z}_2 \times G$ in the phase where the $\mathbb{Z}_2$ is spontaneously broken. To the $\mathbb{Z}_2$ domain walls, we attach a $d-1$ dimensional $G$-SPT, and then condense the bound state to restore the $\mathbb{Z}_2$ symmetry. The result is a nontrivial $\mathbb{Z}_2 \times G$-SPT.

However, for $d=1$, the $d-1$ dimensional $G$-SPTs are just objects carrying nonzero $G$-charge. Naively, it seems that attaching $G$-charge to the $\mathbb{Z}_2$ domain wall and condensing the pair ought to spontaneously break the $G$ symmetry. However, the $1d$ $\mathbb{Z}_2 \times \mathbb{Z}_2$ SPT (the cluster state) seems to be constructed in precisely this way, and doesn't break any symmetries. Why doesn't condensing a bound state involving a symmetry charge lead to symmetry breaking?

I believe there's also a similar construction of the bosonic integer quantum hall state in $d=2$. One starts with a bilayer of bosons with $U(1) \times U(1)$ symmetry, i.e. individually conserved boson numbers. Then, we attach a boson from one layer to a vortex of the other layer and condense the pair, and finally break the symmetry to the diagonal subgroup. Why doesn't the condensation break one of the $U(1)$ symmetries? Is the reason the same as the above case?

Finally, I'd like to ask what is different about the above two scenarios from the situation that occurs at the Neel-VBS transition. My understanding is that the system has a $SO(3) \times C_4$ symmetry, and in the VBS phase where $C_4$ is spontaneously broken, the $C_4$ defect carries a projective $SO(3)$ charge. Condensing the $C_4$ defect to restore $C_4$ then necessarily breaks $SO(3)$, which leads to the Neel phase. This seems consistent with the intuition that condensing symmetry charges leads to symmetry breaking, and I'd like to understand why that intuition doesn't apply in the above two scenarios.


1 Answer 1


Perhaps it helps to remember that one of the consistency conditions for a $\mathbb{Z}_2$ decorated domain wall construction in $d=1$ is that the $G$ charge bound to a domain wall is such that two of these charges fuse to the trivial charge. (Because it has to be consistent with the fusion rules of domain walls). Now, with periodic boundary conditions, the total number of domain walls is always even, so when you imagine the wavefunction as a superposition over domain wall configurations, all the configurations you are superposing have trivial $G$ charge. Therefore, the wavefunction itself has trivial $G$ charge and thus is $G$-invariant.

That's different from what would happen if you just condensed $G$-charges on their own without binding them to domain walls. In that case the wavefunction would be a superposition over all charge sectors, hence it would fail to have definite $G$ charge, i.e. it would break the $G$ symmetry.

The example you mentioned in your final paragraph is different, because the defect is carrying a projective representation of $\mathrm{SO}(3)$, which is different from a regular charge that would correspond to a linear representation (actually it would have to be a one-dimensional linear representation to be admissible in a decorated domain wall construction).

  • $\begingroup$ Thanks, the explanation for why symmetry breaking doesn't occur in the decorated domain wall construction makes sense. In the bosonic IQH construction, I suppose it's similar because the superimposed configurations must have zero total vortex number and therefore zero total charge as well? $\endgroup$
    – pianyon
    Sep 3, 2020 at 6:09
  • $\begingroup$ I'm confused as to why this isn't an obstruction to symmetry breaking in the Neel-VBS case. Is it the size of the representation that allows symmetry breaking to happen? $\endgroup$
    – pianyon
    Sep 3, 2020 at 6:19

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