Abstract

It is well known that symmetry-protected topological (SPT) phases host nontrivial boundaries that cannot be mimicked in a lower-dimensional system with a conventional realization of symmetry. However, for SPT phases of bosons (fermions) within the cohomology (supercohomology) classification the boundary can be recreated without the bulk at the cost of a non-onsite-symmetry action. This raises the following question: Can one also mimic the boundaries of SPT phases which lie outside the (super)cohomology classification? In this paper, we study this question in the context of $(2+1)$-dimensional $[(2+1)\mathrm{D}]$ fermion SPTs. We focus on the root SPT phase for the symmetry group $G={Z}_{2}\ifmmode\times\else\texttimes\fi{}{Z}_{2}^{f}$. Starting with an exactly solvable model for the bulk of this phase constructed by Tarantino and Fidkowski, we derive an effective one-dimensional (1D) lattice model for the boundary. Crucially, the Hilbert space of this 1D model does not have a local tensor product structure, but rather is obtained by placing a local constraint on a local tensor product Hilbert space. We derive the action of the ${Z}_{2}$ symmetry on this Hilbert space and find a simple three-site Hamiltonian that respects this symmetry. We study this Hamiltonian numerically using exact diagonalization and DMRG and find strong evidence that it realizes an Ising conformal field theory where the ${Z}_{2}$ symmetry acts as the Kramers-Wannier duality; this is the expected stable gapless boundary state of the present SPT. A simple modification of our construction realizes the boundary of the $(2+1)\mathrm{D}$ topological superconductor protected by time-reversal symmetry $\mathcal{T}$ with ${\mathcal{T}}^{2}={(\ensuremath{-}1)}^{\mathcal{F}}$.

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