Does the surface topological order on the boundary of 3D topological insulator also have topological ground state degeneracy?

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The boundary of a 3D topological insulator can be fully gapped (under strong interaction) by the surface topological order without breaking the symmetry (see Fidkowski-Chen-Vishwanath, Metlitski-Kane-Fisher, Bonderson-Nayak-Qi, Wang-Potter-Senthil). Suppose we make a solid torus shape of 3D topological insulator, and turn on the interaction to gap out the boundary (which is now a torus) by the surface topological order. For 2D topological ordered state placed on a torus, we expect a topological ground state degeneracy equal to the number of anyon types. However does the surface topological ordered 3D topological insulator also have the topological ground state degeneracy? If so how to understand the topological ground state degeneracy emerging from a symmetry protected trivial state which is not expected to have the degeneracy? Also, is the surface topological ground state degeneracy still related to the number of anyon types?

This post imported from StackExchange Physics at 2015-02-05 10:18 (UTC), posted by SE-user Everett You
There should be topological degeneracy. One can imagine having a 2D topologically ordered state living on a true torus, which embeds in 3D. We would not question about the topological degeneracy. Now instead of having a torus sitting in a vaccum, we insert into the inside of the torus some trivial insulator (meaning no topologically nontrivial excitations in 3D). Obviously the topological degeneracy does not change. This is the same situation we have on the surface of a 3D SPT, because the surface SET is topologically ordered regardless of the symmetry, and once we break all symmetry we just get back to the previous "trivial" case. The symmetry only manifests in the fractionalization of quantum numbers on the surface.

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The brief answer is yes, they do have topological degeneracy. My understanding is that if you only focus on the surface topological order with some symmetry given by the bulk symmetry protected topological (SPT) state, there is nothing wrong with that symmetry enriched topological order (SET) states. That means they also have all the properties as the usual SET states in pure 2D. However, if you try to gauge the surface SET, you will find the obstruction to define a consistent gauged theory. More precisely, the gauged theory won't satisfy the pentagon equation. Those obstructions are classified by the forth cohomology group, $H^{4}(G,U(1))$, where $G$ is the symmetry group of the bulk SPT. Usually, people say those SETs are anomalous. More details can be found in this paper http://arxiv.org/abs/1403.6491

This post imported from StackExchange Physics at 2015-02-05 10:19 (UTC), posted by SE-user Sheng-Jie Huang
answered Feb 5, 2015 by (50 points)

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