Can a nondegenerate fermionic topological Mott insulator (TMI) state support an emergent bosonic topological order?

Based on my recent study and motivated by a recent paper, I have a naive question.

Consider a 2d Hubbard model for electrons at half filling $H=\sum c_k^\dagger h_k c_k+U\sum n_{i\uparrow }n_{i\downarrow }$, assuming the following facts:

(1) the noninteracting part is a Chern insulator (CI)

(2) for large enough $U(>U_c)$ the system is a nonmagnetic Mott insulator and it has a unique ground state

(3) this nonmagnetic Mott insulator has a nonzero Chern number $C=\frac{1}{24\pi^2}\int dk_0dk^2tr(\epsilon^{\mu \nu \lambda }G\partial_\mu G^{-1}G\partial_\nu G^{-1}G\partial_\lambda G^{-1})$ equal to that of the CI, indicating it is a TMI.

On the other hand, in the the slave-rotor mean-field description $c_{i\sigma }=e^{i\theta_i}f_{i\sigma }$, the above TMI phase may be interpreted as a fractionalized CI in terms of the spinons $f_{i\sigma }$ or a chiral spin liquid (CSL, the projected spinon mean-field state).

Question: from the slave-rotor viewpoint, the CSL with spinon band Chern number $C$ seems to have a (bosonic) topological degeneracy, is this possible for our fermionic TMI without ground-state degeneracy?

• thank you for the answer. I think we are interested in the second meaning of TMI (as a fermionic SPT) as you mentioned. In the slave-rotor mean-field description, the TMI state is a direct product of the rotor state and the spinon state, i.e., $\mid \Psi_{TMI}>=\mid \Psi_{\theta }> \otimes \mid \Psi_{f}>$. Without saying topological degeneracy, is it possible that the projected spin state $\mid \Psi_{spin}>=P\mid \Psi_{f}>$ has a nonzero topological entanglement entropy while $\mid \Psi_{TMI}>$ is a fermionic SPT state? – Kai Li Oct 30 '15 at 15:31
• Before answering your actual question, I think $|\Psi_{TMI}\rangle=|\Psi_\theta\rangle\otimes|\Psi_f\rangle$ is not quite right, since you also have to do a projection in front of the tensor product. Anyway, one possible scenario is that the rotors are in a superfluid phase (spontaneous breaking the emergent U(1) symmetry), so there is no dynamical gauge field interacting with the $f$ fermions. In that case, one would end up with a SPT phase. I'm not sure why you need to consider a separate spin state, but naively, that spin state can definitely be a topologically ordered state. – Meng Cheng Oct 30 '15 at 20:38
• thanks for your reminding, I missed some projection for recovering the physical electron state. If the rotor state $\mid \Psi_\theta \rangle$ is a Mott phase such that the corresponding electron state $\mid \Psi_{TMI} \rangle$ is a Mott insulator, does this spin-charge separation of electrons imply that the TMI state $\mid \Psi_{TMI} \rangle$ of the Hubbard model must be degenerate and hence cannot be a fermionic SPT ? – Kai Li Oct 31 '15 at 7:43
• Actually our recent work on a Hubbard model gives the known results '(2)' and '(3)' presented in my question via numerical calculations of Green’s function, but we cannot determine whether the ground-state is degenerate and we assume the uniqueness such that the Green’s function expression for the Chern number makes sense. I also did a slave-rotor mean-field calculation which (of course) yields a fractionalized CI $\mid \Psi_f \rangle$ for spinons, but I'm mot sure the slave-rotor approach is whether compatible with the numerical approach. – Kai Li Oct 31 '15 at 7:57
• If the rotors are in the Mott phase, then the emergent U(1) gauge symmetry is not affected since the rotors are gapped and can be integrated out. The effective theory is just U(1) gauge fields coupled to $f$ fermions with Chern number, so this is most likely some kind of chiral spin liquid with nontrivial topological entanglement entropy. – Meng Cheng Oct 31 '15 at 16:41