# Interacting fermionic SPT phases in 2d with time-reversal symmetry

Interacting fermionic SPT phases in 1d and 3d with $\mathbb{Z}_2^T$ symmetry are classified by $\mathbb{Z}_8$ and $\mathbb{Z}_{16}$ respectively, as shown in the paper by Fidkowski and Kitaev http://arxiv.org/abs/1008.4138, and Wang and Senthil http://arxiv.org/abs/1401.1142. I'd like to know the classification for the 2d case. Can anyone suggest some reference on that?

Here is the classification table of fermionic SPT phases, copied from A. Kapustin et. al., arXiv:1406.7329. In the table, the arrow denotes the reduction of classification by interaction: free fermion classification $\to$ interacting fermion classification. $$\begin{array}{c|ccc} d= & 1 & 2 & 3\\ \hline \text{BDI} & \mathbb{Z}\to\mathbb{Z}_8 & 0\to 0 & 0\to 0 \\ \text{D} & \mathbb{Z}_2\to\mathbb{Z}_2 & \mathbb{Z}\to\mathbb{Z} & 0\to 0 \\ \text{DIII} & \mathbb{Z}_2\to\mathbb{Z}_2 & \mathbb{Z}_2\to\mathbb{Z}_2 & \mathbb{Z}\to\mathbb{Z}_{16} \\ \end{array}$$

BDI class: $\mathbb{Z}_2^T$ symmetry with $\mathcal{T}^2=+1$. DIII class: $\mathbb{Z}_2^T$ symmetry with $\mathcal{T}^2=-1$. D class: no symmetry (apart form fermion parity).

One must note that, the 1d $\mathbb{Z}_8$ and the 3d $\mathbb{Z}_{16}$ classified fermionic SPT phases mention in the question are actually protected by different types of time-reversal symmetries, and belongs to different symmetry classes. Depending on the signature of $\mathcal{T}^2$ being $+1$ or $-1$, the $\mathbb{Z}_2^T$ symmetry is actually ascribed to either BDI or DIII symmetry class. In 1d and 3d, there are $\mathbb{Z}$ classified free fermionic SPT phases (1d BDI or 3d DIII) to start with. The phenomenon of interaction reduced classification only happens if we start from a $\mathbb{Z}$ classified free fermionic SPT state. In 2d, there is no such free fermion SPT state to start with, so there is no analogous interaction reduced classification in 2d with $\mathbb{Z}_2^T$ symmetry protection.

• Hi Everett, I have some confusions:(1).As to the notation $Z_2^T$ with $T^2=-1$, the symmetry group generated by TR symmetry seems to be $Z_2^T=[1,T,T^2,T^3]$, which is in fact a $Z_4$ group, so why don't we use the notation $Z_4^T$ instead of $Z_2^T$? (2). Is the fermion-number-parity symmetry mentioned in your comments same as the particle-hole symmetry in the classification table? And should we distinguish them? Thank you very much! – Kai Li Apr 5 '15 at 10:57
• @KaiLi $T^2=-1$ is a projective representation of the $Z_2^T$ group on the fermions. The symmetry of the entire fermion system is still $T^2=+1$ because fermions must appear in pairs. This is just like in the PSG, when you have some 2-fold symmetry squared to -1, you tend to call it projective rep, other than group extension, although they are mathematically the same. If our universe is built from qubits, then fermion systems actually have topological order, so the -1 sign change of a single fermion can be gauged away, and should not be considered as a symmetry operation. – Everett You Apr 6 '15 at 7:25
• @KaiLi No, fermion number parity is not the particle hole symmetry in the classification table. All fermion systems have the fermion number parity conservation, but not all of them are particle-hole symmetric. – Everett You Apr 6 '15 at 7:27
• Thanks a lot. Naively thinking, if the fermion Hamiltonian contains terms like $c+c^\dagger$, then it seems that the fermion-number-parity symmetry is explicitly broken, does this situation make sense? – Kai Li Apr 6 '15 at 9:41
• @KaiLi It does not make sense to write down a term like $c+c^\dagger$ in the Hamiltonian, because Hamiltonian must be a bosonic operator that corresponds to the energy of the system. – Everett You Apr 6 '15 at 9:49

The non-interacting classification was obtained in the seminal "periodic table" papers by Kitaev and Ryu/Snyder/Furusaki/Ludwig. The interacting classification of 2D fermionic SPT phases with time-reversal symmetry has been considered by several groups already, including:

Gu and Wen's group super-cohomology theory, http://arxiv.org/abs/1201.2648

Kapustin et. al. cobordism group, as already mentioned in the previous answer.

and a recent treatment based on tensor category http://arxiv.org/abs/1501.01313

All approaches agree that there are no 2D nontrivial fermionic SPT with $T^2=1$.