to the current. This is not the only choice, but with this choice the operator $\hat{O}$ when written in $k$-space corresponds to $$\hat{O} = \sum_k ( c^\dagger_{k,A}  c^\dagger_{k,B} )
\left( 
\begin{array}{cc} 0 & 0  \\ 
0 & 1
\end{array}
\right)
\left( 
\begin{array}{c} c_{k,A}  \\ 
c_{k,B} 
\end{array}
\right),$$ which gives equal moments. Clearly, the choice for the spatial coefficients $x_j$ and $y_j$ are due to the fact that in this case the system consists of a set of independent dipoles. When $J$ is taken to zero, and $J'$ kept finite, then the appropriate choice to fix $\hat{O}$ is $x_j=0$ and $y_j=-1$. If instead the sign of $\Delta$ is changed $\hat{O}$ is again defined by the $x_j=0$ and $y_j=-1$. These results are clearly due to the reversal of the direction of the dipole moment within the unit cell. The results presented in Fig. \[fig:mom01\] corroborate our derivation.

Results and Analysis {#sec:results}
====================

We first look at the system with $J'=0$. In this case, the band structure of the system is simply two flat lines in the Brillouin zone. The system can be thought of as a simple two-state system. We calculated the first four GICs, from which we obtained the corresponding GIMs. 