Consider spin-orbit coupling (of strength $\lambda_1$) on lattice, with the below Hamiltonian
$$H = i \lambda_1 \sum_{<ij>} ~\frac{E_{ij} \times R_{ij}}{|E_{ij} \times R_{ij}|} \cdot \sigma ~c_i^\dagger c_j $$
with lattice sites $i, j$, nearest-neighbor connecting sites vector $R_{ij}$, E-field $E_{ij}$ and Paulis matrix $\sigma$.
Consider 2D plane, so $R_{ij} = (R_{ij}^x, R_{ij}^y, 0)$ and choose E-field $E_{ij} = (E_{ij}^x, E_{ij}^y, 0)$, with $E_{ij}^x, E_{ij}^y >0$. Factor in above Hamiltonian is
$$\frac{E_{ij} \times R_{ij}}{|E_{ij} \times R_{ij}|} \cdot \sigma = \sigma_z ~\text{sgn} (E_{ij}^x R_{ij}^y - E_{ij}^y R_{ij}^x)$$
Paper here considers 2d Kagome lattice, with Hamiltonian for spin orbit appearing in 1st line of Eq. (1). Going into k-space, the authors shown in Eq. (2) that spin-orbit Hamiltonian gives terms with cosines, like $\cos (k_x)$.
However, it looks to me like terms should be sines, like $\sin(k_x)$.
Consider the 2-d Kagome lattice shown in Fig 1 of the paper. There will be terms proportional to that below to make the horizontal part of the lattice along x direcion, where $R_{ij}^y = 0$:
$$\sum_n \text{sgn} (- E_{ij}^y R_{ij}^x) c_n^\dagger c_{n+1} \to \sum_k e^{-i k_x} c_k^\dagger c_k $$
and
$$ \sum_n \text{sgn} (- E_{ij}^y R_{ij}^x) c_n^\dagger c_{n-1} \to \sum_k - e^{+i k_x} c_k^\dagger c_k $$
But beacuse of opposite direction of $R_{ij}^x$ in top and bottom cases sgn function will be different, so that exponentials $\exp$ add to make a $\sin(k_x)$ and not a $\cos(k_x)$ as is put in second line of Eq. (2) of the paper.
Where is a gap in my understanding of spin orbit on 2-d lattice?