e^{-i k_x},\\
&& \gamma^{BC}_{y,2~5} = \gamma^{BC}_{y,3~6} = \gamma^{BC}_{y,4~7} = \gamma^{BC}_{y,6~9} = \gamma^{BC}_{y,7~10} = \gamma^{BC}_{y,8~11}=\gamma^{BC}_{y,10~13} = \gamma^{BC}_{y,11~14}=\gamma^{BC}_{y,12~15}=\frac{K_2}{2},\\
&& \gamma^{BC}_{y,4~13} = \frac{K_2}{2}e^{-i (k_x - k_y)},\\
&& \gamma^{BC}_{y, 5~8} = \gamma^{BC}_{y,13~16} = \frac{K_1}{2}e^{-i k_y},\end{aligned}$$ where $\gamma^{BC}_{y, ab} = (\alpha^{BC}_{y, ba})^*$ and rest of the elements are zero.

The diagonal matrix elements of $\alpha^{BC}_z$ for the $z$-flavored bosons in the BC are $\alpha^{BC}_{z,a~a=1\sim16} = 2J_3 - K_1 - K_2 - 2 K_3$, while the off-diagonal matrix elements can be obtained from those of $\alpha^{BC}_y$ with $K_n \rightarrow J_n$. In addition, the matrix elements of $\gamma^{BC}_z$ can be obtained from those of $\gamma^{BC}_y$ by replacing $K_n \rightarrow J_n$.

The matrix elements for the PL are, $$\begin{aligned}
&& \alpha^{PL}_{y,a~a=1\sim16} = \frac{9}{2}J_3 - 2 K_3, \\
\nonumber && \alpha^{PL}_{y,1~2} = \alpha^{PL}_{y,1~5} = \alpha^{PL}_{y,2~6}=\alpha^{PL}_{y,3~4} =\alpha^{PL}_{y,3~7} = \alpha^{PL}_{y,4~8} = \alpha^{PL}_{y,5~6} =\alpha^{PL}_{y,7~8} = \alpha^{PL}_{y,9~10} = \alpha^{PL}_{y,9~13} = \alpha^{PL}_{y,10~14}=\alpha^{PL}_{y,11~12} =\\
&&\hspace{9cm} =\alpha^{PL}_{y,11~15} = \alpha^{PL}_{y,12~16} = \alpha^{PL}_{y,13~14} = \alpha^{PL}_{y,15~16} =\frac{K_1}{2},\\
&& \alpha^{PL}_{y,1~6} = \alpha^{PL}_{y,2~5} =\alpha^{PL}_{y,3~8} =\alpha^{PL}_{y,4~7} = \alpha^{PL}_{y,6~11} = \alpha^{PL}_{y,7~10} = \alpha^{PL}_{y,9~14} =\alpha^{PL}_{y,10~13} = \alpha^{PL}_{y,11~16} = \alpha^{PL}_{y,12~15} =\frac{K_2}{2},\\
&& \alpha^{PL}_{y,1~6} = \frac{K_2}{2}e^{-i(k_x + k_y)},\\
&& \alpha^{PL}_{y,2~15} = \alpha^{PL}_{y,3~14} = \frac{K_2}{2} e^{-i k_x}, \\
&& \alpha^{PL}_{y,4~13} = \frac{K_2}{2} e^{-i(k_x - k_y)},\\
&& \alpha^{PL}_{y,5~12} = \alpha^{PL}_{y,8~9}=\frac{K_2}{2} e^{-i k_y},\end{aligned}$$ where $\alpha^{PL}_{y, ab} = (\alpha^{PL}_{y, ba})^*$ 