$-t_1$ for $T_{111}$ is explicitly written as $$\begin{aligned}
|-t_1 \rangle &=& 
\frac{1}{2}{\textrm e}^{{\textrm i}\left(\theta_{111}-\frac{\pi}{2}\right)}
       \left[ |+ \uparrow \rangle
 + {\textrm e}^{-{\textrm i}\frac{\pi}{4}} |+ \downarrow \rangle
\right] \nonumber \\
& & 
- \frac{1}{2} \left[ |- \uparrow \rangle
 + {\textrm e}^{ {\textrm i}\frac{3 \pi}{4}} |- \downarrow \rangle
\right],\end{aligned}$$ with $\theta_{111}$ being an angle between vector $(1,1,1)$ and the $z$ axis, that is, $\cos \theta_{111} = \sqrt{1/3}$, $\sin \theta_{111} = \sqrt{2/3}$. This state is different from a ground state assumed by Lovesey [*et al*]{}, who considered the state deviating from $\Gamma_8$ quartet.[@Lovesey03] Using the eigenstates as bases, $T_p$ is represented as $$T_p = 
\left(\begin{array}{cccc}
    -t_1 & 0 & 0 & 0 \\
    0 &+t_1 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
\end{array} \right).$$ The ground state is given by assigning either of eigenstates of $\pm t_1$ to each sublattice; which eigenstate is relevant depends on the sign of acting mean field. As shown in Fig. \[fig.struc2\], one longitudinal order and two transverse orders are possible in the triple-**k** ordering; for the longitudinal one, eigenstates of $T_{111}$, $T_{\overline{1}11}$, $T_{1\overline{1}1}$ and $T_{11\overline{1}}$ are assigned to sublattices 1, 3, 4 and 2, 