Y}_{L_1}({{\mbox{\boldmath $x$}}}'_1)
 {\cal Y}_{L_2}({{\mbox{\boldmath $x$}}}'_2)]_{L_{12}}
{\cal Y}_{L_3}({{\mbox{\boldmath $x$}}}'_3)\right]_{LM},
\label{K.basis}$$ where $\widetilde{{{\mbox{\boldmath $x$}}}'}=({{\mbox{\boldmath $x$}}}'_1,{{\mbox{\boldmath $x$}}}'_2,{{\mbox{\boldmath $x$}}}'_3)$ is the K-coordinate set (see Fig. 1(b)) and $A_K$ is a 3$\times$3 diagonal matrix. Noting that ${{\mbox{\boldmath $x$}}}'$ is equal to ${{\mbox{\boldmath $x$}}}'=U_{KH}{{\mbox{\boldmath $x$}}}$, we observe that the basis function (\[K.basis\]) is obtained from Eq. (\[cgtgv\]) by a particular choice of parameters, that is, $\widetilde{u_1}$=(1,0,0), $\widetilde{u_2}$=(0,$-\frac{1}{2}$,$1$) and $\widetilde{u_3}$=(0,$\frac{2}{3}$,$\frac{2}{3}$), and the matrix $A$ is related to $A_K$ by $$\begin{aligned}
A=(u_1 u_2 u_3)A_K \left(
\begin{array}{c}
\widetilde{u_1}\\
\widetilde{u_2}\\
\widetilde{u_3}\\
\end{array}
\right)
=\widetilde{U_{KH}}A_K U_{KH}.\end{aligned}$$ Thus the form of the $F$-function remains unchanged under the transformation of relative coordinates.

Note that $A$ is no longer diagonal. The choice of a different set of coordinates ends up only choosing appropriate parameters for $A$, $u_1$, $u_2$, and $u_3$.

It is also noted that the triple global vectors in Eq. (\[cgtgv\]) are a minimum number of vectors to provide all possible spatial functions with arbitrary $L$ and parity $\pi$. A natural parity state with $\pi=(-1)^L$ can be described by only one global vector, that is, using e.g., $L_1=L$, $L_2=0$, $L_{12}=L$, $L_3=0$ [@vs95; @suzuki98; @GVR]. To describe an unnatural parity state with $\pi=(-1)^{L+1}$ except for $0^-$ case, we need at least two global vectors, 