four vectors ${\bf e}_a$ are chosen to be orthonormal among themselves and orthogonal to all ${\bf e}_\t$. Consequently, we have a basis satisfying $${\bf e}_\t\cdot{\bf e}_{\s} = A_{\t\s}\; ,\qquad
 {\bf e}_\t\cdot{\bf e}_a = 0\; ,\qquad
 {\bf e}_a\cdot{\bf e}_b = \d_{ab}\; .$$ Expanding the fields $\bph$ $$\bph = \sum_A\r_A{\bf e}_A$$ as before one finds for the Lagrangian  $$\cL = \frac{1}{4}\sum_{\t ,\s}A_{\t ,\s}\dot{\r}_\t\dot{\r}_\s
       +\frac{1}{4}\sum_a\dot{\r}_a^2-V$$ with potential $$V = 4\sum_\t\z_\t^2\,\exp\left(\sum_\s A_{\t\s}\r_\s\right)\; .$$ The modes $\r_a$ are decoupled from this potential and their equations of motion immediately lead to the general solution $$\r_a = k_a(\t -\t_a )\; ,$$ where $k_a$ and $\t_a$ are constants. Following the methods of Ref. [@Kostant:qu], the solutions for the other modes are obtained as $$e^{-\r_\t} = \sum_{\bla\in\L_\t}b_\t (\bla )\exp\left(\bla\cdot ({\bf k}
              \t - \btau )\right)\; , \label{sol4}$$ where ${\bf k}$ and $\btau$ are constant vectors. The three sets $\L_\t$ contain the weights of the fundamental representation ${\bf 4}$, the vector representation ${\bf 6}$ and the anti-fundamental representation $\bar{\bf 4}$ of $SU(4)$, respectively, and are explicitly given by $$\begin{aligned}
 \L_1 &=& \{ (100),(-110),(0-11),(00-1)\} \\
 \L_2 &=& \{ (010),(1-11),(-101),(10-1),(-11-1),(0-10)\} \\
 \L_3 &=& \{ (001),(01-1),(1-10),(-100)\} \; .\end{aligned}$$ Finally, $b_\t (\bla )$ represents a set of constants which depends on $\z_\t$ as well 