-\mathbf{1}_{n_{-}} &     & \\
    & \mathbf{0}_{n_0} & \\
    &      & \mathbf{1}_{n_{+}}
\end{pmatrix} \ , \quad
Y_a=
\begin{pmatrix}
Y_a^{(-)} &     & \\
    & Y_a^{(0)} & \\
    &      & Y_a^{(+)}
\end{pmatrix}$$ up to the SU($N_{\rm Y}$) symmetry and $N_{\rm Y}=n_{-} + n_0 + n_{+}$. This implies that $$\begin{aligned}
\frac{N_{\rm Y}-n_0}{N_{\rm Y}} M^3 =M \ , \quad \quad \quad
\label{eq: M} 
{{\left[ M, Y_a \right]}} =0 \ .\end{aligned}$$ Plugging (\[eq: M\]) into the equation of motion (\[eq: EOM\]) with the ansatz , we find that $X_\mu$ and $Y_a$ decouple as $$\begin{aligned}
{{\left[ X^\nu, {{\left[ X_\nu, X_0 \right]}} \right]}} - \xi X_0 &= 0  \ , \nonumber \\
{{\left[ X^\nu, {{\left[ X_\nu, X_i \right]}} \right]}} - \zeta X_i &= 0 \quad (i=1,2,3) \ ,
\label{eq: X_mu}
\\
\label{eq: Y_a}
{{\left[ Y^b, {{\left[ Y_b, Y_a \right]}} \right]}} - \zeta Y_a&= 0  \quad (a=4, \ldots , 9) \ . $$

In this work, we set the parameters $\xi$ and $\zeta$ to be equal in order to respect the SO(3,1) symmetry and take them to be positive. Note that once we impose this condition on the initial values of $\xi$ and $\zeta$, it is satisfied automatically at each step. The value of $\kappa$ turns out to be 