(\phi_i \times \phi_{i-1}) -(\phi_i^\dagger \times \phi_{i-1}^\dagger)
- (\phi_i^\dagger \times \phi_{i-1}) -(\phi_i \times \phi_{i-1}^\dagger)
\right)
\ ,\end{aligned}$$ which we can rewrite as, $${\cal L}\supset -{{\textstyle\frac{1}{2}}}\,
(\phi_{i}^{\alpha\beta} \phi_{i}^{*\alpha\beta}) \
\mathcal{M}^2_{ij\, \alpha\alpha' \, \beta\beta'} \
\left( \begin{array}{r} \phi_j^{*\alpha'\beta'} \\ \phi_j^{\alpha'\beta'} \end{array} \right)$$ and the mass matrix is again a direct product of two matrices in the gauge and lattice index spaces $$\mathcal{M}^2_{ij\, \alpha\alpha' \, \beta\beta'}=
g^2 v^2\
\Omega_{ij}
\left( \begin{array}{rrrr}
A_{\alpha\alpha' \, \beta\beta'}  & B_{\alpha\alpha' \, \beta\beta'} \\
B_{\alpha\alpha' \, \beta\beta'} &  A_{\alpha\alpha' \, \beta\beta'}
\end{array} \right)
\ .$$ The lattice matrix is the same as the one appearing in the gauge boson mass matrix while now the gauge matrices are non diagonal and are given by $$\begin{aligned}
    \label{Ascalar}
A_{\alpha\alpha' \, \beta\beta'} & = &
\delta_{\alpha\alpha'} \delta_{\beta\beta'}
-{{\textstyle\frac{1}{M}}}\, \delta_{\alpha\beta} \delta_{\alpha'\beta'}\ , \\
    \label{Bscalar}
B_{\alpha\alpha' \, \beta\beta'} & = &
\delta_{\alpha\beta'} \delta_{\beta\alpha'}
-{{\textstyle\frac{1}{M}}}\, \delta_{\alpha\beta} \delta_{\alpha'\beta'}
\ .\end{aligned}$$ The second terms in (\[Ascalar\]) and (\[Bscalar\]) are due to the projection out of the trace in the Fierz transformations (\[Fierz\]), as a consequence of the tracelessness of the generators of $SU(M)$.

We already know the eigenvalues of $\Omega$ from (\[eq:mkomega\]) and it is easy to check that the gauge matrix has only two eigenvalues: 0, with 