a canonically normalized kinetic term). This leads to a complex mass matrix for the fermions that is once again a direct product of a lattice and a gauge structure $${{\textstyle\frac{1}{2}}} \, (\lambda_{i\, \alpha\beta} \left|\, q_{i\, \alpha\beta}) \right.
\ \mathcal{M}_{ij\, \alpha\alpha'\, \beta \beta'} \
\left( \begin{array}{r} \lambda_{j\, \alpha'\beta'} \\\hline  q_{j\, \alpha'\beta'} \end{array} \right)
+ {\rm h.c.}$$ with $$\mathcal{M}_{ij\, \alpha\alpha'\, \beta \beta'} = 
i\sqrt{2} gv \
B_{\alpha\alpha'\, \beta\beta'} \
\left( \begin{array}{r|r}  & \Theta^t \\\hline \Theta &  \end{array} \right)_{ij}$$ where the gauge matrix $B$ has been defined in (\[Bscalar\]) and the lattice matrix $\Theta$ is given by $$\Theta = \left( \begin{array}{rrrr}
1&-1&&\\
&\ddots&\ddots&\\
&&\ddots&-1\\
-1&&&1
\end{array} \right)$$ It is then easy to derive the fermionic spectrum. The square of the gauge matrix $B^\dagger B=B B^\dagger$ has one zero eigenvalue and $M^2-1$ degenerate eigenvalues equal to 1. On the other hand, $\Theta^\dagger\Theta= \Theta \Theta^\dagger =\Omega$, showing that the fermionic mass levels agree with those computed before for the vectors and the scalars. Concerning the zero modes, as a result of the tracelessness of the generators of $SU(M)_i$ the trace of $q_i$ does not acquire a mass from the Kähler potential. Instead it combines with the fermionic component of the singlet $S_i$ to form a 