scalar and has a negative eigenvalue, therefore lacking a massless zero mode and does not survive at low energies.

In the case with general $P$, it must comply with $P^2=1$ from which we know that it only has eigenvalues $\pm 1$. This condition separates the generators $\{T_a\}=\{t_b,\tilde{t}_{\hat{b}}\}$ where $$P t_b P= t_b, \ \ \ P \tilde{t}_{\hat{b}} P=-\tilde{t}_{\hat{b}}.
\label{eq:hatb}$$ Therefore only the vector fields $A_\mu^b$ corresponding to parity even generators have a zero mode and, at low energies, the gauge group is broken into the subgroup generated by $t_b$, i.e. the ones that commute with $P$. Although the vector fields $A_\mu^{\hat{b}}$ corresponding to parity odd generators are massive, the 4d scalars $A_5^{\hat{b}}$ corresponding to the broken generators now have a zero mode, as we now discuss.

Since the orbifold operation $y=-y$ implies a parity transformation, there are two sources of a negative eigenvalue for the fields. The first one coming from the Lorentz representation, separating the extra dimensional component ($m=5$) from the 4d ones ($m=\mu$). The second one comes from the gauge representation, separating the broken generators ($\tilde{t}_{\hat{b}}$) from the unbroken ones ($t_b$). Each contribution is shown separately for each component of the gauge field, in the table \[tab:1\]. Only 