the fields with final combined $+1$ eigenvalue have zero mode.

   $A_m^b T_b$   $T_b=t_b$   $T_b=\tilde{t}_{\hat{b}}$
  ------------- ----------- ---------------------------
    $m=\mu $     $(1)(1)$            $(1)(-1)$
     $m=5 $      $(-1)(1)$          $(-1)(-1)$

  : Eigenvalues of $A_m^b$ under $(P)(\mathcal{P}_5)$ respectively for each component. The components $A_\mu^b, A_5^{\hat{b}}$ have a combined $+1$ eigenvalue and hence have massless modes.[]{data-label="tab:1"}

Then we conclude that the gauge transformation $P$ breaks the gauge symmetry at low energies into the subgroup that commutes with it. Furthermore it allows the 4d scalars corresponding to the broken generators to have zero modes.

Note that the $P$ by itself breaks the gauge symmetry at low energies but it can’t reduce the rank of the remaining gauge group. The effective 4d scalars with zero modes come from the broken generators, and if they obtain a VEV, they would reduce the rank. However they come from the extra dimensional gauge vector component and they do not have a potential that gives them a VEV. As we discuss in the next subsection, a mechanism capable of giving the 4d scalars with zero modes a VEV may be provided by Wilson lines. In this way, the combination of Wilson lines and orbifolds may break the rank of the gauge group.

Wilson 