condition can always be rotated to be $$P=e^{i\pi T_{3}}=\left(\begin{array}{ccc}-1&0&0 \\ 0 & -1 & 0\\ 0 & 0 &1 \end{array}\right),
\label{eq:pso3}$$ which breaks $SO(3)\to U(1)$ (and also does not reduce rank).

However, as discussed in the previous section, the combination of Wilson lines and orbifold boundary conditions can reduce the rank of the remaining symmetry group. The boundary condition $P$ leaves an $U(1)$ invariant and the Wilson Line can give a VEV in that remaining $U(1)$.

Imposing the $U$ from Eq. \[eq:uso3\] together with the $P$ from Eq. \[eq:pso3\], they must comply with the consistency condition from Eq. \[eq:pupu\] $$\begin{split}
PUPU&=1\\
&=(Pe^{i\beta^aT_a}P)e^{i\beta^aT_a}=e^{i\beta^aPT_aP}e^{i\beta^aT_a}\\
&=e^{i(-\beta^1T_1-\beta^2T_2+\beta^3T_3)}e^{i(\beta^1T_1+\beta^2T_2+\beta^3T_3)},
\end{split}$$ which has 2 nontrivial solutions. The first one is where $\beta^3=0$ and general real $\beta^{1,2}$. This is equivalent to a flavon alignment $$\braket{\phi}=\frac{1}{2\pi R}\left(\begin{array}{c} \beta^1\\ \beta^2 \\ 0
\end{array}\right),$$ which breaks the remaining $U(1)$. However in this example there is also a rank preserving solution. [^7]

In the $SO(3)$ gauge theory with 6 dimensions where the orbifold is $T^2/\mathbb{Z}_2$, there would be two flavons obtained. There would also be two possible Wilson lines associated with each extra dimensions $$U_5=e^{i\beta^a_5T_a},\ \ \ U_6=e^{i\beta^a_6T_a}.$$ Coming from translations of independent circles, these phases must commute $[U_5,U_6]=0$, and this fixes 