choice of the Wilson line phase.

In the presence of both Wilson lines and orbifold boundary conditions, we need to consider the generators of the $SO(3)$ algebra which are $$T_{1}=\left(\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0\end{array} \right),\ \ T_{2}=\left(\begin{array}{ccc}0 & 0 & -1 \\ 0 & 0 & 0 \\ 1 & 0 & 0\end{array} \right),\ \ T_{3}=\left(\begin{array}{ccc}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{array} \right),$$ Any element of the algebra of $SO(3)$, and therefore any phase from a Wilson line as in Eq. \[eq:psi2pir\], can be written as $$U=e^{i\beta^aT_a},  \ \ \ \ 
T_a\beta^a=\left(\begin{array}{ccc}
0&\beta^3&\beta^2\\
-\beta^3&0&\beta^1\\
-\beta^2&-\beta^1&0
\end{array}\right),
\label{eq:uso3}$$ with real $\beta^a$. This can be rewritten in the usual flavon notation as a triplet column vector $$\braket{\phi}=\frac{1}{2\pi R}\left(\begin{array}{c}\beta^1\\ \beta^2 \\ \beta^3\end{array}\right),
\label{eq:uso31}$$ so that we can obtain any desired real alignment by choosing the phase.[^6]

If we only have the Wilson line, without any $SO(3)$ orbifold breaking condition, it can always be rotated to the $(0,0,1)^T$ direction, which breaks $SO(3)\to U(1)$, preserving the rank.

On the other hand if we consider orbifold breaking without Wilson lines, the parity $P$ boundary 