$$\beta^a_5\propto\beta^a_6,$$ which means that the two flavon VEVs must be aligned with each other so nothing is gained by having two flavons with identically aligned VEVs. However, we can override this condition assuming an $SO(3)$ gauge theory, where there is a non-commutative twist in the fibre bundle as described in Sec. \[sec:noncom\], leading to non-aligned VEVs in that case. However for the moment we shall continue to consider the usual commutative case, and go on to consider another example.

$SU(4)$ Gauge-Flavon Unification {#su4}
================================

Flavon sector
-------------

As a second example, let us assume that we have a $SU(4)$ gauge family symmetry model in $5$ dimensions with an orbifold $S^1/Z_2$. Since this group allows complex representations, it will allow two non-aligned flavons in conjugate representations. In this case, let us assume that the extra dimensions are orbifolded with a twist on the ED parity $$P_M\sim{\mathrm{diag}}(1,1,1,-1),$$ that breaks $SU(4)\to SU(3)\times U(1)$. The generators $T_a$ with $a=1,...,15$ decompose as $$\begin{split}
T_a&=\lambda_a,\ \ \ {\rm for}\ \ \ a=1,...,8,\ \ \ T_{15}={\mathrm{diag}}(-1/3,-1/3,-1/3,1)\\
T_9&=
\left(\begin{array}{cccc} 0&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&0
\end{array}\right),\ \ \ T_{11}=
\left(\begin{array}{cccc} 0&0&0&0\\0&0&0&1\\0&0&0&0\\0&1&0&0
\end{array}\right),\ \ \ T_{13}=
\left(\begin{array}{cccc} 0&0&0&0\\0&0&0&0\\0&0&0&1\\0&0&1&0
\end{array}\right),\\
T_{10}&=
\left(\begin{array}{cccc} 0&0&0&-i\\0&0&0&0\\0&0&0&0\\i&0&0&0
\end{array}\right),\ \ \ T_{12}=
\left(\begin{array}{cccc} 0&0&0&0\\0&0&0&-i\\0&0&0&0\\0&i&0&0
\end{array}\right),\ \ \ T_{14}=
\left(\begin{array}{cccc} 0&0&0&0\\0&0&0&0\\0&0&0&-i\\0&0&i&0
