 Solution

In this part the earlier variable $n$ specializes to six. A group representation of $U(1)\times SU(6)$ factors through the quotient group $U(6)$ exactly when the central subgroup $K$ in solution acts trivially. For an irreducible SU(6) tensor representation represented by a Young diagram with $b$ boxes, the scalar matrix ${\zeta }^{-1}{I}_6$ acts as ${\zeta }^{-b}$, where $\zeta =e^{2\pi i/6}$.

The prescribed representation $\tilde{R}(z,U)=R(U)$ gives no compensating phase from the first factor. Its value on the generator of $K$ is therefore $R({\zeta }^{-1}{I}_6)={\zeta }^{-b}I$. The box counts in solution giveConsequentlyConcretely, conjugation $T\mapsto WT{W}^{-1}$ defines it for every $W\in U(6)$, and the scalar phase cancels.

The qualification “prescribed lifts” matters. The defining representation of $U(6)$ certainly exists, but its pullback is $(z,U)\mapsto zU$, not $(z,U)\mapsto U$. For instance $(\zeta ,{\zeta }^{-1}{I}_6)$ and $(1,I_6)$ represent the same quotient element, while the latter prescription assigns them different matrices in the defining representation. More generally, for an integer $q$, a lift $z^{q}R(U)$ descends precisely whenThis descent criterion for representations of U(n) explains how all the listed SU(6) representations can be extended after choosing a suitable first-factor U(1) charge, even though only one descends with that charge fixed to zero.