= 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 <3/b/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> $\widetilde R(z,U)=R(U)$ 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 <3/c/solution> give
$$
\begin{array}{c|ccccc}
R&\mathbf6&\overline{\mathbf6}&\mathbf{21}&\mathbf{15}&\mathbf{35}\\\hline
b\pmod6&1&5&2&2&0\\
\widetilde R(\zeta,\zeta^{-1}I_6)&\zeta^{-1}I&\zeta I&\zeta^{-2}I&\zeta^{-2}I&I
\end{array}
$$
Consequently
$$
\boxed{\text{Among these prescribed lifts, only the adjoint }\mathbf{35}\text{ descends to }U(6).}
$$
Concretely, conjugation $T\mapsto WTW^{-1}$ 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 $z^qR(U)$ descends precisely when
$$
\boxed{q\equiv b\pmod6.}
$$
This <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.
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