Descent criterion for representations of U(n) ($q\equiv b(\mathrm{mod}n)$)

The map $(z,U)\mapsto zU$ identifies the unitary group with $(U(1)\times SU(n))/K$, where $K=\{({\zeta }^k,{\zeta }^{-k}{I}_n):\zeta =e^{2\pi i/n}\}$. If an irreducible SU(n) tensor transformation has $b$ boxes in its Young diagram, the central scalar $\zeta I_n$ acts as ${\zeta }^b$. Therefore, for integer $q$, $(z,U)\mapsto z^{q}R(U)$ is trivial on $K$ precisely when ${\zeta }^{q-b}=1$, proving the criterion. Adding a full $n$-box column changes $b$ by $n$ and leaves the condition unchanged. In particular a lift with trivial first-factor U(1) charge descends only if its central action is trivial; this does not prevent other lifts of the same special unitary group representation from descending.