Past exam of the groups and representations course of the University of Oxford / 2020 / 3 / d / Solution

An element commuting with the full first SU(3) factor has its first component in that factor's center of a group, which is discrete. In the second factor, commutation with every $\mathrm{diag}(U,1)$ forces the doublet block to be scalar and the mixed blocks to vanish, by the Schur lemma. Unitarity and unit determinant then give the connected commuting subgroupThis is the centralizer of the standard SU2 subgroup in SU3 in the second factor; the full centralizer in $G$ also includes the first factor's discrete center of a group ${\mathbb{Z}}_3$. The Lie algebra generator is $iX$ in the anti-Hermitian convention. The opposite sign or a common nonzero rescaling changes the charge convention, not the commuting direction.

In the defining representation, the U(1) charges are $1,1,-2$. In the dual representation they have the opposite signs. In the Adjoint representation a component $T^i{}_j$ has U(1) charge $X_i-X_j$. Thus its upper-right doublet has charge $3$, its lower-left doublet has charge $-3$, and its diagonal blocks have charge zero. Adding the charges to solution givesEach subscript denotes the eigenvalue of $X$, so a state of charge $q$ acquires the phase $e^{iq\theta }$. The connected subgroup generated by this circle and the second SU(2) is $(SU(2)\times U(1{)}_X)/{\mathbb{Z}}_2$: $\theta =\pi$ gives $\mathrm{diag}(-I_2,1)$, also the central element of the embedded SU(2). This finite intersection does not change the displayed Lie algebra branching of a representation rules.