= 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 $\operatorname{diag}(U,1)$ 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 <subgroup>
$$
\boxed{U(1)_X=\{(I_3,\operatorname{diag}(e^{i\theta},e^{i\theta},e^{-2i\theta})):\theta\in\mathbb R\},\qquad X=\operatorname{diag}(1,1,-2).}
$$
This 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 <3/c/solution> gives
$$
\boxed{\begin{aligned}
(\mathbf3,\mathbf3)&\downarrow=(\mathbf3,\mathbf2)_{1}\oplus(\mathbf3,\mathbf1)_{-2},\\
(\overline{\mathbf3},\mathbf1)&\downarrow=(\overline{\mathbf3},\mathbf1)_{0},\\
(\mathbf1,\mathbf3)&\downarrow=(\mathbf1,\mathbf2)_{1}\oplus(\mathbf1,\mathbf1)_{-2},\\
(\mathbf1,\mathbf8)&\downarrow=(\mathbf1,\mathbf3)_{0}\oplus(\mathbf1,\mathbf2)_{3}\oplus(\mathbf1,\mathbf2)_{-3}\oplus(\mathbf1,\mathbf1)_{0}.
\end{aligned}}
$$
Each 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 $\operatorname{diag}(-I_2,1)$, also the central element of the embedded <SU(2)>. This finite intersection does not change the displayed <Lie algebra> <branching of a representation> rules.
Back to article page
