 Solution

Work entirely with left-handed Weyl fermions; a right-handed field is counted through its left-handed charge conjugate. For a simple non-Abelian group $G$, defineThe two orientations of a chiral triangle supply the symmetrized trace. Spectator representation dimensions multiply each trace. The perturbative gauge anomaly conditions for $G\times U(1)$ are thereforeThe first concerns $G^3$, the second $G^{2}U(1)$, and the third $U(1{)}^3$. A $GU(1{)}^2$ triangle has group factor ${\sum }_i{Q}_i^2{\mathrm{tr}}_{R_i}{T}^a=0$, because non-Abelian generators of a semisimple group are traceless. If coupling to gravity is required, also impose the mixed gauge-gravitational condition ${\sum }_i\dim (R_i)Q_i=0$. Cancellation of these local triangles is not a universal substitute for checking global anomalies; the Witten SU(2) anomaly is the relevant additional test here.^[ref]

For the Standard Model, one generation consists of $Q_L:(3,2{)}_{1/6}$, $u^c:(\bar{3},1{)}_{-2/3}$, $d^c:(\bar{3},1{)}_{1/3}$, $L_L:(1,2{)}_{-1/2}$ and $e^c:(1,1{)}_1$. These are precisely the all-left-handed representations; conjugating the singlets reverses their hypercharges relative to right-handed particles. Check the requested triangles separately:

For $3$–$3$–$3$, normalize the cubic coefficient of a fundamental to $+1$. Its conjugate has $-1$, so the weak-doublet multiplicity givesFor $2$–$2$–$2$, the doublet generators obey $\{T^b,T^c\}={\delta }^{bc}I/2$, so $\mathrm{tr}(T^a\{T^b,T^c\})=0$. Thus every doublet separately has zero local cubic weak anomaly. For the global Witten SU(2) anomaly, each generation has three coloured quark doublets and one lepton doublet, an even total of four. All three generations give twelve, so the global obstruction also cancels.

For $3$–$3$–$1$, use $T(3)=T(\bar{3})=1/2$. The two conjugate triplets have the same quadratic index, not its negative:For $2$–$2$–$1$, the three quark colours and one lepton doublet giveFor $1$–$1$–$1$, count every coloured and weak component:The mixed gauge-gravitational coefficient also vanishes:Unlisted gauge combinations contain a single generator of at least one simple factor. For example, $3$–$2$–$1$ has ${\mathrm{tr}}_3{T}^a{\mathrm{tr}}_2{t}^{b}Y$, and $3$–$3$–$2$ has ${\mathrm{tr}}_3(T^a{T}^b){\mathrm{tr}}_2{t}^c$; both vanish. The same reasoning covers $3$–$1$–$1$, $2$–$1$–$1$ and $3$–$2$–$2$, including permutations of the legs. Scalars contribute no chiral triangle anomaly. Therefore