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Use the electroweak hypercharge normalization $Q=T_3+Y$, rather than $Q=T_3+\hat{Y}/2$. In the latter convention all tabulated hypercharges would be doubled. The electroweak representation and hypercharge table for one minimal Standard Model generation isThe quarks are colour triplets, the leptons are colour singlets, and all listed right-handed fields are weak singlets. A sterile right-handed neutrino, if added, has representation $(1,1{)}_0$ and changes none of the anomaly traces.

On each left weak doublet, $T_3=\mathrm{diag}(1/2,-1/2)$. The electric charges are therefore $1/6\pm 1/2=(2/3,-1/3)$ for $Q_L$ and $-1/2\pm 1/2=(0,-1)$ for $L_L$. On each right singlet $T_3=0$, so its electric charge equals its hypercharge. These are the observed quark and lepton charge assignments.

For Standard Model anomaly cancellation, traces must include both components of each doublet and all three colours. Every right weak singlet has $T_3^2=0$, givingOn the left doublets, $T_3^2=I/4$, while ${\mathrm{tr}}_L{T}_3=0$ separately in each doublet. ConsequentlyThe linear trace isEquivalently, ${\mathrm{tr}}_{L}Q=3(2/3-1/3)+(0-1)=0$. This proves cancellation of the mixed weak-hypercharge gauge anomaly within one generation.

The one-generation hypercharge anomaly traces for the cubic condition areThus both required anomaly differences vanish generation by generation. The right-handed subtraction is important: counting all physical fields with the same chirality sign would not give the cubic cancellation.