= Solution
Count every right-handed field as a left-handed conjugate, which reverses its $U(1)$ charge and conjugates its non-Abelian representation. The nontrivial local anomaly-cancellation conditions are
$$
[SU(3)]^2U(1):\qquad \boxed{2q-u-d=0},
$$
$$
[SU(2)]^2U(1):\qquad \boxed{3q+l=0},
$$
$$
[U(1)]^3:\qquad
\boxed{6q^3+2l^3-3u^3-3d^3-x^3=0}.
$$
The pure $SU(3)^3$ anomaly cancels because the two triplets in $Q_L$ balance the conjugates of $u_R$ and $d_R$. There is no perturbative $SU(2)^3$ anomaly, and the <Witten SU(2) anomaly> also vanishes because the three colored quark doublets plus one lepton doublet make four. The <mixed gauge-gravitational anomaly> is excluded at this stage as requested.
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