= Solution
Write every fermion as a left-handed Weyl field. One generation is
$$
Q:(\mathbf3,\mathbf2)_{1/6},\quad
u^c:(\bar{\mathbf3},\mathbf1)_{-2/3},\quad
d^c:(\bar{\mathbf3},\mathbf1)_{1/3},\quad
L:(\mathbf1,\mathbf2)_{-1/2},\quad
e^c:(\mathbf1,\mathbf1)_1,\quad
\nu^c:(\mathbf1,\mathbf1)_0.
$$
The perturbative anomaly coefficients vanish as follows:
$$
\begin{aligned}
[SU(3)_c]^3 &: &&2-1-1=0,\\
[SU(3)_c]^2U(1)_Y &: &&2\!\left(\frac16\right)-\frac23+\frac13=0,\\
[SU(2)_L]^2U(1)_Y &: &&3\!\left(\frac16\right)-\frac12=0,\\
[U(1)_Y]^3 &: &&6\!\left(\frac16\right)^3+3\!\left(-\frac23\right)^3
+3\!\left(\frac13\right)^3+2\!\left(-\frac12\right)^3+1=0,\\
[\mathrm{gravity}]^2U(1)_Y &: &&6\!\left(\frac16\right)+3\!\left(-\frac23\right)
+3\!\left(\frac13\right)+2\!\left(-\frac12\right)+1=0.
\end{aligned}
$$
The common Dynkin-index factors have been suppressed in the mixed non-Abelian lines. An $SU(2)^3$ perturbative anomaly vanishes because the doublet is pseudoreal, while mixed anomalies containing one non-Abelian generator vanish because that generator is traceless. There are $3+1=4$ left-handed $SU(2)$ doublets after color multiplicity, so the nonperturbative <Witten SU(2) anomaly> also cancels. The gauge-singlet right-handed neutrino changes none of these sums.
Back to article page
