= Solution
Use the <electroweak hypercharge> normalization $Q=T_3+Y$, rather than $Q=T_3+\widehat Y/2$. In the latter convention all tabulated hypercharges would be doubled. The <electroweak representation and hypercharge table> for one minimal <Standard Model> generation is
$$
\begin{array}{c|c|c|c|c}
\text{field}&SU(3)_{\rm colour}&SU(2)_T&Y&Q\\ \hline
Q_L=(u_L,d_L)^T&\mathbf3&\mathbf2&1/6&(2/3,-1/3)\\
u_R&\mathbf3&\mathbf1&2/3&2/3\\
d_R&\mathbf3&\mathbf1&-1/3&-1/3\\
L_L=(\nu_L,e_L)^T&\mathbf1&\mathbf2&-1/2&(0,-1)\\
e_R&\mathbf1&\mathbf1&-1&-1
\end{array}
$$
The <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 $(\mathbf1,\mathbf1)_0$ and changes none of the anomaly traces.
On each left weak doublet, $T_3=\operatorname{diag}(1/2,-1/2)$. The electric charges are therefore $1/6\pm1/2=(2/3,-1/3)$ for $Q_L$ and $-1/2\pm1/2=(0,-1)$ 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 $T_3^2=0$, giving
$$
\boxed{\operatorname{tr}_R(T_3^2Y)=0.}
$$
On the left doublets, $T_3^2=I/4$, while $\operatorname{tr}_LT_3=0$ separately in each doublet. Consequently
$$
\boxed{\operatorname{tr}_L(T_3^2Y)=\tfrac14\operatorname{tr}_LY=\tfrac14\operatorname{tr}_LQ.}
$$
The linear trace is
$$
\operatorname{tr}_LY=3\cdot2\cdot\frac16+2\left(-\frac12\right)=1-1=0.
$$
Equivalently, $\operatorname{tr}_LQ=3(2/3-1/3)+(0-1)=0$. This proves cancellation of the mixed weak-hypercharge <gauge anomaly> within one generation.
The <one-generation hypercharge anomaly traces> for the cubic condition are
$$
\begin{aligned}
\operatorname{tr}_LY^3&=6\left(\frac16\right)^3+2\left(-\frac12\right)^3=\frac1{36}-\frac14=-\frac29,\\
\operatorname{tr}_RY^3&=3\left(\frac23\right)^3+3\left(-\frac13\right)^3+(-1)^3=\frac89-\frac19-1=-\frac29.
\end{aligned}
$$
Thus \b[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.
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