= Cubic anomaly coefficient
{title2=$A(R)$}
For a <representation> $R$ of $SU(N)$ with $N\ge3$, choose the <Hermitian quantum generator convention> and normalize the cubic anomaly coefficient by
$$
\operatorname{tr}_R\bigl(T^a\{T^b,T^c\}\bigr)=A(R)\operatorname{tr}_{\mathbf N}\bigl(T^a\{T^b,T^c\}\bigr),\qquad A(\mathbf N)=1.
$$
Here the <anticommutator> symmetrizes the last two generators; <cyclic property of the trace> makes the whole tensor symmetric. This coefficient multiplies the local four-dimensional <gauge anomaly> from a chiral fermion in $R$. It adds under <direct sums>.
The <conjugate representation> has generators $-(T^a)^T$, so transposition and the three minus signs give $A(\overline R)=-A(R)$. Hence a <real representation> or <pseudoreal representation> has zero cubic anomaly coefficient. For $SU(3)$, the <two-index antisymmetric representation> is the conjugate triplet and has coefficient $-1$. For $SU(2)$ the symmetric cubic invariant itself vanishes, so normalization to a nonzero fundamental cubic coefficient is unavailable; this statement concerns the local cubic anomaly and does not settle global anomalies.<Lectures on Anomalies by Adel Bilal>[Section 7.2, equations (7.13)–(7.15), representation conjugation and vanishing cubic traces]{ref}
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