= Decomposition of three SU(2) spinors
{title2=$\mathbf2\otimes\mathbf2\otimes\mathbf2=\mathbf4\oplus\mathbf2\oplus\mathbf2$}
For $Q^{\alpha\beta\gamma}$, put $S=Q^{(\alpha\beta\gamma)}$, $d_1^\alpha=\epsilon_{\beta\gamma}Q^{\alpha\beta\gamma}$, $d_2^\alpha=\epsilon_{\beta\gamma}Q^{\beta\alpha\gamma}$ and $d_3^\alpha=\epsilon_{\beta\gamma}Q^{\beta\gamma\alpha}$. With $\epsilon_{12}=1$ and inverse $\epsilon^{12}=-1$, the two-dimensional alternating identity gives $d_1-d_2+d_3=0$. The inverse decomposition is $Q^{\alpha\beta\gamma}=S^{\alpha\beta\gamma}+\epsilon^{\alpha\beta}(2d_1^\gamma-d_2^\gamma)/3-\epsilon^{\alpha\gamma}(d_1^\beta+d_2^\beta)/3$. The symmetric part is the spin-$3/2$ <SU(2) representation>, and the two independent contractions are spin-$1/2$ representations. This is the spin coupling underlying ground-state three-quark <baryons>, before the <Pauli constraint on three-quark flavour multiplets> is imposed.
// Target: quantum-field-theory.bigb
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