Clebsch-Gordan decomposition for SU2

For $m,n\ge 0$,The identity follows by multiplying the weight characters and comparing their nested weight strings.

Realize $V_n={\mathrm{Sym}}^n{\mathbb{C}}^2$. Contract $r$ pairs of indices, one from each symmetric tensor, with the alternating invariant tensor $ϵ$, and symmetrize the remaining indices. This is a nonzero intertwining operator into $V_{m+n-2r}$. In the bihomogeneous polynomial realization, $(z_1{w}_2-z_2{w}_1{)}^r{z}_1^{m-r}{w}_1^{n-r}$ is a highest-weight vector of weight $m+n-2r$. Its lowering string has length $m+n-2r+1$. These inequivalent irreducible summands exhaust the tensor product, since their dimensions sum to $(m+1)(n+1)$.

For $Q^{\alpha \beta \gamma }$, put $S=Q^{(\alpha \beta \gamma )}$, $d_1^{\alpha }={ϵ}_{\beta \gamma }{Q}^{\alpha \beta \gamma }$, $d_2^{\alpha }={ϵ}_{\beta \gamma }{Q}^{\beta \alpha \gamma }$ and $d_3^{\alpha }={ϵ}_{\beta \gamma }{Q}^{\beta \gamma \alpha }$. With ${ϵ}_{12}=1$ and inverse ${ϵ}^{12}=-1$, the two-dimensional alternating identity gives $d_1-d_2+d_3=0$. The inverse decomposition is $Q^{\alpha \beta \gamma }=S^{\alpha \beta \gamma }+{ϵ}^{\alpha \beta }(2d_1^{\gamma }-d_2^{\gamma })/3-{ϵ}^{\alpha \gamma }(d_1^{\beta }+d_2^{\beta })/3$. 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

On the multiplicity-one summand $V_{2n-2j}\subset V_n\otimes V_n$, interchange of tensor factors acts by $(-1{)}^j$. A highest-weight vector exhibiting the sign is

The odd-parity summands give

The alternating summands in the tensor square give ${\bigwedge }^2{V}_4\cong V_6\oplus V_2$. Since $V_4$ is self-dual and has trivial determinant,