 Solution

Write $|m_1,m_2⟩$ for product spin states, and $J_{+}=J_{1+}+J_{2+}$. A highest-weight state is killed by $J_{+}$. For $s=2$ the unique $m=2$ state is $|1,1⟩$. In the $m=1$ subspace, $J_{+}$ kills $a|1,0⟩+b|0,1⟩$ exactly when $a+b=0$. In the $m=0$ subspace it kills $a|1,-1⟩+b|0,0⟩+c|-1,1⟩$ exactly when $a+b=b+c=0$. Normalizing givesThe left states are coupled states and the right states are product states. Lowering commutes with particle interchange, so the whole spin-$s$ multiplet has the exchange sign of its highest-weight state: $+1$ for $s=0,2$ and $-1$ for $s=1$, or $(-1{)}^s$. Integer-spin identical particles are bosons; their spatial partial wave changes by $(-1{)}^{\ell }$ under interchange. Symmetry of the full state therefore requires $\ell +s$ even.

For decay from total angular momentum one, coupling requires $|\ell -s|\le 1\le \ell +s$. The case $s=0$ forces $\ell =1$ and violates Bose symmetry. For $s=1$, the allowed range is $\ell =0,1,2$ and symmetry leaves $\ell =1$. For $s=2$, the range is $\ell =1,2,3$ and symmetry leaves $\ell =2$. The identical particles' intrinsic parity product is $+1$, so total final parity is $(-1{)}^{\ell }$. HenceThis uses the stated massive-spin angular-momentum model; no additional restrictions for massless transverse vector particles have been assumed.