sl3 decomposition of the symmetric-square dual tensor product ($(S^{2}V)\otimes V^{\ast }\cong {\Gamma }_{2,1}\oplus {\Gamma }_{1,0}$)

For the defining three-dimensional special linear Lie algebra representation $V$, contraction $(uv)\otimes f\mapsto f(u)v+f(v)u$ maps $(S^{2}V)\otimes V^{\ast }$ onto $V$. The kernel contains the highest-weight vector $e_1^2\otimes e_3^{\ast }$ of Dynkin labels $(2,1)$. The Weyl dimension formula gives its irreducible module dimension $15$, the entire kernel dimension; the Weyl complete reducibility theorem supplies the displayed splitting.

Let $t_{ij}=(e_i\otimes e_j+e_j\otimes e_i)/2$, and let $f_i$ be the dual basis. The equivariant contraction sends $t_{ij}\otimes f_k$ to ${\delta }_{ik}{e}_j+{\delta }_{jk}{e}_i$. Its kernel has a weight basis consisting of $t_{ii}\otimes f_k$ for $i\ne k$, $t_{ij}\otimes f_k$ for three distinct indices with $i<j$, and $t_{ij}\otimes f_j-\frac{1}{2}{t}_{ii}\otimes f_i$ for $i\ne j$. The three families have sizes six, three and six. This is the irreducible fifteen-dimensional highest-weight representation with highest weight $2L_1-L_3$.

The formal character identity ${\chi }_{{\wedge }^{2}U}(x)=\frac{1}{2}({\chi }_U(x{)}^2-{\chi }_U(x^2))$ counts unordered pairs of distinct basis vectors. Substituting ${\chi }_{2,1}=h_2(x)h_1(x^{-1})-h_1(x)$ and the Weyl character formula gives the displayed multiplicity-free decomposition. Its dimensions sum to $21+42+24+15+3=105=\left(\genfrac{}{}{0ex}{}{15}{2}\right)$. The sl3 interlacing character formula provides an exact weight-enumeration check.