= Solution
Use the <contraction splitting of a defining module tensor a dual symmetric square>. For a quadratic <polynomial> $p$ in the dual variables, define
$$
C(e_i\otimes p)=\partial_{y_i}p,\qquad
\iota(y_j)=Q_j:=\sum_{i=1}^3e_i\otimes y_i y_j.
$$
Both maps are <intertwiners>. The first is the natural <tensor contraction>; the second uses the <invariant tensor> $\sum_i e_i\otimes y_i$. Directly, $C(Q_j)=4y_j$, so $C\iota=4I$ and
$$
W=\ker C\oplus\operatorname{im}\iota,\qquad
\operatorname{im}\iota\cong V^*,\qquad \dim\ker C=18-3=15.
$$
The vector $T_{1;33}=e_1\otimes y_3^2$ lies in $\ker C$ and is killed by both positive simple-root operators $E_{12},E_{23}$. It is a <highest-weight vector> of <weight> $L_1-2L_3=\omega_1+2\omega_2$. By <Weyl complete reducibility theorem>, it generates the corresponding irreducible summand. The <Weyl dimension formula> for $\mathfrak{sl}_3$ gives $\dim\Gamma_{a,b}=(a+1)(b+1)(a+b+2)/2$, hence $\dim\Gamma_{1,2}=15$, exhausting the <kernel>. Therefore
$$
\boxed{V\otimes\operatorname{Sym}^2V^*\cong\Gamma_{1,2}\oplus\Gamma_{0,1},
\qquad \Gamma_{0,1}=V^*.}
$$
The <kernel> has the nine outer <weights> of multiplicity one and each inner <weight> $-L_j$ of multiplicity two; the dual summand has just the three <weights> $-L_j$, each once.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-2-sl3-submodule-weights.png]
{title=Weight diagrams and basis labels of the fifteen-dimensional kernel and the dual summand}
Back to article page
