Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 2 / 1 / b

Let $e_1,e_2,e_3$ be the basis of the defining representation and $y_1,y_2,y_3$ its dual. On a diagonal element of the Cartan subalgebra their weights are $L_i$ and $-L_i$, with $L_1+L_2+L_3=0$. Use the tensor product and symmetric square basisIts weight is $L_i-L_j-L_k$. For the three repeated weights putIf $\{i,l\}=\{1,2,3\}\setminus \{j\}$, the complete weight-space decomposition isThere are six spaces of the first type, three of the second, and three of the third. Their dimensions sum to $6+3+9=18$. The weight diagram below labels every space with all of its basis vectors.

Use the contraction splitting of a defining module tensor a dual symmetric square. For a quadratic polynomial $p$ in the dual variables, defineBoth 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 $C\iota =4I$ andThe 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. ThereforeThe 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.

Use the basis labels in the preceding diagrams. For each $j$, write $\{i,l\}=\{1,2,3\}\setminus \{j\}$ with $i<l$, and defineThe contraction satisfies $C(A_j)=2y_j$ and $C(B_{i;j})=y_j$, so both $K$ vectors lie in $\ker C$. They are independent and form a basis of its weight space at $-L_j$.

Thus the bases marked on the weight diagram of ${\Gamma }_{1,2}$ areand the diagram of ${\Gamma }_{0,1}$ uses the basis $Q_j=A_j+B_{i;j}+B_{l;j}$ at $-L_j$. Each $Q_j$ is complementary to the two kernel vectors because its contraction is $4y_j\ne 0$. All vectors shown are therefore genuine bases, not merely weight counts. The highest-weight vectors areThe latter is highest because $\iota$ intertwines the dual Lie algebra representation and $y_3$ is its highest-weight vector.