Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 3 / 1 / b

Let $x_1,x_2,x_3$ be the coordinate basis of the defining representation $W$ of ${\mathfrak{sl}}_3$. Write $s_{ij}=x_i{x}_j$ for $1\le i\le j\le 3$. The six $s_{ij}$ form a basis of the symmetric power ${\mathrm{Sym}}^{2}W$, with weight $L_i+L_j$. A basis for the tensor product of Lie algebra representations $T$ consists of all 36 ordered tensorsThe order of the two factors matters; there is no quotient identifying their interchange.

Each weight of a representation has the form $k_1{L}_1+k_2{L}_2+k_3{L}_3$, where the nonnegative integers satisfy $k_1+k_2+k_3=4$. Its weight multiplicity is the number of ways to split this triple into two nonnegative triples, each summing to two. The possible patterns, including every permutation, areFor example, $(2,2,0)$ splits as $(2,0,0)+(0,2,0)$, the reverse split, or $(1,1,0)+(1,1,0)$, giving multiplicity three. For $(2,1,1)$, the four splits pair $(2,0,0)$ with $(0,1,1)$ and $(1,1,0)$ with $(1,0,1)$, each in both orders. The multiplicities sum to $3+12+9+12=36$, verifying the dimension.

The following weight diagram uses Dynkin labels $(p,q)=(k_1-k_2,k_2-k_3)$; the circle numbers give multiplicities and orange points mark the Closed dominant Weyl chamber.

Take the simple roots $L_1-L_2$ and $L_2-L_3$. A weight represented by $(k_1,k_2,k_3)$ is in the Closed dominant Weyl chamber exactly when $k_1\ge k_2\ge k_3$. Among nonnegative triples of total four, the possibilities are $(4,0,0),(3,1,0),(2,2,0),(2,1,1)$. Using $L_1+L_2+L_3=0$, their explicit weight space bases areEach listed tensor has the required summed weight, and the monomial basis from part (i) shows linear independence and exhausts all possible tensors of that weight. Thus these are all the requested bases.

Use the positive simple-root raising operators $E_{12}=x_1{\partial }_{x_2}$ and $E_{23}=x_2{\partial }_{x_3}$. On the weight space of $2L_1-L_3$, the first maps both $s_{11}\otimes s_{12}$ and $s_{12}\otimes s_{11}$ to $s_{11}\otimes s_{11}$, while the second annihilates both. Consequentlyis a nonzero highest-weight vector; the remaining positive-root operator $E_{13}=[E_{12},E_{23}]$ also annihilates it. Its Dynkin labels are $(2,1)$, and the generated submodule $Z$ is the irreducible highest-weight representation with that highest weight.

Put $F_1=E_{21}=x_2{\partial }_{x_1}$ and $F_2=E_{32}=x_3{\partial }_{x_2}$, acting on both tensor factors. Direct calculation givesDefine the linearly independent tensorsThey both have weight $L_1$, andTherefore $A,B\in Z_{L_1}$. To prove that they exhaust this weight space, its weight differs from the highest weight by $(L_1-L_2)+(L_2-L_3)$. The Poincaré-Birkhoff-Witt theorem expresses all descendants at that difference using $F_2{F}_{1}v$ and $F_1{F}_{2}v$: the third possible operator $E_{31}=[F_2,F_1]$ is their difference. Hence the space has dimension at most two, and the two independent vectors already found give