Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 1 / 1 / b

Let $e_1=x,e_2=y,e_3=z$, and let ${ϵ}_i$ be the weight of a representation defined by ${ϵ}_i(\mathrm{diag}(h_1,h_2,h_3))=h_i$. Since the trace is zero, ${ϵ}_1+{ϵ}_2+{ϵ}_3=0$. The dual representation gives weight $-{ϵ}_i$ to $e_i^{\ast }$. PutTaking $j\le k$ gives an eighteen-vector tensor product basis. Its weight is ${ϵ}_j+{ϵ}_k-{ϵ}_i$.

For explicit coordinates use the usual simple coroots $H_1=\mathrm{diag}(1,-1,0)$ and $H_2=\mathrm{diag}(0,1,-1)$, and write each weight as $(\mu (H_1),\mu (H_2))$. Then ${ϵ}_1=(1,0)$, ${ϵ}_2=(-1,1)$ and ${ϵ}_3=(0,-1)$. Collecting equal weights gives every weight space:The first six weights are $2{ϵ}_j-{ϵ}_i$ with $i\ne j$; the next three are $-2{ϵ}_i$; the last three are the defining weights ${ϵ}_i$. Thus there are nine multiplicity-one weights and three multiplicity-three weights, giving total dimension $9+9=18$.

The planar coordinates are $a{\omega }_1+b{\omega }_2$ with ${\omega }_1=(1,0)$ and ${\omega }_2=(1/2,\sqrt{3}/2)$, so the weight diagram uses an equilateral root geometry rather than orthogonal Dynkin-coordinate axes. Every point is labelled with its complete weight-space basis.

Define an equivariant contraction $C:V^{\ast }\otimes {\mathrm{Sym}}^{2}V\to V$ byThis uses the natural dual pairing, so it commutes with the Lie algebra representation. Also define $s(v)={\sum }_i{e}_i^{\ast }\otimes e_{i}v$. The identity tensor ${\sum }_i{e}_i^{\ast }\otimes e_i$ is invariant, making $s$ equivariant. Direct contraction gives $C(s(v))=3v+v=4v$. Thus $s$ is injective and $C$ is surjective, andis a direct decomposition into $\ker C$ and $s(V)$. In particular $\dim \ker C=18-3=15$ and $s(V)$ is a defining irreducible module of highest weight ${\omega }_1$.

For the usual upper-triangular choice of positive roots, $T_3^{11}=z^{\ast }\otimes x^2$ lies in the kernel and is killed by both simple raising operators $E_{12},E_{23}$. Its highest weight is $2{ϵ}_1-{ϵ}_3=2{\omega }_1+{\omega }_2$, with Dynkin labels $(2,1)$. The highest-weight classification and complete reducibility give an irreducible copy $L(2,1)$ generated by this vector. The Weyl dimension formula for ${\mathfrak{sl}}_3$ isso $\dim L(2,1)=15$. It fills the entire kernel. Therefore the sl3 decomposition of the symmetric-square dual tensor product, with its tensor factors exchanged, isThe weight diagrams below record both summands. Their labels use $K_{ij}=2T_j^{ij}-T_i^{ii}$ for $j\ne i$ and $S_i={\sum }_j{T}_j^{ij}$; these are the weight bases detailed in part (iii).

The nine one-dimensional weight spaces from part (i) all lie in $\ker C$, because the dual index differs from every polynomial index. Their bases remain $T_i^{jj}$ at weight $2{ϵ}_j-{ϵ}_i$ for $i\ne j$, and $T_i^{jk}$ at weight $-2{ϵ}_i$ for three distinct indices $i,j,k$.

At weight ${ϵ}_i$, the full space has basis $T_i^{ii},T_j^{ij},T_k^{ik}$, where $j,k$ are the two other indices. Their contractions are $2e_i,e_i,e_i$. Thus the contraction kernel has the independent basis $K_{ij}=2T_j^{ij}-T_i^{ii}$ and $K_{ik}=2T_k^{ik}-T_i^{ii}$. Explicitly the weight basis of the trace-free symmetric-square dual tensor module is completed byTogether with the nine single vectors, these give fifteen independent vectors. They are precisely the $T$ and $K$ labels on the fifteen-dimensional weight diagram.

The defining summand has only the following three weight spaces:Each $S_i=s(e_i)$ is nonzero, has its indicated weight and satisfies $C(S_i)=4e_i$. These are the labels on the three-dimensional diagram. The highest-weight vectors in the two summands areAt each repeated weight, the two kernel vectors together with $S_i$ span the original three-dimensional weight space, since the kernel has dimension two and $C(S_i)\ne 0$. Thus the displayed diagrams specify all submodule weight-space bases.