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