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

Realize the defining sl2 Lie algebra Lie algebra representation as $\mathbb{C}x\oplus \mathbb{C}y$, with $H=x{\partial }_x-y{\partial }_y$, $E=x{\partial }_y$, $F=y{\partial }_x$. A basis of the symmetric power is $e_i=x^{4-i}{y}^i$ for $0\le i\le 4$, with $H{e}_i=(4-2i)e_i$. Hence the exterior square basis $w_{ij}=e_i\wedge e_j$, $i<j$, has weight $8-2(i+j)$.

Collecting equal weights gives the complete weight-space decomposition:Every one of the ten exterior basis vectors appears exactly once. Thus $U={⨁}_{m=6,4,2,0,-2,-4,-6}{U}_m$, with the displayed bases and dimensions.

The weight diagram has weights $6,4,2,0,-2,-4,-6$ with multiplicities $1,1,2,2,2,1,1$. An Irreducible Lie algebra representation of the sl2 Lie algebra with highest weight $n$ has weights $n,n-2,\dots ,-n$, all of multiplicity one. The top weight therefore forces one $L(6)$ summand. Subtracting its formal character leaves weights $2,0,-2$, each once, which is $L(2)$. By Weyl complete reducibility theorem,Their dimensions $7+3=10$ account for the entire module.

Use the exterior-power Lie algebra representation:Reorder wedge factors when necessary and interpret repeated factors as zero. The explicit weight basis of the exterior square of Sym4 for sl2 isfor $L(6)$, andfor $L(2)$. Their weights are $6-2j$ and $2-2j$, respectively.

Direct application of the displayed operators gives $E{v}_0=E{t}_0=0$ and the lowering chainsThe raising coefficients along the first chain are $2,5,12,1,2,3$; along the second, $E{t}_1=2t_0$ and $E{t}_2=t_1$. Thus each span is stable and is the required irreducible module. At the common weights $2,0,-2$, the $v$ and $t$ vectors are independent; the other weight spaces occur only in the first span. This proves that the ten vectors are a basis of $U$ and that the two spans are complementary. The highest-weight vectors are