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

Take a basis $x,y$ of the defining representation of the sl2 Lie algebra, with $Hx=x$, $Hy=-y$, $Ey=x$, $Ex=0$, $Fx=y$ and $Fy=0$. A monomial basis of the symmetric power ${\mathrm{Sym}}^{4}V$ isFor the exterior power $U={\bigwedge }^2{\mathrm{Sym}}^{4}V$, put $w_{ij}=v_i\wedge v_j$ for $i<j$. These ten elements form a basis, and their weights of a representation add:Thus the complete weight-space decomposition, with explicit bases, isThe weight diagram therefore has multiplicities $1,1,2,2,2,1,1$ at weights $6,4,2,0,-2,-4,-6$, respectively.

The induced raising operator and lowering operator act bySince $E{w}_{01}=0$ and $w_{01}$ has weight $6$, it generates the seven-dimensional irreducible sl2 Lie algebra module ${\mathrm{Sym}}^{6}V$. By complete reducibility of semisimple Lie algebra representations, subtracting its weights $6,4,2,0,-2,-4,-6$, each of multiplicity one, from the weight diagram leaves weights $2,0,-2$ once each. HenceThese are the only two irreducible isomorphism types occurring as submodules.

For the three-dimensional summand, calculate $E{w}_{03}=3w_{02}$ and $E{w}_{12}=w_{02}$. Thereforeis a nonzero highest-weight vector of weight $2$. The classification of finite-dimensional sl2 representations shows that the submodule $W$ it generates is the irreducible module ${\mathrm{Sym}}^{2}V$, of dimension three.

Apply the lowering operator $F$ to the highest-weight vector from part (ii). Since $F{w}_{03}=4w_{13}+w_{04}$ and $F{w}_{12}=2w_{13}$, we obtainThis is nonzero because $w_{04},w_{13}$ are distinct elements of the exterior power basis. Both have weight zero. As a consistency check, the raising operator sends this vector to $2v_{+}$, as required by the sl2 highest-weight lowering formula.