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

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.