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

Table of contents
- Solution iii

Solution

Apply the lowering operator

F

to the highest-weight vector from part (ii). Since

F w_{03} = 4 w_{13} + w_{04}

and

F w_{12} = 2 w_{13}

, we obtain

F v_{+} = w_{04} - 2 w_{13} \in W_{0} .

This 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

2 v_{+}

, as required by the sl2 highest-weight lowering formula.

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