Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 102 / 4

For a dominant integral highest weight $\lambda$, the Weyl dimension formula iswhere ${\Phi }^{+}$ is the chosen positive system of a root system, ${\alpha }^{\vee }$ is a coroot, and $\rho$ is the Weyl vector.

For the B2 root system with ${\alpha }_1$ short, the positive roots areSubstituting $\lambda =a{\omega }_1+b{\omega }_2$ into the Weyl dimension formula for B2 gives

Let $V=V({\omega }_2)$ be the five-dimensional defining representation of the so5 Lie algebra. The tensor square splits into its symmetric square and exterior square:The invariant symmetric form spans a trivial subrepresentation of $S^{2}V$, while its traceless complement is the irreducible $V(2{\omega }_2)$ of dimension $14$. The identification ${\Lambda }^{2}V\cong {\mathfrak{so}}_5$ makes the exterior square the ten-dimensional Adjoint representation, whose highest weight is the highest root $2{\omega }_1$. Therefore the Tensor-square decomposition of the defining so5 representation is

The Poincare-Birkhoff-Witt theorem shows that the weights of the Verma module $M({\omega }_2)$ arewith multiplicities given by the corresponding Kostant partition function.

The Dynkin labels of ${\omega }_2$ are $(0,1)$. Hence the two simple-root singular vectors are $f_1{v}_{{\omega }_2}$, of weight ${\omega }_2-{\alpha }_1$, and $f_2^2{v}_{{\omega }_2}$, of weight ${\omega }_2-2{\alpha }_2$. They generate the Maximal proper submodule of a dominant integral Verma module. Its set of weights is consequently