 Solution

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