Traceless symmetric square of the defining so5 representation ($V(2{\omega }_2)$)

For the five-dimensional defining representation $V=V({\omega }_2)$ of ${\mathfrak{so}}_5$, the invariant quadratic form splitsThus the traceless symmetric square is irreducible of dimension $15-1=14$.

For the five-dimensional defining representation of ${\mathfrak{so}}_5$, the symmetric square is the sum of its invariant trace and the traceless representation $V(2{\omega }_2)$, while the exterior square is the Adjoint representation $V(2{\omega }_1)$. Hence