 Solution

The weights of the defining five-dimensional representation areIts highest weight is ${\epsilon }_1={\omega }_2$, so irreducibility identifies it as

If $v_{+}$ is a highest-weight vector, then $v_{+}\otimes v_{+}$ is a highest-weight vector of weight $2{\omega }_2$ in $V\otimes V$. The subrepresentation it generates is therefore $V(2{\omega }_2)$. Equivalently, it is the Traceless symmetric square of the defining so5 representation; the invariant quadratic form supplies the complementary trivial line in $S^{2}V$.

Putting $a=0$ and $b=2$ into the formula from part i gives