The weights of the defining five-dimensional representation are
$$
\varepsilon_1,\ \varepsilon_2,\ 0,\ -\varepsilon_2,\ -\varepsilon_1.
$$
Its highest weight is $\varepsilon_1=\omega_2$, so irreducibility identifies it as
$$
V\cong V(\omega_2).
$$

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 $S^2V$.

Putting $a=0$ and $b=2$ into the formula from part i gives
$$
\boxed{
\dim V(2\omega_2)
=\frac{(1)(3)(4)(7)}6=14}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 102 / 5 / c / ii / Solution

{c}
{title2=$V(2\omega_2)$}

For the five-dimensional defining representation $V=V(\omega_2)$ of $\mathfrak{so}_5$, the invariant quadratic form splits
$$
S^2V\cong\mathbb C\oplus V(2\omega_2).
$$
Thus the traceless symmetric square is irreducible of dimension $15-1=14$.


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 Incoming links: Traceless symmetric square of the defining so5 representation