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>:
$$
V\otimes V=S^2V\oplus\Lambda^2V.
$$
The invariant symmetric form spans a trivial subrepresentation of $S^2V$, while its traceless complement is the irreducible $V(2\omega_2)$ of dimension $14$. The identification $\Lambda^2V\cong\mathfrak{so}_5$ 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
$$
\boxed{V\otimes V\cong V(0)\oplus V(2\omega_1)\oplus V(2\omega_2).}
$$

Solution

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

{c}
{title2=$V(\omega_2)^{\otimes2}=V(0)\oplus V(2\omega_1)\oplus V(2\omega_2)$}

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
$$
V(\omega_2)\otimes V(\omega_2)
\cong V(0)\oplus V(2\omega_1)\oplus V(2\omega_2).
$$

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 Incoming links: Tensor-square decomposition of the defining so5 representation