Let $V=V(\omega_1)$, so $V^*=V(\omega_2)$. First,
$$
V\otimes V
=\operatorname{Sym}^2V\oplus\Lambda^2V
\cong V(2\omega_1)\oplus V(\omega_2).
$$
The highest-weight tensor-product rule gives
$$
V(2\omega_1)\otimes V(\omega_2)
\cong V(2\omega_1+\omega_2)\oplus V(\omega_1)
$$
and
$$
V(\omega_2)\otimes V(\omega_2)
\cong V(2\omega_2)\oplus V(\omega_1).
$$
The dimensions $15+3+6+3=27$ check the decomposition. Therefore the <Triple tensor decomposition for the defining sl3 representation> is
$$
\boxed{V\otimes V\otimes V^*
\cong V(2\omega_1+\omega_2)\oplus
V(2\omega_2)\oplus V(\omega_1)^{\oplus2}}.
$$
Thus one may take
$$
(\lambda_1,\lambda_2,\lambda_3,\lambda_4)
=(2\omega_1+\omega_2,\,2\omega_2,\,\omega_1,\,\omega_1).
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 102 / 4 / e / Solution

{c}
{title2=$V\otimes V\otimes V^*$}

For the defining representation $V=V(\omega_1)$ of $\mathfrak{sl}_3$,
$$
V\otimes V\otimes V^*
\cong
V(2\omega_1+\omega_2)\oplus
V(2\omega_2)\oplus
V(\omega_1)^{\oplus2}.
$$


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