The weights of the seven-dimensional irreducible $G_2$ representation are zero and the six short roots. In the $A_2$ <weight lattice>, the three positive short roots are
$$
\alpha=\omega_2-\omega_1,\qquad
\alpha+\beta=\omega_1,\qquad
2\alpha+\beta=\omega_2.
$$
Together with their negatives, they split into the weight sets of the two dual three-dimensional <Fundamental representations of sl3>; the zero weight supplies a trivial representation. By the <Restriction of the seven-dimensional G2 representation to long-root A2>,
$$
\boxed{V_7\!\downarrow_{\mathfrak{sl}_3}
\cong V(\omega_1)\oplus V(\omega_2)\oplus V(0).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 102 / 5 / c / Solution

{c}
{title2=$\mathbf7\downarrow_{A_2}=\mathbf3\oplus\overline{\mathbf3}\oplus\mathbf1$}

The seven-dimensional fundamental $G_2$ representation has weights zero and the six short roots. Under its long-root $A_2$ subalgebra, these are the weights of the defining three-dimensional representation, its dual, and the trivial representation:
$$
V_7\downarrow_{A_2}\cong V(\omega_1)\oplus V(\omega_2)\oplus V(0).
$$


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 Incoming links: Restriction of the seven-dimensional G2 representation to long-root A2