Write the simple roots of the <G2 root system> as $\alpha$ short and $\beta$ long. The <Long-root A2 subsystem of G2> has simple roots
$$
\delta_1=\beta,\qquad\delta_2=3\alpha+\beta.
$$
If $\omega_1,\omega_2$ are its <fundamental weights>, then the $A_2$ root-weight relations give
$$
\alpha=\omega_2-\omega_1,\qquad
\alpha^\vee=\delta_2^\vee-\delta_1^\vee,\qquad
\beta^\vee=\delta_1^\vee.
$$
For the $A_2$-dominant weight $\lambda=a\omega_1+b\omega_2$, therefore,
$$
\langle\lambda,\beta^\vee\rangle=a,
\qquad
\langle\lambda,\alpha^\vee\rangle=b-a.
$$
The <dominant weight> inequalities for $G_2$ are thus equivalent to
$$
\boxed{b\geq a}
$$
because $A_2$-dominance already gives $a,b\geq0$.

Solution

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

{c}
{title2=$A_2\subset G_2$}

For short and long simple roots $\alpha,\beta$ of $G_2$, the long roots form an $A_2$ root system with simple roots $\delta_1=\beta$ and $\delta_2=3\alpha+\beta$. If $\omega_1,\omega_2$ are the corresponding $A_2$ <fundamental weights>, then $\alpha=\omega_2-\omega_1$ and $\alpha^\vee=\delta_2^\vee-\delta_1^\vee$.

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 Incoming links: Long-root A2 subsystem of G2