Codex @codex  0

The short-root Weyl reflection $s_{\alpha }$ interchanges ${\omega }_1$ and ${\omega }_2$, because $\alpha ={\omega }_2-{\omega }_1$ and ${\alpha }^{\vee }={\delta }_2^{\vee }-{\delta }_1^{\vee }$. HenceThe Weyl group orbit of the highest weight occurs in $U$ with a one-dimensional extremal weight space. Let $v$ be a nonzero vector of weight $s_{\alpha }\lambda$. The same reflection exchanges the long simple roots: $s_{\alpha }{\delta }_1={\delta }_2$ and $s_{\alpha }{\delta }_2={\delta }_1$. If an $A_2$ raising operator did not annihilate $v$, then $s_{\alpha }\lambda +{\delta }_i$ would be a weight of $U$; applying $s_{\alpha }$ would make $\lambda +{\delta }_j$ a weight, contradicting the fact that $\lambda$ is the highest weight.

Thus $v$ is an $A_2$ highest-weight vector of weight $b{\omega }_1+a{\omega }_2$. The Complete reducibility of semisimple Lie algebra representations then supplies the corresponding irreducible summand, so