laced, then ${V({\eta})}\simeq{\mathfrak g}$ is the adjoint representation: in this case the claim follows straightforward by considering the map ${\mathfrak g}\otimes {V({{\lambda}})} \to {V({{\lambda}})}$ induced by the ${\mathfrak g}$-module structure on ${V({{\lambda}})}$, which is non-zero since ${\lambda}$ is non-zero.

Suppose now that ${\Delta}$ is not simply laced. If ${\lambda}$ satisfies condition $(\star)$, then by Lemma \[lem:traslazione\] it is enough to study the case ${\lambda}= \omega_{\alpha}$ where ${\alpha}$ is a short simple root:

*Type ${\mathsf B}_r$*: ${V({\omega_r})}\subset{V({\omega_1})}\otimes {V({\omega_r})}$.

*Type ${\mathsf C}_r$*: ${V({\omega_i})}\subset{V({\omega_2})}\otimes {V({\omega_i})}$, with $i<r$.

*Type ${\mathsf F}_4$*: ${V({\omega_3})}\subset{V({\omega_4})}\otimes {V({\omega_3})}$ and ${V({\omega_4})}\subset{V({\omega_4})}\otimes {V({\omega_4})}$.

*Type ${\mathsf G}_2$*: ${V({\omega_1})}\subset{V({\omega_1})}\otimes {V({\omega_1})}$.

If ${\lambda}$ does not satisfy condition $(\star)$, by Lemma \[lem:traslazione\] we can assume that ${\lambda}={\omega}_{\alpha}$ with ${\alpha}$ a long root:

*Type ${\mathsf B}_r$*: ${V({\omega_i})}\subset{V({\omega_1})}\otimes {V({\omega_{i-1}})}$, if $1<i<r$, and ${V({\omega_1})}\subset{V({\omega_1})}\otimes {V({0})}$.

*Type ${\mathsf C}_r$*: ${V({\omega_r})}\subset{V({\omega_2})}\otimes {V({\omega_{r-2}})}$.

*Type ${\mathsf F}_4$*: ${V({\omega_1})}\subset{V({\omega_4})}\otimes {V({\omega_4})}$ and ${V({\omega_2})}\subset{V({\omega_4})}\otimes {V({\omega_1+\omega_4})}$.

*Type ${\mathsf G}_2$*: ${V({\omega_2})}\subset{V({\omega_1})}\otimes {V({\omega_1})}$.

The above mentioned inclusion relations for tensor products are essentially known: let us treat the case of type ${\mathsf C}_r$ with ${\lambda}=\omega_i$ and $i<r$, the other cases are easier or can be checked directly.

Let $v_0$ be a highest weight vector of ${V({\omega_2})}$ and $w_0$ be a highest weight vector of ${V({\omega_i})}$. Let $f$ be 