and if ${\lambda}$ does not satisfy $(\star)$ then $${V({{\lambda}+\omega_1})} \subset {V({\omega_2})} \otimes {V({{\lambda}^{\mathrm{lb}}})}.$$

3.  If $G$ is of type ${\mathsf G}_2$ and if ${\alpha}_S \in \operatorname{Supp}({\lambda})$ then $${V({{\lambda}+\omega_1})} \subset {V({\omega_2})} \otimes {V({{\lambda}})}.$$

By Lemma \[lem:traslazione\] it is enough to check the statements for ${\lambda}=\omega_{\alpha}$ with ${\alpha}$ a long root in the first two cases and ${\alpha}={\alpha}_S$ in the last case.

*Type ${\mathsf C}_r$*: by Lemma \[lem:traslazione\] it is enough to check that ${V({\omega_{r-1}})}\subset {V({\omega_1})} \otimes {V({\omega_r})}$.

*Type ${\mathsf F}_4$*: we have ${\lambda}=\omega_1$ or ${\lambda}=\omega_2$ and $\omega_S+\zeta=\omega_1+\omega_4$.

*Type ${\mathsf G}_2$*: we have ${\lambda}=\omega_2$ and ${\lambda}^{\mathrm{lb}}=\omega_1$ in point (2) and ${\lambda}=\omega_1$ in point (3).

Normality and non-normality of $X_\Sigma$ {#ssez:normalita}
-----------------------------------------

We are now able to state the main theorem.

\[teo:normalita\]Let $\Sigma$ be a simple set of dominant weights and let ${\lambda}$ be its maximal element. The variety $X_\Sigma$ is normal if and only if $\Sigma \supset {\mathrm{LB}}({\lambda})$.

Theorem A stated in the introduction follows immediately by considering the case $\Sigma=\{{\lambda}\}$. The remaining part of this section will be devoted to the proof of Theorem \[teo:normalita\]. The general strategy will be based on Proposition \[prp:normalita\] and will proceed by induction on the dominance order of weights. The ingredients of this induction will 