make use of following lemma.

\[lem:BGno\]

1.  Let $G$ be of type ${\sf B}_r$. Then, for any $n$, ${V({(n-1){\omega}_1})} \not \subset {V({{\omega}_1})}^{\otimes n}$.

2.  Let $G$ be of type ${\sf G}_2$. Then, for any $n$, ${V({{\omega}_1+(n-1){\omega}_2})} \not \subset {V({{\omega}_2})}^{\otimes n}$.

We consider only the first case, the second is similar. Fix a highest weight vector $v_1 \in {V({{\omega}_1})}$. If ${\alpha}$ is any simple root and if $1 {\leqslant}s {\leqslant}r$, notice that $f_{\alpha}$ acts non-trivially on $f_{{\alpha}_{s-1}}\cdots f_{{\alpha}_1}v_1$ if and only if ${\alpha}=
{\alpha}_s$. The $T$-eigenspace of weight 0 in ${V({{\omega}_1})}$ is spanned by $v_0=f_{{\alpha}_r}\cdots f_{{\alpha}_1}v_1$, and similarly the $T$-eigenspace of weight $(n-1){\omega}_1$ in ${V({{\omega}_1})}^{\otimes n}$ is spanned by $v_1^{\otimes i-1} \otimes v_0 \otimes v_1^{\otimes n-i}$, where $1 {\leqslant}i {\leqslant}n$. Since the vectors $$e_{{\alpha}_r} (v_1^{\otimes i-1} \otimes v_0 \otimes v_1^{\otimes
      n-i}) = v_1^{\otimes i-1} \otimes (e_{{\alpha}_r} v_0) \otimes
    v_1^{\otimes n-i}$$ are linearly independent, there exists no maximal vector of weight $(n-1){\omega}_1$ in ${V({{\omega}_1})}^{\otimes n}$.

Dual results will be needed to describe the subsets ${\Sigma}$ which give rise to a normal variety $X_{\Sigma}$, but before we need to introduce some further notation.

If $\Phi$ is an irreducible root system and ${\Delta}$ is a basis for $\Phi$ we will denote by $\eta$ the 