^\circ_{\lambda}} : {\widetilde}X^\circ_{\lambda}\to X^\circ_\Sigma$ is an isomorphism. Denote by $\bar B(\Sigma)$ the coordinate ring of $X^\circ_\Sigma$ and by $\bar A({\lambda})$ the coordinate ring of ${\widetilde}X^\circ_{\lambda}$; then we have $$\bar A({\lambda})= \{\frac{{\varphi}}{{\varphi}_{\lambda}^n}{\, : \,}{\varphi}\in A_n({\lambda})\}
\supset \{\frac{{\varphi}}{{\varphi}_{\lambda}^n}{\, : \,}{\varphi}\in B_n(\Sigma)\} = \bar B(\Sigma)$$ and $X_\Sigma$ is normal if and only if $\bar A({\lambda})=\bar B
(\Sigma)$. The rings $\bar A({\lambda})$ and $\bar B(\Sigma)$ are not $G\times G$-modules, however since $X^\circ_\Sigma$ is an open subset of $X_\Sigma$ we still have an action of the Lie algebra ${\mathfrak g}\oplus{\mathfrak g}$ on them.

By [@Ka], $\bar A({\lambda})$ is generated by the elements of the form ${\varphi}/{\varphi}_{\lambda}$ with ${\varphi}\in A_1({\lambda})$. In particular we have the following lemma.

\[lem:general-normality\] The variety $X_\Sigma$ is normal if and only if for all $\mu\in{\Lambda}^+$ such that $\mu{\leqslant}{\lambda}$ there exists $n>0$ such that $$s^{{\lambda}-\mu}{E({\mu+(n-1){\lambda}})}^*
\subset B_n(\Sigma).$$

Let ${\varphi}_\mu \in s^{{\lambda}-\mu}{E({\mu})}^*$ be a highest weight vector and suppose that $X_\Sigma$ is normal. Then, by the descriptions of $\bar A({\lambda})$ and $\bar B(\Sigma)$, for every dominant weight $\mu{\leqslant}{\lambda}$ there exist $n>0$ and $\varphi\in B_n(\Sigma)$ such that ${{\varphi}}/{{\varphi}_{\lambda}^n} = {{\varphi}_\mu}/{{\varphi}_{\lambda}}$ or equivalently ${\varphi}={\varphi}_\mu {\varphi}_{\lambda}^{n-1}\in B_n(\Sigma)$. Since ${\varphi}$ is a highest weight vector of the module $s^{{\lambda}-\mu}{E({\mu+(n-1){\lambda}})}^*$ the claim follows.

Conversely assume that for every 