and $\eta + {\alpha}\in {\Pi({{\lambda}+ \mu})}{\smallsetminus}\{{\lambda}+\mu\}$, against the maximality of $\eta$. Thus $\eta - \mu$ is maximal in ${\Pi({{\lambda}})}
{\smallsetminus}\{{\lambda}\}$ and similarly $\eta - {\lambda}$ is maximal in ${\Pi({\mu})}
{\smallsetminus}\{\mu\}$. Therefore, on one hand it must be $$\eta - \mu = {\lambda}- {\alpha}$$ with ${\alpha}\in \operatorname{Supp}({\lambda})$, while on the other hand it must be $$\eta - {\lambda}= \mu - {\beta}$$ with ${\beta}\in \operatorname{Supp}(\mu)$. Since $\operatorname{Supp}({\lambda}) \cap \operatorname{Supp}(\mu) =
{\varnothing}$, this is impossible and shows that, if $(u, v) \in \ker d
\psi_x$, then it must be $u = 0$ or $v = 0$. Suppose now that $(u,0)
\in \ker d\psi_x$: then $u v_\mu = 0$ and by the irreducibility of $G/B$ also $u=0$. A similar argument applies if $v=0$.

$ii).$ Suppose that $v,w \in V({\lambda})$ are such that $v^{n} = w^{n}$: then $v=tw$ for some $t\in {\Bbbk}$. Thus $\psi_{\lambda}^n$ is injective. Let us show now that $\psi_{\lambda}^n$ is smooth; it is enough to show it in $x = [v_{\lambda}]$ where $v_{\lambda}\in V({\lambda})$ is a highest weight vector. Let $V \subset V({\lambda})$ be the $T$-stable complement of ${\Bbbk}v_{\lambda}$, identified with the tangent space $T_{x} {\mathbb P}(V({\lambda}))$. If $v
\in V$, the differential $d (\psi_{\lambda}^n)_{x}$ is described as follows $$d (\psi^n_{\lambda})_{x} (v) 