Singular dot weights have zero line-bundle cohomology

Put ${\eta }_0=\lambda +\rho$ and $n_{-}(\eta )=\mathrm{\#}\{\gamma \in {\mathrm{\Phi }}^{+}:⟨\eta ,{\gamma }^{\vee }⟩<0\}$. Starting from ${\lambda }_0=\lambda$, write ${\eta }_j={\lambda }_j+\rho$. Whenever ${\eta }_j$ is not dominant, choose a simple root $\alpha$ with $⟨{\eta }_j,{\alpha }^{\vee }⟩<0$ and set ${\lambda }_{j+1}=s_{\alpha }\cdot {\lambda }_j$, so ${\eta }_{j+1}=s_{\alpha }{\eta }_j$. The reflection $s_{\alpha }$ permutes ${\mathrm{\Phi }}^{+}\setminus \{\alpha \}$ and sends $\alpha$ to $-\alpha$; by [F3], $n_{-}({\eta }_{j+1})=n_{-}({\eta }_j)-1$. Continue whenever a negative simple pairing remains, including when other simple pairings vanish. After exactly $m=n_{-}({\eta }_0)$ steps the count is zero, so all simple pairings are nonnegative and ${\eta }_m$ is dominant. Every ${\eta }_j$ remains singular because the Weyl group preserves root hyperplanes.

At each of these $m$ steps, $⟨{\lambda }_j,{\alpha }^{\vee }⟩=⟨{\eta }_j,{\alpha }^{\vee }⟩-1\le -2$. The reverse form of [F1] gives $H^{i+1}(X,{\mathcal{L}}_{{\lambda }_j})\cong H^i(X,{\mathcal{L}}_{{\lambda }_{j+1}})$ for every $i\ge 0$, and composing yields $H^{i+m}(X,{\mathcal{L}}_{\lambda })\cong H^i(X,{\mathcal{L}}_{{\lambda }_m})$. Because ${\eta }_m$ is dominant and singular, its stabilizer is nontrivial; [F2] therefore gives a simple root ${\alpha }_i$ with $⟨{\eta }_m,{\alpha }_i^{\vee }⟩=0$. Thus $⟨{\lambda }_m,{\alpha }_i^{\vee }⟩=-1$, and [F1] makes every cohomology group of ${\mathcal{L}}_{{\lambda }_m}$ vanish. Conversely, a zero simple-coroot pairing puts a Weyl translate on a root hyperplane, so it implies that ${\eta }_0$ is singular. Therefore the wall condition is equivalent to the stated condition on the unique dominant orbit point. We obtain $H^q(X,{\mathcal{L}}_{\lambda })=0$ for every $q\ge m$.

Put $\mu =-\lambda -2\rho$, so $\mu +\rho =-{\eta }_0$ is singular. Applying steps 1.1-2.1 with $\mu$ in place of $\lambda$ gives $m^{\mathrm{\prime }}=n_{-}(-{\eta }_0)$ and $H^q(X,{\mathcal{L}}_{\mu })=0$ for every $q\ge m^{\mathrm{\prime }}$.

Let $N=\mathrm{\mid }{\mathrm{\Phi }}^{+}\mathrm{\mid }=\dim X$. Since ${\eta }_0$ is singular, at least one positive-root coroot pairs to zero; each other positive root contributes to exactly one of $n_{-}({\eta }_0)$ and $n_{-}(-{\eta }_0)$, so $m+m^{\mathrm{\prime }}\le N-1$. For $0\le i<m$, this implies $N-i\ge N-m+1>m^{\mathrm{\prime }}$. By Serre duality [F5] and the line-bundle identification [F6], $H^i(X,{\mathcal{L}}_{\lambda }{)}^{\vee }\cong H^{N-i}(X,{\mathcal{L}}_{\lambda }^{\vee }\otimes {\omega }_X)\cong H^{N-i}(X,{\mathcal{L}}_{\mu })=0$ by step 3.1. Hence the remaining low-degree groups $H^i(X,{\mathcal{L}}_{\lambda })$ also vanish. Together with step 2.1, which covers every degree $q\ge m$, this proves vanishing for all $i\ge 0$. In particular $H^0(X,{\mathcal{L}}_{\lambda })=0$, so the line bundle does not geometrically realise a nonzero irreducible representation.