To fix (albeit standard) notation, let $G$ be a complex semisimple algebraic group, and $T \subset B \subset G$ choices of maximal torus and Borel subgroup, respectively. Let $X^\ast(T)$ be the character lattice of $T$, and $W=N(T)/T$ the Weyl group.
To each $\lambda \in X^\ast(T)$, we can associate a line bundle $L_\lambda$ on the flag variety $G/B$ such that when $\lambda$ is dominant, we have $H^0(G/B, L_\lambda) \neq 0$ and $H^k(G/B, L_\lambda) = 0 $ for all $k \geq 1$. In fact, these two cohomological features are equivalent to $\lambda$ being dominant, by the Borel-Weil theorem (and really even just first condition suffices!)
Now fixing some element $w \in W$, we can consider the Schubert variety $X_w:= \overline{BwB/B} \subset G/B$, and abusing notation we restrict $L_\lambda$ to a line bundle on $X_w$ (we can assume that $\lambda$ is zero on all simple coroots $\alpha_i^\vee$ where the simple reflection $s_i$ does not appear in a reduced expression for $w$). Again when $\lambda$ is dominant, we have the same cohomology (non)vanishings for $L_\lambda$ as for the $G/B$ setting. However, it need not be the case that $H^0(X_w, L_\lambda) \neq 0 \implies \lambda$ dominant; there are nice characterizations of when global sections exist that are weaker than dominance.
Question: If we assume both that
- $H^0(X_w, L_\lambda) \neq 0$, and
- $H^k(X_w, L_\lambda) = 0$ for all $k \geq 1$,
can we recover that $\lambda$ is dominant? (where again, we really mean dominant on the "support" of $w$). I know that there are very few results in the direction of Borel-Weil-Bott for Schubert varieties, but one might hope for these two cohomological conditions together to force dominance as a soft generalization of the $G/B$ case.
