The Borel-Weil-Bott theorem

Suppose first that $\lambda +\rho$ is not regular. Then [F1] gives $H^i(X,{\mathcal{L}}_{\lambda })=0$ for all $i\ge 0$, which is the first clause of the Statement.

Now suppose that $\lambda +\rho$ is regular, and let $w$ be the unique element with $w\cdot \lambda$ dominant, which is the complementary case to step 1.1 and exists uniquely by [F3]. Put ${\lambda }_j=w_j\cdot \lambda$ along the reduced chain of [F3], so that ${\lambda }_j+\rho =w_j(\lambda +\rho )$ and ${\lambda }_{\mathrm{\ell }}=w\cdot \lambda$. At the step from $j$ to $j+1$ the chosen simple root satisfies $⟨w_j(\lambda +\rho ),{\alpha }^{\vee }⟩\le -1$ by [F3], hence $⟨{\lambda }_j,{\alpha }^{\vee }⟩=⟨w_j(\lambda +\rho ),{\alpha }^{\vee }⟩-1\le -2\le -1$, so the second clause of [F2] applies and gives $H^{i+1}(X,{\mathcal{L}}_{{\lambda }_j})\cong H^i(X,{\mathcal{L}}_{{\lambda }_{j+1}})$ for all $i\ge 0$. Composing over the $\mathrm{\ell }=\mathrm{\ell }(w)$ steps yields $H^{i+\mathrm{\ell }}(X,{\mathcal{L}}_{\lambda })\cong H^i(X,{\mathcal{L}}_{w\cdot \lambda })$ for all $i\ge 0$. Since $w\cdot \lambda$ is dominant integral, [F4] gives $H^{i+\mathrm{\ell }}(X,{\mathcal{L}}_{\lambda })=0$ for every $i>0$, and for $i=0$ gives $H^{\mathrm{\ell }}(X,{\mathcal{L}}_{\lambda })\cong H^0(X,{\mathcal{L}}_{w\cdot \lambda })\cong L(w\cdot \lambda {)}^{\ast }$. Thus the cohomology vanishes above degree $\mathrm{\ell }$ and has the asserted value in degree $\mathrm{\ell }$.

It remains to exclude nonzero cohomology in degrees $i<\mathrm{\ell }$. Consider the weight $\mu =-\lambda -2\rho$. Since $\mu +\rho =-(\lambda +\rho )$ is regular, [F3] applies to $\mu$; let $v$ be the unique element with $v\cdot \mu$ dominant. Then $v\cdot \mu +\rho =v(\mu +\rho )=-v(\lambda +\rho )$, and for $v=w_{0}w$ this equals $-w_0(w(\lambda +\rho ))$: since $w(\lambda +\rho )$ is dominant, $w_0(w(\lambda +\rho ))$ is antidominant, so its negative $-w_0(w(\lambda +\rho ))$ is dominant; hence $(w_{0}w)\cdot \mu$ is dominant and uniqueness gives $v=w_{0}w$. By [F6] the Borel-Weil-Bott degree of $\mu$ is therefore $\mathrm{\ell }(w_{0}w)=N-\mathrm{\ell }$. Applying step 2.1 to $\mu$ in place of $\lambda$ gives $H^q(X,{\mathcal{L}}_{\mu })=0$ for every $q>N-\mathrm{\ell }$. By Serre duality [F5], for $0\le i<\mathrm{\ell }$ one has $H^i(X,{\mathcal{L}}_{\lambda }{)}^{\ast }\cong H^{N-i}(X,{\mathcal{L}}_{\mu })$ with $N-i>N-\mathrm{\ell }$, so $H^i(X,{\mathcal{L}}_{\lambda })=0$.

Combining steps 1.1, 2.1 and 3.1: if $\lambda +\rho$ is not regular all cohomology vanishes; if it is regular, then $H^{\mathrm{\ell }(w)}(X,{\mathcal{L}}_{\lambda })\cong L(w\cdot \lambda {)}^{\ast }$ and $H^i(X,{\mathcal{L}}_{\lambda })=0$ for every $i\mathrm{\ne }\mathrm{\ell }(w)$, since degrees above $\mathrm{\ell }(w)$ vanish by step 2.1 and degrees below vanish by step 3.1. These are the only nonvanishing cohomology groups, so the theorem is proved.