The dominant Borel-Weil section extends from the big cell

Define $v_0:\mathrm{\Omega }\to \mathbb{C}$ by $v_0(u^{-}b)=\lambda (b)$, where $\lambda$ also denotes the character of $B$ trivial on $U$. This is well-defined and regular by the isomorphism $U^{-}\times B\to \mathrm{\Omega }$ of [F1], it satisfies $v_0(g{b}^{\mathrm{\prime }})=\lambda (b^{\mathrm{\prime }})v_0(g)$ for $g\in \mathrm{\Omega }$, $b^{\mathrm{\prime }}\in B$, and $v_0(1)=1$. Since $\mathrm{\Omega }$ is dense open in the irreducible $G$, the function $v_0$ is a rational function on $G$, regular on $\mathrm{\Omega }$.

In the mixed decomposition [F3], the cell $C_w=U^{-}{n}_{w}B$ is the left translate by $n_{w_0}$ of $B{n}_{w_0^{-1}w}B$. Its dimension is therefore $\dim B+\mathrm{\ell }(w_{0}w)=\dim G-\mathrm{\ell }(w)$: $w_0$ reverses all root signs, so the inversion definition gives $\mathrm{\ell }(w_{0}w)=N-\mathrm{\ell }(w)$, where $N=\mathrm{\mid }{\mathrm{\Phi }}^{+}\mathrm{\mid }$. The cell $C_1$ is $\mathrm{\Omega }$, and length one means a simple reflection by [F4]. Since this is a finite decomposition into irreducible locally closed cells, the prime-divisor components of $G\setminus \mathrm{\Omega }$ are exactly $D_{\alpha }=\overline{C_{s_{\alpha }}}$ for simple $\alpha$.

Fix a simple $\alpha$, and let $m$ be the order of $v_0$ along $D_{\alpha }$. By [F8], at its generic point $v_0=z^{m}h$ for a uniformizer $z$ and a unit $h$. Shrink an open neighborhood $V$ of that point so that $z,h,h^{-1}$ are regular on $V$, the equality holds as rational functions, and the zero set of $z$ on $V$ is precisely $D_{\alpha }\cap V$. Such a shrink is possible by clearing the finitely many denominators and removing the other irreducible components of the zero set of $z$. Choose $p\in C_{s_{\alpha }}\cap V$; this intersection is nonempty because the cell is dense in its closure. Write $p=u^{-}{n}_{s_{\alpha }}b$ and let $T(g)=u^{-}gb$. On $\mathrm{\Omega }$ the defining formula gives $v_0(T(g))=\lambda (b)v_0(g)$, hence the same equality holds rationally on $G$. Pulling the local expression back along $T$ gives $v_0=\lambda (b{)}^{-1}(z\circ T{)}^m(h\circ T)$ on a neighborhood of $n_{s_{\alpha }}$, with $h\circ T$ a unit.

In $S{L}_2$, the open set $a\mathrm{\ne }0$ consists of matrices $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=u_{-}(c\mathrm{/}a)\mathrm{diag}(a,a^{-1})u_{+}(b\mathrm{/}a)$. By [F5], its image lies in $\mathrm{\Omega }$, and ${\phi }_{\alpha }^{\ast }{v}_0=a^k$ there, where $k=⟨\lambda ,{\alpha }^{\vee }⟩$. In particular the curve $\gamma (t)={\phi }_{\alpha }\left(\begin{array}{cc}t& -1\\ 1& 0\end{array}\right)$ satisfies $\gamma (0)=n_{s_{\alpha }}$, $\gamma (t)\in \mathrm{\Omega }$ for $t\mathrm{\ne }0$, and $v_0(\gamma (t))=t^k$ for $t\mathrm{\ne }0$.

Pull the expression of step 2.1 back along $\gamma$. The regular germ $z\circ T\circ \gamma$ vanishes at $0$ and is not identically zero, because $T\gamma (t)\in \mathrm{\Omega }$ for $t\mathrm{\ne }0$, whereas its local zero set is $D_{\alpha }$. Its order is therefore some positive integer $e$. The germ $h\circ T\circ \gamma$ is a unit, so the order of the pulled-back rational function is $me$. Step 2.2 identifies this order with $k$, hence $me=k$. Dominance [F7] gives $k\ge 0$, and $e>0$ implies $m\ge 0$. Thus $v_0$ is regular at the generic point of every $D_{\alpha }$. No equality $e=1$ or transversality of the curve is needed.

If $\lambda$ is dominant integral, then $⟨\lambda ,{\alpha }^{\vee }⟩\ge 0$ for every simple root $\alpha$ by [F7], so by step 3.1 it is regular at the generic point of every boundary divisor; every other prime divisor meets $\mathrm{\Omega }$, where $v_0$ is regular by step 1.1; by the pole criterion of [F6] it extends to a regular function $v\in \mathcal{O}(G)$. The functional equation $v(gb)=\lambda (b)v(g)$ holds on the dense open set $\mathrm{\Omega }$ by step 1.1 and hence on all of $G$, since both sides are regular in $g$ for fixed $b$; likewise $v(1)=v_0(1)=1$ because $1\in \mathrm{\Omega }$. By [F2] it gives $H^0(X,{\mathcal{L}}_{\lambda })\mathrm{\ne }0$. Conversely, a nonzero function $f$ in the model of [F2] has $f(g_0)\mathrm{\ne }0$ somewhere, and $g\mapsto f(g_{0}g)\mathrm{/}f(g_0)$ has the same functional equation and value $1$ at the identity.