Borel Weil and Borel Weil Bott

By [F1] the map $\pi$ is a Zariski-locally trivial $B$-torsor, so $X$ has a finite open cover by charts $V_i$ on which $\pi$ admits regular sections ${\sigma }_i:V_i\to G$, with $\pi \circ {\sigma }_i=\mathrm{i}\mathrm{d}$; on an overlap $V_i\cap V_j$ the two sections satisfy ${\sigma }_j(x)={\sigma }_i(x)b_{ij}(x)$ for a morphism $b_{ij}:V_i\cap V_j\to B$, because both points lie in the same fibre, which is a right $B$-coset by [F3], and the two trivializations of the associated bundle over the overlap differ by $b_{ij}$, as [F1] records.

Let $s\in H^0(X,{\mathcal{L}}_{\lambda })$. Its pullback ${\pi }^{\ast }s$ is a regular section of ${\pi }^{\ast }{\mathcal{L}}_{\lambda }$ over $G$, and the pullback of an associated bundle along the torsor projection is canonically trivial: the map $\tau :G\times \mathbb{C}\to {\pi }^{\ast }{\mathcal{L}}_{\lambda }$, $\tau (g,v)=(g,[g,v])$, is an isomorphism over $G$, since it is bijective on the fibre ${\pi }^{-1}(\pi (g))=gB$ by the relation $[gb,v]=[g,\lambda (b{)}^{-1}v]$ of [F2] and is the identity trivialization over each chart of [F1]. Write ${\tau }^{-1}({\pi }^{\ast }s)(g)=(g,f(g))$ for a regular function $f\in \mathcal{O}(G)$. For $b\in B$ one has $s(\pi (gb))=s(\pi (g))$ by [F3], while $s(\pi (gb))=[gb,f(gb)]$ and $s(\pi (g))=[g,f(g)]=[gb,\lambda (b)f(g)]$ by [F2]; the second coordinate of the class over $gb$ is unique, so $f(gb)=\lambda (b)f(g)$. This defines the map from sections to functions.

Conversely, let $f\in \mathcal{O}(G)$ satisfy $f(gb)=\lambda (b)f(g)$ for all $g,b$, and define $s(x)=[g,f(g)]$ for any $g$ with $\pi (g)=x$. This is well-defined: every other representative of $x$ is $gb$ with $b\in B$ by [F3], and $[gb,f(gb)]=[gb,\lambda (b)f(g)]=[g,f(g)]$ by [F2]. It is a section: the projection sends $s(x)=[g,f(g)]$ to $gB=x$. It is regular: on a chart $V_i$ of step 1.1 the formula $s(x)=[{\sigma }_i(x),f({\sigma }_i(x))]$ exhibits $s$ as a composition of regular maps into the locally trivial bundle ${\mathcal{L}}_{\lambda }$, and regularity is local on the cover $\{V_i\}$, which is finite by [F1]; since $s$ is a section, the compatibility of the local formulae on overlaps is automatic from well-definedness.

The two constructions are inverse: a function $f$ recovered from $s$ satisfies $[g,f(g)]={\pi }^{\ast }s(g)=s(\pi (g))$, so reconstructing $s$ from $f$ returns the original section, and starting from $f$ the recovered function is $g\mapsto$ second coordinate of $[g,f(g)]$, which is $f(g)$. The identification is natural in $\lambda$: the canonical isomorphism ${\mathcal{L}}_{\lambda }\otimes {\mathcal{L}}_{\mu }\to {\mathcal{L}}_{\lambda +\mu }$ of [F2] sends $[g,u]\otimes [g,v]$ to $[g,uv]$, so on sections it corresponds to pointwise multiplication of the functions assigned to $\lambda$ and $\mu$. The dual bundle isomorphism is the fibrewise dual construction; evaluation ${\mathcal{L}}_{\lambda }\otimes {\mathcal{L}}_{-\lambda }\to {\mathcal{O}}_X$ corresponds to pointwise multiplication of functions with opposite $B$-equivariance. In particular, the dictionary does not identify a dual section with the pointwise reciprocal of an arbitrary section. These constructions are compatible with the trivialization $\tau$ used above.

For left translation: the equivariant structure acts on a section $s$ by $(g_0\cdot s)(x)=g_0\cdot s(g_0^{-1}x)$ and $g_0\cdot [g,v]=[g_{0}g,v]$ by [F2], so the function of $g_0\cdot s$ is $g\mapsto$ second coordinate of $g_0\cdot [g_0^{-1}g,f(g_0^{-1}g)]$, which is $f(g_0^{-1}g)$; this is the stated left translation action. For the right $B$-action on sections $(b\cdot s)(x)=s(x)\cdot b$, the associated function is $(b\cdot f)(g)=\lambda (b{)}^{-1}f(g)=f(g{b}^{-1})$ by [F2], so ${\mathrm{e}\mathrm{v}}_1(b\cdot f)=(b\cdot f)(1)=f(b^{-1})=\lambda (b{)}^{-1}f(1)=b\cdot ({\mathrm{e}\mathrm{v}}_{1}f)$ for the action $v\mapsto \lambda (b{)}^{-1}v$ on the fibre ${\mathbb{C}}_{-\lambda }$ at $eB$; hence ${\mathrm{e}\mathrm{v}}_1$ is $B$-equivariant into that fibre. Nothing here asserts that ${\mathrm{e}\mathrm{v}}_1$ is surjective: it is the zero map whenever the space of such functions is zero, and the present lemma only identifies that space with $H^0(X,{\mathcal{L}}_{\lambda })$.

For $g\in G$ let $a_g:X\to X$ be $x\mapsto gx$. The left action on the total bundle gives ${\mathrm{\Phi }}_g:{\mathcal{L}}_{\lambda }\to a_g^{\ast }{\mathcal{L}}_{\lambda }$, whose fibre map at $x$ sends $v\in ({\mathcal{L}}_{\lambda }{)}_x$ to $g\cdot v\in ({\mathcal{L}}_{\lambda }{)}_{gx}$. These maps are algebraic in $g$ and satisfy ${\mathrm{\Phi }}_{gh}=a_h^{\ast }{\mathrm{\Phi }}_g\circ {\mathrm{\Phi }}_h$: at $x$ the composite first applies $h$ to the fibre over $x$, then $g$ to the fibre over $hx$. Define ${\rho }_i(g):=H^i(a_{g^{-1}}^{\ast }{\mathrm{\Phi }}_g)\circ a_{g^{-1}}^{\ast }:H^i(X,{\mathcal{L}}_{\lambda })⟶H^i(X,{\mathcal{L}}_{\lambda })\mathrm{.}$ Here $a_{g^{-1}}^{\ast }$ first pulls cohomology to $H^i(X,a_{g^{-1}}^{\ast }{\mathcal{L}}_{\lambda })$, and the pulled-back bundle map $a_{g^{-1}}^{\ast }{\mathrm{\Phi }}_g:a_{g^{-1}}^{\ast }{\mathcal{L}}_{\lambda }\to {\mathcal{L}}_{\lambda }$ then returns to the original cohomology group. The cocycle gives ${\rho }_i(gh)={\rho }_i(g){\rho }_i(h)$ and ${\rho }_i(e)=\mathrm{i}\mathrm{d}$, so these are linear actions. They are natural in $\lambda$ because the bundle maps are.

The groups $V_i:=H^i(X,{\mathcal{L}}_{\lambda })$ are finite-dimensional: $X$ is smooth projective, hence proper over $\mathbb{C}$ by [F3], and ${\mathcal{L}}_{\lambda }$ is coherent by [F2], so [F4] applies.

On degree zero, the formula in step 1.1 sends a section to $x\mapsto g\cdot s(g^{-1}x)$, which corresponds by [F1] to left translation $(g\cdot f)(g^{\mathrm{\prime }})=f(g^{-1}{g}^{\mathrm{\prime }})$. If $\varphi :{\mathcal{L}}_{\lambda }\to {\mathcal{L}}_{\mu }$ is $G$-equivariant, its compatibility with the total-space action makes the induced family maps commute with $\varphi$; by functoriality [F4], every $H^i(\varphi )$ intertwines the actions ${\rho }_i$.

To prove rationality, put $R=\mathcal{O}(G)$, let $q:G\times X\to X$ be projection and define the algebraic automorphism $A:G\times X\to G\times X$ by $A(g,x)=(g,g^{-1}x)$. The product cover $G\times U_I$ and [F5] compute $H^i(G\times X,q^{\ast }{\mathcal{L}}_{\lambda })$ by the Cech complex whose terms, by [F6], are $R{\otimes }_{\mathbb{C}}\mathrm{\Gamma }(U_I,{\mathcal{L}}_{\lambda })$. Its differentials are $1_R\otimes d$, so exactness of tensoring over the field $\mathbb{C}$ identifies this cohomology with $R{\otimes }_{\mathbb{C}}{V}_i$. The left action on the line bundle gives an algebraic isomorphism $A^{\ast }{q}^{\ast }{\mathcal{L}}_{\lambda }\to q^{\ast }{\mathcal{L}}_{\lambda }$ over $G\times X$, sending the fibre over $(g,x)$ by the action of $g$ from $({\mathcal{L}}_{\lambda }{)}_{g^{-1}x}$ to $({\mathcal{L}}_{\lambda }{)}_x$. Pullback by $A$ followed by this isomorphism induces an $R$-linear endomorphism of $R\otimes V_i$. Under evaluation at each $g\in G$, the same product Cech computation identifies its fibre map with ${\rho }_i(g)$ of step 1.1. Thus the matrix entries of ${\rho }_i$ are regular functions on $G$; the action law makes ${\rho }_i:G\to \mathrm{GL}(V_i)$ an algebraic group homomorphism.

The identifications ${\mathcal{L}}_{\lambda }\otimes {\mathcal{L}}_{\mu }\cong {\mathcal{L}}_{\lambda +\mu }$ and ${\mathcal{L}}_{\lambda }^{\vee }\cong {\mathcal{L}}_{-\lambda }$ are morphisms of equivariant bundles by [F2], so the induced cohomology maps intertwine the actions from step 1.1. Each algebraic representation of step 2.2 differentiates to a $\mathfrak{g}$-module, and the tensor/dual identifications remain equivariant for both actions.

Fix a root parametrization $u_{-\alpha }(t)={\exp }_G(tx)$ with $0\mathrm{\ne }x\in {\mathfrak{g}}_{-\alpha }$. The derived left action is $(x\cdot f)(g)=\frac{d}{dt}{\mathrm{\mid }}_{0}f(u_{-\alpha }(-t)g)$. If every $x\in {\mathfrak{n}}^{-}$ annihilates $f$, put $p_g(t)=f(u_{-\alpha }(-t)g)$. For every $t_0$, the group law gives $p_g^{\mathrm{\prime }}(t_0)=(x\cdot f)(u_{-\alpha }(-t_0)g)=0$. Thus this polynomial has zero derivative everywhere and is constant over $\mathbb{C}$. Each root subgroup fixes $f$, and their product is $U^{-}$ by [F3], so $U^{-}$ fixes $f$. Conversely, differentiating a $U^{-}$-invariant function gives ${\mathfrak{n}}^{-}f=0$.

Assume now that $f$ is invariant under $U^{-}$. For $u^{-}\in U^{-}$ and $b\in B$ the invariance gives $f(u^{-}b)=f(b)$, and the functional equation of [F1] with $g=1$ gives $f(b)=f(1\cdot b)=\lambda (b)f(1)$. Hence $f(u^{-}b)=\lambda (b)f(1)$ for all $u^{-}\in U^{-}$, $b\in B$.

Let $f,f^{\mathrm{\prime }}$ be two $U^{-}$-invariant sections of ${\mathcal{L}}_{\lambda }$ with $f(1)=f^{\mathrm{\prime }}(1)$. By step 2.1, $f$ and $f^{\mathrm{\prime }}$ agree on $\mathrm{\Omega }=U^{-}B$, which is dense open in the irreducible variety $G$ by [F2]; two regular functions on $G$ agreeing on a dense open subset agree everywhere, so $f=f^{\mathrm{\prime }}$. The linear map $f\mapsto f(1)$ is therefore injective on the space of $U^{-}$-invariant sections, which has dimension at most $1$.

Finally let $t\in T$ and put $g=t\cdot f-\lambda (t{)}^{-1}f$. By [F3] the torus normalizes every $U_{-\alpha }$, so $t\cdot f$ is again $U^{-}$-invariant: for $u^{-}\in U^{-}$ there is $u^{\mathrm{\prime }-}=t{u}^{-}{t}^{-1}\in U^{-}$ with $u^{-}\cdot (t\cdot f)=t\cdot ((t^{-1}{u}^{-}t)\cdot f)=t\cdot f$. Hence $g$ is a $U^{-}$-invariant section, and $g(1)=(t\cdot f)(1)-\lambda (t{)}^{-1}f(1)=f(t^{-1})-\lambda (t{)}^{-1}f(1)=\lambda (t^{-1})f(1)-\lambda (t{)}^{-1}f(1)=0$ by the functional equation. By step 3.1 the vanishing of $g(1)$ forces $g=0$, that is $t\cdot f=\lambda (t{)}^{-1}f$. If $f$ is nonzero, it therefore has weight $-\lambda$.

Let $h$ be any section of $T$-weight $-\lambda$. Left translation and right $B$-equivariance [F1] give $h(t{u}^{-}{t}^{-1})=h(u^{-})$ for $t\in T$. In the polynomial root coordinates on $U^{-}$ of [F3], conjugation scales each coordinate by the character $-\beta$ for a positive root $\beta$. A nonconstant monomial has character $-{\sum }_{\beta }{m}_{\beta }\beta \mathrm{\ne }0$: every positive root has nonnegative simple-root coefficients, and some $m_{\beta }>0$. Since distinct torus characters are linearly independent (restrict a finite list to a one-parameter subgroup separating their exponents), a conjugation-invariant polynomial is constant. Thus $h$ is constant on $U^{-}$ and $h(u^{-}b)=\lambda (b)h(1)$ on $\mathrm{\Omega }$. Density [F2] makes $h$ $U^{-}$-invariant on $G$. Step 3.1 now shows that the entire weight-$(-\lambda )$ space has dimension at most one; any nonzero invariant section spans it.

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.