Borel, opposite unipotent groups and root coordinates

By [F7] fix a faithful rational closed immersion $G\hookrightarrow GL(V)$. Restrict the rational representation to $T\cong ({\mathbb{G}}_m{)}^r$. Its coaction is a finite Laurent-polynomial sum, so comparison of coefficients in the coaction identity decomposes $V$ as the direct sum of finitely many character weight spaces $V_{\mu }$. For $e_{\alpha }\in {\mathfrak{g}}_{\alpha }$ and $v\in V_{\mu }$, differentiating $t{e}_{\alpha }{t}^{-1}=\alpha (t)e_{\alpha }$ from [F1] gives $e_{\alpha }v\in V_{\mu +\alpha }$. Choose a real linear functional on the character lattice that is positive on every simple root, hence every positive root, and order the finitely many weights of $V$ by its value. Every $X\in {\mathfrak{n}}^{+}={⨁}_{\alpha >0}{\mathfrak{g}}_{\alpha }$ strictly raises this common filtration, so $X^{\dim V}=0$. This proves nilpotence of every sum of positive-root operators, not merely of the individual root vectors.

The solvable subalgebra $\mathfrak{b}$ is maximal solvable: if $\mathfrak{s}\supseteq \mathfrak{b}$ is a solvable subalgebra and $\mathfrak{s}\mathrm{\ne }\mathfrak{b}$, then $\mathfrak{g}={\mathfrak{n}}^{-}\oplus \mathfrak{b}$ and $\mathfrak{h}\subseteq \mathfrak{b}$ is $\mathrm{ad}$-stable, so $\mathfrak{s}$ contains a nonzero weight component $\mathfrak{s}\cap {\mathfrak{g}}_{{\alpha }_0}$ with ${\alpha }_0\in {\mathrm{\Phi }}^{-}$; write ${\mathfrak{g}}_{-}=-{\alpha }_0\in {\mathrm{\Phi }}^{+}$, so that ${\mathfrak{g}}_{{\alpha }_0}\subseteq \mathfrak{s}$ and ${\mathfrak{g}}_{-{\alpha }_0}\subseteq {\mathfrak{n}}^{+}\subseteq \mathfrak{s}$; by [F2] applied to the root $-{\alpha }_0$ there are $e_{-{\alpha }_0}\in {\mathfrak{g}}_{-{\alpha }_0}$ and $f_{-{\alpha }_0}\in {\mathfrak{g}}_{{\alpha }_0}$ with $[e_{-{\alpha }_0},f_{-{\alpha }_0}]=h_{-{\alpha }_0}\mathrm{\ne }0$, so $\mathfrak{s}$ contains the copy of ${\mathfrak{s}\mathfrak{l}}_2$ spanned by these three elements, contradicting solvability of $\mathfrak{s}$ because ${\mathfrak{s}\mathfrak{l}}_2$ is not solvable.

By [F3] the Lie algebra ${\mathfrak{n}}^{+}$ is nilpotent, say ${\gamma }_{c+1}({\mathfrak{n}}^{+})=0$, so the Lie subalgebra generated by any two elements of ${\mathfrak{n}}^{+}$ is nilpotent of class at most $c$ and every Dynkin term of [F5] with more than $c$ nested brackets vanishes identically on ${\mathfrak{n}}^{+}\times {\mathfrak{n}}^{+}$; hence the series of [F4] truncates to a polynomial map $P:{\mathfrak{n}}^{+}\times {\mathfrak{n}}^{+}\to {\mathfrak{n}}^{+}$. Applying [F4] to the real Lie group $GL(V)$ with nilpotent elements $X,Y\in {\mathfrak{n}}^{+}\subseteq \mathfrak{g}\mathfrak{l}(V)$ gives $\exp (X)\exp (Y)=\exp (P(X,Y))$ on a neighborhood of $(0,0)$; both sides are holomorphic functions of $(X,Y)$ on the complex vector space ${\mathfrak{n}}^{+}\times {\mathfrak{n}}^{+}$, so by the identity theorem the identity holds for all $X,Y\in {\mathfrak{n}}^{+}$.

Fix nonzero $e_{{\alpha }_i}\in {\mathfrak{g}}_{{\alpha }_i}$ and define $F:{\mathbb{A}}^m\to {\mathfrak{n}}^{+}$ by $F(z)=z_1{e}_{{\alpha }_1}\ast \cdots \ast z_m{e}_{{\alpha }_m}$, the iterated polynomial group law $X\ast Y=P(X,Y)$ of step 2.1, so that $u_{{\alpha }_1}(z_1)\cdots u_{{\alpha }_m}(z_m)=\exp (F(z))$ by step 2.1 and [F1]. In the basis $e_{{\alpha }_1},\dots ,e_{{\alpha }_m}$ of ${\mathfrak{n}}^{+}$ ordered by increasing height, the bracket of two basis elements is a combination of basis elements of strictly larger height by [F3], so expanding $P$ and the iterated product gives $F_k(z)=z_k+Q_k(z_1,\dots ,z_{k-1})$ with $Q_k$ polynomial; such a map is a bijection with polynomial inverse, defined recursively by $z_1=F_1$, $z_k=F_k-Q_k(z_1,\dots ,z_{k-1})$.

Define $\theta :{\mathbb{A}}^m\to G$ by $\theta (z)=\exp (F(z))$, which by the preceding step equals $u_{{\alpha }_1}(z_1)\cdots u_{{\alpha }_m}(z_m)$ and is therefore a morphism of varieties into $G$. Since $F$ is a bijection, $\theta$ is injective, and $\theta ({\mathbb{A}}^m)=U$ is an abstract subgroup of $G$: because $F$ is a bijection, for $x,y\in {\mathfrak{n}}^{+}$ the elements $\exp (x)$ and $\exp (y)$ satisfy $\exp (x)\exp (y)=\exp (x\ast y)=\exp (P(x,y))$ with $P(x,y)\in {\mathfrak{n}}^{+}$ by step 2.1, and $\exp (x{)}^{-1}=\exp (-x)$ follows from the same identity with $y=-x$.

$U$ is a closed subgroup of $G$. The morphism $\theta :{\mathbb{A}}^m\to G$ has irreducible image and its closure $K$ is an irreducible closed subgroup, since multiplication and inverse carry the dense subgroup $U$ into itself. By [F6], $U$ is constructible, so its density in $K$ gives a nonempty open subset $O\subseteq U$. For any $k\in K$, the two nonempty opens $k{O}^{-1}$ and $O$ of the irreducible variety $K$ meet; writing $k{o}_1^{-1}=o_2$ yields $k=o_2{o}_1\in U$. Thus $K=U$. The coordinate inverse is established separately below.

The common filtration of step 1.1 bounds the nilpotence index of every $X\in {\mathfrak{n}}^{+}$ by $D=\dim V$, so the matrix logarithm $L(g)={\sum }_{j=1}^{D-1}(-1{)}^{j+1}(g-I{)}^j\mathrm{/}j$ is a regular polynomial map $G\to \mathrm{End}(V)$, even though its value need not be a Lie-algebra element for arbitrary $g$. Choose a linear projection $p:\mathrm{End}(V)\to {\mathfrak{n}}^{+}$ that is the identity on ${\mathfrak{n}}^{+}$, and define $r(g)=F^{-1}(p(L(g)))\in {\mathbb{A}}^m$, using the polynomial inverse from step 3.1. For $g=\theta (z)=\exp (F(z))$, finite formal logarithm and exponential are inverse in the nilpotent algebra generated by $F(z)$, so $r(\theta (z))=z$ as a polynomial identity. Therefore $\theta$ is a section of the separated morphism $r:G\to {\mathbb{A}}^m$, hence a closed immersion with regular inverse $r{\mathrm{\mid }}_U$. Its differential at zero is the identity ${\mathfrak{n}}^{+}\to {\mathfrak{n}}^{+}$, so $\mathrm{Lie}U={\mathfrak{n}}^{+}$ and $\dim U=m$. The source ${\mathbb{A}}^m$ is connected, and every $\exp X\in U$ is unipotent by step 1.1. Thus the product map in the statement is an algebraic isomorphism onto the closed connected unipotent subgroup $U$.

The torus $T$ normalizes $U$: for $t\in T$, conjugation by $t$ is an automorphism of $G$ with $t{U}_{{\alpha }_i}{t}^{-1}=U_{{\alpha }_i}$ by [F1], and since the product map of step 6.1 is onto $U$, $tU{t}^{-1}={\prod }_{i}t{U}_{{\alpha }_i}{t}^{-1}={\prod }_i{U}_{{\alpha }_i}=U$. Moreover $T\cap U=1$: an element of $T$ is diagonalisable as an endomorphism of $V$ by step 1.1, an element of $U=\exp ({\mathfrak{n}}^{+})$ is unipotent by step 6.1, and an endomorphism that is both diagonalisable and unipotent is the identity, so $T\cap U\subseteq \{1\}$.

Hence $B=TU$ is a semidirect product on complex points: $T$ normalizes $U$ and $T\cap U=1$ by step 7.1. The multiplication morphism $m:T\times U\to G$ has constructible image by [F6], an abstract subgroup because $T$ normalizes $U$, and irreducible source. The same dense-open subgroup argument as step 5.1 makes its image a closed irreducible algebraic subgroup $B$; over $\mathbb{C}$ it is smooth. Its dimension is $\dim T+\dim U$ because the point fibres of $m$ are singletons by $T\cap U=1$, so its Lie algebra is $\mathfrak{h}\oplus {\mathfrak{n}}^{+}=\mathfrak{b}$. At every point the differential of $m$ is an isomorphism onto $T_{b}B$: at the identity this is the direct sum of $\mathfrak{h}$ and ${\mathfrak{n}}^{+}$, and translations handle the other points. Thus $m$ is étale. It is injective on complex points; the off-diagonal of $(T\times U){\times }_B(T\times U)$ is an open finite-type complex scheme with no complex points and hence empty, so $m$ is a monomorphism. A surjective étale monomorphism is an isomorphism by fppf descent, proving that the inverse $B\to T\times U$ is regular. Since $T$ is abelian and $U$ is normal unipotent, $B$ is solvable with unipotent radical $U$.

Every connected solvable closed subgroup $S\subseteq G$ with $S\supseteq B$ equals $B$: its Lie algebra $\mathrm{Lie}S$ is a solvable subalgebra of $\mathfrak{g}$ containing $\mathfrak{b}$ (the Lie algebra of a closed subgroup is a subalgebra, and the derived series of $\mathrm{Lie}S$ is contained in the Lie algebra of the derived series of $S$, which terminates), so $\mathrm{Lie}S=\mathfrak{b}$ by step 1.2, whence $\dim S=\dim \mathrm{Lie}S=\dim \mathfrak{b}=\dim B$ because connected algebraic groups over $\mathbb{C}$ of characteristic zero are smooth; an inclusion of irreducible closed subvarieties of the same dimension is an equality, so $S=B$.

The whole construction applied to the negative root system ${\mathrm{\Phi }}^{-}$ produces $U^{-}=\exp ({\mathfrak{n}}^{-})$ with $\mathrm{Lie}{U}^{-}={\mathfrak{n}}^{-}$, its coordinate isomorphism, and $B^{-}=T⋉U^{-}$ with $\mathrm{Lie}{B}^{-}=\mathfrak{h}\oplus {\mathfrak{n}}^{-}$; the argument uses only the height function of [F3], which is defined on all roots, and the corresponding root subgroups $U_{\alpha }$ for $\alpha \in {\mathrm{\Phi }}^{-}$ supplied by [F1].

Finally $X^{\ast }(B)\to X^{\ast }(T)$ is bijective: a character $\chi$ of $T$ extends to $B=T⋉U$ by making it trivial on the normal subgroup $U$, so the restriction map is surjective, and it is injective because a character $\psi$ of $B$ with $\psi {\mathrm{\mid }}_T=1$ is trivial on every $U_{\alpha }$ (a morphism ${\mathbb{G}}_a\to {\mathbb{G}}_m$ is given by a unit of $\mathbb{C}[z]$, hence is constant) and $U={\prod }_i{U}_{{\alpha }_i}$ by step 6.1, so $\psi$ is trivial on $U$ and on $BT$. The Axiom of Choice is assumed; [F4] uses the countable-choice BCH interface, while the faithful-representation supplier [F7] is choice-free. The root suppliers [F2] and [F3] retain their stated hypotheses; steps 3.1 to 8.1 then select only finitely many data.