Minimal parabolic from one negative simple root

The Gauss decomposition $S{L}_2=B_2\bigsqcup B_{2}w{B}_2$ holds, where $B_2$ is the upper triangular subgroup of determinant one and $w=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$: a matrix $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ with $c=0$ lies in $B_2$, and one with $c\mathrm{\ne }0$ equals $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\left(\begin{array}{cc}{c}^{-1}& a\\ 0& c\end{array}\right)\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}1& c^{-1}d\\ 0& 1\end{array}\right)$. Since ${\phi }_{\alpha }$ maps the diagonal torus $D$ onto ${\alpha }^{\vee }({\mathbb{C}}^{\times })$ and the standard unipotent subgroups onto $U_{\pm \alpha }$ by [F1], it maps $B_2=D⋉U_2$ onto $B_{\alpha }:=T_{\alpha }{U}_{\alpha }\subseteq B$ with $T_{\alpha }={\alpha }^{\vee }({\mathbb{C}}^{\times })$ and $U_{\alpha }\subseteq U$ by [F2]; hence $V_{\alpha }:={\phi }_{\alpha }(S{L}_2)=B_{\alpha }\bigsqcup B_{\alpha }{n}_{\alpha }{B}_{\alpha }$, and $V_{\alpha }\subseteq B\cup B{n}_{\alpha }B$ because $B_{\alpha }\subseteq B$ and $B_{\alpha }{n}_{\alpha }{B}_{\alpha }\subseteq B{n}_{\alpha }B$. In particular $U_{-\alpha }\subseteq B\cup B{n}_{\alpha }B$ and $w\mathrm{\notin }{B}_2$.

For every root $\beta$ one has $\mathrm{Ad}(n_{\alpha }){\mathfrak{g}}_{\beta }={\mathfrak{g}}_{s_{\alpha }(\beta )}$. Indeed by [F8] $\mathrm{Ad}(n_{\alpha })$ is a Lie algebra automorphism, so for $x\in {\mathfrak{g}}_{\beta }$ and $H\in \mathfrak{h}$ one has $[H,\mathrm{Ad}(n_{\alpha })x]=\mathrm{Ad}(n_{\alpha })[\mathrm{Ad}(n_{\alpha }{)}^{-1}H,x]=\beta (\mathrm{Ad}(n_{\alpha }{)}^{-1}H)\mathrm{Ad}(n_{\alpha })x=(s_{\alpha }\beta )(H)\mathrm{Ad}(n_{\alpha })x$, because $\mathrm{Ad}(n_{\alpha })$ preserves $\mathfrak{h}$ and acts there as the involutive reflection $s_{\alpha }$, so that $\mathrm{Ad}(n_{\alpha }{)}^{-1}H=s_{\alpha }H$, and because the dual reflection satisfies $(s_{\alpha }\beta )(H)=\beta (s_{\alpha }H)$. By [F5] the functional $s_{\alpha }(\beta )$ is a root, so by [F4] and [F7] the target space ${\mathfrak{g}}_{s_{\alpha }(\beta )}$ is the one-dimensional root space of that functional; as $\mathrm{Ad}(n_{\alpha })$ is invertible and ${\mathfrak{g}}_{\beta }\mathrm{\ne }0$, the image is all of it.

For every $\beta \in {\mathrm{\Phi }}^{+}\setminus \{\alpha \}$ one has $s_{\alpha }(\beta )\in {\mathrm{\Phi }}^{+}\setminus \{\alpha \}$. Write $\beta ={\sum }_{\gamma \in \mathrm{\Delta }}{n}_{\gamma }\gamma$ with $n_{\gamma }\ge 0$ integers, using that $\beta$ is positive and [F6]; as $\beta \mathrm{\ne }\alpha$ and the root system is reduced, some $n_{\gamma }$ with $\gamma \mathrm{\ne }\alpha$ is nonzero. The reflected functional $s_{\alpha }(\beta )=\beta -⟨\beta ,{\alpha }^{\vee }⟩\alpha$ has the same coefficient $n_{\gamma }$ at every simple root $\gamma \mathrm{\ne }\alpha$, hence has the positive coefficient $n_{\gamma }>0$ at $\gamma$ and is a root, so all its simple coefficients are nonnegative and it is a positive root; it is different from $\alpha$, whose coefficient at $\gamma$ is $0$.

The two double cosets $B$ and $B{n}_{\alpha }B$ are disjoint. It suffices to show $n_{\alpha }\mathrm{\notin }B$, since $B\cap B{n}_{\alpha }B\mathrm{\ne }\mathrm{\varnothing }$ would give $n_{\alpha }\in B$. Suppose $n_{\alpha }=tu$ with $t\in T$ and $u\in U$, using the decomposition $B=T⋉U$ of [F2]. Then $\mathrm{Ad}(u){\mathrm{\mid }}_{\mathfrak{h}}=\mathrm{Ad}(t{)}^{-1}\mathrm{Ad}(n_{\alpha }){\mathrm{\mid }}_{\mathfrak{h}}=s_{\alpha }$, because $t$ acts trivially on $\mathfrak{h}$; but writing $u=u_{{\beta }_1}(z_1)\cdots u_{{\beta }_m}(z_m)$ in the height-compatible order of [F2] and using $\mathrm{Ad}(u_{\beta }(z))=e^{z{\mathrm{ad}}_{e_{\beta }}}$ from [F8], each factor maps $\mathfrak{h}$ into $\mathfrak{h}+{\mathfrak{n}}^{+}$: indeed $[e_{\beta },H]=-\beta (H)e_{\beta }\in {\mathfrak{n}}^{+}$ and ${\mathfrak{n}}^{+}$ is a Lie subalgebra, so every term $({\mathrm{ad}}_{e_{\beta }}{)}^{k}H$ with $k\ge 1$ lies in ${\mathfrak{n}}^{+}$. A composition of maps of the form $\mathrm{i}\mathrm{d}+N$ with $N$ mapping $\mathfrak{h}$ into ${\mathfrak{n}}^{+}$ again maps $\mathfrak{h}$ into $\mathfrak{h}+{\mathfrak{n}}^{+}$ and induces the identity on the $\mathfrak{h}$-component, so $(\mathrm{Ad}(u)-\mathrm{i}\mathrm{d})(\mathfrak{h})\subseteq {\mathfrak{n}}^{+}$; comparing with $s_{\alpha }$ gives $(s_{\alpha }-\mathrm{i}\mathrm{d})(\mathfrak{h})\subseteq \mathfrak{h}\cap {\mathfrak{n}}^{+}=0$ by [F4], hence $s_{\alpha }=\mathrm{i}\mathrm{d}$ on $\mathfrak{h}$, contradicting $s_{\alpha }(h_{\alpha })=-h_{\alpha }\mathrm{\ne }{h}_{\alpha }$. Therefore $n_{\alpha }\mathrm{\notin }B$.

Fix a height-compatible order of ${\mathrm{\Phi }}^{+}$ with $\alpha$ first and write $U^{\mathrm{\prime }\mathrm{\prime }}={\prod }_{\beta \in {\mathrm{\Phi }}^{+}\setminus \{\alpha \}}{U}_{\beta }$ in the induced order, so that $U=U_{\alpha }\cdot U^{\mathrm{\prime }\mathrm{\prime }}$ by [F2]. For $u\in U$, written as $u=u_{\alpha }(z)u^{\mathrm{\prime }\mathrm{\prime }}$ with $u^{\mathrm{\prime }\mathrm{\prime }}\in U^{\mathrm{\prime }\mathrm{\prime }}$, one has $n_{\alpha }u{n}_{\alpha }=C_{n_{\alpha }}(u)n_{\alpha }^2$ with $n_{\alpha }^2\in T$ by [F1]; by [F9] applied to the automorphism $C_{n_{\alpha }}$ of [F8] and the parametrization $u_{\beta }(s)={\exp }_G(s{e}_{\beta })$ of [F3], step 1.2 gives $C_{n_{\alpha }}(u_{\alpha }(z))\in U_{-\alpha }$ and $C_{n_{\alpha }}(u_{\beta })\subseteq U_{s_{\alpha }\beta }$ for each factor $u_{\beta }$ of $u^{\mathrm{\prime }\mathrm{\prime }}$, so by step 1.3 the element $C_{n_{\alpha }}(u^{\mathrm{\prime }\mathrm{\prime }})$ is a product of elements of the subgroups $U_{s_{\alpha }\beta }\subseteq U$ and therefore lies in $U$; hence $n_{\alpha }U{n}_{\alpha }\subseteq U_{-\alpha }\cdot U\cdot T\subseteq U_{-\alpha }\cdot B$.

Moreover $B\cap V_{\alpha }=B_{\alpha }$. The preimage $H:={\phi }_{\alpha }^{-1}(B)$ is a closed subgroup of $S{L}_2$ containing $B_2$, and $H\mathrm{\ne }S{L}_2$ since $n_{\alpha }={\phi }_{\alpha }(w)\mathrm{\notin }B$ by step 1.4. If $H$ contained an element $g\mathrm{\notin }{B}_2$, then $g\in B_{2}w{B}_2$ by step 1.1, so $w\in B_{2}g{B}_2\subseteq H$ and $H$ contains $⟨B_2,w⟩=S{L}_2$, since every element of $B_{2}w{B}_2$ is a product of two elements of $B_2$ and one $w$ and $S{L}_2=B_2\cup B_{2}w{B}_2$; this contradicts $H\mathrm{\ne }S{L}_2$, so $H=B_2$ and $B\cap V_{\alpha }={\phi }_{\alpha }(B_2)=B_{\alpha }$.

Consequently $n_{\alpha }B{n}_{\alpha }=(n_{\alpha }T{n}_{\alpha })(n_{\alpha }U{n}_{\alpha })=T\cdot n_{\alpha }U{n}_{\alpha }\subseteq T\cdot U_{-\alpha }\cdot B\subseteq B\cdot U_{-\alpha }\cdot B\subseteq B\cdot (B\cup B{n}_{\alpha }B)\cdot B=B\cup B{n}_{\alpha }B$, where the middle inclusion uses step 2.1, the next uses $U_{-\alpha }\subseteq B\cup B{n}_{\alpha }B$ from step 1.1, and the last uses $B\cdot B\subseteq B$ and $B\cdot B{n}_{\alpha }B\cdot B=B{n}_{\alpha }B$.

The simple double coset has a one-root chart: $B{n}_{\alpha }B=U_{\alpha }{n}_{\alpha }B$. Indeed $B=T{U}_{\alpha }{U}^{\mathrm{\prime }\mathrm{\prime }}$ by [F2] and step 2.1, while $n_{\alpha }^{-1}{U}^{\mathrm{\prime }\mathrm{\prime }}{n}_{\alpha }\subseteq U\subseteq B$ by steps 1.2–1.3, so the $U^{\mathrm{\prime }\mathrm{\prime }}$ factor moves through $n_{\alpha }$ into the right $B$ factor and $T$ is absorbed there. The multiplication $U_{\alpha }\times B\to B{n}_{\alpha }B$, $(u,b)\mapsto u{n}_{\alpha }b$, has singleton complex-point fibres: equality of two outputs would put a nontrivial element of $U_{\alpha }$ in $n_{\alpha }B{n}_{\alpha }^{-1}$, whereas conjugation by $n_{\alpha }^{-1}$ sends it into $U_{-\alpha }\cap B=1$ by step 2.2. Thus the image is irreducible of dimension $\dim B+1$ by the fibre-dimension formula [F10].

Define $Q=B\cup B{n}_{\alpha }B$. Then $Q\cdot Q\subseteq Q$: one has $B\cdot B\subseteq B$ and $B\cdot B{n}_{\alpha }B\cdot B\subseteq B{n}_{\alpha }B$, so the only non-formal product is $(B{n}_{\alpha }B)(B{n}_{\alpha }B)=B(n_{\alpha }B{n}_{\alpha })B\subseteq B(B\cup B{n}_{\alpha }B)B=Q$ by step 3.1. Also $Q^{-1}=Q$, because $B^{-1}=B$ and $(B{n}_{\alpha }B{)}^{-1}=B{n}_{\alpha }^{-1}B\subseteq B{n}_{\alpha }B$: indeed $n_{\alpha }^{-1}=n_{\alpha }{n}_{\alpha }^{-2}\in n_{\alpha }T\subseteq B{n}_{\alpha }B$ by [F1], using $T\subseteq B$. Hence $Q$ is a subgroup of $G$ containing $B$ and, by step 1.1, containing $U_{-\alpha }$.

Therefore $P_{\alpha }=Q$: since $Q$ is a subgroup containing the two generators $B$ and $U_{-\alpha }$ of $P_{\alpha }$, one has $P_{\alpha }\subseteq Q$; conversely $B\subseteq P_{\alpha }$, and $n_{\alpha }={\phi }_{\alpha }(w)\in V_{\alpha }={\phi }_{\alpha }(S{L}_2)$ is a product of elements of $U_{\alpha }\subseteq B$, of $U_{-\alpha }$ and of $T_{\alpha }\subseteq T\subseteq B$, so $n_{\alpha }\in P_{\alpha }$ and $B{n}_{\alpha }B\subseteq P_{\alpha }$, whence $Q\subseteq P_{\alpha }$. Thus $P_{\alpha }=B\cup B{n}_{\alpha }B$, and with step 1.4 this is the disjoint union $B\bigsqcup B{n}_{\alpha }B$.

The set $P_{\alpha }=B\bigsqcup B{n}_{\alpha }B$ is constructible because $B$ is closed and $B{n}_{\alpha }B$ is the morphic image of $U_{\alpha }\times B$ of step 3.2. The latter image is irreducible. For every $z\mathrm{\ne }0$, the rank-one Gauss decomposition of step 1.1 puts $u_{-\alpha }(z)$ in $B{n}_{\alpha }B$, while $u_{-\alpha }(z)\to 1$ algebraically as $z\to 0$; hence $1$ and, by left $B$-translation, all of $B$ lie in the closure of $B{n}_{\alpha }B$. Consequently $Z=\overline{P_{\alpha }}=\overline{B{n}_{\alpha }B}$ is irreducible of dimension $\dim B+1$ by step 3.2. Since $P_{\alpha }$ is a dense constructible subgroup of $Z$ by step 5.1, it contains a nonempty open $O$ of $Z$. For any $z\in Z$, the two nonempty opens $z{O}^{-1}$ and $O$ meet, so $z\in OO\subseteq P_{\alpha }$; thus $P_{\alpha }=Z$ is closed and irreducible, hence connected, and has dimension $\dim B+1$. As a closed algebraic subgroup over $\mathbb{C}$ it is smooth.

Construct the quotient as a scheme using two product charts. Put $s_0(z)=u_{-\alpha }(z)$ and $s_{\mathrm{\infty }}(t)=u_{\alpha }(t)n_{\alpha }$, and let $m_0:{\mathbb{A}}^1\times B\to P_{\alpha }$ and $m_{\mathrm{\infty }}:{\mathbb{A}}^1\times B\to P_{\alpha }$ send $(z,b)$ to $s_0(z)b$ and $(t,b)$ to $s_{\mathrm{\infty }}(t)b$. The maps are injective on complex points: for $m_0$ this uses $U_{-\alpha }\cap B=1$ from step 2.2, and for $m_{\mathrm{\infty }}$ it uses $n_{\alpha }^{-1}{U}_{\alpha }{n}_{\alpha }=U_{-\alpha }$ and the same intersection. Their differential at every point is an isomorphism onto $T{P}_{\alpha }$: at $1$ and $n_{\alpha }$, after left translation, its two summands are ${\mathfrak{g}}_{-\alpha }$ and $\mathfrak{b}$, which sum directly to $\mathrm{Lie}{P}_{\alpha }$: that Lie algebra contains $\mathfrak{b}\oplus {\mathfrak{g}}_{-\alpha }$ because $P_{\alpha }$ contains $B$ and $U_{-\alpha }$, while $\dim P_{\alpha }=\dim B+1$ by step 6.1; translations handle all other points. Since source and target are smooth finite-type schemes over $\mathbb{C}$, the maps are étale. Their injectivity on closed points makes each an étale monomorphism: the complement of the open diagonal in its finite-type fibre product is closed and has no $\mathbb{C}$-point, hence is empty. Thus both $m_0$ and $m_{\mathrm{\infty }}$ are open immersions. The image of $m_{\mathrm{\infty }}$ is $B{n}_{\alpha }B$ by step 3.2, and $B$ lies in the image of $m_0$, so the two open images cover $P_{\alpha }$ on closed points and therefore as schemes. In $S{L}_2$, with $s_0(z)=\left(\begin{array}{cc}1& 0\\ z& 1\end{array}\right)$ and $s_{\mathrm{\infty }}(t)=\left(\begin{array}{cc}t& -1\\ 1& 0\end{array}\right)$, one calculates $s_0(z{)}^{-1}{s}_{\mathrm{\infty }}(t)=\left(\begin{array}{cc}t& -1\\ 1-zt& z\end{array}\right)$, which is upper triangular exactly when $zt=1$; the overlap transition is the upper-triangular matrix at $t=z^{-1}$, carried by ${\phi }_{\alpha }$ into $B$. Therefore the local first-coordinate maps glue to a morphism $q:P_{\alpha }\to {\mathbb{A}}^1{\cup }_{z=t^{-1}}{\mathbb{A}}^1={\mathbb{P}}^1$, and each preimage is ${\mathbb{A}}^1\times B$ with right $B$ acting on its second factor. These local products show directly on every test scheme that $q$ represents the fppf sheaf quotient $P_{\alpha }\mathrm{/}B$ and is a Zariski-locally trivial right $B$-torsor. This proves (iii), including the $t=0$ point that the earlier calculation dividing by a matrix entry omitted.

$\mathrm{Lie}{P}_{\alpha }=\mathfrak{b}\oplus {\mathfrak{g}}_{-\alpha }$: since $P_{\alpha }$ is a closed subgroup containing $B$ and $U_{-\alpha }$, its Lie algebra contains $\mathrm{Lie}B+\mathrm{Lie}{U}_{-\alpha }=\mathfrak{b}+{\mathfrak{g}}_{-\alpha }$ by [F2] and [F3]; here $\mathfrak{b}\cap {\mathfrak{g}}_{-\alpha }=0$ by [F4], since $-\alpha \mathrm{\notin }{\mathrm{\Phi }}^{+}$, so $\mathfrak{b}+{\mathfrak{g}}_{-\alpha }=\mathfrak{b}\oplus {\mathfrak{g}}_{-\alpha }$ has dimension $\dim \mathfrak{b}+1=\dim B+1$. Connected algebraic groups over $\mathbb{C}$ of characteristic zero are smooth, so $\dim \mathrm{Lie}{P}_{\alpha }=\dim P_{\alpha }=\dim B+1$ by step 6.1, and the inclusion of vector spaces of the same dimension is an equality.

Collecting the steps proves (i), (ii) and the two-chart description of (iii): $P_{\alpha }=B\bigsqcup B{n}_{\alpha }B$ is closed and connected by steps 5.1 and 6.1, $\mathrm{Lie}{P}_{\alpha }=\mathfrak{b}\oplus {\mathfrak{g}}_{-\alpha }$ with $\dim P_{\alpha }=\dim B+1$ by steps 6.1 and 7.2, and the two affine charts and their fppf quotient gluing $t=z^{-1}$ are established in step 7.1. The Axiom of Choice enters through [F11]: full AC is assumed by the root-subgroup suppliers [F1]–[F3], root results [F5], [F7], and constructibility and fibre dimension in [F10]; the Lie-group suppliers [F8] and [F9] require only countable choice; the argument selects only the fixed simple root $\alpha$, the fixed root vectors and the fixed height order of ${\mathrm{\Phi }}^{+}$.