Root subgroups of a split reductive group

The subtorus $T_{\alpha }$ has codimension one in $T$ and is central in $G_{\alpha }$. Thus the semisimple rank of the split reductive group $G_{\alpha }$ is at most one. It is nonzero because its Lie algebra contains the nonzero root space ${\mathfrak{g}}_{\alpha }$, whereas a reductive group of semisimple rank zero is a torus. Apply [F2]: the central cover of $G_{\alpha }\mathrm{\prime }$ has the two standard root groups of ${\mathrm{S}\mathrm{L}}_2$, with one-dimensional root spaces. The central torus $Z(G_{\alpha }{)}_t$ contributes only zero weights, by the exact central-torus product clause of [F1], so $G_{\alpha }$ has exactly two nonzero weights, $\alpha$ and $-\alpha$. By [F1], the roots of $G$ restricting trivially to $T_{\alpha }$ are precisely the rational multiples of $\alpha$; hence those multiples are only $\pm \alpha$ and $\mathrm{Lie}(G_{\alpha })=\mathfrak{t}\oplus {\mathfrak{g}}_{\alpha }\oplus {\mathfrak{g}}_{-\alpha }$. This proves (a).

Choose the central cover $v:{\mathrm{S}\mathrm{L}}_2\to G_{\alpha }\mathrm{\prime }$ from [F2], taking its diagonal torus onto $T_1=(T\cap G_{\alpha }\mathrm{\prime }{)}_t$. Its standard root groups map isomorphically onto $U_{\pm \alpha }$, giving (b), including the Lie algebras and the smooth $T$-stable subgroup containment criterion of [F1]. The element $v(n_{\alpha })$ normalizes $T_1$ and centralizes the central torus of $G_{\alpha }$, hence normalizes $T=T_{1}Z(G_{\alpha }{)}_t$. The Weyl group therefore has the two standard rank-one elements. Choose the sign of the image under $v$ of the standard coroot so that its pairing with $\alpha$ is $2$. On $T_1$ conjugation by $v(n_{\alpha })$ acts by inversion, and on the central torus it acts trivially. These two subtori generate $T$ up to finite intersection, so the rank-one reflection formula holds on $X(T)\otimes \mathbb{Q}$, and hence on $X(T)$: $s_{\alpha }(x)=x-⟨x,{\alpha }^{\vee }⟩\alpha$. This formula determines the pairing of ${\alpha }^{\vee }$ with every character, proving its uniqueness in $X_{\ast }(T)$. This proves (c).

For (d): each root space ${\mathfrak{g}}_{\alpha }$ is contained in $\mathrm{Lie}(T\cdot U_{\alpha })$, so $\mathrm{Lie}$ of the subgroup $H$ generated by $T$ and the $U_{\alpha }$ contains $\mathfrak{t}$ and every root space, i.e. all of $\mathfrak{g}$; since $H$ is smooth (a generated subgroup of a reductive group by smooth subgroups is smooth, by the same generation criterion [F2]) and $G$ is connected, $H=G$ by the Lie-generation criterion [F2]. For (e): (rd1) is (c), (rd2) follows on roots from conjugation by $n_{\alpha }$, which sends $U_{\beta }$ to $U_{s_{\alpha }\beta }$; the uniqueness characterization of each coroot in(c) also sends ${\beta }^{\vee }$ to $(s_{\alpha }\beta {)}^{\vee }$ under the dual conjugation action, giving the required coroot-reflection invariance, and (rd3) holds because $W(G,T)$ is finite (it is a quotient of the normalizer of a maximal torus, finite over $k$); the reflection subgroup acts simply transitively on the Weyl chambers, while $W(G,T)$ acts faithfully and simply transitively on the Borel subgroups containing $T$. The Borel–chamber bijection identifies these two sets, so the reflection subgroup has the same order as $W(G,T)$ and equals it (Milne's Weyl identification theorem).

For (f), choose a cocharacter strictly positive on ${\mathrm{\Phi }}^{+}$. Its conjugation contracts $U=B_u$ and each positive root group to the identity. For any ordering, multiplication ${\prod }_{\alpha \in {\mathrm{\Phi }}^{+}}{U}_{\alpha }\to U$ is equivariant and its differential at the identity is the direct-sum isomorphism of the positive root spaces. The strictly contracting smooth affine comparison in [F3] therefore makes it a global variety isomorphism; this is the argument of [F4] and uses no commutativity of the root groups. If $H\subseteq U$ is smooth and $T$-stable, the root-group containment criterion puts precisely the root groups corresponding to the weights of $\mathrm{Lie}H$ in $H^{\circ }$. The same contracting argument makes their ordered product isomorphic to $H^{\circ }$. To prove $H=H^{\circ }$, order those groups first in the coordinates of $U$. Those coordinates decompose $H$ as $H^{\circ }$ times a finite residual locus whose points have only the omitted root coordinates. The connected torus acts trivially on this finite component set, so that residual locus is fixed by $T$. But $U^T=U\cap C_G(T)=U\cap T=1$. Thus $H$ is connected and is exactly the ordered product of its root groups, proving the asserted subgroup description.

Let $\mathrm{\Psi }\subseteq {\mathrm{\Phi }}^{+}$ be the weights of such an $H$. The Chevalley formula in [F4] and uniqueness of ordered root coordinates show that if $\alpha ,\beta \in \mathrm{\Psi }$ and $N_{\alpha ,\beta ;i,j}$ is nonzero in $k$, the root $i\alpha +j\beta$ must be in $\mathrm{\Psi }$: the corresponding coordinate of the commutator is a nonzero polynomial in $x,y$ and cannot be an omitted coordinate of $H$. Conversely, for quasi-closed $\mathrm{\Psi }$, take the coordinate subvariety $H={\prod }_{\alpha \in \mathrm{\Psi }}{U}_{\alpha }$ in the coordinates of $U$. Collecting products and inverses using [F4] creates only root coordinates still in $\mathrm{\Psi }$, so $H$ is a closed $T$-stable subgroup. Its product coordinates make it smooth connected with precisely those weights. Ordinary root closure implies quasi-closure by the root-string combinatorics. If $\mathrm{char}k=0$ or $\mathrm{char}k=p>3$, every structure coefficient that occurs in [F4] remains nonzero; in particular a root sum of two members of $\mathrm{\Psi }$ must remain in $\mathrm{\Psi }$. Thus quasi-closure and ordinary root closure coincide in those characteristics. This proves the corrected criterion.