Fixed loci and centralizers of torus actions are connected

Smoothness of $G^S$ follows from [F1], without affineness. The dual-number fixed condition at the identity is precisely invariance in the tangent representation, so $\mathrm{Lie}(G^S)={\mathfrak{g}}^S$. This also gives the centralizer formula when $S$ acts by subgroup conjugation.

We establish a rigidity consequence for Borels used below. Over an algebraically closed field, if two group homomorphisms ${\phi }_1,{\phi }_2:G_R\to K_R$ agree on $B_R$, the map $g\mapsto {\phi }_1(g){\phi }_2(g{)}^{-1}$ is right $B_R$-invariant and descends to $(G\mathrm{/}B{)}_R$. The quotient is smooth connected complete, hence integral with global functions $k$; by [F7] its base change has global functions $R$. Since $K_R$ is affine, the descended map is constant, and its value at $eB$ is the identity. Thus ${\phi }_1={\phi }_2$. Applying this to conjugation by $c\in C_G(B)(R)$ and the identity, for every $R$, proves $C_G(B)=Z(G)$; in particular $Z(B)\subseteq Z(G)$.

For connectedness of a subgroup centralizer in affine $G$, work over an algebraic closure and choose a faithful representation $W$ by [F2]. Its finite torus weights have finitely many nonzero differences. Choose an integral cocharacter $\lambda$ avoiding their pairing-zero hyperplanes; then two weights have equal $\lambda$-weight exactly when they have equal $S$-weight. Matrix block comparison gives $C_G(S)=C_G(\lambda ({\mathbf{G}}_m))$. This smooth centralizer occurs as the middle factor of the open immersion in [F3]. If it had two connected components, the products of each component with the identity components of the two other factors would give disjoint nonempty open subsets of the irreducible smooth connected group $G$, a contradiction. It is therefore connected. For an external action form the smooth connected affine semidirect product $G⋊S$. Its subgroup centralizer of $S$ is $G^S\times S$ as a scheme, so the preceding conclusion implies connectedness of $G^S$. Descent proves the assertion over $k$.

If a smooth connected affine group $H$ has a nilpotent Borel $D$, then $H=D$, by induction on $\dim D$. When $\dim D=0$, $D=1$ and $H=H\mathrm{/}D$ is affine and complete, hence has dimension zero and is trivial by [F7]. Otherwise [F5] gives a positive-dimensional smooth connected $N\subseteq Z(D)$, which is central in $H$ by step 1.2. The quotient $D\mathrm{/}N$ is a nilpotent Borel of $H\mathrm{/}N$: a larger smooth connected solvable subgroup of $H\mathrm{/}N$ pulls back to a larger smooth connected solvable subgroup of $H$, contradicting maximality of $D$. The affine smooth quotient exists by [F6]. Induction gives $H\mathrm{/}N=D\mathrm{/}N$, hence $H=D$.

Let $C=C_G(T)$ for maximal $T$. By steps 1.1–1.2 it is smooth connected affine, and $T\subseteq Z(C)$. Choose a Borel $D$ of $C$ containing $T$; it decomposes as $D=D_u⋊T$ by [F4], since $T$ is already maximal in $G$. Centrality of $T$ makes this a direct product, so $D$ is nilpotent. Step 2.2 gives $C=D$, proving nilpotence. Its maximal torus $T$ is unique by [F5]. Therefore $N_G(C)$ preserves $T$; the connected group $N_G(C{)}^{\circ }$ acts trivially on $T$ by [F6], so it is contained in $C_G(T)=C$. The reverse inclusion follows from connectedness of $C$, proving $N_G(C{)}^{\circ }=C$.

Put $C=C_G(S)$ and let $B\supseteq S$ be a Borel. The group $C\cap B=C_B(S)$ is smooth connected by steps 1.1–1.2 applied to affine $B$, and solvable as a subgroup of $B$. Let $Y$ be the reduced closure of $CB$ in $G$. It is irreducible, hence connected, as the closure of the image of connected smooth $C\times B$, and is stable under right multiplication by $B$. The transporter of $S$ into $B$ is closed, contains $CB$, and hence contains $Y$. Thus $(y,s)\mapsto y^{-1}sy$ defines a morphism $Y\times S\to B$. If $q:B\to B\mathrm{/}{B}_u$ is the torus quotient, the maps $s\mapsto q(y^{-1}sy)$ form a family of torus homomorphisms parametrized by connected $Y$. Rigidity [F6] makes them equal to $q{\mathrm{\mid }}_S$, their value at $y=e$.

Choose a maximal torus $T\subseteq B$ containing $S$. For $y\in Y(k)$ the torus $y^{-1}Sy\subseteq B$ can be conjugated into $T$ by some $u\in B_u(k)$, by [F4]. Since $yu\in Y$, step 3.2 gives $q((yu{)}^{-1}s(yu))=q(s)$ for every $s\in S$. Both arguments lie in $T$, where $q{\mathrm{\mid }}_T$ is an isomorphism, so $(yu{)}^{-1}s(yu)=s$ scheme-theoretically. Thus $yu\in C(k)$ and $y\in CB(k)$. The multiplication image $CB$ is constructible by [F4] and contains every closed point of its closure. Its constructible complement in $Y$ is therefore empty, since a nonempty constructible subset of a variety over an algebraically closed field contains a closed point. Hence $CB$ is closed as a subset. The quotient map $G\to G\mathrm{/}B$ is open and surjective, so its image $q(C)$ is closed because its inverse image is $CB$. Its image in the complete quotient $G\mathrm{/}B$ is a closed orbit of $C$, identified scheme-theoretically with $C\mathrm{/}(C\cap B)$ by the homogeneous-space theorem; it is complete. To prove maximality, suppose a smooth connected solvable $D\subseteq C$ contains $C\cap B$. Its action on this complete quotient has a fixed point by the Borel fixed-point theorem, so $D\subseteq c(C\cap B)c^{-1}$ for some $c$, implying $\dim D\le \dim (C\cap B)$ and hence equality of the two smooth connected subgroups. Thus $C\cap B$ is a Borel of $C$.