Cartan subgroups: conjugacy, density and normalizers

By [F1] the Cartan $C=C_G(T)$ is smooth connected nilpotent and satisfies $N_G(C{)}^{\circ }=C$. It lies in some Borel $B^{\mathrm{\prime }}$ because it is connected solvable, and $T\subseteq B^{\mathrm{\prime }}$. For any other Borel $B\supseteq T$, [F2] gives $B=n{B}^{\mathrm{\prime }}{n}^{-1}$ with $n\in N_G(T)(k)$. Such $n$ preserves $C$, so $C\subseteq B$. In the reductive case [F7] gives $C=T$. This proves (a).

Conjugacy of maximal tori in [F2] implies conjugacy of their centralizers, proving the first assertion of (b). To prove density, decompose $\mathfrak{g}$ into $T$-weights. There are finitely many nonzero weights; choose $t\in T(k)$ with $\chi (t)\mathrm{\ne }1$ for each of them, possible because their kernels are proper closed subsets of the irreducible torus. On the zero-weight space $\mathfrak{c}$ the differential of $G\times C\to G$, $(g,c)\mapsto gc{g}^{-1}$, at $(e,t)$ receives all of $\mathfrak{c}$ from its second factor. On every other weight space the first-factor differential is multiplication by $1-\chi (t)$, hence invertible. The differential is surjective. A morphism between smooth varieties with surjective differential is smooth on a neighbourhood of this point and hence has open image there. This nonempty open subset consists of conjugates of points of $C$; algebraically closed rational points in each nonempty fibre give the asserted dense open union of Cartan subgroups.

To prove the full neutral-normalizer clause for possibly nonsmooth $H\supseteq C$, put $X=G\mathrm{/}H$, smooth by [F4]. The connected normalizer of $T$ acts trivially on $T$ by rigidity, so $N_G(T{)}^{\circ }=C$. It has finitely many connected components; therefore $N_G(T)(k)\mathrm{/}C(k)$ is finite. Every $T$-fixed coset $gH\in X(k)$ satisfies $g^{-1}Tg\subseteq H$. Both $T$ and $g^{-1}Tg$ lie in the smooth connected group $(H_{\mathrm{r}\mathrm{e}\mathrm{d}}{)}^{\circ }$ and are maximal tori there, since they are maximal in $G$. By [F2] some $h\in H(k)$ makes $gh\in N_G(T)(k)$. Thus the fixed cosets are represented by this finite set of normalizer components. Since $X^T$ is smooth by [F1] and has finitely many geometric points, it is finite étale.

Let $B\supseteq T$ be a Borel. For every $R$ and $z\in C_G(B)(R)$, the morphism $g\mapsto zg{z}^{-1}{g}^{-1}$ is right $B_R$-invariant, since $z$ commutes with $B_R$, and descends to $(G\mathrm{/}B{)}_R$. By [F5] any such morphism to affine $G_R$ is constant, with its identity value at $eB$. Therefore $C_G(B)=Z(G)$ as group schemes. Step 1.1 gives $Z(G)\subseteq C_G(T)\subseteq B$, so $Z(B)=C_G(B)\cap B=Z(G)$. This proves (c).

The quotient $N_G(H)\mathrm{/}H$ is a closed subgroup scheme of $X$: pullback along the faithfully flat cover $G\to X$ identifies it with the closed subgroup $N_G(H)\subseteq G$, so closed immersion descends. It lies in $X^T$ because $T\subseteq H$ and each normalizer point transports $T$ into $H$. A closed subscheme of a finite étale scheme over algebraically closed $k$ is finite étale. Therefore $N_G(H)\mathrm{/}H$ is finite étale and its identity fibre is $H$. The connected component of $N_G(H)$ lies in that fibre; together with the reverse inclusion this proves $N_G(H{)}^{\circ }=H^{\circ }$, without claiming $H$ smooth. For smooth $H$, [F3] also gives $\mathrm{Lie}{N}_G(H)=\mathrm{Lie}H$: the zero-weight space ${\mathfrak{g}}^T=\mathfrak{c}$ lies in $\mathfrak{h}$, so $(\mathfrak{g}\mathrm{/}\mathfrak{h}{)}^T=0$, hence its $H$-invariants vanish. Thus the full normalizer of such $H$ is smooth.

We prove $N_G(B)=B$ by induction on $\dim G$, with the zero-dimensional case immediate. The normalizer is smooth by step 2.2 because $B$ contains $C$. For $x\in N_G(B)(k)$, conjugate by a point of $B$ so that $x$ normalizes $T$, using [F2]. Then $\phi :T\to T$, $t\mapsto xt{x}^{-1}{t}^{-1}$, is a homomorphism. If it is not surjective, its kernel contains a positive-dimensional torus $S$. Thus $x\in C_G(S)$ and normalizes $C_G(S)\cap B$, a Borel by [F1]. If $C_G(S)\mathrm{\ne }G$, induction on this smaller smooth connected group puts $x$ in $B$. If $C_G(S)=G$, then $S$ is central; induction on $G\mathrm{/}S$ and its Borel $B\mathrm{/}S$ again puts $x$ in $B$. The quotient Borel assertion follows by pulling back any larger smooth connected solvable subgroup along the smooth central-torus quotient.

If $\phi$ is surjective, choose by [F5] a line $L=kv$ whose scheme-theoretic stabilizer is $N_G(B)$. Its character on $T$ is trivial, because it is trivial on commutators $[x,t]$ and these exhaust $T$ as a group scheme. The unipotent subgroup $B_u$ also fixes $v$, by the fixed-vector criterion on this one-dimensional representation. Hence $B$ fixes $v$ and the orbit morphism descends to $G\mathrm{/}B\to V$. By [F5] it is the constant $v$, so $G$ fixes $v$ and therefore $G=N_G(B)$. This makes $B$ normal. All conjugate Borels then equal $B$ and, by step 1.1, all Cartans lie in $B$. Their dense union in step 1.2 forces the closed subgroup $B$ to equal $G$. Thus $x\in B$ in this case as well. Since $N_G(B)$ and $B$ are smooth with identical algebraically closed points, they are equal as group schemes.

Every subgroup variety $P\supseteq B$ is connected: for $p\in P(k)$, the two Borels $B$ and $pB{p}^{-1}$ of $P^{\circ }$ are conjugate by $P^{\circ }(k)$, so after multiplying $p$ by such a point it normalizes $B$. Step 4.1 puts that product in $B\subseteq P^{\circ }$, hence $p\in P^{\circ }$. Let $P=(N_G(B_u){)}_{\mathrm{r}\mathrm{e}\mathrm{d}}$, a subgroup variety by [F4], containing $B$. The subgroup $B_u$ is maximal among smooth connected unipotent subgroups of $G$: any larger such group lies in a Borel, and its dimension is at most that Borel's unipotent radical, which has the same dimension as $B_u$ by conjugacy. The smooth connected affine quotient $P\mathrm{/}{B}_u$ therefore has no nontrivial smooth connected unipotent subgroup, since its inverse image would be a larger such subgroup of $G$. By [F6] it is a torus, so $P$ is solvable and Borel maximality gives $P=B$. Thus $(N_G(B_u){)}_{\mathrm{r}\mathrm{e}\mathrm{d}}=B$, completing (d). The unreduced equality can fail: in characteristic $2$ and $G={\mathrm{P}\mathrm{G}\mathrm{L}}_2$, $I+\epsilon E_{21}$ normalizes upper $B_u$ over $k[\epsilon ]\mathrm{/}{\epsilon }^2$ but is outside upper $B$.