Exposé XIII. Regular elements of algebraic groups and Lie algebras

by A. Grothendieck

1. An auxiliary lemma on varieties with operators

Let $S$ be a prescheme, $G$ an $S$-prescheme in groups acting on the left on an $S$-prescheme $V$, $W$ a closed $S$-subprescheme of $V$, $N$ its stabilizer in $G$ — the subgroup of $G$ whose points, with values in an $S^{\prime }$ over $S$, are the $g\in G(S^{\prime })$ such that $g\cdot W_{S^{\prime }}=W_{S^{\prime }}$. We endow $(Sch)/S$ with the faithfully flat quasi-compact topology, and identify $G$, $V$, $W$ with the corresponding sheaves (cf. IV). We shall therefore argue in the category of sheaves on $(Sch)/S$, and in this section the term "locally" refers to the topology we have just specified on $(Sch)/S$. Note that $N$ is a sheaf; consider the quotient sheaf $G/N$. One sees at once that it is isomorphic to the following functor: to every $S^{\prime }$ over $S$, one associates the set of subsheaves $W^{\prime }$ of $V_{S^{\prime }}$ that are locally conjugate to $W_{S^{\prime }}$ by the group $G$. Let $X$ be the subsheaf of $G/N{\times }_{S}V$ whose value, for every $S^{\prime }$ over $S$, is the set of pairs $(W^{\prime },v)$, where $W^{\prime }$ is as above and $v$ is a section of $W^{\prime }$ over $S^{\prime }$ (hence a section of $V_{S^{\prime }}$ over $S^{\prime }$). Let $Z$ be the inverse image of $X$ in $G{\times }_{S}V$, so that we have the cartesian diagram

              Z  ─────────→  X
              │              │
        (✱)   │              │ i
              ↓              ↓
           G ×_S V ─→ G/N ×_S V,

where $i$ is the canonical immersion, and the second horizontal arrow comes from the canonical morphism $G\to G/N$.

sees that $Z(S^{\prime })$ is the set of pairs $(g,v)\in G(S^{\prime })\times V(S^{\prime })$ such that $v\in g\cdot W_{S^{\prime }}(S^{\prime })$. Consequently, $Z$ is isomorphic to the sheaf $G{\times }_{S}W$, by means of the isomorphism

G ×_S W ─⥲→ Z

defined by $(g,w)\mapsto (g,g\cdot w)$. Thus the preceding cartesian diagram gives the cartesian diagram

                 q
              G ×_S W ─────→  X
                │              │
       (✱✱)    λ│              │ i
                ↓              ↓
              G ×_S V ─→ G/N ×_S V

where $\lambda (g,w)=(g,g\cdot w)$, hence $q(g,w)=(\bar{g},g\cdot w)$, where ḡ denotes the image of $g$ under the canonical map $G(S^{\prime })\to (G/N)(S^{\prime })$. One sees finally from diagram (✱) that $Z\to X$ makes $Z$ into a principal bundle with base $X$ and group $N$ acting on the right by $(g,v)\cdot n=(gn,v)$, so that in (✱✱), $q:G{\times }_{S}W\to X$ makes $G{\times }_{S}W$ into a principal bundle with base $X$ and group $N$ acting on the right by

(g, w) · n = (gn, n⁻¹ · w).

We summarize the principal morphisms above in the following diagram:

                       G ×_S W
                      ╱       ╲
                     q         ϕ
                    ╱           ╲
                   ↓             ↘
        G/N ←── p─ X ──ψ──→ V
                   │
                pr₁│  i↓    ↗pr₂
                   ↓
              G/N ×_S V
(D)

where $\psi =p{r}_2\circ i$ and $\phi =\psi \circ q$, i.e. $\phi (g,w)=g\cdot w$. If $v$ is a section of $V$ over $S$, the subsheaf $X_v$ of $X$, inverse image of this section by $\psi$, is given by $X_v(S^{\prime })=$ set of subsheaves $W^{\prime }$ of $V_{S^{\prime }}$ which are locally conjugate to $W_{S^{\prime }}$ by $G$ and which contain the section $v_{S^{\prime }}$ of $V_{S^{\prime }}$, while the subsheaf of $G{\times }_{S}W$ inverse image of $v$ by $\phi$ is isomorphic to the subsheaf $M_v$ of $G$ given by $M_v(S^{\prime })=$ set of $g\in G(S^{\prime })$ such that $v_{S^{\prime }}\in g\cdot W(S^{\prime })$, i.e. such that $g^{-1}{v}_{S^{\prime }}\in W(S^{\prime })$. If $v$ is a section of $W$ and not merely of $V$, then $M_v$ evidently contains $N$.

In these explicit descriptions we have not used the fact that $G$, $V$, $W$ are representable (nor that the site over which we work is defined in terms of preschemes!). But suppose now that $N$ is representable and faithfully flat and quasi-compact over $S$, and that $G/N$ is representable. When $S$ is the spectrum of a field, and $G$ is of finite type over $k$, one knows that this hypothesis is necessarily satisfied (VIII 6 and VI_B.11.18). One then sees from the cartesian diagram (✱✱), using the theory of faithfully flat quasi-compact descent and the fact that $Z\to G{\times }_{S}V$ is a closed immersion, that $X$ is representable (it is obtained by descent of the closed subprescheme $Z$ of $G{\times }_{S}V$ by the faithfully flat quasi-compact morphism $G{\times }_{S}V\to G/N{\times }_{S}V$). Hence diagram (D) is a diagram of morphisms of preschemes over $S$.

We henceforth assume that $S$ is the spectrum of a field $k$, and that $G$, $V$, $W$ are of finite type over $k$. Let $\mathfrak{n}$ be the Lie algebra of $N$, so that $\dim N⩽rank\mathfrak{n}$, with equality if and only if $N$ is smooth over $k$ (Exp VI ¹). Let $a\in W(k)$, and consider the subscheme $M_a$ of $G$ defined above, containing $N$, and isomorphic to ${\phi }^{-1}(a)$; we shall denote by ${\mathfrak{m}}_a$ its Zariski tangent space at the identity element $e$ of $N$, so that one has

(1)    𝔫 ⊂ 𝔪_a,    dim N ⩽ rank 𝔫 ⩽ rank 𝔪_a.

Lemma 1.1. With the preceding notation:

a) Consider the following conditions:

• (i) $\mathfrak{n}={\mathfrak{m}}_a$ and $N$ is smooth over $k$.
• (i bis) $\dim N=ran{k}_k{\mathfrak{m}}_a$.
• (ii) The morphism $\psi :X\to V$ is unramified at $(\bar{e},a)$.
• (iii) $M_a$ and $N$ coincide in a neighborhood of $e$.

Then one has the implications (i) ⇔ (i bis) ⇒ (ii) ⇔ (iii).

b) Suppose $\phi :G{\times }_{S}W\to V$ is smooth at $(e,a)$. Then $M_a$ is smooth over $k$ at $e$, and $\psi$ is smooth at $(\bar{e},a)$.

Proof. a) The equivalence of (i) and (i bis) follows at once from the relations (1) and from the fact already noted that $N$ is smooth over $k$ if and only if $\dim N=ran{k}_k\mathfrak{n}$. On the other hand, consider the inclusion morphism $N\to M_a$; it is well known ² that if $N$ is smooth over $k$ at $e$ and the tangent map at $e$ is surjective, then $N\to M_a$ is smooth at $e$, hence (being an immersion) is an isomorphism at $e$, which shows that (i) implies (iii). To prove the equivalence of (ii) and (iii), consider as above $X_a={\psi }^{-1}(a)$ and use the isomorphism $M_a\simeq {\phi }^{-1}(a)\simeq q^{-1}(X_a)$ to obtain a morphism $p_a:M_a\to X_a$ which makes $M_a$ a principal homogeneous bundle with group $N_{X_a}$. Consider the following diagram

                  j_a
       M_a  ←──────────── N
        │                  │
      p_a│                 │
        ↓        j_a′      ↓
       X_a  ←──────  S = Spec(k),

where $\mathrm{Spec}(k)\to X_a$ is defined by the point $(\bar{e},a)$ of $X$, and $j_a:N\to M_a$ is the canonical immersion. To say that $\psi$ is unramified at $(\bar{e},a)$ means that $j_a^{\prime }$ is an open immersion, or equivalently that it induces an isomorphism $S\stackrel{\sim }{\to }\mathrm{Spec}({\mathcal{O}}_{X_a,\bar{a}})$. Since $p_a$ is flat, it is equivalent to say that the morphism deduced from the preceding by the base change $\mathrm{Spec}({\mathcal{O}}_{M_a,e})\to \mathrm{Spec}({\mathcal{O}}_{X_a,\bar{a}})$ is an isomorphism; now this deduced morphism is none other than the morphism $\mathrm{Spec}({\mathcal{O}}_{N,e})\to \mathrm{Spec}({\mathcal{O}}_{M_a,e})$, which proves the equivalence of (ii) and (iii).

One will note moreover that the proof shows that conditions (ii), (iii) imply the following condition, apparently stronger than (iii):

• (iii bis) $N$ is an open and closed subscheme of $M_a$.

b) The first assertion follows from the fact that $M_a$ is isomorphic to ${\phi }^{-1}(a)$, the second from the fact that $q$ is flat and $q(e,a)=(\bar{e},a)$.

2. Density theorem and theory of the regular points of G

We shall apply the constructions and notation of the preceding section to the case where $G$ is a connected smooth algebraic group over $k$, where $V=G$ on which $G$ acts by inner automorphisms, and where $W$ is a connected smooth algebraic subgroup $H$ of $G$. We shall denote by $\mathfrak{g}$ the Lie algebra of $G$, by $\mathfrak{h}$ that of $H$, by $N$ the normalizer of $H$ in $G$, and by $\mathfrak{n}$ the Lie algebra of $N$. If $a\in G(k)$, we shall denote, as in section 1, by $M_a$ the symmetric of its transporter into $H$, so that if $a\in H(k)$, one has $N\subset M_a$; in this case, we denote by ${\mathfrak{m}}_a$ the Zariski tangent space of $M_a$ at the identity element $e$ of $G$. Note that

𝔥 ⊂ 𝔫 ⊂ 𝔪_a    for a ∈ H(k).

We shall need the following:

Lemma 2.0. In order that $H=N^0$, it is necessary and sufficient that $(\mathfrak{g}/\mathfrak{h}{)}^H=0$ (where the first member denotes the subspace of invariants under the action of $H$ deduced from the adjoint representation). When this condition is satisfied, $N$ is smooth and $\dim X=\dim G$. In any case, $\dim X⩽\dim G$, and this inequality is an equality if and only if $H$ is of finite index in $N$.

Indeed, one has seen (II 5.2.3 (i)) that $\mathfrak{n}$ equals the inverse image of $(\mathfrak{g}/\mathfrak{h}{)}^H$ under the morphism $\mathfrak{g}\to \mathfrak{g}/\mathfrak{h}$, hence $(\mathfrak{g}/\mathfrak{h}{)}^H=0$ is equivalent to $\mathfrak{h}=\mathfrak{n}$, which is also equivalent (since $H$ is a smooth connected algebraic subgroup of the algebraic group $N$) to $H=N^0$ (cf. VI.2). This evidently implies that $N$ is smooth. On the other hand, one has

dim X = (dim G − dim N) + dim H = dim G − (dim N − dim H),

hence

dim X ⩽ dim G,

with equality if and only if $\dim H=\dim N$, i.e. if and only if $H$ is of finite index in $N$. This is the case in particular if $H=N^0$, which completes the proof of 2.0.

Theorem 2.1. Let $G$ be a smooth connected algebraic group over the algebraically closed field $k$, $H$ a connected smooth algebraic subgroup, $N$ its normalizer, $\mathfrak{g}$, $\mathfrak{h}$, $\mathfrak{n}$ the Lie algebras, $X=G{\times }^{N}H$ the scheme (fibered over $G/N$ with typical fiber $H$) introduced in section 1, $\psi :X\to G$ the canonical morphism (whose image is also the image of $\phi :G\times H\to G$ defined by $\phi (g,h)=int(g)h=gh{g}^{-1}$). The following conditions are all equivalent:

• (i) $H$ contains a Cartan subgroup (XII 1) $C$ of $G$.
• (i bis) $H$ has the same reductive rank and the same nilpotent rank (XII 1) as $G$.
• (ii) $H$ contains a maximal torus $T$ of $G$, and $(\mathfrak{g}/\mathfrak{h}{)}^T=0$.

• (iii) The set of conjugates of $H$ containing a given maximal torus is finite and non-empty, and $H$ is of finite index in its normalizer.
• (iv) There exists $a\in H(k)$ contained only in finitely many conjugates of $H$ (or merely such that ${\psi }^{-1}(a)$ has an isolated point), and $H$ is of finite index in its normalizer.
• (iv bis) The morphism $\psi :X\to G$ is generically quasi-finite (i.e. there exists a dense open of $X$ on which $\psi$ is quasi-finite), and $H$ is of finite index in its normalizer.
• (v) There exists a dense open $U$ in $G$ such that for every $x\in U(k)$, the set of conjugates of $H$ containing $x$ is finite and non-empty, i.e. $\psi :X\to G$ is dominant and generically quasi-finite.
• (vi) There exists a dense open $U$ of $G$ such that every $x\in U(k)$ is contained in a conjugate of $H$, i.e. $\psi :X\to G$ is dominant.
• (vii) There exists $a\in H(k)$ such that the subspace of $\mathfrak{g}/\mathfrak{h}$ of fixed points of $a{d}_{\mathfrak{g}/\mathfrak{h}}(a)$ is zero.

Furthermore, these conditions imply that $H$ is its own connected normalizer, i.e. $N$ is smooth and $\dim H=\dim N$, and that $\psi :X\to G$ is generically étale.

Proof. By 2.0, one has $\dim X⩽\dim G$, with equality if and only if $\dim H=\dim N$, i.e. $H$ of finite index in $N$. From the inequality $\dim X⩽\dim G$ it follows that $\psi$ is dominant if and only if it is dominant and generically quasi-finite, or again if and only if $\psi$ is generically quasi-finite and $\dim X=\dim G$. As this latter equality means also, by 2.0, that $H$ is of finite index in $N$, we have proved the equivalence of (vi), (v), (iv bis). The equivalence of (iv) and (iv bis) is immediate.

The equivalence of (i) and (i bis) is immediate from the definitions, and is left to the reader. On the other hand, if $H$ contains a Cartan subgroup $C$ of $G$, then it contains the maximal torus $T$ of $C$, which is a maximal torus of $G$. Since $C$ is the centralizer of $T$, its Lie algebra $\mathfrak{c}$ is given by

$$\mathfrak{c}={\mathfrak{g}}^T$$

(II, 5.2.3 (ii)). Hence as $H\supset C$, so that $\mathfrak{h}\supset \mathfrak{c}$, it follows that $\mathfrak{c}\subset {\mathfrak{g}}^T$, which is equivalent (by I, 4.7.3) to the relation

(✱)    (𝔤/𝔥)^T = 0.

Conversely, suppose that $H$ contains the maximal torus $T$ and that the preceding relation holds, i.e. that $\mathfrak{h}\supset \mathfrak{c}$; I claim that $H$ contains the centralizer $C$ of $T$ (which will establish (i) ⇔ (ii)). This follows from the:

Lemma 2.1.1. Let $G$ be a smooth algebraic group over the field $k$, $T$ a subgroup of multiplicative type of $G$, $C$ its connected centralizer (equal to the centralizer of $T$ if $G$ is connected and $T$ is a torus, (XII 6.6 b))), $H$ a smooth subgroup of $G$ containing $T$. In order that $H$ contain $C$, it is necessary and sufficient that its Lie algebra $\mathfrak{h}$ contain that $\mathfrak{c}$ of $C$.

Indeed, one knows (XI 2.4) that $Cent{r}_G(T)$ is smooth over $k$, hence $C$ is smooth over $k$; similarly $Cent{r}_H(T)$ is smooth over $k$, and $Cent{r}_H(T)=Cent{r}_G(T)\cap H$ has $\mathfrak{h}\cap \mathfrak{c}$ as Lie algebra; hence the hypothesis implies that the smooth subgroup $Cent{r}_H(T)$ of the smooth group $Cent{r}_G(T)$ has the same Lie algebra, so it contains the connected component $C$ of the latter, and thus $H$ contains $C$. QED.

Let us prove the equivalence of (i) and (iii), which amounts to proving that if $H$ contains the maximal torus $T$ of $G$, then the condition $H\supset C$ (which is also equivalent to (✱) above, as we have just seen) is equivalent to the fact that $H$ is of finite index in its normalizer and that the set of conjugates of $H$ containing $T$ is finite. If $H$ contains $C$, hence if one has (✱), then a fortiori

(✱✱)   (𝔤/𝔥)^H = 0,

and we know that $\mathfrak{n}$ is the inverse image of the first member of the preceding relation under the canonical homomorphism $\mathfrak{g}\to \mathfrak{g}/\mathfrak{h}$ (II 5.2.3 (i)), so that the preceding relation means, by 2.0, that $H=N^0$, and a fortiori $H$ is of finite index in its normalizer. Now consider the diagram of subgroups

   T  ───→  N(T) ∩ H  ───→  H
              │              │
              ↓              ↓
       N(T) ∩ N(H)  ───→  N(H)
              │
              ↓
            N(T).

Using the theorem of conjugacy of maximal tori in $H$ (XII 6.6 a)), one sees that every conjugate of $H$ containing $T$ is a conjugate of $H$ by an element of $N(T)(k)$, so the set of conjugates of $H$ containing $T$ is in bijective correspondence with the set of points of $N(T)/N(T)\cap N(H)$ with values in $k$; but as $H\supset C$, one has $N(T)\cap H\supset C$, hence the preceding set is a quotient of $(N(T)/C)(k)$, which is a finite set, hence is finite. This proves (i) ⇒ (iii). Conversely, suppose (iii), i.e. $N(T)/N(T)\cap N(H)$ finite and $N(H)/H$ finite. Using again the conjugacy theorem in $H$, one sees again that the homomorphism

N(T) ∩ N(H)/N(T) ∩ H ─→ N(H)/H

induced by the preceding diagram is bijective on points with values in $k$ (in fact, it is an isomorphism), hence as the latter is finite, so is the former, so $N(T)\cap H$ is of finite index in $N(T)$, hence contains $C=N(T{)}^0$, hence $H\supset C$. Thus (i), (i bis), (ii), (iii) are equivalent conditions.

Let us prove that (ii) ⇒ (vii). One sees at once that conditions (ii) and (vii) are each invariant under an extension $k\to k^{\prime }$ of the base field, with $k^{\prime }$ algebraically closed, which allows us to assume that $k$ has infinite transcendence degree over its prime subfield. Then it is well known (and easily verified) that there exists an element $a$ of $T(k)$ such that the subgroup of $T(k)$ it generates is dense in $T$ for the Zariski topology. One easily concludes that $(\mathfrak{g}/\mathfrak{h}{)}^T=(\mathfrak{g}/\mathfrak{h}{)}^{ad(a)}$, and as the first member is zero by hypothesis, one concludes (vii).

Let us prove (vii) ⇒ (vi). This implication is contained in the following result, which sharpens 2.1:

Corollary 2.2. Let $G$ be a smooth algebraic group over a field $k$, $H$ a smooth algebraic subgroup, $N$ its normalizer in $G$, $\phi :G\times H\to G$ the morphism defined by $\phi (g,h)=ad(g)h=gh{g}^{-1}$, $\psi :X=G{\times }^{N}H\to G$ the morphism deduced from $\phi$ by passage to the quotient (cf. section 1), $a\in H(k)$. The following conditions are equivalent:

• (i) $\phi$ is smooth at $(e,a)$.
• (ii) $\psi$ is étale at $(\bar{e},a)$, and $N$ is smooth over $k$.
• (iii) $(\mathfrak{g}/\mathfrak{h}{)}^{ad(a)}=0$ (where $\mathfrak{g}$, $\mathfrak{h}$ are the Lie algebras of $G$, $H$).

These conditions imply $H^0=N^0$.

We know that the smoothness of $\phi$ (which is a morphism between smooth $k$-preschemes) at a $k$-rational point is equivalent to the surjectivity of the tangent map at that point. Now an immediate calculation shows that this tangent map is written (using the usual identifications of the tangent spaces at points of $G$ and $H$ with the Lie algebras of $G$ and $H$)

dϕ(ξ, η) = (id − ad(a)) · ξ + η,

regarded as a map from $\mathfrak{g}\times \mathfrak{h}$ to $\mathfrak{g}$. Surjectivity therefore amounts to the surjectivity of $(id-ad(a))$ on $\mathfrak{g}/\mathfrak{h}$, i.e. to (iii). Now (iii) implies a fortiori $(\mathfrak{g}/\mathfrak{h}{)}^H=0$, i.e. (cf. 2.0, where the connectedness hypothesis at the beginning of the section is unnecessary) $H^0=N^0$. From this we deduce that $N$ is smooth, and $\dim H=\dim N$, whence $\dim X=\dim G$. Now since $q:G\times H\to X$ is flat and $\psi \circ q=\phi$ is smooth at $(e,a)$, it follows that $\psi$ is smooth at $q(e,a)=(\bar{e},a)$, hence étale at this point by reasons of dimension. So we have proved (i) ⇔ (iii) ⇒ (ii); on the other hand (ii) ⇒ (i), because the smoothness of $N$ implies that of $q$. QED.

Let us finally prove (vi) ⇒ (i), which, together with the implications already established, will prove the theorem. Suppose first that $G$ is affine. Let $U$ be a non-empty open of $G$ such that $x\in U(k)$ implies that $x$ is contained in a conjugate of $H$. Let $C$ be a Cartan subgroup of $G$. Using the implication (i) ⇒ (iv) for $C$ in place of $H$ (this is Borel's "density theorem"), it follows that one can find a conjugate of $C$ which meets $U$, so one can assume $U\cap C\ne \varnothing$, i.e. that there exists a non-empty open $V$ in $C$ such that for every $x\in V(k)$, $x$ is contained in a conjugate of $H$. Write $C$ as a product

$$C=T\cdot C_u$$

where $T$ is the maximal torus of $C$ (which is a maximal torus of $G$) and $C_u$ is the unipotent part of $C$, $T$ being in the center of $C$ (Bible 6 th. 2). We can again assume that $k$ has infinite transcendence degree over its prime subfield, which allows us to find an element $t$ of $T(k)$ that is in the projection of $V$ onto $T$ (which is a non-empty open of $T$), i.e. $t\cdot C_u\cap V\ne \varnothing$, and such that $t$ "generates" $T$. Since every algebraic subgroup of $G$ containing a product $t\cdot u$ ($t\in T(k)$, $u\in C_u(k)$) contains the two factors (Bible 4 th. 3), it follows, with the preceding choice of $t$, and taking $t\cdot u\in V(k)$, that there exists a conjugate of $H$ containing $t$, hence $T$. So we may already assume that one has

$$T\subset H.$$

If $W$ is the open of $C_u$ inverse image of $V$ by $u\mapsto t\cdot u$, one sees therefore that for every element $x$ of $(T\cdot W)(k)$, there exists a conjugate of $H$ containing $T$ and $x$. As we have already noted, such a conjugate is of the form $int(g)\cdot H$, where $g\in N(T)(k)$. Consider then the morphism

f : N(T) × H → G

defined by $f(g,h)=int(g)\cdot h=gh{g}^{-1}$; then the image of $f$ contains $T\cdot W$, so as $N(T)$ is a finite union of translates $C\cdot g_i$ (where $g_i\in N(T)(k)$) since $C$ is of finite index in $N(T)$, it follows that there exists a dense open $V_0$ of $C=T\cdot C_u$ which is contained in the image of $(C\cdot g_i)\times H$ by $f$. Replacing $H$ by $int(g_i)\cdot H$ if necessary, we may assume $g_i=e$, i.e. $f(C\times H)\supset V_0$. So for every $u\in V_0(k)$, there exist $v\in C(k)$ and $h\in H(k)$ such that

v⁻¹ h v = u, whence vuv⁻¹ ∈ H(k),

hence, setting

C₀ = C ∩ H = Centr_H(T),

$vu{v}^{-1}\in C_0$, whence $int(v)\cdot C_0\supset u$. This proves that the union of the conjugates of $C_0$ in $C$ (by elements of $C(k)$) is dense, which implies (as we have seen for the pair $(G,H)$ instead of $(C,C_0)$) that $C_0$ is of finite index in its normalizer in $C$. By Bible 7, lemma of section 1, it follows that $C_0=C$, hence $C=H$, which proves (vi) ⇒ (i) when $G$ is affine.

In the general case, we proceed by induction on $n=\dim G$, the assertion being trivial if $n=0$. Let $Z$ be the center of $G$, and distinguish two cases:

1°) $\dim Z\cap H>0$: then setting $G^{\prime }=G/(Z\cap H)$, one has $\dim G^{\prime }<n$; on the other hand the hypothesis (vi) on $H$ implies the same condition for the image $H^{\prime }$ of $H$ in $G^{\prime }$, so $H^{\prime }$ contains a Cartan subgroup $C^{\prime }$ of $G^{\prime }$, hence $H$ contains the inverse image $C$ of $C^{\prime }$, which is a Cartan subgroup by XII 6.6 e).

2°) $\dim Z\cap H=0$: then the canonical morphism $H\to G/Z$ is a finite morphism, and as $G/Z$ is affine by virtue of XII 6.1, it follows that $H$ is affine, hence every homomorphism from $H$ into an abelian variety is zero (and even every morphism of preschemes from $H$ to an abelian variety is zero): this follows from the fact that an affine smooth connected algebraic group over an algebraically closed field is a rational variety, or simply that it is the union of its Borel subgroups (Bible 6 th. 5 b)), and it follows very easily from the structure theorems Bible 6.2 and 6.3 that an affine smooth connected solvable group is a rational variety. Now use Chevalley's structure theorem for $G$, according to which $G$ is an extension of an abelian variety $A$ by a smooth affine group. Then the image of $H$ in $A$ is zero, $H$ being affine; on the other hand it is identical to $A$, because the union of its conjugates in $A$ must be dense. So $A=0$, hence $G$ is affine, and we are reduced to the case already treated. This completes the proof of 2.1.

Corollary 2.3. Assume that the equivalent conditions of 2.1 hold.

a) Let $k(X)$ (resp. $k(G)$) be the field of rational functions of $X$ (resp. $G$); then $k(X)$ is a finite separable extension of $k(G)$. Denote its degree by $d$.

b) Let $T$ be a maximal torus of $G$ contained in $H$ (which exists by form 2.1 (ii)), and let $C$ be the corresponding Cartan subgroup of $G$. Then $C\subset H$. On the other hand, $Nor{m}_G(T)$ is a smooth subgroup of $G$ and Norm_G(T) ∩ Norm_G(H) = Norm_{Norm_G(H)}(T) is a smooth subgroup of it, of finite index equal to $d$ (defined in a)). The number of conjugates of $H$ containing a given maximal torus or a given Cartan subgroup is equal to $d$.

c) Let $U$ be the largest open of $G$ such that $\psi :X\to G$ induces a morphism ${\psi }^{-1}(U)\to U$ that is finite and étale. Then $U$ is a dense open, and for $g\in G(k)$, one has $g\in U(k)$ if and only if there exist exactly $d$ conjugates of $H$ containing $g$, or again if and only if there exist at least $d$ distinct conjugates $H_i$ of $H$ containing $g$ such that for every $i$, $(\mathfrak{g}/{\mathfrak{h}}_i{)}^{ad(g)}=0$ (where ${\mathfrak{h}}_i=Lie(H_i)$).

Assertion a) comes from the fact that $\psi$ is generically étale (which was stated at the end of 2.1); this also implies that the open $U$ introduced in c) is non-empty, i.e. dense, and the two characterizations stated for the elements of $U(k)$ (taking into account that $\psi$ is separated, $X$ integral and $G$ integral normal, SGA 1 I 10.11, and that $(\mathfrak{g}/{\mathfrak{h}}_i{)}^{ad(g)}=0$ means that $\psi$ is étale at the point $x_i$ of ${\psi }^{-1}(g)$ corresponding to $H_i$). If $H$ contains the maximal torus $T$ of $G$, then the centralizers of $T$ in $H$ and $G$ have the same dimension, and are smooth and connected (XII 6.6 b)), hence are equal, which proves that $C\subset H$. Moreover, one knows that the normalizer of $T$ in a smooth group containing it is smooth (XI, 2.4 bis), so $Nor{m}_G(T)$ and $Nor{m}_{Nor{m}_G(H)}(T)$ are smooth (N.B. we have noted that $N=Nor{m}_G(H)$ is smooth at the end of the statement of 2.1); also $Nor{m}_N(T)$ contains $C$, which is of finite index in $N(T)$, so it is of finite index in $N(T)$. Using the theorem of conjugacy for maximal tori in $H$, one sees that the index in question equals the number of conjugates of $H$ containing $T$, or what amounts to the same, containing $C$. Now since the union of the conjugates of $C$ in $G$ is dense (by 2.1 (i) ⇒ (vi) applied to $C$ instead of $H$), and the open $U$ defined in c) is evidently stable under inner automorphisms, one sees that $C\cap U\ne \varnothing$. Proceeding as in the proof of the implication (vi) ⇒ (ii) of 2.1, one concludes that (up to a harmless change of base field) there exists $g\in (C\cap U)(k)$ such that every conjugate of $H$ containing $g$ contains $T$, and hence also $C$. So the conjugates of $H$ containing $C$ are those containing $g$, and as $g\in U(k)$, their number equals $d$, which completes the proof of b). One will note that we have in fact shown that the set of conjugates of $H$ containing $T$ is a homogeneous set under the group of rational points of

$$W_G(T)=Nor{m}_G(T)/Cent{r}_G(T),$$

which proves in particular that

d ⩽ order of the Weyl group of G.

Corollary 2.4. With the notation of 2.1, the following conditions are equivalent:

• (i) $\psi :X\to G$ is a birational morphism.
• (ii) There is exactly one conjugate of $H$ containing a given Cartan subgroup of $G$.
• (iii) $H$ contains a Cartan subgroup $C$ of $G$, and $Nor{m}_G(H)\supset Nor{m}_G(C)$.
• (iv) There exists a non-empty open $V$ of $G$ such that $g\in V(k)$ implies that $g$ is contained in exactly one conjugate of $H$.

This is clear from 2.1 and 2.3.

Corollary 2.5. Assume that the conditions of 2.4 hold, and let $g\in G(k)$. The following conditions are equivalent:

• (i) $g\in U(k)$, where $U$ is defined in 2.3 c), i.e. $g$ is contained in exactly one conjugate of $H$.
• (ii) The set of conjugates of $H$ containing $g$ is finite and non-empty.
• (iii) The scheme ${\psi }^{-1}(g)$ "of conjugates of $H$ containing $g$" has an isolated point.
• (iv) There exists a conjugate $H^{\prime }$ of $H$ containing $g$, and one has $(\mathfrak{g}/{\mathfrak{h}}^{\prime }{)}^{ad(g)}=0$, where ${\mathfrak{h}}^{\prime }=Lie(H^{\prime })$.

Finally, $U$ is also the largest open of $G$ such that $\psi$ induces an isomorphism ${\psi }^{-1}(U)\stackrel{\sim }{\to }U$.

The equivalence of (i), (ii), (iii) and the last assertion follow from the "Main Theorem"³ applied to the birational morphism $\psi :X\to G$, given that $G$ is normal. The equivalence of these conditions with (iv) follows at once from the last assertion of 2.1 characterizing the set of points of $X$ at which $\psi$ is étale.

Theorem 2.6. Let $G$ be a smooth connected algebraic group over an algebraically closed field $k$, $C$ a Cartan subgroup, associated to a maximal torus $T$, $N=Nor{m}_G(C)=Nor{m}_G(T)$ (cf. XII 8.4), let $X=G{\times }^{N}C$, where $N$ acts on the left factor $G$ by right translations, and on the right factor $C$ by inner automorphisms, and let $\psi :X\to G$ be the canonical morphism.

a) The morphism $\psi$ is birational.

b) Let $U$ be the largest open of $G$ such that $\psi$ induces an isomorphism ${\psi }^{-1}(U)\stackrel{\sim }{\to }U$ (cf. 2.5). Let

ρ = ρ_ν(G) = dim C

be the nilpotent rank of $G$. Then for every $g\in G(k)$, the multiplicity of the eigenvalue 1 in $ad(g)$ acting on $\mathfrak{g}$ is at least equal to $\rho$, and for it to equal $\rho$, it is necessary and sufficient that $g\in U(k)$.

Proof. Since condition 2.1 (i) is satisfied, one can apply 2.4 (iii) ⇒ (i), which establishes a). In Bible 7 (in the case where $G$ is affine) the points of $U(k)$ are called the regular points of $G(k)$, and we shall follow this terminology, calling $U$ the open of regular points of $G$. (N.B. The proof given in the Bible that the set in question is itself open is incorrect, but we have obtained it in the present Exposé under more general conditions.)

Let us prove b); for this, introduce for every $g\in G(k)$ the characteristic polynomial

P(ad(g), t) = tⁿ + c₁(g) tⁿ⁻¹ + ⋯ + c_n(g);

one sees at once (replacing $k$ by any algebra over $k$) that the $c_i(g)$ come from well-determined sections

$$c_i\in \Gamma (G,{\mathcal{O}}_G).$$

When $g\in G(k)$ is an element contained in a Cartan subgroup (for example a regular element), which we may assume to be $C$, then by 2.5 (iv) one sees that $(\mathfrak{g}/\mathfrak{c}{)}^{ad(g)}=0$ if and only if $g$ is regular (where $\mathfrak{c}$ denotes the Lie algebra of $C$); on the other hand, since $C$ is nilpotent, one sees at once that $a{d}_{\mathfrak{c}}(g)$ has only the eigenvalue 1, which proves that the multiplicity of the eigenvalue 1 in $a{d}_{\mathfrak{g}}(g)$ is $⩾\rho$, and equal exactly to $\dim C=\rho$ if and only if $g$ is regular. In particular, the polynomial above is divisible by $(t-1{)}^{\rho }$. Since the relation of divisibility by $(t-1{)}^{\rho }$ is expressed by linear relations (with integer coefficients) among the coefficients of the polynomial, and these relations are satisfied for $g\in U(k)$, $U$ being a dense open, it follows ($G$ being reduced) that they hold for all $g$; in fact one has a relation

(†)    tⁿ + c₁ tⁿ⁻¹ + ⋯ + c_n = (t − 1)^ρ (tⁿ⁻ρ + b₁ tⁿ⁻ρ⁻¹ + ⋯ + b_{n−ρ})

in the ring of polynomials over $\Gamma (G,{\mathcal{O}}_G)$; in particular for every $g\in G(K)$, $ad(g)$ has the eigenvalue 1 with multiplicity at least $\rho$. Moreover, we have seen that equality holds if $g$ is regular; let us prove the converse. To this end, suppose first $G$ affine, and write $g$ as a product

$$g=g_s{g}_u$$

of its semisimple part and its unipotent part (Bible 4 section 4); then

$$ad(g)=ad(g_s)ad(g_u)$$

is the analogous decomposition of $ad(g)$ (loc. cit. cor. to th. 3), so $ad(g)$ and $ad(g_s)$ have the same eigenvalues (counted with multiplicity); in particular the eigenvalue 1 appears with the same multiplicity in $ad(g)$ and in $ad(g_s)$.

Moreover, by Bible 7 th. 2 cor. 1, $g$ is regular if and only if $g_s$ is. Hence to prove b), we may assume $g=g_s$, i.e. $g$ semisimple, so contained in a maximal torus by Bible 6 th. 5 c), and a fortiori in a Cartan subgroup — the case already treated. This proves b) in the case $G$ affine. In the general case, let $Z=Centr(G{)}_{red}$; then by XII 6.6 e) the Cartan subgroups of $G$ are the inverse images of those of $G^{\prime }=G/Z$, hence $g$ is regular in $G$ if and only if its image $g^{\prime }$ in $G^{\prime }$ is regular in $G^{\prime }$. On the other hand, since $Z$ is smooth, the Lie algebra ${\mathfrak{g}}^{\prime }$ of $G^{\prime }$ is none other than $\mathfrak{g}/\mathfrak{z}$, where $\mathfrak{z}=Lie(Z)$, and $ad(g^{\prime })$ is none other than $ad(g{)}_{\mathfrak{g}/\mathfrak{z}}$; hence the multiplicity of the eigenvalue 1 in $ad(g)$ equals $d=\dim Z$ plus the multiplicity of the eigenvalue 1 in $ad(g^{\prime })$, whence at once the first is equal to the nilpotent rank of $G$ if and only if the second is equal to the nilpotent rank of $G^{\prime }$. Thus we are reduced to the case of $G^{\prime }$; now $G^{\prime }$ being affine by XII 6.1, this case has already been treated. This completes the proof of 2.6.

Corollary 2.7. With the notation of the preceding proof,⁴ let

b = 1 + b₁ + ⋯ + b_{n−ρ} ∈ Γ(G, 𝒪_G).

Then the open of regular points of $G$ is given by

$$U=G_b$$

(set of points of $G$ at which $b$ is invertible); in particular $U$ is an affine open if $G$ is affine.

Corollary 2.8. Let $H$ be a smooth connected algebraic subgroup of $G$ containing a Cartan subgroup of $G$.

a) Let $C$ be an algebraic subgroup of $H$. In order that $C$ be a Cartan subgroup of $H$, it is necessary and sufficient that it be a Cartan subgroup of $G$.

b) Let $g\in G(k)$, and let $d$ be the integer introduced in 2.3. In order that $g$ be a regular point of $G$, it is necessary and sufficient that there exist exactly $d$ conjugates $H_i$ of $H$ containing $g$, and that for each $i$, $g$ be a regular element of $H_i$; or again, that there be at most $d$ conjugates of $H$ containing $g$, and that $g$ be regular in one of them. If so, and if $C$ is the unique Cartan subgroup of $G$ containing $g$, then the conjugates of $H$ containing $g$ are the conjugates of $H$ containing $C$.

c) Let $g\in H(k)$; in order that $g$ be regular in $G$, it is necessary and sufficient that it be regular in $H$, and that one have $(\mathfrak{g}/\mathfrak{h}{)}^{ad(g)}=0$.

Let us prove a). Under either hypothesis on $C$, the unique maximal torus $T$ of $C$ is a maximal torus of $H$ and of $G$ ($H$ having the same reductive rank as $G$); hence as $Cent{r}_H(T)\subset Cent{r}_G(T)$ are smooth connected groups of the same dimension, they are equal, so it amounts to the same to say that $C$ is equal to one or the other of these two groups, which proves a).

Let us prove b). Suppose first $g$ regular in $G$, let $C$ be the unique Cartan subgroup of $G$ containing $g$. Then by 2.3 b) there exist exactly $d$ conjugates $H_i$ of $H$ containing $C$. Since $(\mathfrak{g}/\mathfrak{c}{)}^{ad(g)}=0$, i.e. $ad(g)$ has no eigenvalue +1 on $\mathfrak{g}/\mathfrak{c}$, one has a fortiori $(\mathfrak{g}/{\mathfrak{h}}_i{)}^{ad(g)}=0$, hence by 2.3 c) there are exactly $d$ conjugates of $H$ containing $g$, namely the $H_i$. For such an $H_i$, a Cartan subgroup of $H_i$ containing $g$ is a Cartan subgroup of $G$ containing $g$ by a), hence equals $C$, which proves that $g$ is regular in $H_i$. Conversely, suppose that there are at most $d$ conjugates $H_i$ of $H$ containing $g$, and that $g$ is regular in one of them, which we may assume to be $H$. Let us prove that $g$ is regular in $G$. Since $g$ is regular in $H$, it is contained in a unique Cartan subgroup $C$ of $H$; by a) this is a Cartan subgroup of $G$. Let $C^{\prime }$ be a Cartan subgroup of $G$ containing $g$; let us prove $C^{\prime }=C$ (which will prove that $g$ is regular in $G$). Indeed, by 2.3 b) there exist exactly $d$ conjugates of $H$ containing $C^{\prime }$, and as these latter contain $g$, they are necessarily the $H_i$, hence the $H_i$ and in particular $H$ contain $C^{\prime }$. Hence $C$, $C^{\prime }$ are two Cartan subgroups of $H$ (by a)) that contain the same regular element $g$ of $H$, so they are equal. QED.

Let us prove c). Denoting by $\nu (u)$ the nullity⁵ of $id-u$, for an endomorphism of a finite-dimensional vector space, one has

$$\nu (ad(g{)}_{\mathfrak{g}})=\nu (ad(g{)}_{\mathfrak{h}})+\nu (ad(g{)}_{\mathfrak{g}/\mathfrak{h}}),$$

and the two terms on the right-hand side are respectively $⩾$ nilpotent rank of $H$ (equal to the nilpotent rank $\rho$ of $G$ by a)) and $⩾0$; hence one has $\nu (ad(g{)}_{\mathfrak{g}})=\rho$ if and only if $\nu (ad(g{)}_{\mathfrak{h}})=\rho$ and $\nu (ad(g{)}_{\mathfrak{g}/\mathfrak{h}})=0$, i.e. $g$ is regular in $G$ if and only if $g$ is regular in $H$ and $ad(g{)}_{\mathfrak{g}/\mathfrak{h}}$ has no non-zero invariants. QED.

Remarks 2.9. In the statement of 2.1, one cannot weaken condition (iii) by assuming only that $H$ contains a maximal torus and is of finite index in its normalizer, even if one further requires that this normalizer be smooth, i.e. that $H=N^0$, and even when $G$ is affine solvable. An example is furnished by the group $G$ of matrices of the form

       ⎛ t a c ⎞
   g = ⎜ 0 1 b ⎟,
       ⎝ 0 0 1 ⎠

and the subgroup $H$ of matrices of the preceding form with $b=c=0$ (N.B. the Cartan subgroup of $G$ is here the matrices $g$ with $a=c=0$).

Remarks 2.10. Let $G$ be a smooth algebraic group over $k$, $H$ a smooth algebraic subgroup, but we no longer assume $H$ and $G$ connected. Suppose that $H^0$ contains a Cartan subgroup of $G^0$. Then $(\mathfrak{g}/\mathfrak{h}{)}^H\subset (\mathfrak{g}/\mathfrak{h}{)}^{H^0}=0$, hence $H^0=N^0$ ($N$ is the normalizer of $H$), in particular $N$ is smooth. However, one easily constructs examples, with $G$ connected, where $H$ has a connected component $H_i$ such that for no $h\in H_i(k)$ one has $(\mathfrak{g}/\mathfrak{h}{)}^{ad(h)}=0$, i.e. the morphism $(g,h)\mapsto ad(g)\cdot h$ from $G\times H_i$ to $G$ is étale (nor even quasi-finite) at no point (take for example for $H$ the normalizer of the maximal torus in $SL(2{)}_k$). Similarly, even if the image $H^{\prime \prime }$ of $H$ in the finite group $G^{\prime \prime }=G/G^0$ equals $G^{\prime \prime }$, it is not necessarily true that the union of conjugates of $H$ in $G$ is dense (take for $G$ the semidirect product of $\mathbb{Z}/2\mathbb{Z}$ with $G^0=SL(2{)}_k$ on which it acts by "symmetry", and for $H$ the semidirect product $\mathbb{Z}/2\mathbb{Z}\cdot T$, where $T$ is a maximal torus of $G^0$). On the other hand, if one does not assume a priori that $H^0$ contains a Cartan subgroup of $G^0$, but that the union of conjugates of $H$ in $G$ is dense, then $H^0$ necessarily contains a Cartan subgroup of $G^0$: to verify this, one may evidently suppose $G$ connected, and it suffices to redo the proof of 2.1 (vi) ⇒ (i), which is valid without assuming $H$ connected.

3. The case of an arbitrary base prescheme

Suppose first that we are over a base field $k$, not necessarily algebraically closed. Since conditions 2.1 (i bis), (iv bis), (v), (vi) are invariant under extension of the base field, one sees by passing to the algebraic closure $\bar{k}$ of $k$ that they are equivalent to each other, and equivalent to the fact that $H_{\bar{k}}$ contains a Cartan subgroup of $G_{\bar{k}}$. When this condition is satisfied, then (with the notation of 2.3) it will still be true that $k(X)$ is a finite separable extension of $k(G)$, of degree $d$ independent of any extension of the base. If $U$ is the largest open of $G$ such that $\psi$ induces a morphism ${\psi }^{-1}(U)\to U$ that is finite and étale, then the formation of $U$ commutes with extension of the base field. If $\psi$ is birational, then $U$ is also the largest open of $G$ such that $\psi$ induces an isomorphism ${\psi }^{-1}(U)\stackrel{\sim }{\to }U$, and if then $g\in U(k)$, there exists a subgroup $H^{\prime }$ of $G$ and only one, conjugate to $H$ over the algebraic closure $\bar{k}$ of $k$, such that $H^{\prime }\ni g$.

A point $g\in G(k)$ is said to be regular if it is regular as an element of $G(\bar{k})=G_{\bar{k}}(\bar{k})$. More generally, the construction of 2.7 gives us an open of $G$, whose formation commutes with any extension of the base field, called the open of regular points of $G$, which is also characterized by the fact that for every algebraically closed extension $K$ of $k$ and every point $g\in G(K)$, $g$ is a regular point of G_K if and only if $g\in U(K)$. If $g\in U(k)$,⁶ one sees that $g$ is contained in one and only one Cartan subgroup of $G$, as we shall show under more general conditions below.

Let $G$ be a smooth, separated, finite-type prescheme in groups with connected fibers over the prescheme $S$; consider the functor $\mathcal{C}:(Sch{)}^{\circ }/S\to (Ens)$ defined by

𝓒(S′) = set of Cartan subgroups (XII 3.1) of G_{S′}.

Suppose this functor is representable by a smooth prescheme over $S$; we give in XV a necessary and sufficient condition for this to be so, but we already know that this hypothesis is satisfied if $G$ is affine over $S$ with locally constant reductive rank (XII 3.3), or more generally if $G$ admits locally for the fpqc topology a maximal torus (XII 7.1 a)), for example if $S$ is the spectrum of a field. Let $X$ be the Cartan subgroup of the $\mathcal{C}$-prescheme in groups $G_{\mathcal{C}}$, the "universal Cartan subgroup" of $G$. As a prescheme over $S$, $X$ therefore represents the functor

X(S′) = set of pairs (C, g), C a Cartan subgroup of G_{S′} and g a section of C over S′.

Consider the canonical projection morphism $(C,g)\mapsto g$

$$\psi :X\to G.$$

One then has the:

Theorem 3.1. Under the preceding conditions on $G$, and with the preceding notation, let $U$ be the set of $g\in G$ such that $g$ is a regular element of its fiber $G_s$.

Then $U$ is open, and it is also the largest open $U$ of $G$ such that $\psi$ induces an isomorphism ${\psi }^{-1}(U)\stackrel{\sim }{\to }U$.

Let us first prove that $U$ is open. From the hypothesis of representability of $\mathcal{C}$ as a smooth prescheme over $S$, since its structural morphism is evidently surjective, one concludes at once that $G$ admits locally for the étale topology a Cartan subgroup, and that the nilpotent rank of the fibers of $G$ is locally constant. The same holds for the dimension of the fibers of $G$, and up to localizing on $S$, one may assume that both are constant, say $\rho$ and $n$. Consider then the Killing polynomial

P_G(t) = tⁿ + c₁ tⁿ⁻¹ + ⋯ + c_n ∈ A[t],   where A = Γ(G, 𝒪_G).

The restriction of this polynomial to the fibers $G_s$ of $G$, and in particular to the fibers at the maximal points of $S$, is divisible by $(t-1{)}^{\rho }$, which is expressed by the fact that certain linear combinations with integer coefficients of the $c_i$ vanish on the fibers $G_s$. When $S$ is reduced (which we may assume in order to establish that $U$ is open), it follows that they are themselves zero, hence the Killing polynomial itself is divisible by $(t-1{)}^{\rho }$, say

P_G(t) = (t − 1)^ρ (tⁿ⁻ρ + b₁ tⁿ⁻ρ⁻¹ + ⋯ + b_{n−ρ}).

Let $b$ be the sum of the coefficients $b_0=1,b_1,\cdots ,b_{n-\rho }$ of the second factor; then by 2.7 applied to the fibers of $G$, one sees that

$$U=G_b,$$

which indeed proves that $U$ is open.

To prove that ${\psi }^{-1}(U)\to U$ is an isomorphism, we are reduced by SGA 1, I 5.7 to verifying it fiber by fiber, which reduces to the case of a base field, which we may assume algebraically closed. Then there exists a Cartan subgroup $C$ of $G$, and if $N$ is its normalizer, $\mathcal{C}$ is identified by the conjugacy theorem XII 7.1 a) and b) with $G/N$, and the morphism $\psi :X\to G$ considered here is none other than that defined in section 2. One concludes then by 2.6 b). The same reasoning also shows that $U$ is the largest open of $G$ such that $\psi$ induces an isomorphism ${\psi }^{-1}(U)\to U$.

Corollary 3.2. Under the conditions of 3.1, let $g$ be a regular section of $G$, i.e. such that for every $s\in S$, $g(s)$ is a regular point of $G_s$. Then there exists one and only one Cartan subgroup $C$ of $G$ such that $g$ is a section of $C$.

Indeed, the hypothesis on $g$ means that $g$ is a section of $U$, and the conclusion means that there exists a unique section of $X$ lifting it, which is just another way of expressing that ${\psi }^{-1}(U)\to U$ is an isomorphism.

Note now that the open ${\psi }^{-1}(U)$ of the Cartan subgroup $X$ of $G_{\mathcal{C}}$ is none other than the open of $X$ consisting of the points of $X$ that are regular in $G_{\mathcal{C}}$ (by which we mean: regular in their fiber). One thus obtains a natural "fibration" of the dense open $U$ of regular points of $G$ over the prescheme $\mathcal{C}$, the fibers being dense opens of Cartan subgroups of the fibers of $G_{\mathcal{C}}$ (namely the opens of regular points in $G$). One finds for example the following result (which will be considerably refined in the following Exposé):

Corollary 3.3. Let $G$ be a smooth connected algebraic group over the field $k$, $\mathcal{T}$ the scheme of maximal tori of $G$ (≃ the scheme of Cartan subgroups of $G$). Then the field of functions $k(G)$ of $G$ is isomorphic to the field of functions of a smooth connected affine nilpotent algebraic group $C$ over the field of functions $k(\mathcal{T})$ of $\mathcal{T}$, namely $C=$ "the generic Cartan subgroup of $G$". If $G$ is affine of zero unipotent rank, i.e. if the Cartan subgroups of G_K are tori, then $k(G)$ is a unirational extension of $k(\mathcal{T})$.

Of course, by generic Cartan subgroup of $G$, we mean (by abuse of language) the Cartan subgroup of $G_{k(\mathcal{T})}$ generic fiber of $X$ over $\mathcal{T}$. It remains only to prove the last assertion of 3.3, which is contained in the following well-known result (due to Chevalley):

Lemma 3.4. Let $k$ be a field, $T$ a torus over $k$, $k(T)$ the field of rational functions on $T$; then $k(T)$ is a unirational extension of $k$, i.e. is contained in a pure transcendental extension of $k$.

Indeed, let $k^{\prime }$ be a finite separable extension of $k$ that splits⁷ $T$ (X 1.4); then $T{\otimes }_k{k}^{\prime }$ is a rational variety, i.e. admits a dense open isomorphic to a dense open of affine space ${\mathbb{A}}_{k^{\prime }}^n$; hence T′ = ∏_{Spec(k′)/Spec(k)} T_{k′}/Spec(k′) is a rational variety (because it admits a dense open isomorphic to a dense open of ${\prod }_{\mathrm{Spec}(k^{\prime })/\mathrm{Spec}(k)}{\mathbb{A}}_{k^{\prime }}^n$, which is isomorphic to affine space of dimension mn over $k$, where $m=[k^{\prime }:k]$). Consider the norm homomorphism from $T^{\prime }$ to $T$ (defined whenever $T$ is a commutative group scheme over $k$); the composite $T\to T^{\prime }\to T$ is the $m$-th power in $T$, hence dominant, hence $T^{\prime }\to T$ is dominant, which proves that $T$ is unirational.

Let us return to the conditions of 3.1, but assume further that $G$ admits, locally for the fpqc topology, a maximal torus (XII 7.1). Let $Y$ be the maximal torus of the Cartan subgroup $X$ of $G_{\mathcal{C}}$, so that the morphism $\psi :X\to G$ induces a morphism $Y\to G$ whose image is set-theoretically the semisimple elements of the fibers of $G$ (XII 8). Finally, it follows from 3.1 that the restriction of $\psi$ to the open $Y_{reg}$ of regular points of $Y$ induces a closed immersion

$$Y_{reg}\to U=G_{reg}.$$

Making explicit the meaning of $Z=Y_{reg}$ as a functor on $S$, one finds:

Corollary 3.5. Let $G$ be a smooth, separated, finite-type $S$-prescheme in groups with connected fibers over the prescheme $S$, admitting locally for the fpqc topology a maximal torus. Let $Z:(Sch{)}^{\circ }/S\to (Ens)$ be the functor defined by

Z(S′) = set of regular sections of G_{S′} over S′ that are contained in
        a maximal torus of G_{S′}.

Then $Z$ is representable by a closed sub-prescheme, smooth over $S$, of the open $U=G_{reg}$ of $G$ introduced in 3.1.

To finish, let us note the following result, which sharpens the density theorem 2.1 (i) ⇒ (vi):

Corollary 3.6. Under the conditions of 3.5, let $C$ be a Cartan subgroup of $G$, and consider the morphism

ϕ : Z × C → G

defined by $\phi (g,h)=ad(g)h=gh{g}^{-1}$. Then $\phi$ is dominant.

It evidently suffices to prove this fiber by fiber, which reduces us to the case where $S$ is the spectrum of an algebraically closed field. Let $T$ be the maximal torus of $C$, $t_0$ an element of $T(k)$ regular in $G$, $c_0$ an element of $C(k)$ regular in $G$; consider ${\phi }^{-1}(\phi (t_0,c_0))$, whose $k$-rational points are the pairs $(t,c)$ with $t\in Z(k)$, $c\in C(k)$, such that

ad(t) c = ad(t₀) c₀   i.e.   c = ad(t⁻¹ t₀) c₀,

which are therefore in bijective correspondence with the $t\in Z(k)$ such that $ad(t^{-1}{t}_0)c_0\in C$, or equivalently (since $c_0$ is regular) such that $t^{-1}{t}_0\in N$ (normalizer of $C$), i.e. $t\in N$. One obtains an open and closed part of this fiber by restricting to the $t\in Z(k)$ such that $t\in C(k)$. So we have found a connected component of ${\phi }^{-1}\phi (t_0,c_0)$ isomorphic to $T$ (N.B. that the preceding set-theoretic reasoning indeed gives an isomorphism of schemes is seen by replacing points with values in $k$ by points with values in an arbitrary $k$-prescheme); hence the generic fiber of $\phi$ is of dimension $⩽\dim T$, so $Im(\phi )$ is of dimension ⩾ dim Z × C − dim T = dim Z + dim C − dim T; now one has dim Z = dim Y = dim 𝓒 + dim T = dim G − dim C + dim T, whence finally $\dim Im(\phi )⩾\dim G$, so $\phi$ is dominant. QED.

Remarks 3.7. One will note that the reasoning shows moreover that the connected component at $(t_0,c_0)$ of the fiber ${\phi }^{-1}(\phi (t_0,c_0))$ is isomorphic to $T$, in particular is smooth over $k$, and has the same dimension as the generic fiber, which implies that $\phi$ is in fact smooth at $(t_0,c_0)$ (which one ought to be able to verify also by calculating the tangent map). It follows that under the conditions of 3.6 the induced morphism $Z{\times }_S{C}_{reg}\to G_{reg}$ (where one has set $C_{reg}=C\cap G_{reg}$) is a smooth morphism. One sees similarly that the analogous morphism $Z\times T_{reg}\to Z$ (where $T$ is a maximal torus of $G$) is smooth; more generally, for every smooth connected invariant algebraic subgroup $H$ of $C$ containing a regular element $c_0$ of $G(k)$, the image of $Z\times H\to G$ is dense in that of $G\times H\to G$.

4. Lie algebras over a field: rank, regular elements, Cartan subalgebras

In what follows of this Exposé, we take up the theory developed by Chevalley in his book Théorie des Groupes de Lie III (Act. Sc. Ind. 1226, Paris 1955), the technique of schemes allowing us to eliminate the hypothesis of characteristic zero. We begin by recalling in the present section certain well-known notions and results.

Let $\mathfrak{g}$ be a Lie algebra over a ring $k$. For every $a\in \mathfrak{g}$, one denotes by $ad(a)$ the endomorphism

ad(a) · x = [a, x]

of $\mathfrak{g}$, which is a derivation of the Lie algebra $\mathfrak{g}$. Now for any derivation $D$ of the Lie algebra $\mathfrak{g}$, the nil-space of $D$, i.e. the union of the kernels of the iterates of $D$, is a Lie subalgebra of $\mathfrak{g}$, as one sees from the Leibniz formula

Dᵐ([x, y]) = Σ_{0 ⩽ p ⩽ m} (m choose p) [Dᵖ x, Dᵐ⁻ᵖ y].

We shall set

Nil(a, 𝔤) = nil-space of ad(a) = ⋃_{m ⩾ 0} Ker ad(a)ᵐ;

when no confusion need be feared, we shall denote it simply $Nil(a)$, and we shall call it the nil-space of $a$ (in $\mathfrak{g}$).

Proposition 4.1. For every $a\in \mathfrak{g}$, its nil-space $Nil(a,\mathfrak{g})$ is a Lie subalgebra of $\mathfrak{g}$, equal to its own normalizer.

It remains to prove that it is its own normalizer, i.e. that every element of $\mathfrak{g}/Nil$ annihilated by the adjoint representation of Nil on $\mathfrak{g}/Nil$ is zero, which is trivial (since every element in this quotient annihilated by $Ad(a)$ is zero).

In what follows of this section, we suppose that $k$ is a field, and that $\mathfrak{g}$ is of finite dimension over $k$. We shall denote by $W(\mathfrak{g})$ the scheme over $k$ defined by $\mathfrak{g}$, whose points in the $k$-algebra $A$ are the elements of $\mathfrak{g}{\otimes }_{k}A$. If $a\in \mathfrak{g}$, the characteristic polynomial of $ad(a)$ is also called the characteristic polynomial or Killing polynomial of $a$ in $\mathfrak{g}$, namely

P_𝔤(a, t) = tⁿ + c₁(a) tⁿ⁻¹ + ⋯ + c_n(a),

where $n=ran{k}_k\mathfrak{g}$, the $c_i(a)\in k$. Taking this polynomial also for $a\in \mathfrak{g}{\otimes }_{k}A$,⁸ where $A$ is any $k$-algebra, one sees that the $c_i(a)$ come from well-determined sections $c_i$ of the structural sheaf of $W(\mathfrak{g})$, i.e. from elements of the symmetric algebra $A=Sy{m}_k({\mathfrak{g}}^{\vee })$, where ${\mathfrak{g}}^{\vee }$ is the dual of the $k$-module $\mathfrak{g}$. (When $k$ is an infinite field, the $c_i$ are determined by knowing the corresponding polynomial functions $\mathfrak{g}\to k$, but this is no longer the case if $k$ is a finite field.) Let $r$ be the largest integer such that the Killing polynomial

P_𝔤(t) = tⁿ + c₁ tⁿ⁻¹ + ⋯ + c_n ∈ A[t]

is divisible by tʳ, i.e. one has:

P_𝔤(t) = tⁿ + c₁ tⁿ⁻¹ + ⋯ + c_{n−r} tʳ,    c_{n−r} ≠ 0.

The integer $r$ is called the nilpotent rank of the Lie algebra $\mathfrak{g}$. It is invariant under extension of the base field.

Proposition 4.2. Let $r$ be the nilpotent rank of $\mathfrak{g}$, and $a\in \mathfrak{g}$. Then one has

rank_k Nil(a, 𝔤) ⩾ r,

with equality if and only if

$$c_{n-r}(a)\ne 0.$$

In this case, $Nil(a,\mathfrak{g})$ is a nilpotent Lie algebra (and we shall see in 5.7 b) the converse, when $\mathfrak{g}$ is the Lie algebra of an algebraic group $G$ smooth over $k$).

The first assertion is trivial, since by definition $ran{k}_{k}Nil(a,\mathfrak{g})=$ multiplicity of the zero root in $P_{\mathfrak{g}}(a,t)$. Let us prove that if $c_{n-r}(a)\ne 0$, then $Nil(a)$ is nilpotent, which also means that for every $x\in Nil(a)$, $a{d}_{Nil(a)}(x)$ is a nilpotent endomorphism. One may assume $k$ algebraically closed; then as $ad(a{)}_{\mathfrak{g}/Nil(a)}$ is injective, there exists a non-empty open $U$ of $W(Nil(a))$ such that for every $x\in U(k)$, $ad(x{)}_{\mathfrak{g}/Nil(a)}$ is injective, hence $Nil(x)\subset Nil(a)$; one may further suppose $U$ contained in the open of points where $c_{n-r}$ does not vanish (since this open is non-empty by $c_{n-r}(a)\ne 0$), and then $Nil(x)$ having the same dimension as $Nil(a)$, one will have $Nil(x)=Nil(a)$. Consequently, for every $x\in U(k)$, $ad(x{)}_{Nil(a)}$ is nilpotent, and by the principle of extension of algebraic identities, this remains true for every $x\in Nil(a)$, hence $Nil(a)$ is nilpotent.

One says that the element $a$ of $\mathfrak{g}$ is regular if $c_{n-r}(a)\ne 0$, i.e. if $ran{k}_{k}Nil(a,\mathfrak{g})=r$. When $k$ is infinite, this also means that $ran{k}_{k}Nil(a,\mathfrak{g})$ is the smallest possible (for $a$ varying in $\mathfrak{g}$). In any case, the notion of regular element of $\mathfrak{g}$ is invariant under extension of the base field, and the set of points of $W(\mathfrak{g})$ that are regular (i.e. that come from regular points of $W(\mathfrak{g})$ with values in a suitable extension of $k$) is open, since identical to $W(\mathfrak{g}{)}_{c_{n-r}}$ (set of points where $c_{n-r}$ is invertible).

Corollary 4.3. Let $a$ be a regular element of $\mathfrak{g}$, and $\mathfrak{h}$ a Lie subalgebra of $\mathfrak{g}$ containing $a$. Then $\mathfrak{h}$ is nilpotent if and only if $\mathfrak{h}\subset Nil(a,\mathfrak{g})$; in particular, $Nil(a,\mathfrak{g})$ is a maximal nilpotent subalgebra of $\mathfrak{g}$.

Since $Nil(a)$ is nilpotent, the relation $\mathfrak{h}\subset Nil(a)$ indeed implies that $\mathfrak{h}$ is nilpotent; and conversely, if $\mathfrak{h}$ is nilpotent, it is contained in the nil-space of its element $a$, i.e. $\mathfrak{h}\subset Nil(a)$.

Proposition 4.4. Suppose $k$ infinite. Let $\mathfrak{d}$ be a Lie subalgebra of $\mathfrak{g}$. Consider the following conditions:

• (i) $\mathfrak{d}$ is maximal nilpotent and contains a regular element of $\mathfrak{g}$.
• (i bis) $\mathfrak{d}$ is of the form $Nil(a,\mathfrak{g})$, where $a$ is a regular element of $\mathfrak{g}$.
• (ii) $\mathfrak{d}$ is nilpotent and of the form $Nil(a,\mathfrak{g})$, where $a\in \mathfrak{g}$.
• (ii bis) $\mathfrak{d}$ is nilpotent, and there exists $a\in \mathfrak{d}$ such that $ad(a{)}_{\mathfrak{g}/\mathfrak{d}}$ is injective.
• (iii) $\mathfrak{d}$ is nilpotent and identical to its own normalizer.

One has the implications:

(i) ⇔ (i bis) ⇒ (ii) ⇔ (ii bis) ⇔ (iii)

(and we shall see in 5.7 a) that if $\mathfrak{g}$ is the Lie algebra of a smooth algebraic group, then all the preceding conditions are equivalent).

The equivalence of (i) and (i bis) is trivial by 4.3, and these conditions trivially imply (ii). The equivalence of (ii) and (ii bis) is also trivial, as is (ii bis) ⇒ (iii) (cf. 4.1). It remains to prove the implication (iii) ⇒ (ii bis), the only one moreover that uses the fact that $k$ is infinite, and which follows at once from:

Lemma 4.5. Let $\mathfrak{d}$ be a nilpotent Lie algebra over an infinite field $k$, acting on a finite-dimensional vector space $V$. Suppose that for every $x\in \mathfrak{d}$, the endomorphism $u(x)$ is non-injective. Then there exists a non-zero element $v$ of $V$ annihilated by $\mathfrak{d}$.

One may suppose $k$ algebraically closed and $\mathfrak{d}$ of finite dimension. One knows then that $V$ is a direct sum of finitely many non-zero stable subspaces $V_i$ ($1⩽i⩽n$), such that for every $i$, and every $x\in \mathfrak{d}$, $u(x)|V_i$ has a single eigenvalue ${\lambda }_i(x)$ (cf. Bourbaki, Groupes et Algèbres de Lie, Chap. I, §4, Exercise 22). Let $c_i(x)$ be the constant term of the characteristic polynomial of $u(x)|V_i$, so that ${\lambda }_i(x)=0$ if and only if $c_i(x)=0$. Then $c_i$ is a polynomial function on $\mathfrak{d}$, and the hypothesis means that $\mathfrak{d}$ is the union of the sets of zeros of the $c_i$. So one of the $c_i$ is zero, which reduces us (replacing $V$ by $V_i$) to the case where $V$ is such that the $u(x)$ ($x\in \mathfrak{d}$) are nilpotent. But then Engel's theorem (Bourbaki loc. cit. th. 1) implies that there exists a non-zero $v$ in $V$ annihilated by $\mathfrak{d}$. QED.

One sees easily that ($k$ being always an infinite field) conditions (i) (i bis) of 4.4 are invariant under any extension of the base field. If they are satisfied, one will say that $\mathfrak{d}$ is a Cartan subalgebra of $\mathfrak{g}$; in the general case ($k$ not necessarily infinite) one will say that $\mathfrak{d}$ is a Cartan subalgebra of $\mathfrak{g}$ if it becomes a Cartan subalgebra under one (and hence any) extension of the base field $k\to k^{\prime }$, with $k^{\prime }$ infinite. This thus implies that $\mathfrak{d}$ is nilpotent and equal to its own normalizer.

Proposition 4.6. a) Let $a$ be an element of $\mathfrak{g}$. If $a$ is regular, it is contained in one and only one Cartan subalgebra of $\mathfrak{g}$ (and we shall see in 6.1 d) the converse when $k$ is algebraically closed and $\mathfrak{g}$ is the Lie algebra of a smooth algebraic group).

b) Let $\mathfrak{d}$ be a Cartan subalgebra of $\mathfrak{g}$, $a$ an element of $\mathfrak{d}$; then $a$ is regular in $\mathfrak{g}$ if and only if $ad(a{)}_{\mathfrak{g}/\mathfrak{d}}$ is injective.

Indeed, for a) one notes that if $a$ is regular, then $Nil(a,\mathfrak{g})$ is a Cartan subalgebra of $\mathfrak{g}$ (because this is true over an infinite extension $k^{\prime }$ of $k$), and it follows then at once from 4.3 that every Cartan subalgebra of $\mathfrak{g}$ containing $a$ is identical to the preceding one. For b) one notes that the nullity⁹ of $ad(a{)}_{\mathfrak{g}}$ is equal to the sum of the nullities of $ad(a{)}_{\mathfrak{d}}$ and of $ad(a{)}_{\mathfrak{g}/\mathfrak{d}}$, and as the first equals $r$, the sum is equal to $r$ if and only if $ad(a{)}_{\mathfrak{g}/\mathfrak{d}}$ is injective. QED.

Corollary 4.7. Let $a$ be a regular element of $\mathfrak{g}$, $\mathfrak{d}$ a Cartan subalgebra of $\mathfrak{g}$ containing $a$, $A$ an algebra over $k$, ${\mathfrak{g}}_A$ and ${\mathfrak{d}}_A\subset {\mathfrak{g}}_A$ the $A$-Lie algebras deduced from $\mathfrak{g}$, $\mathfrak{d}$ by base change, $a_A$ the image of $a$ in ${\mathfrak{g}}_A$. Let $u$ be an automorphism of ${\mathfrak{g}}_A$. In order that $u({\mathfrak{d}}_A)={\mathfrak{d}}_A$, it is necessary and sufficient that one have $u(a_A)\in {\mathfrak{d}}_A$.

The condition is trivially necessary; let us prove that it is also sufficient. If it is satisfied, then ${\mathfrak{d}}^{\prime }=u({\mathfrak{d}}_A)$ is a Lie subalgebra containing $a_A$, and every element $b$ of which is such that $ad(b{)}_{{\mathfrak{d}}^{\prime }}$ is nilpotent (because ${\mathfrak{d}}^{\prime }$ is isomorphic to ${\mathfrak{d}}_A$, which has this property, as follows at once from the definition of "nilpotent" in Bourbaki, Groupes et Algèbres de Lie, Chap. I, §4, def. 1). Taking $b=a_A$, one sees that the nil-space $Nil(b,{\mathfrak{g}}_A)$ contains ${\mathfrak{d}}^{\prime }$; on the other hand it is equal to ${\mathfrak{d}}_A$, and as ${\mathfrak{d}}^{\prime }$ is locally a direct factor in the module ${\mathfrak{g}}_A$ ($\mathfrak{d}$ being so), hence in ${\mathfrak{d}}_A$, and as it is a projective module of the same rank $r$ as the latter, one concludes that it equals it. QED.

Proposition 4.8. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$.

a) The following conditions are equivalent if $k$ is infinite:

• (i) $\mathfrak{h}$ contains a Cartan subalgebra $\mathfrak{d}$ of $\mathfrak{g}$.
• (ii) $\mathfrak{h}$ contains a regular element $a$ of $\mathfrak{g}$, and an element $b$ such that $ad(b{)}_{\mathfrak{g}/\mathfrak{h}}$ is injective.
• (iii) $\mathfrak{h}$ has the same nilpotent rank as $\mathfrak{g}$, and contains a regular element of $\mathfrak{g}$.

These conditions are invariant under extension of the base field $k$.

b) Suppose these conditions are satisfied over a suitable infinite extension $k^{\prime }$ of $k$. Let $a\in \mathfrak{h}$; then $a$ is regular in $\mathfrak{g}$ if and only if it is regular in $\mathfrak{h}$ and $ad(a{)}_{\mathfrak{g}/\mathfrak{h}}$ is injective, i.e. if and only if $Nil(a,\mathfrak{h})=Nil(a,\mathfrak{g})$ and this is a Cartan subalgebra of $\mathfrak{h}$.

c) Under the conditions of b), let $\mathfrak{d}$ be a Lie subalgebra of $\mathfrak{h}$; in order that it be a Cartan subalgebra of $\mathfrak{h}$, it suffices that it be a Cartan subalgebra of $\mathfrak{g}$ (and we shall see in 5.8 that the condition is also necessary if $\mathfrak{g}$ is the Lie algebra of a smooth algebraic group $G$, and $\mathfrak{h}$ the Lie algebra of a smooth algebraic subgroup $H$ of $G$).

One sees at once that conditions (ii) and (iii) of a) are invariant under extension of the base field $k$ (assumed infinite), and that in statements b) and c), one may assume $k$ infinite, which we shall do. If $\mathfrak{h}$ contains the Cartan subalgebra $\mathfrak{d}=Nil(a,\mathfrak{g})$, then $a$ is a regular element of $\mathfrak{g}$ such that $ad(a{)}_{\mathfrak{g}/\mathfrak{h}}$ is injective, hence (i) ⇒ (ii). Conversely, if (ii) is satisfied, then for an element $a$ "sufficiently general" of $\mathfrak{h}$, $a$ satisfies simultaneously the two conditions of (ii), hence $Nil(a,\mathfrak{g})$ is a Cartan subalgebra of $\mathfrak{g}$ and is contained in $\mathfrak{h}$, so one has (i). So (i) and (ii) are equivalent. Suppose them satisfied, let $a$ be a variable element of $\mathfrak{h}$; then

(✱)   rank_k Nil(a, 𝔤) = rank_k Nil(a, 𝔥) + rank_k Nil(ad(a)_{𝔤/𝔥}),

on the other hand, the two terms of the right-hand side are respectively $⩾r_0=$ nilpotent rank of $\mathfrak{h}$, and $⩾0$, the equalities being moreover attained¹⁰ for an element "sufficiently general" of $\mathfrak{h}$. Moreover, one has also $ran{k}_{k}Nil(a,\mathfrak{g})⩾r=$ nilpotent rank of $\mathfrak{g}$, with equality attained for an element "sufficiently general" of $\mathfrak{h}$, and attained if and only if $a$ is regular in $\mathfrak{g}$. One concludes from this that one has $r=r_0$, and that $a$ is regular if and only if the two terms of the right-hand side of (✱) equal respectively $r_0$ and 0, i.e. if and only if $a$ is regular in $\mathfrak{h}$ and $ad(a{)}_{\mathfrak{g}/\mathfrak{h}}$ is injective, which proves b), and c) follows trivially by taking an element $a$ in $\mathfrak{d}$ regular in $\mathfrak{g}$, so that $Nil(a,\mathfrak{g})=\mathfrak{d}$. Moreover, the preceding result shows that (i) ⇒ (iii); finally (iii) ⇒ (i), because under (iii), an element sufficiently general $a$ of $\mathfrak{h}$ is regular in $\mathfrak{h}$ and in $\mathfrak{g}$, hence $Nil(a,\mathfrak{h})\subset Nil(a,\mathfrak{g})$ are respectively Cartan subalgebras of $\mathfrak{h}$ and of $\mathfrak{g}$, and as they have the same rank over $k$, they are identical, which proves (i). This completes the proof of 4.8.

5. The case of the Lie algebra of a smooth algebraic group: density theorem

Let $G$ be a smooth algebraic group over the field $k$, and $\mathfrak{g}$ its Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $G$ act on $W(\mathfrak{g})$ by the adjoint representation, and consider the subscheme $W(\mathfrak{h})$. The construction of section 1 leads us to introduce

$$N=Nor{m}_G(\mathfrak{h})=Nor{m}_G(W(\mathfrak{h})),$$

which is an algebraic subgroup of $G$ (not necessarily smooth),

$$\mathfrak{n}=Lie(N),$$

and the scheme

X = G ×^N W(𝔥)

quotient of $\tilde{X}=G\times W(\mathfrak{h})$ by $N$ acting on the right by $(g,x)\cdot n=(gn,Ad(n^{-1})x)$. We consider the canonical morphisms

                      G × W(𝔥)
                       │     ╲
                      q│      ╲ ϕ
                       │       ╲
                       ↓        ↘
                       X ──ψ──→ W(𝔤).

Theorem 5.1. With the preceding notation, suppose $k$ infinite. Consider the following conditions:

• (i) $\mathfrak{h}$ contains a Cartan subalgebra $\mathfrak{d}$ of $\mathfrak{g}$.
• (ii) There exists $a\in \mathfrak{h}$ such that $ad(a{)}_{\mathfrak{g}/\mathfrak{h}}$ is injective.
• (iii) $\phi :G\times W(\mathfrak{h})\to W(\mathfrak{g})$ is generically smooth.
• (iv) The preceding morphism $\phi$ (or also $\psi :X\to W(\mathfrak{g})$) is dominant, and $\mathfrak{h}$ has the same nilpotent rank as $\mathfrak{g}$.
• (v) $\psi :X\to W(\mathfrak{g})$ is generically smooth and $\mathfrak{h}=\mathfrak{n}$.
• (vi) $\psi :X\to W(\mathfrak{g})$ is dominant, and $\mathfrak{h}=\mathfrak{n}$.
• (vii) $\psi :X\to W(\mathfrak{g})$ is dominant.
• (viii) $\psi :X\to W(\mathfrak{g})$ is dominant, and $N$ is smooth.
• (ix) $\psi :X\to W(\mathfrak{g})$ is dominant, and $\mathfrak{h}=\mathfrak{n}$, and $\mathfrak{h}$ is the Lie algebra of a smooth algebraic subgroup $H$ of $G$.
• (x) $\psi :X\to W(\mathfrak{g})$ is generically quasi-finite, and $\mathfrak{n}=\mathfrak{h}$.
• (xi) $\psi :X\to W(\mathfrak{g})$ is generically étale, and $\mathfrak{n}=\mathfrak{h}$.
• (xii) There exists a smooth algebraic subgroup $H$ of $G$ with Lie algebra $\mathfrak{h}$, and $\mathfrak{h}$ contains a Cartan subalgebra $\mathfrak{d}$ of $\mathfrak{g}$.
• *(xiii) There exists $a\in \mathfrak{h}$ such that $N$ and the transporter $M_a$ of $a$ to $\mathfrak{h}$ (cf. section 1) coincide in a neighborhood of `e`, and `𝔫 = 𝔥`.*