RATIONAL REGULAR NILPOTENT ELEMENTS OF A REDUCTIVE LIE ALGEBRA

DAVID G. A. JACKSON

Abstract. The aim of this paper is give an exposition of some of the main properties of regular nilpotent elements of the Lie algebra of a reductive algebraic group (with not necessarily connected centre) defined over a finite field. The approach presented here will be similar to that of Steinberg [10] and Digne & Michel [4] for regular unipotent elements.

1. Introduction and Notation

We will assume the reader is familiar with the basic theory of linear algebraic groups. Most of what is needed here can be found in Borel [1] or Borel & Tits [2]. Throughout this paper G will denote a connected reductive algebraic group defined over the finite field 𝔽_{q} with corresponding Frobenius endomorphism F : G → G. We will suppose here that G is F -split. We will in general denote Lie algebras by lower case gothic letters (so, for example, 𝔤 = Lie(G)). We get an induced action of Frobenius on 𝔤, which we shall also denote by F. Given an F -stable Borel subgroup B and an F -stable Levi decomposition B = T U we will let Φ be the root system of G with respect to T and Π the system of simple roots determined by the pair T ⊆ B. Let the root subgroup (resp. root subspace) corresponding to root α be denoted by X_{α} (resp. 𝔤_{α}). Let U′ = ∏_{α∈Φ^{+}−Π} X_{α} and Z be the centre of G. We will repeatedly use the fact that ∩_{α∈Π} ker α = Z (cf. Digne & Michel [4, 0.35]). Let W = N_{G}(T)/T be the Weyl group of G with respect to T. We will denote by s_{β} the reflection in root β and by ẇ a representative in N_{G}(T) of w ∈ W.

2. Some Preparatory Lemmas

Lemma 1. Let B, T, U and U′ be as above. The derived group of U is U′ and

(i) If u ∈ U, Y ∈ 𝔲 then Ad u(Y) − Y ∈ 𝔲′. (ii) If g ∈ B, Y ∈ 𝔲 then Ad g(Y) ∈ 𝔲. (iii) If g ∈ B, Y′ ∈ 𝔲′ then Ad g(Y′) ∈ 𝔲′. (iv) If g ∈ B, Y ∈ 𝔟 then Ad g(Y) − Y ∈ 𝔲.

Proof. The first statement is Digne & Michel [4, 14.17]. For (i), the map v ↦ uvu^{−1}v^{−1} sends U to U′. According to Humphreys [5, 10.1(c)], its differential is Y ↦ Ad u(Y)−

2 DAVID G. A. JACKSON

Y and sends 𝔲 to 𝔲′. In a similar fashion, if g ∈ B, differentiation of the map U → U given by x ↦ gxg^{−1} shows that Ad g(Y) ∈ 𝔲 proving (ii). If t ∈ T then differentiating the map U′ → U′ given by x ↦ txt^{−1} shows that Ad t(Y′) ∈ 𝔲′. Now let u ∈ U. Since Y′ ∈ 𝔲, if follows from (i) that Ad u(Y′)−Y′ ∈ 𝔲′ and hence Ad u(Y′) ∈ 𝔲′. Statement (iii) then follows since B = T U. The last statement follows by differentiating the map b ↦ gbg^{−1}b^{−1} from B to U.

Lemma 2. If B is a Borel subgroup of G then B = {g ∈ G | Ad g(𝔟) = 𝔟}.

Proof. Clearly B ⊆ {g ∈ G | Ad g(𝔟) = 𝔟}. Conversely, suppose that Ad g(𝔟) = 𝔟. Let T be any maximal torus in B and let Π be the simple system of roots determined by the pair T ⊆ B. For α ∈ Π let Y_{α} be a non-zero element of 𝔤_{α}. Then Ad g(∑_{α∈Π} Y_{α}) ∈ 𝔟 and hence g ∈ B by Borel [1, 14.23].

The proofs of the following two results are based on Digne & Michel [4, 14.2, 14.4].

Proposition 3 (Springer [8, Remark following 5.6]). If G is a linear algebraic group then dim C_{G}(Y) ⩾ rk(G) for any element Y ∈ 𝔤.

Proof. By Borel [1, 14.25], Y is contained in 𝔟 = Lie(B) for some Borel subgroup B of G. Let U be its unipotent radical. By Lemma 1, Ad B(Y) ⊆ Y + 𝔲 and hence dim Ad B(Y) ⩽ dim U = dim B − rk(G). But dim Ad B(Y) = dim B − dim C_{B}(Y) and hence dim C_{B}(Y) ⩾ rk(G). A fortiori, we must have dim C_{G}(Y) ⩾ rk(G).

Definition. An element Y ∈ 𝔤 is regular in 𝔤 if dim C_{G}(Y) = rk(G).

Note. In the case of the group, one can show the existence of regular elements by finding t ∈ T such that C_{G}(t) = T (cf. Digne & Michel [4, 14.2(ii)]). The corresponding result for the Lie algebra is not true however. For example, take 𝔰𝔩_{2} over a field of characteristic 2. Every semisimple element is then central. We will prove the existence of regular elements of the Lie algebra in Theorem 5.

Proposition 4 (cf. Steinberg [10, 3.7]). Let Y = Y_{s}+Y_{n} be the Jordan decomposition of an element of 𝔤. Then Y is regular in 𝔤 if and only if Y_{n} is regular in C_{𝔤}(Y_{s}).

Proof. By Humphreys [5, 4.2 Proposition (b)] we have C_{G}(Y) = C_{G}(Y_{s}) ∩ C_{G}(Y_{n}), and hence dim C_{G}(Y) = dim C_{C_{G}(Y_{s})}(Y_{n}). Now let T be a maximal torus of G whose Lie algebra contains Y_{s}. Then T ⊆ C_{G}(Y_{s}) and hence rk(G) = rk(C_{G}(Y_{s})). Since C_{𝔤}(Y_{s}) = Lie C_{G}(Y_{s}) (cf. Borel [1, 9.1]), the result follows according to the definition of regular applied to the group C_{G}(Y_{s}).

RATIONAL REGULAR NILPOTENT ELEMENTS OF A REDUCTIVE LIE ALGEBRA 3

3. The main theorem on regular nilpotent elements

Theorem 5 (cf. Steinberg [10, 3.2, 3.3], Digne & Michel [4, 14.14]). Let G be a connected reductive group. Then regular nilpotent elements exist and any two regular nilpotent elements are conjugate under Ad G. Let Y be a nilpotent element of 𝔤. The following properties are equivalent:

(i) Y is regular. (ii) There are only a finite number of Borel subalgebras containing Y. (iii) There is only one Borel subalgebra containing Y. (iv) There exists a Borel subalgebra 𝔟 = Lie B containing Y and a maximal torus T of

B such that in the decomposition Y = ∑_{α∈Φ^{+}} Y_{α} where α runs over the positive roots relative to the pair T ⊂ B and Y_{α} ∈ 𝔤_{α}, none of the Y_{α} for α simple is equal to zero.

(v) Property (iv) is true for any Borel subalgebra 𝔟 = Lie B containing Y and any maximal torus of B.

Proof. We will first show that the statements (i)-(v) are equivalent. In the course of this the first assertion will be proved. (iv) ⟹ (iii). Let B′ be a second Borel subgroup such that Y ∈ 𝔟′ = Lie(B′). By the conjugacy theorem for Borel subgroups (cf. Borel [1, 11.1]) and the Bruhat decomposition, B′ = uẇBẇ^{−1}u^{−1} for some w ∈ W, u ∈ U_{w^{−1}}. Then 𝔟 contains Ad(ẇ^{−1}u^{−1})(Y). According to Lemma 1, Ad u^{−1}(Y) − Y ∈ 𝔲′ and hence Ad u^{−1}(Y) = ∑_{α∈Φ^{+}} Y_{α} where Y_{α} ∈ 𝔤_{α} and Y_{α} ≠ 0 for α ∈ Π. Since Ad ẇ^{−1}(Ad u^{−1}(Y)) ∈ 𝔟 we must have wα > 0 for every α ∈ Π and hence w = 1. Thus B′ = B and (iii) holds. (ii) ⟹ (v). Suppose (v) is false; that is, for some Borel subgroup B whose Lie algebra contains Y and some maximal torus T of B, Y = ∑_{α∈Φ^{+}} Y_{α} with Y_{α} ∈ 𝔤_{α} and Y_{β} = 0 for some β ∈ Π. Then for any u ∈ X_{β}, Ad u^{−1}(Y) − Y ∈ 𝔲′ by Lemma 1 and hence Ad u^{−1}(Y) = ∑_{α∈Φ^{+}} Y′_{α} with Y′_{α} ∈ 𝔤_{α} and Y′_{β} = 0. Hence Ad ṡ_{β}^{−1}u^{−1}(Y) ∈ 𝔟 since the only positive root made negative by s_{β} is β, and hence Y ∈ Ad(uṡ_{β})(𝔟) for any u ∈ X_{β}. These Borel subalgebras are distinct. For if Ad ṡ_{β}(𝔟) = Ad(uṡ_{β})(𝔟) for some u ∈ X_{β} then by applying Lemma 2 in the case of the Borel subalgebra Ad ṡ_{β}(𝔟) = Lie(ṡ_{β}B) we see that u ∈ ṡ_{β}B ∩ X_{β} = ṡ_{β}(B ∩ ṡ_{β}X_{β}) = ṡ_{β}(B ∩ X_{−β}) and hence u = 1. Thus the Borel subalgebras Ad(uṡ_{β})(𝔟) for u ∈ X_{β} constitute an infinite number of Borel subalgebras containing Y, contradicting (ii). (iii) ⟹ (ii). This implication follows trivially.

(v) ⟹ (iv). According to Borel [1, 14.25], each nilpotent element of 𝔤 lies in some Borel subalgebra. The implication immediately follows. We have thus proved the equivalence of (ii), (iii), (iv) and (v).

4 DAVID G. A. JACKSON

(i) ⟹ (v). Let Y be a regular nilpotent element. Let Y ∈ 𝔟 for some Borel subalgebra 𝔟 = Lie(B). Let T be a maximal torus of B and let Y = ∑_{α∈Φ^{+}} Y_{α} where the sum is over positive roots relative to T ⊂ B and Y_{α} ∈ 𝔤_{α}. Suppose that Y_{β} = 0 for some β ∈ Π. Then Y ∈ 𝔭 = Lie(P) where P is the parabolic subgroup B ∪ Bṡ_{β}B. We first show that if g ∈ P then Ad g(Y) = ∑_{α∈Φ^{+}} Y′_{α} with Y′_{α} ∈ 𝔤_{α} and Y′_{β} = 0. First suppose that g ∈ B. Write g = tu with t ∈ T, u ∈ U. Then Ad u(Y) − Y ∈ 𝔲′ and hence Ad g(Y) = Ad t Ad u(Y) = ∑_{α∈Φ^{+}} Y′_{α} for some Y′_{α} ∈ 𝔤_{α} with Y′_{β} = 0. Since Y_{β} = 0 and the only positive root made negative by s_{β} is β, if g = ṡ_{β} then Ad ṡ_{b}(Y) = ∑_{α∈Φ^{+}} Y′_{α} for some Y′_{α} ∈ 𝔤_{α} with Y′_{β} = 0. Hence the Ad P -orbit of Y contains only elements of U of the form ∑_{α∈Φ^{+}} Y′_{α} with Y′_{α} ∈ 𝔤_{α} and Y′_{β} = 0 and thus dim Ad P (Y) ⩽ dim(U) − 1. As dim P = dim U + dim T + 1,

dim C_{P} (Y) = dim P − dim Ad P (Y)

⩾ dim P − (dim U − 1)

= dim U + dim T + 1 − dim U + 1 = dim T + 2 = rk(G) + 2,

a contradiction, as then rk(G) = dim C_{G}(Y) ⩾ rk(G) + 2. (iv) ⟹ (i) and proof of the initial assertion. We now show that if Y is a regular nilpotent element of 𝔤 and Y′ satisfies (iv) then Y and Y′ are conjugate under Ad G. Since regular nilpotent elements exist (cf. Springer [8, 5.9(b)] and Keny [7]), this will show that any element Y′ of 𝔤 satisfying (iv) is regular and hence satisfies condition (i). Note that if Y and Y′ are any two regular nilpotent elements of 𝔤 then this will also show that they are conjugate (since we have shown that Y′ satisfies condition (iv)), thus proving the initial assertion. Replacing Y by an Ad G conjugate we may assume Y ∈ 𝔲. We have shown that Y satisfies (v) and hence, according to Humphreys [6, 16.2(C)] and Digne & Michel [4, 0.31(i)], we may replace Y′ by an Ad T conjugate in such a way that if Y = ∑_{α∈Φ^{+}} Y_{α} and Y′ = ∑_{α∈Φ^{+}} Y′_{α} with Y_{α}, Y′_{α} ∈ 𝔤_{α} then Y_{α} = Y′_{α} for α ∈ Π. Thus Y′ − Y ∈ 𝔲′. As the orbits of a unipotent affine group acting on a variety are closed (cf. Borel [1, 4.10]), the set {Ad u(Y) | u ∈ U} is a closed subvariety of 𝔲 and hence {Ad u(Y) − Y | u ∈ U} is also a closed subvariety of 𝔲. Since Ad u(Y) − Y ∈ 𝔲′ for all u ∈ U and 𝔲′ is a closed subvariety of 𝔲, this implies that {Ad u(Y) − Y | u ∈ U} is a closed subvariety of 𝔲′. Its codimension in 𝔲 is dim C_{U}(Y). We will show that this equals the codimension of 𝔲′ so that 𝔲′ = {Ad u(Y) − Y | u ∈ U}. For the codimension of 𝔲′ is the semisimple rank of G, equal to rk(G) − dim Z. But, by the following lemma, C_{G}(Y) = Z · C_{U}(Y) and hence rk(G) − dim Z = dim C_{G}(Y) − dim Z = dim C_{U}(Y). Hence there exists a u ∈ U such that Ad u(Y) − Y = Y′ − Y. Hence Ad u(Y) = Y′ and Y is Ad G-conjugate to Y′.

Lemma 6 (cf. Springer [8, 5.9(b)], Springer & Steinberg [9, III, 3.7(a)]). Let Y be a regular nilpotent element of 𝔤 and let U be the unipotent radical of the unique Borel subgroup B whose Lie algebra contains Y. Then C_{G}(Y) = Z · C_{U}(Y) as a direct product.

Proof. If Ad g(Y) = Y then, since (i) implies (iii) in Theorem 5, we have Ad g(𝔟) = 𝔟 and hence g ∈ B by Lemma 2. Thus g ∈ C_{B}(Y). Since C_{B}(Y) is a linear algebraic group (being a closed subgroup of B), the Jordan decomposition of g in G is of the form g = tv where t, v ∈ C_{B}(Y), t semisimple and v unipotent. Being unipotent, v must lie in U and hence in C_{U}(Y). By property (v), if T is a maximal torus of B containing t and Π is the set of simple roots with respect to T ⊂ B then α(t) = 1 for all α ∈ Π and hence t ∈ Z. The result follows.

4. Rational conjugacy of regular nilpotent elements

Proposition 7 (cf. Springer & Steinberg [9, III, 1.19(b)]). The regular nilpotent Ad G^{F} -orbits of 𝔤^{F} are parametrized by H^{1}(F, Z/Z^{∘}), the set of F -conjugacy classes in Z/Z^{∘}.

Proof. By Theorem 5, all the regular nilpotent elements of 𝔤 are conjugate under Ad G. Hence, according to Digne & Michel [4, 3.21], the Ad G^{F} -orbits of regular nilpotent elements in 𝔤^{F} are parametrized by H^{1}(F, A(Y)) where Y is any rational regular nilpotent element and A(Y) = C_{G}(Y)/C_{G}(Y)^{∘}. But according to Lemma 6, C_{G}(Y) = Z · C_{U}(Y). By Springer & Steinberg [9, III, 3.7(b)] and Keny [7, p. 197, line 8], C_{U}(Y) is connected regardless of the characteristic and hence C_{G}(Y)/C_{G}(Y)^{∘} ≅ Z/Z^{∘}. Thus H^{1}(F, A(Y)) ≅ H^{1}(F, Z/Z^{∘}) giving the result.

Lemma 8. Let A be a finite abelian group. If F is an automorphism of A then |H^{1}(F, A)| = |A^{F} |.

Proof. Since A is abelian, if L : A → A is the map x ↦ x^{−1}F (x) then two elements of A are F -conjugate if and only if they represent the same coset in A/L(A). Also, as the kernel of L is A^{F}, L(A) ≅ A/A^{F}. Since A is finite we find that |H^{1}(F, A)| = |A/L(A)| = |A|/|A/A^{F} | = |A^{F} |.

Corollary 9. We have |H^{1}(F, Z)| = |Z^{F} |/|Z^{∘F} |

6 DAVID G. A. JACKSON

Proof. We have |H^{1}(F, Z)| = |H^{1}(F, Z/Z^{∘})| by Springer & Steinberg [9, I, 2.6]. Then by the above lemma, |H^{1}(F, Z/Z^{∘})| = |(Z/Z^{∘})^{F} |. This last quantity equals the desired result by Digne & Michel [4, 3.13] since Z^{∘} is connected.

It is possible to be more precise with the parametrization of the rational regular nilpotent orbits. For this we need to extend the definition of the map L as follows.

Definition (cf. Digne, Lehrer & Michel [3, 1.1]). If H is any F -stable algebraic subgroup of G, the Lang map L_{H} : H → H is defined by L_{H}(h) = h^{−1}F (h). If H = G then we will write L for L_{G}.

Lemma 10 (Digne, Lehrer & Michel [3, 1.3]). Let H be an F -stable connected closed subgroup of G such that H ⊇ Z. Then H^{1}(F, Z) and L_{H}^{−1}(Z) are groups, ZH^{F} is a normal subgroup of L_{H}^{−1}(Z) and H^{1}(F, Z) ≅ L_{H}^{−1}(Z)/ZH^{F}.

Proof. It is instructive to present a proof which does not explicitly use the methods of Galois cohomology.

First we need to show that H^{1}(F, Z) and L_{H}^{−1}(Z) are groups. If z_{1}, z_{2}, a, b ∈ Z then z_{1}^{−1}aF (z_{1})z_{2}^{−1}bF (z_{2}) = z^{−1}(ab)F (z) where z = z_{1}z_{2} and hence F -conjugacy classes of Z multiply, showing that H^{1}(F, Z) is a group. If a, b ∈ L_{H}^{−1}(Z) then L_{H}(ab^{−1}) = ba^{−1}F (a)F (b^{−1}) = a^{−1}F (a)bF (b^{−1}) ∈ Z and hence L_{H}^{−1}(Z) is a group. We must now show that ZH^{F} ⊴ L_{H}^{−1}(Z). If g ∈ L_{H}^{−1}(Z), z ∈ Z and h ∈ H^{F} then gzhg^{−1} = zghg^{−1} and

F (ghg^{−1}) = F (g)hF (g)^{−1} = gg^{−1}F (g)h(g^{−1}F (g))^{−1}g^{−1} = ghg^{−1}

since g^{−1}F (g) ∈ Z. Hence ghg^{−1} ∈ H^{F} and ZH^{F} ⊴ L_{H}^{−1}(Z).

Define α : L_{H}^{−1}(Z) → H^{1}(F, Z) by sending h ∈ L_{H}^{−1}(Z) to the F -conjugacy class of L_{H}(h) in Z. We first show that α is surjective. Let z ∈ Z. Since H is connected and z ∈ H, by the Lang-Steinberg Theorem (cf. Springer-Steinberg [9, I, 2.2]) there exists h ∈ H such that L_{H}(h) = z. Then α(h) is the class represented by z. Now h ∈ L_{H}^{−1}(Z) lies in ker α if and only if h^{−1}F (h) = a^{−1}F (a) for some a ∈ Z, which occurs if and only if ha^{−1} ∈ H^{F} for some a ∈ Z, that is, if and only if h ∈ ZH^{F}. Thus ker α = ZH^{F} and α induces an isomorphism L_{H}^{−1}(Z)/ZH^{F} ≅ H^{1}(F, Z).

Lemma 11 (cf. Steinberg [11, 10.10]). Let 𝔟_{1} and 𝔟_{2} be two F -stable Borel subalgebras of G. Then 𝔟_{1} and 𝔟_{2} are conjugate under Ad G^{F}.

Proof. Let 𝔟_{i} = Lie(B_{i}). Then there exists g ∈ G such that B_{2} = ^{g}B_{1} and hence 𝔟_{2} = Ad g(𝔟_{1}). Since 𝔟_{2} is F -stable, Ad(g^{−1}F (g))(𝔟_{1}) = 𝔟_{1} and hence g^{−1}F (g) ∈ B_{1} by Lemma 2. By the Lang-Steinberg Theorem (cf. Springer-Steinberg [9, I, 2.2]) there exists h ∈ B_{1} such that g^{−1}F (g) = h^{−1}F (h). Then gh^{−1} ∈ G^{F} and 𝔟_{2} = Ad(gh^{−1})(𝔟_{1}).

Theorem 12 (cf. Digne, Lehrer & Michel [3, 3.2]). Let G be a connected reductive group. Fix an F -stable Borel subgroup B of G with unipotent radical U and an F - stable maximal torus T of B. Let L_{T} : T → T be the map t ↦ t^{−1}F (t) and let H^{1}(F, Z) be the set of F -conjugacy classes in Z = Z. For each z ∈ H^{1}(F, Z), let t_{z} ∈ T be such that t_{z}^{−1}F (t_{z}) lies in z. Define

𝔲_{z} = {Ad g(Ad t_{z}(Y_{0}) + Y′) | g ∈ G^{F} , Y′ ∈ 𝔲′^{F} }

where Y_{0} is a fixed rational regular nilpotent element. Then the following hold:

(i) The sets 𝔲_{z} for z ∈ H^{1}(F, Z) are disjoint. (ii) The set of regular nilpotent elements of 𝔤^{F} is the union of the 𝔲_{z}. (iii) 𝔲_{z} = Ad t_{z}(𝔲_{1}) and hence each 𝔲_{z} has the same cardinality. (iv) The 𝔲_{z} are precisely the regular nilpotent orbits in 𝔤^{F}.

Proof. Note that t_{z} is a representative of z in L_{T}^{−1}(Z) according to the isomorphism H^{1}(F, Z) ≅ L_{T}^{−1}(Z)/ZT^{F} of Lemma 10. Also note that the sets 𝔲_{z} are well-defined. For suppose that t_{z} and t′_{z} ∈ L_{T}^{−1}(Z) both represent z ∈ H^{1}(F, Z). Then t′_{z} = t_{z}ah for some a ∈ Z, h ∈ T^{F} and

Ad t′_{z}(Y_{0}) + Y′ = Ad ah(Ad t_{z}(Y_{0}) + Ad(ah)^{−1}(Y′))
= Ad h(Ad t_{z}(Y_{0}) + Ad h^{−1}(Y′))

since Ad a(Y) = Y for a ∈ Z, Y ∈ 𝔤. Since Ad h^{−1}(Y′) ∈ 𝔲′^{F} and h ∈ G^{F} it follows that the definition of 𝔲_{z} does not depend on the choice of representative for z in L_{T}^{−1}(Z).

(i) Suppose that 𝔲_{z} ∩ 𝔲_{z′} ≠ ∅. Then

Ad t_{z}(Y_{0}) + Y′ = Ad g(Ad t_{z′}(Y_{0}) + Y′′)

for some g ∈ G^{F}, Y′, Y′′ ∈ 𝔲′^{F}. Now Ad t_{z}(Y_{0}) + Y′ ∈ 𝔟 and Ad t_{z′}(Y_{0}) + Y′′ ∈ 𝔟. By Theorem 5, Ad t_{z′}(Y_{0}) + Y′′ is regular and hence 𝔟 = Ad g(𝔟) and consequently, by Lemma 2, g ∈ B^{F}. Write g = su with s ∈ T^{F}, u ∈ U^{F}. Since Ad g(Y′′) ∈ 𝔲′^{F} (by Lemma 1), we see that

Ad(s^{−1}t_{z})(Y_{0}) ≡ Ad(ut_{z′})(Y_{0}) mod 𝔲′^{F} .

But ut_{z′} = t_{z′}t_{z′}^{−1}ut_{z′} and

Ad(ut_{z′})(Y_{0}) = Ad t_{z′}(Ad(t_{z′}^{−1}ut_{z′})(Y_{0}))
= Ad t_{z′}(Y_{0} + Ad(t_{z′}^{−1}ut_{z′})(Y_{0}) − Y_{0})
≡ Ad t_{z′}(Y_{0}) mod 𝔲′^{F}

since t_{z′}^{−1}ut_{z′} ∈ U and hence, by Lemma 1, Ad(t_{z′}^{−1}ut_{z′})(Y_{0}) − Y_{0} ∈ 𝔲′. Thus

Ad(s^{−1}t_{z})(Y_{0}) ≡ Ad t_{z′}(Y_{0}) mod 𝔲′^{F}

which implies that

Ad(t_{z′}^{−1}s^{−1}t_{z})(Y_{0}) ≡ Y_{0} mod 𝔲′^{F} .

Since Y_{0} is regular and t_{z′}^{−1}s^{−1}t_{z} ∈ T this implies that α(t_{z′}^{−1}s^{−1}t_{z}) = 1 for all α ∈ Π and hence that t_{z′}^{−1}s^{−1}t_{z} ∈ Z. Since T is abelian, we see that t_{z′}ZT^{F} = t_{z}ZT^{F} and hence z = z′. (ii) By Theorem 5, every regular nilpotent element Y of 𝔤^{F} lies in some Borel subalgebra 𝔟_{1} of G. Since Y is rational and regular 𝔟_{1} = F (𝔟_{1}). Thus, by Lemma 11, there exists g ∈ G^{F} such that Ad g(𝔟_{1}) = 𝔟 and hence Ad g(Y) ∈ 𝔟^{F}. Since Y is nilpotent, this implies that Ad g(Y) ∈ 𝔲^{F} (cf. Borel [1, 10.6(4)]) and we may assume that Y ∈ 𝔲^{F}. Let T be any rational maximal torus contained in B and let Φ and Π be defined as usual. Now write Y = ∑_{α∈Φ^{+}} Y_{α} where Y_{α} ∈ 𝔤_{α}. Then Y is regular if and only if Y_{α} ≠ 0 for all α ∈ Π. Thus, by Humphreys [6, 16.2(C)], there exists a t ∈ T such that Ad t(Y_{0}) ≡ Y mod 𝔲′. Since Y is rational we see that Ad t(Y_{0}) ≡ Ad F (t)(Y_{0}) mod 𝔲′ and hence that Ad(t^{−1}F (t))(Y_{0}) ≡ Y_{0} mod 𝔲′. But this implies that α(t^{−1}F (t)) = 1 for all α ∈ Π and hence that t^{−1}F (t) ∈ Z and t ∈ L_{T}^{−1}(Z). Thus Y ∈ 𝔲_{z} for some z ∈ H^{1}(F, Z). (iii) Since t_{z}^{−1}F (t_{z}) ∈ Z, Ad(t_{z}^{−1}F (t_{z}))(Y′) = Y′ for Y′ ∈ 𝔲′ and thus Y′ ∈ 𝔲′^{F} if and only if Ad t_{z}(Y′) ∈ 𝔲′^{F}. Hence

𝔲_{z} = {Ad g(Ad t_{z}(Y_{0}) + Y′) | g ∈ G^{F} , Y′ ∈ 𝔲′^{F} }
= {Ad g(Ad t_{z}(Y_{0} + Y′)) | g ∈ G^{F} , Y′ ∈ 𝔲′^{F} }
= {Ad t_{z} Ad(t_{z}^{−1}gt_{z})(Y_{0} + Y′) | g ∈ G^{F} , Y′ ∈ 𝔲′^{F} }.

Since t_{z} ∈ L_{T}^{−1}(Z), t_{z}^{−1}F (t_{z}) ∈ Z and hence t_{z}^{−1}gt_{z} ∈ G^{F} if and only if g ∈ G^{F}. Thus

\begin{aligned}𝔲_{z} & = \{Ad t_{z}(Ad g(Y_{0} + Y′)) | g ∈ G^{F} , Y′ ∈ 𝔲′^{F} \} \\ & = Ad t_{z}(𝔲_{1}). \end{aligned}

RATIONAL REGULAR NILPOTENT ELEMENTS OF A REDUCTIVE LIE ALGEBRA 9

(iv) From Proposition 7 we know that there are |H^{1}(F, Z)| regular nilpotent Ad G^{F} - orbits on 𝔤^{F}. On the other hand, by definition, each of the 𝔲_{z} is a union of Ad G^{F} - orbits. Since there are also |H^{1}(F, Z)| distinct sets 𝔲_{z}, these must be precisely the regular nilpotent Ad G^{F} -orbits.

Note. This result is valid for all characteristics. The corresponding result for regular unipotent elements (Digne, Lehrer & Michel [3, 3.2]), however, is only true for good characteristic.

Lemma 13 (cf. Digne & Michel [4, 14.23]). The number of regular nilpotent elements of 𝔤^{F} is

\frac{|G^{F} |}{|Z^{∘F} |q^{l}}

where l is the semi-simple rank of G.

Proof. According to Theorem 5, if Y is regular nilpotent then it is contained in a unique Borel subalgebra and hence the number of rational regular nilpotent elements is equal to the product of the number of rational Borel subalgebras (which, by Lemma 2 and the conjugacy of Borel subgroups, is |G^{F} /B^{F} | for any rational Borel subgroup B) and the number of such elements in a given rational Borel subalgebra. By Borel [1, 10.6(4)] and Theorem 5, if B = T U is any F -stable Levi decomposition with Φ and Π defined as usual then, writing

𝔟 = Lie T ⊕ ⨁_{α∈Φ^{+}} 𝔤_{α},

we see that Y is regular nilpotent in 𝔟 if and only if Y ∈ 𝔲 and the 𝔤_{α} components are nonzero for α a simple root. Hence the number of regular nilpotents in 𝔟^{F} is |𝔲′^{F} | times the number of elements of 𝔲^{F} /𝔲′^{F} for which every component is nontrivial in the decomposition 𝔲/𝔲′ ≅ ⨁_{α∈Π} 𝔤_{α}. We shall call such elements regular also. By Humphreys [6, 16.2(C)] T acts transitively on the set of regular elements of 𝔲/𝔲′ and by Digne & Michel [4, 0.35] the stabilizer of any given regular element is Z. Hence, according to Digne & Michel [4, 3.21], the number of T^{F} -orbits of rational regular elements of 𝔲^{F} /𝔲′^{F} is |H^{1}(F, Z)|. In each orbit however there are |T^{F} /Z^{F} | elements. Hence the number of regular nilpotent elements in 𝔤^{F} is

\frac{|G^{F} |}{|B^{F} |} · |𝔲′^{F} | · |H^{1}(F, Z)| · \frac{|T^{F} |}{|Z^{F} |}.

Now |B^{F} | = |T^{F} ||U^{F} |, |𝔲^{F} /𝔲′^{F} | = q^{l} and according to Corollary 9,

|H^{1}(F, Z)| = \frac{|Z^{F} |}{|Z^{∘F} |}.

Substituting these values in gives the desired number.

Corollary 14. If Y ∈ 𝔤^{F} is regular nilpotent then

|C_{G^{F}} (Y)| = q^{l}|Z^{∘F} ||H^{1}(F, Z)| = q^{l}|Z^{F} |.

Proof. According to Theorem 12, the regular nilpotent Ad G^{F} -orbits are the sets 𝔲_{z} where z ∈ H^{1}(F, Z) and they have the same cardinality. Hence each regular nilpotent Ad G^{F} -orbit has size |G^{F} |/(q^{l}|Z^{∘F} ||H^{1}(F, Z)|). This gives the first equality. The second equality follows from Corollary 9.

References

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University of Sydney Australia 2006