The index of centralizers of elements of reductive Lie algebras

same as arXiv:0904.1778


Introduction
In this note k is an algebraically closed field of characteristic 0.
1.1.Let g be a finite dimensional Lie algebra over k and consider the coadjoint representation of g.By definition, the index of g is the minimal dimension of stabilizers g x , x ∈ g * , for the coadjoint representation: indg := min{dim g x ; x ∈ g * } .
The definition of the index goes back to Dixmier [Di74].It is a very important notion in representation theory and in invariant theory.By Rosenlicht's theorem [Ro63], generic orbits of an arbitrary action of a linear algebraic group on an irreducible algebraic variety are separated by rational invariants; in particular, if g is an algebraic Lie algebra, indg = deg tr k(g * ) g , where k(g * ) g is the field of g-invariant rational functions on g * .The index of a reductive algebra equals its rank.For an arbitrary Lie algebra, computing its index seems to be a wild problem.However, there is a large number of interesting results for several classes of nonreductive subalgebras of reductive Lie algebras.For instance, parabolic subalgebras and their relatives as nilpotent radicals, seaweeds, are considered in [Pa03a], [TY04], [J07].The centralizers, or normalizers of centralizers, of elements form another interesting class of such subalgebras, [E85a], [Pa03a], [Mo06b].The last topic is closely related to the theory of integrable Hamiltonian systems [Bol91].Let us precise this link.
From now on, g is supposed to be reductive.Denote by G the adjoint group of g.The symmetric algebra S(g) carries a natural Poisson structure.By the so-called argument shift method, for x in g * , we can construct a Poisson-commutative family F x in S(g) = k[g * ]; see [MF78] or Remark 1.3.It is generated by the derivatives of all orders in the direction x ∈ g * of all elements of the algebra S(g) g of g-invariants of S(g).Moreover, if G.x denotes the coadjoint orbit of x ∈ g * : Theorem 1.1 ([Bol91], Theorems 2.1 and 3.2).There is a Poisson-commutative family of polynomial functions on g * , constructed by the argument shift method, such that its restriction to G.x contains 1  2 dim (G.x) algebraically independent functions if and only if indg x = ind g.
Motivated by the preceding result of Bolsinov, A.G. Elashvili formulated the conjecture: ind g x = rkg for all x ∈ g * , where rkg is the rank of g.Elashvili's conjecture also appears in the following problem: Is the algebra S(g x ) g x of invariants in S(g x ) under the adjoint action a polynomial algebra?This question was formulated by A. Premet in [PPY07, Conjecture 0.1].After that, O. Yakomiva discovered a counterexample [Y07], but the question remains very interesting.As an example, under certain hypothesis, and under the condition that Elashvili's conjecture holds, the algebra of invariants S(g x ) g x is polynomial in rkg variables, [PPY07, Theorem 0.3].
In this paper, we give a proof of Elashvili's conjecture.Hence we claim: Theorem 1.2.Let g be a reductive Lie algebra.Then ind g x = rkg for all x ∈ g * .
During the last decade, Elashvili's conjecture caught attention of many invariant theorists [Pa03a], [Ch04], [Y06a], [De08].To begin with, describe some easy but useful reductions.Since the g-modules g and g * are isomorphic, it is equivalent to prove Theorem 1.2 for centralizers of elements of g.On the other hand, by a result due to E.B. Vinberg, pointed out in [Pa03a], the inequality indg x ≥ rkg holds for all x ∈ g.So only remains to prove the opposite one.Given x ∈ g, let x = x s + x n be its Jordan decomposition.Then g x = (g xs ) xn .The subalgebra g xs is reductive of rank rkg.Thus, the verification of the Elashvili's conjecture reduces to the case of nilpotent elements.At last, one can clearly restrict itself to the case of simple g.
Review now the main results obtained so far on Elashvili's conjecture.If x is regular, then g x is a commutative Lie algebra of dimension rkg.So, Elashvili's conjecture is obviously true in that case.Further, the conjecture is known for subregular nilpotent elements and nilpotent elements of height 2 and 3, [Pa03a], [Pa03b].Remind that the height of a nilpotent element e is the maximal integer m such that (ad e) m = 0.More recently, O. Yakimova proved the conjecture in the classical case [Y06a].To valid the conjecture in the exceptional type, W.A. DeGraaf used the computer programme GAP (cf.[De08]).Since there are many nilpotent orbits in the Lie algebras of exceptional type, it is difficult to present the results of such computations in a concise way.In 2004, the first author published a case-free proof of Elashvili's conjecture applicable to all simple Lie algebras; see [Ch04].Unfortunately, the argument in [Ch04] has a gap in the final part of the proof, which was pointed out by L. Rybnikov.
Because of the importance of Elashvili's conjecture in invariant theory, it would be very appreciated to find a conceptual proof of Elashvili's conjecture applicable to all finite-dimensional simple Lie algebras.The proof we propose in this paper is fresh and almost general.More precisely, it remains 7 isolated cases; one nilpotent orbit in type E 7 and six nilpotent orbits in type E 8 have to be considered separately.For these 7 orbits, the use of GAP is unfortunately necessary.1.2.Description of the paper.Let us briefly explain our approach.Denote by N(g) the nilpotent cone of g.As noticed previously, it suffices to prove indg e = rkg for all e in N(g).If the equality holds for e, it does for all elements of G.e; we shortly say that G.e satisfies the Elashvili's conjecture.
From a nilpotent orbit O l of a reductive factor l of a parabolic subalgebra of g, we can construct a nilpotent orbit of g having the same codimension in g as O l in l and having other remarkable properties.The nilpotent orbits obtained in such a way are called induced; the other ones are called rigid.Richardson orbits are induced nilpotent orbits from a zero orbit.We refer the reader to Subsection 2.3 for more precisions about this topic.Using Bolsinov's criterion of Theorem 1.1, we first prove Theorem 1.2 for all Richardson nilpotent orbits and then, by an induction process, we show that the conjecture reduces to the case of rigid nilpotent orbits.To deal with rigid nilpotent orbits, we use methods developed in [Ch04] by the first author, and resumed in [Mo06a] by the second author, based on nice properties of Slodowy slices of nilpotent orbits.
In more details, the paper is organized as follows: We state in Section 2 the necessary preliminary results.In particular, we investigate in Subsection 2.2 extensions of Bolsinov's criterion and we establish an important result (Theorem 2.7) which will be used repeatedly in the sequel.We prove in Section 3 the conjecture for all Richardson nilpotent orbits.Starting with Section 4, we develop a reduction method: we introduce in Section 4 a property (P 1 ) given by Definition 4.3 which turns out to be equivalent to Elashvili's conjecture.By using this equivalence and results of Section 2, we show in Section 4 that Elashvili's conjecture reduces to the case of rigid nilpotent orbits.From Section 5, we handle the rigid nilpotent orbits: we introduce and study in Section 5 a property (P 2 ) given by Definition 5.2.Then, in Section 6, we are able to deal with almost all rigid nilpotent orbits.The remaining cases are dealt with set-apart by using a different approach, still in Section 6.

Notations.
• If E is a subset of a vector space V , we denote by span(E) the vector subspace of V generated by E. The grassmanian of all d-dimensional subspaces of V is denoted by Gr d (V ).By cone of V , we mean a subset of V invariant under the natural action of k * := k \ {0} and by a bicone of V × V we mean a subset of V × V invariant under the natural action of k * × k * on V × V .
• From now on, we assume that g is semisimple of rank ℓ and we denote by ., .the Killing form of g.We identify g to g * through ., . .Unless otherwise specified, the notion of orthogonality refers to the bilinear form ., . .
• Denote by S(g) g the algebra of g-invariant elements of S(g).Let f 1 , . . ., f ℓ be homogeneous generators of S(g) g of degree d 1 , . . .,d ℓ respectively.We choose the polynomials f 1 , . . .,f ℓ so that d 1 ≤ • • • ≤d ℓ .For i = 1, . . ., d i and (x, y) ∈ g × g, we may consider a shift of f i in direction y: f i (x + ty) where t ∈ k.Expanding f i (x + ty) as a polynomial in t, we obtain where y → (m!)f (m) i (x, y) is the differential at x of f i at the order m in the direction y.The elements f (m) i as defined by (1) are invariant elements of S(g) ⊗ k S(g) under the diagonal action of G in g × g.Note that f .One says that the family F x is constructed by the argument shift method.
• Let i ∈ {1, . . ., ℓ}.For x in g, we denote by ϕ i (x) the element of g satisfying (df i ) x (y) = f (1) i (x, y) = ϕ i (x), y , for all y in g.Thereby, ϕ i is an invariant element of S(g) ⊗ k g under the canonical action of G.We denote by ϕ (m) i , for 0 ≤ m ≤ d i − 1, the elements of S(g) ⊗ k S(g) ⊗ k g defined by the equality: • For x ∈ g, we denote by g x = {y ∈ g | [y, x] = 0} the centralizer of x in g and by z(g x ) the center of g x .The set of regular elements of g is and we denote by g reg,ss the set of regular semisimple elements of g.Both g reg and g reg,ss are G-invariant open dense subsets of g.
• The nilpotent cone of g is N(g).As a rule, for e ∈ N(g), we choose an sl 2 -triple (e, h, f ) in g given by the Jacobson-Morozov theorem [CMa93, Theorem 3.3.1].In particular, it satisfies the equalities: The action of ad h on g induces a Z-grading: Recall that e, or G.e, is said to be even if g(i) = 0 for odd i.Note that e ∈ g(2), f ∈ g(−2) and that g e , z(g e ) and g f are all ad h-stable.
• All topological terms refer to the Zariski topology.If Y is a subset of a topological space X, we denote by Y the closure of Y in X.
1.4.Acknowledgments.We would like to thank O. Yakimova for her interest and useful discussions; and more particularly for bringing Bolsinov's paper to our attention.We also thank A.G. Elashvili for suggesting Lawther-Testerman's paper [LT08] about the centers of centralizers of nilpotent elements.

Preliminary results
We start in this section by reviewing some facts about the differentials of generators of S(g) g .Then, the goal of Subsection 2.2 is Theorem 2.7.We collect in Subsection 2.3 basics facts about induced nilpotent orbits.
We turn now to the elements ϕ (m) i , for i = 1, . . ., ℓ and 0 ≤ m ≤ d i −1, defined in Subsection 1.3 by (2).Recall that d i is the degree of the homogeneous polynomial f i , for i = 1, . . ., ℓ.The integers d 1 − 1, . . ., d ℓ − 1 are thus the exponents of g.By a classical result [Bou02, Ch.V, §5, Proposition 3], we have d i = b g where b g is the dimension of Borel subalgebras of g.For (x, y) in g × g, we set: The subspaces V x,y will play a central role throughout the note.
Remark 2.2.(1) For (x, y) ∈ g×g, the dimension of V x,y is at most b g since d i = b g .Moreover, for all (x, y) in a nonempty open subset of g × g, the equality holds [Bol91].Actually, in this note, we do not need this observation.
(2) By Lemma 2.1(ii), if x is regular, then g x is contained in V x,y for all y ∈ g.In particular, if so, dim [x, V x,y ] = dim V x,y − ℓ.
The subspaces V x,y were introduced and studied by Bolsinov in [Bol91], motivated by the maximality of Poisson-commutative families in S(g).These subspaces have been recently exploited in [PY08] and [CMo08].The following results are mostly due to Bosinov, [Bol91].We refer to [PY08] for a more recent account about this topic.
(i)[PY08, Lemma A1] The subspace V x,y of g is the sum of the subspaces g x+ty where t runs through any nonempty open subset of k.
(ii)[PY08, Lemma A4] The subspace g y + V x,y is a totally isotropic subspace of g with respect to the Kirillov form Let σ and σ i , for i = 1, . . ., ℓ, be the maps respectively, and denote by σ ′ (x, y) and σ ′ i (x, y) the tangent map at (x, y) of σ and σ i respectively.Then σ ′ i (x, y) is given by the differentials of the f (m) i 's at (x, y) and σ ′ (x, y) is given by the elements σ ′ i (x, y).
Proof.(i) The verifications are easy and left to the reader.
(iii) Suppose that x is regular and suppose that σ ′ (x, y)(v, w ′ ) = 0 for some w ′ ∈ g.Then by (i), v is orthogonal to the elements ϕ 1 (x), . . .,ϕ ℓ (x).So by Lemma 2.1(ii), v is orthogonal to g x .Since g x is the orthogonal complement of [x, g] in g, we deduce that v lies in [x, g].Conversely, since σ(x, y) = σ(g(x), g(y)) for all g in G, the element ([u, x], [u, y]) belongs to the kernel of σ ′ (x, y) for all u ∈ g.So, the converse implication follows.
2.2.On Bolsinov's criterion.Let a be in g and denote by π the map Remark 2.5.Recall that the family (F x ) x∈g constructed by the argument shift method consists of all elements f (m) i (x, .)for i = 1, . . ., ℓ, 1 ≤ m ≤ d i , see Remark 1.3.By definition of the morphism π, there is a family constructed by the argument shift method whose restriction to G.a contains 1 2 dim G.a algebraically independent functions if and only if π has a fiber of dimension 1 2 dim G.a.
In view of Theorem 1.1 and the above remark, we now concentrate on the fibers of π.For (x, y) ∈ g × G.a, denote by F x,y the fiber of π at π(x, y): Lemma 2.6.Let (x, y) be in g × G.a.
(i) The irreducible components of F x,y have dimension at least 1 2 dim G.a. (ii) The fiber F x,y has dimension 1 2 dim G.a if and only if any irreducible component of F x,y contains an element (x, y ′ ) such that (g y ′ + V x,y ′ ) ⊥ has dimension 1 2 dim G.a. Proof.We prove (i) and (ii) all together.The tangent space T x,y ′ (F x,y ) of F x,y at (x, y ′ ) in F x,y identifies to the subspace of elements w of [y ′ , g] such that σ ′ (x, y ′ )(0, w) = 0. Hence, by Lemma 2.4(ii), But by Lemma 2.3(ii), (g y ′ + V x,y ′ ) ⊥ has dimension at least 1 2 dim G.a; so does dim T x,y ′ (F x,y ).This proves (i).Moreover, the equality holds if and only if (g y ′ + V x,y ′ ) ⊥ has dimension 1 2 dim G.a, whence the statement (ii).
Let p be a proper parabolic subalgebra of g and let l be a reductive factor of p.We denote by p u the nilpotent radical of p. Denote by L the connected closed subgroup of G whose Lie algebra is ad l and denote by P the normalizer of p in G.We shall say that e ∈ N(g) is an induced (respectively rigid, Richardson) nilpotent element of g if the G-orbit of e is an induced (respectively, rigid, Richardson) nilpotent orbit of g.Proposition 2.9 ([CMa93], Proposition 7.1.4).If l 1 and l 2 are two Levi subalgebras of g with In other words, Proposition 2.9 says that the induction of orbits is transitive.
Remark 2.10.As a consequence of Proposition 2.9, a nilpotent orbits is always induced, not necessarily in a unique way, from a rigid nilpotent orbit.
We use now classical results about polarizable elements and sheets to obtain interesting properties of Richardson elements useful for Section 4. All results we need to state Theorem 2.11 are contained in [TY05, §33 and §39].For any x in g, we denote by C(x) the G-invariant cone generated by x and we denote by C(x) the closure of C(x) in g.
Theorem 2.11.Let e ∈ N(g) be a Richardson element.There is a semisimple element s of g satisfying the following two conditions: (1) dim G.e = dim G.s; (2) the subset of nilpotent elements contained in C(s) is the closure of G.e. Since s is a nonzero semisimple element, there exists i in {1, . . ., ℓ} such that f i (s) = 0.For any x in G.s, for j = 1, . . ., ℓ. Hence C(s) is contained in the nullvariety in g of the functions: Recall that N(g) is the nullvariety in g of the f j 's.As a consequence, the set In addition, it is a finite union of nilpotent orbits.Hence the subset of nilpotent elements belonging to C(s) is exactly the closure of G.e. Indeed, a nilpotent orbit contained in C(s) different from G.e has dimension strictly smaller than dim G.e as we saw before.In conclusion, s satisfies the desired conditions (1) and (2).
Remark 2.12.The above proof shows that, for semisimple s as in Theorem 2.11,

Proof of Theorem 1.2 for Richardson nilpotent orbits
Let e be a Richardson nilpotent element of g.Our goal is to prove: ind g e = ℓ.Let s be a semisimple element of g verifying the two conditions of Theorem 2.11 and let x * be a regular element of g such that f j (s) = f j (x * ) for all j = 1, . . ., ℓ.Let i ∈ {1, . . ., ℓ} such that f i (s) = 0.
is the union of the nilpotent cone and of the cone generated by the closure of G.x * . Proof.
Then τ is a proper morphism and its fiber are finite.So τ is surjective since D(f i ) * and k * × G.x * have the same dimension, whence the statement.
As an abbreviation, we set: For j = 1, . . ., ℓ, recall that σ j is the map: Let B j be the nullvariety of σ j in g × g and let B be the union of B 1 , . . .,B ℓ ; it is a bicone of g × g.Denote by ρ and τ the canonical maps: Let σ 0 be the restriction to (g × g)\B of σ; it has values in k d× .As σ j (sx, sy) = s d j σ j (x, y) for all (x, y) ∈ g × g and j = 1, . . ., ℓ, the map τ •σ 0 factors through ρ.Denote by σ 0 the map from ρ(g × g\B) to P d making the following diagram commutative: and let Γ be the graph of the restriction to ρ(g × C(s)\B) of σ 0 .
Lemma 3.2.The set Γ is a closed subset of P(g × g) × P d .
Proof.Let Γ be the inverse image of Γ by the map ρ × τ .Then Γ is the intersection of the graph of σ and (g . But for all x in g such that f j (x) = 0 for any j, σ(x, e) belongs to k d× .Thus, the open subset U = Z ∩ k d× of Z is convenient and the proposition follows.
Theorem 3.4.Let e be a Richardson nilpotent element of g.Then the index of g e is equal to ℓ.
Proof.Let s be as in Theorem 2.11.Since s is semisimple, g s is a reductive Lie algebra of rank ℓ.So the index of g s is equal to ℓ. Besides, by Theorem 2.7, (1)⇒(6), applied to s, Hence for all z in a dense subset of σ(g × G.s), the fiber of the restriction of σ to g × G.s at z has minimal dimension Recall that Z is the closure of σ(g × C(s)) in k d .We deduce from the above equality that Z has dimension By Proposition 3.3, there exists an open subset U of Z contained in σ(g × C(s)) having a nonempty intersection with σ(g × G.e).Let i be in {1, . . ., ℓ} such that f i (s) = 0.For z ∈ k d , we write z = (z i,j ) 1≤i≤ℓ 0≤j≤d i its coordinates.Let V i be the nullvariety in U of the coordinate z i,d i .
Then V i is not empty by the choice of U .Since U is irreducible and since z i,d i is not identically zero on U , V i is equidimensional of dimension 1 2 dim G.e + ℓ.By Theorem 2.11 and Remark 2.12, the nullvariety of is an open subset of g × G.e.So σ(g × G.e) has dimension 1 2 dim G.e + ℓ.Then by Theorem 2.7, (6)⇒(1), the index of g e is equal to ℓ.

The property (P 1 ) and reduction to rigid nilpotent orbits
Let p be a proper parabolic subalgebra of g, let l be a Levi factor of p and denote by p u the nilpotent radical of p.We denote by L the connected subgroup of G whose Lie algebra is ad l.We consider a nilpotent orbit L. e of l and we choose an element e of Ind g l (L.e) which belongs to ẽ + p u .Note that rkl = ℓ and that dim g e = dim l e e by Theorem 2.8.It will be shown in this Section that under the assumption indl e e = ℓ, the index of g e is equal to ℓ (Theorem 4.10).This will enable us to reduce the proof of Elashvili's conjecture to the case of rigid nilpotent orbits.4.1.A preliminary result.Let f 1 , . . ., f ℓ be homogeneous generators of S(l) l of degrees d 1 , . . ., d ℓ respectively, with As a rule, we will write down with a tilde all relative notions to the f i 's.For example, we denote by σ the map where b l is the dimension of Borel subalgebras of l.Recall that the morphism σ was defined in Subsection 2.1.For (x, y) ∈ l × l, denote by Γ x,y the fiber at σ(x, y) of σ and denote by Γ x,y the fiber at σ(x, y) of the restriction of σ to l × l.Lemma 4.1.Let (x, y) ∈ l × l.The fiber Γ x,y is a union of irreducible components of Γ x,y and σ is constant on each irreducible component of Γ x,y .
Proof.Let d ≥ max(d ℓ , d ℓ ) and let us choose t 1 , . . .,t d pairwise different elements in k.Denote by κ and κ the maps from l to k ℓ , respectively.For j = 1, . . ., d, we denote by τ j the map l × l → l, (x ′ , y ′ ) → x ′ + t j y ′ and we denote by χ and χ the maps from l × l to k dℓ , j for all (x ′ , y ′ ) in l × l and all j = 1, . . ., d. Conversely, if F is the fiber of χ at χ(x, y), then for all (x ′ , y ′ ) ∈ F , all j = 1, . . ., d and all i = 1, . . ., ℓ, we have: So, for i = 1, . . ., ℓ, the elements f (m) i (x ′ , y ′ ) are solutions of a linear system whose determinant is a Van der Monde determinant defined by some elements among the t j 's.As these elements are pairwise different, this system has a unique solution.As a consequence, σ is constant on F .In conclusion, F = Γ x,y .The same argument shows that Γ x,y is the fiber of χ at χ(x, y).
The subalgebra S(l) l is a finite extension of the subalgebra generated by the restrictions to l of f 1 , . . .,f ℓ since l contains a Cartan subalgebra of g.So there exists a finite morphism ρ : Then ρ d is a finite morphism and χ = ρ d • χ.As a result, Γ x,y is contained in Γ x,y .Moreover, χ is constant on each irreducible component of Γ x,y since χ(Y ) is finite for any irreducible component Y of Γ x,y .This completes the proof of the lemma.Proposition 4.2.Assume that indl e e = ℓ.Then σ(p × {e}) has dimension 1 2 dim G.e + ℓ − dim p u .Proof.By Theorem 2.7, (1)⇒(6), applied to l and e, we deduce from our hypothesis that σ(l×{ e}) has dimension 1 2 dim L. e + ℓ.On the other hand, Lemma 4.1 tells us that the minimal dimension of fibers of the restriction of σ to l × { e} is the minimal dimension of fibers of the restriction of σ to l × { e}.Hence σ(l × { e}) has dimension 1 2 dim L. e + ℓ as well.For (x, y) ∈ l×l and (v, w) ∈ p u ×p u , we have σ(x+v, y+w) = σ(x, y); so σ(p×{e}) = σ(l×{ e}).In conclusion σ(p × {e}) has dimension 1 2 L. e + ℓ.Now, since dim l e e = dim g e , we have and the proposition follows.
4.2.The property (P 1 ).Let p − be the opposite parabolic subalgebra to p. Denote by p u,− the nilpotent radical of p − , by P − the normalizer of p − in G and by P u,− the unipotent radical of P − .We introduce now a property (P 1 ).It will turn out to be equivalent to the equality indg e = ℓ under the assumption ind l e e = ℓ (Corollary 4.7).
Definition 4.3.We say that e has Property (P 1 ) if for all (g, x) in a nonempty open subset of P u,− × p, we have [g(x), V g(x),e ] ⊥ ∩ p u,− = 0.
Remark 4.4.The image of P u,− × p by the map (g, x) → g(x) is dense in g.Hence, e has Property (P 1 ) if and only if for all y in a nonempty open subset of g, we have [y, V y,e ] ⊥ ∩ p u,− = 0.
We start with a preliminary result.For (g, x) ∈ P u,− × p, set: Note that Σ g,x is a closed subset of P u,− containing g.We denote by T g (Σ g,x ) the tangent space of Σ g,x at g.
Proof.Let v be in p u,− .By Lemma 2.4, (ad v)g belongs to T g (Σ x ) if and only if [v, g(x)] is orthogonal to V g(x),e whence the lemma.
Proposition 4.6.If e has Property (P 1 ) then Moreover, if the following equality holds: then e has Property (P 1 ).
By Lemma 2.4, for v in g, v belongs to the tangent space at x of S g,x if and only if g(v) is orthogonal to V g(x),e or, equivalently, if and only if v is orthogonal to V x,g −1 (e) .Since S 1,x is the fiber at σ(x, e) of the map x → σ(x, e) from p to σ(p × {e} we deduce that for all x in a nonempty open subset of p. So, for all (g, x) in a dense open subset of P u,− × p, we have dim (V ⊥ x,g −1 (e) ∩ p) ≤ dim p − dim σ(p × {e}).Hence, there exists a dense open subset U of P u,− × p satisfying the following condition: Suppose now that e has Property (P 1 ) and show dim ψ(P u,− × p) ≥ dim σ(p × {e}) + dim p u .Since e has Property (P 1 ), we can assume furthermore that U satisfies the following condition: (C 2 ) for all (g, x) ∈ U , we have [g(x), V g(x),e ] ⊥ ∩ p u,− = 0.
Let F be an irreducible component of a fiber of the restriction of ψ to U and let τ be the restriction to F of the projection map P u,− × p → p.By (C 2 ) and Lemma 4.5, the fibers of τ are finite.So there exists a dense open subset O of τ (F ) such that the restriction of τ to τ −1 (O) is an étale covering.Therefore, for x ∈ O and for g ∈ P u,− such that (g, x) ∈ τ −1 (O), {g} × S g,x is a neighborhood of (g, x) in F .Hence for all (g, x) ∈ F , F has dimension dim S g,x .Then, by (C 1 ), we get dim ψ(U ) = ψ(P u,− × p) ≥ dim σ(p × {e}) + dim p u .
Suppose now that the image of ψ has dimension dim σ(p × {e}) + p u and prove that e has Property (P 1 ).In this case, we can assume that the dense open subset U of P u,− × p satisfies (C 1 ) and also the following condition: (C 3 ) for all (g, x) in U , the fiber at ψ(g, x) of the restriction ψ 0 of ψ to U has dimension dim p − dim σ(p × {e}).
As a result, for all (g, x) ∈ U , the subset Σ g,x is finite.Let (g, x) be in U and let v be in p u,− such that v is orthogonal to [g(x), V g(x),e ].Then [v, g(x)] is orthogonal to V g(x),e .So by Lemma 4.5, (ad v)g belongs to the tangent space at g of Σ g,x .Hence v = 0.In other words, e has Property (P 1 ).
The following corollary yields the expected equivalence: Corollary 4.7.Assume that ind l e e = ℓ.Then, e has property (P 1 ) if and only if indg e = ℓ.
Proof.By Proposition 4.2, σ(p × {e}) has dimension 1 2 dim G.e + ℓ − dim p u and by Lemma 2.6(i), dim σ(g × {e}) ≤ 1 2 dim G.e + ℓ.Hence by Proposition 4.6, σ(g × {e}) has dimension 1 2 dim G.e + ℓ if and only if e has property (P 1 ).But Theorem 2.7, (1)⇔(6), says that indg e = ℓ if and only if e has property (P 1 ).4.3.Reduction to rigid nilpotent orbits.The aim of this section is Theorem 4.10.Let p, p u , l, e, e be as in Subsections 4.1 and 4.2.Recall that e is a nilpotent element which belongs to e + p u .For x in p and for y in N(l), we set: Since the map (x, y) → V x,y is G-equivariant, s x only depends on G.x.As an abbreviation, we denote by Ω y the intersection Ind g l (L.y) ∩ (L.y + p u ).By Theorem 2.8, remind that Ω y is a P -orbit.
(i) If x ∈ Ω y has Property (P 1 ), then so does any element of Ω y .
(ii) For all (z, x) in a nonempty open subset of g reg × Ω y , [z, V z,x ] ⊥ has dimension s x .
(ii) For x in Ω y , set: . Then there are s ′ x elements η 1 , . . .,η s ′ x among the elements ϕ (m) i 's, such that η 1 (z, x), . . .,η s ′ x (z, x) is a basis of V z,x .By continuity, for all (z ′ , x ′ ) in an open subset U of g reg × Ω y containing (z, x), the elements η 1 (z ′ , x ′ ), . . .,η s ′ x (z ′ , x ′ ) are linearly independent.Hence by maximality of s Let y be in N(l).The integer s x does not depend on the choice of an element x in Ω y .So one is justified to set s y := s x for any x in Ω y .Lemma 4.9.Let y, y ′ ∈ N(l) such that L.y is contained in the closure of L.y ′ in l.If x ∈ Ω y has Property (P 1 ) then any element of Ω y ′ has Property (P 1 ) too.
Proof.Let x be in Ω y and assume that x has Property (P 1 ).Suppose that there exists x ′ ∈ Ω y ′ such that x ′ has not Property (P 1 ).We are going to prove that, for all z in a nonempty open subset of g reg , there exists a subspace E of g of dimension s x ′ , contained in [z, V z,x ] ⊥ and such that E ∩ p u,− = 0. Once proved, this will lead us to a contradiction.Indeed, by Remark 4.4, Property (P 1 ) for x means that, for any z in a nonempty open subset of g reg , we have [z, V y,z ] ⊥ ∩ p u,− = 0.
So, let us prove the above statement.From our assumption and by Lemma 4.8(i), no element of Ω y ′ has Property (P 1 ).Remind that s y ′ = s x ′ for all x ′ ∈ Ω y ′ by definition of s y ′ .By Lemma 4.8(ii), for all (z, x ′ ) in a nonempty open subset U of Let z be in the projection of U to g reg and set: Then U z is a dense open subset of Ω y ′ .Denote by T the subset of elements (E, D) of Gr s y ′ (g) × P(p u,− ) such that D is contained in E. Then T is a projective variety.Set: Denote by T 0 the closure of T 0 in g × T .Since T is a projective variety, the projection of T 0 to g is closed.In particular, it contains the closure of Ω y ′ in g.Since L.y is contained in the closure of L.y ′ in l, there exists (E, D) in T such that (x, E, D) belongs to T 0 .For any i = 1, . . ., ℓ, 0 ≤ m ≤ d i − 1, and for all x ′ in U z , ϕ is the subspace generated by the ϕ (m) i (z, x)'s.So, E is convenient since its intersection with p u,− contains the line D and since the projection of U to g reg is an open subset of g reg .
We are now ready to prove the main result of this section: Theorem 4.10.Assume that ind a x = rka for all reductive subalgebra a strictly contained in g and for all x in a. Then for all induced nilpotent orbit O g in g and for all e in O g , ind g e = ℓ.
Proof.Let l be a Levi subalgebra of g and let O g the induced orbit from a nilpotent orbit L. e of l.From our assumption, indl e e = ℓ.Choose now e in O g ∩ p; we wish to prove ind g e = ℓ.Let O 0 be the induced nilpotent orbit of g from the zero orbit of l.It is a Richardson orbit by definition.Therefore by Theorem 3.4, for all x in O 0 , ind g x = ℓ.So by Corollary 4.7, all elements of Ω 0 have Property (P 1 ).Since 0 belongs to the closure of L.ẽ in l, any element of Ω e e has Property (P 1 ) by Lemma 4.9.So by Corollary 4.7, indg e = ℓ since e belongs to Ω e e by the choice of e.
From that point, our goal is to prove Theorem 1.2 for rigid nilpotent element; Theorem 4.10 tells us that this is enough to complete the proof.

The Slodowy slice and the property (P 2 )
In this section, we introduce a property (P 2 ) in Definition 5.2 and we prove that e ∈ N(g) has Property (P 2 ) if and only if indg e = ℓ.Then, for most rigid nilpotent orbits of g, we will show next section that they do have Property (P 2 ).5.1.Blowing up of S. Let e be a nilpotent element of g and consider an sl 2 -triple (e, h, f ) containing e as in Subsection 1.3.The Slodowy slice is the affine subspace S := e + g f of g which is a transverse variety to the adjoint orbit G.e. Denote by B e (S) the blowing up of S centered at e and let p : B e (S) → S be the canonical morphism.The variety S is smooth and p −1 (e) is a smooth, irreducible, hypersurface of B e (S).The use of the blowing-up B e (S) for the computation of the index was initiated by the first author in [Ch04] and resumed by the second author in [Mo06a].Here, we use again this technique to study the index of g e .Describe first the main tools extracted from [Ch04] we need.
For  is a nonempty set.In addition, α(x) ⊂ g p(x) for all x ∈ Ω. Definition 5.2.We say that e has Property (P 2 ) if z(g e ) ⊂ α(x) for all x in Ω ∩ p −1 (e).
Remark 5.3.Suppose that e is regular.Then g e is a commutative algebra, i.e. z(g e ) = g e .If x ∈ Ω ∩ p −1 (e), then α(x) = g e since p(x) = e is regular in this case.On the other hand, ind g e = dim g e = ℓ since e is regular.So e has Property (P 2 ) and ind g e = ℓ.
We aim to prove that e has Property (P 2 ) if and only if indg e = ℓ.As a consequence of Remark 5.3, we will assume that e is a nonregular nilpotent element of g.

5.
2. On the property (P 2 ).This subsection aims to show: Property (P 2 ) holds for e if and only if ind g e = ℓ.We start by stating that if (P 2 ) holds, then so does the assertion (B) of Theorem 5.1.
Let L g be the S(g)-submodule ϕ ∈ S(g) ⊗ k g satisfying [ϕ(x), x] = 0 for all x in g.It is known that L g is a free module of basis ϕ 1 , . . ., ϕ ℓ , cf. [Di79].We investigate an analogous property for the Slodowy slice S = e + g f .We denote by S reg the intersection of S and g reg .As e is nonregular, the set (S \ S reg ) contains e.
Lemma 5.4.The set S \ S reg has codimension 3 in S and each irreducible component of S \ S reg contains e.

Proof. Let us consider the morphism
By a Slodowy's result [Sl80], this morphism is a smooth morphism.So its fibers are equidimensional of dimension dim g f .In addition, by [V72], g \ g reg is a G-invariant equidimensional closed subset of g of codimension 3. Hence S \ S reg is an equidimensional closed subset of S of codimension 3.
Denoting by t → g(t) the one parameter subgroup of G generated by ad h, S and S\S reg are stable under the action of t −2 g(t) for all t in k * .Furthermore, for all x in S, t −2 g(t)(x) goes to e when t goes to ∞, whence the lemma.

Denote by k[S] the algebra of regular functions on S and denote by L
Lemma 5.5.The module L S is a free module of basis ϕ 1 | S , . . ., ϕ ℓ | S where ϕ i | S is the restriction to S of ϕ i for i = 1, . . ., ℓ.
The following proposition accounts for an important step to interpret Assertion (B) of Theorem 5.1: Proof.Since g x is the orthogonal complement of [x, g] in g, our hypothesis says that ϕ(x) is orthogonal to g x for all x in a nonempty open subset S ′ of S. The intersection S ′ ∩ S reg is not empty; so by Lemma 2.1(ii), ϕ(x), ϕ i | S (x) = 0 for all i = 1, . . ., ℓ and for all x ∈ S ′ ∩ S reg .Therefore, by continuity, ϕ(x), ϕ i | S (x) = 0 for all i = 1, . . ., ℓ and all x ∈ S. Hence ϕ(x) ∈ [x, g] for all x ∈ S reg by Lemma 2.1(ii) again.Consequently by Lemma 5.4, Lemma 5.5 and the proof of the main theorem of [Di79], there exists an element ψ ∈ k[S] ⊗ k g which satisfies the condition of the proposition.
Let u 1 , . . .,u m be a basis of g f and let u * 1 , . . ., u * m be the corresponding coordinate system of S = e + g f .There is an affine open subset Y ⊂ B e (S) with Y ∩ p −1 (e) = ∅ such that k[Y ] is the set of linear combinations of monomials in (u * 1 ) −1 , u * 1 , . . ., u * m whose total degree is nonnegative.In particular, we have a global coordinates system u * 1 , v * 2 , . . ., v * m on Y satisfying the relations: Note that, for x ∈ Y , we so have: So, the image of Y by p is the union of {e} and the complementary in S of the nullvariety of u * 1 .Let Y ′ be an affine open subset of Y contained in Ω and having a nonempty intersection with p −1 (e).Denote by L Y ′ the set of regular maps ϕ from Y ′ to g satisfying [ϕ(x), p(x)] = 0 for all x ∈ Y ′ .Lemma 5.7.Suppose that e has Property (P 2 ).For each z ∈ z(g e ), there exists Proof.Let z be in z(g e ).Since Y ′ ⊂ Ω, for each y ∈ Y ′ , there exists an affine open subset U y of Y ′ containing y and regular maps ν 1 , . . .,ν ℓ from U y to g such that ν 1 (x), . . .,ν ℓ (x) is a basis of α(x) for all x ∈ U y .Let y be in Y ′ .We consider two cases: (1) Suppose p(y) = e.Since e has Property (P 2 ), there exist regular functions a 1 , . . .,a ℓ on U y satisfying for all x ∈ U y ∩ p −1 (e).The intersection U y ∩ p −1 (e) is the set of zeroes of u * 1 in U y .So there exists a regular map ψ from U y to g which satisfies the equality: ] = 0 for all x ∈ U y since α(x) is contained in g p(x) for all x ∈ Ω.
(2) Suppose p(y) = e.Then we can assume that U y ∩ p −1 (e) = ∅ and the map ψ = (u * 1 ) −1 z satisfies the condition: [z − u * 1 (x)ψ(x), p(x)] = 0 for all x ∈ U y .In both cases (1) or (2), we have found a regular map ψ y from U y to g satisfying: [z − (u * 1 ψ y )(x), p(x)] = 0 for all x ∈ U y .Let y 1 , . . .,y k be in Y ′ such that the open subsets U y 1 , . . .,U y k cover Y ′ .For i = 1, . . ., k, we denote by ψ i a regular map from for all i, j.Then there exists a well-defined map ψ z from Y ′ to g whose restriction to U y i is equal to ψ i − ψ i for all i, and such that z − u * 1 ψ z belongs to L Y ′ .Finally, the map ψ z verifies the required property.
Let z be in z(g e ).We denote by ϕ z the regular map from Y to g defined by: Corollary 5.8.Suppose that e has Property (P 2 ) and let z be in z(g e ).There exists Proof.By Lemma 5.7, there exists ], for all x ∈ Y ′ .So the map ψ z is convenient, since u * 1 is not identically zero on Y ′ .The following lemma is easy but helpful for Proposition 5.10: Lemma 5.9.Let v be in g e .Then, v belongs to z(g e ) if and only if [v, g f ] ⊂ [e, g].
Proof.As [x, g] is the orthogonal complement of g x in g for all x ∈ g, we have: But g is the direct sum of g e and [f, g] and [v, g e ] is contained in g e since v ∈ g e .Hence [v, g f ] is contained in [e, g] if and only if v is in z(g e ).
Proposition 5.10.Suppose that e has Property (P 2 ) and let ϕ be in k Proof.Since ϕ is a regular map from Y to g, there is a nonnegative integer d and ϕ and ϕ is a linear combination of monomials in u * 1 , . . ., u * m whose total degree is at least d.By hypothesis on ϕ, we deduce that for all x ∈ S such that u * 1 (x) = 0, ϕ(x) is in [g, x].Hence by Proposition 5.6, there exists ψ in k[S] ⊗ k g such that ϕ(x) = [ ψ(x), x] for all x ∈ S.
For |i| equal to l, the term in of [ ψ(e + τ l+2 v), e + τ l+2 v] is equal to [ψ i(j) , e] + [ψ i , u j ].Since (u * 1 ) d vanishes on the set of k[τ l+2 ]-points of Y whose source is a zero of u * 1 , this term is equal to 0, whence the claim.Recall that Y ′ is an affine open subset of Y contained in Ω and having a nonempty intersection with p −1 (e).
Corollary 5.12.Suppose that e has Property (P 2 ).Let ϕ be in .Since ϕ is a regular map from Y ′ to g, there is m i ≥ 0 such that a m i i ϕ is the restriction to Y ′ of some regular map ϕ i from Y to g.For m i big enough, ϕ i vanishes on Y \ D(a i ); hence ϕ i (x) ∈ [g, p(x)] for all x ∈ Y .So, by Proposition 5.6, there is a regular map ψ i from Y ′ to g such that ϕ i (x) = [ψ i (x), p(x)] for all x ∈ Y ′ .Then for all x ∈ D(a i ), we have ϕ an affine open subset of Y , there exists a regular map ψ from Y ′ to g which satisfies the condition of the corollary.
We are now in position to prove the main result of this section: Theorem 5.13.The equality indg e = ℓ holds if and only if e has Property (P 2 ).
Proof.By Corollary 5.12, if e has Property (P 2 ), then Assertion (B) of Theorem 5.1 is satisfied.Conversely, suppose that indg e = ℓ and show that e has Property (P 2 ).By Theorem 5.1, (A)⇒(B), Assertion (B) is satisfied.We choose an affine open subset Y ′ of Y , contained in Ω, such that Y ′ ∩ p −1 (e) = and verifying the condition of the assertion (B).Let z ∈ z(g e ).Recall that the map ϕ z is defined by (5).Let x be in Y ′ .If u * 1 (x) = 0, then ϕ z (x) belongs to [g, p(x)] by (5).If u * 1 (x) = 0 , then by Lemma 5.9, ϕ z (x) belongs to [e, g].So there exists a regular map ψ from Y ′ to g such that ϕ z (x) = [ψ(x), p(x)] for all x ∈ Y ′ by Assertion (B).Hence we have ] for all x ∈ Y .So α(x) contains z for all x in Ω ∩ Y ′ ∩ p −1 (e).Since p −1 (e) is irreducible, we deduce that e has Property (P 2 ).

5.3.
A new formulation of the property (P 2 ).Recall that Property (P 2 ) is introduced in Definition 5.2.As has been noticed in the proof of Lemma 5.4, the morphism G × S → g, (g, x) → g(x) is smooth.As a consequence, the set S reg of v ∈ S such that v is regular is a nonempty open subset of S. For x in S reg , g e+t(x−e) has dimension ℓ for all t in a nonempty open subset of k since x = e + (x − e) is regular.Furthermore, since k has dimension 1, [Sh94, Ch.VI, Theorem 1] asserts that there is a unique regular map β x : k → Gr ℓ (g) satisfying β x (t) = g e+t(x−e) for all t in a nonempty open subset of k.
Recall that Y is an affine open subset of B e (S) with Y ∩ p −1 (e) = ∅ and that u * 1 , v * 2 , . . ., v * m is a global coordinates system of Y , cf. (4).Let S ′ reg be the subset of x in S reg such that u * 1 (x) = 0.For x in S ′ reg , we denote by x the element of Y whose coordinates are 0, v * 2 (x), . . ., v * m (x).
Lemma 5.14.Let x be in Proof.(i) The map β x is a regular map and [β x (t), e + t(x − e)] = 0 for all t in a nonempty open subset of k.So, β x (0) is contained in g e .
(ii) Since S ′ reg has an empty intersection with the nullvariety of u * 1 in S, the restriction of p to p −1 (S ′ reg ) is an isomorphism from p −1 (S ′ reg ) to S ′ reg .Furthermore, β x (t) = α(p −1 (e + tx − te)) for any t in k such that e + t(x − e) belongs to S ′ reg and p −1 (e + tx − te) goes to x when t goes to 0. Hence β x (0) is equal to α( x) since α and β are regular maps.
Corollary 5.15.The element e has Property (P 2 ) if and only if z(g e ) ⊂ β x (0) for all x in a nonempty subset of S = e + g f .Proof.The map x → x from S ′ reg to Y is well-defined and its image is an open subset of Y ∩p −1 (e).Let S ′′ reg be the set of x ∈ S ′ reg such that x ∈ Ω and let Y ′′ be the image of S ′′ reg by the map x → x.

Then S ′′
reg is open in S reg and Y ′′ is dense in Ω ∩ p −1 (e) since p −1 (e) is irreducible.Since α is regular, e has property (P 2 ) if and only if α(x) contains z(g e ) for all x in Y ′′ .By Lemma 5.14(ii), the latter property is equivalent to the fact that β x (0) contains z(g e ) for all x in S ′′ reg .
(ii) If z(g e ) has dimension 1, then e has Property (P 2 ).
(ii) is an immediate consequence of (i) since ϕ 1 (e) = e by our choice of d 1 .
Remark 5.17.When g is simple of classical type, z(g e ) is generated by ϕ 1 (e), . . .,ϕ ℓ (e) means that z(g e ) is generated by powers of e.For example, this situation always occurs when g has type A or C and occurs in most cases when g has type B or D (cf.[Y06b] and [Mo06c, Théorème 1.1.8]).

Proof of Theorem 1.2 for rigid nilpotent orbits
We intend to prove in this section the following theorem: Theorem 6.1.Suppose that g is reductive and let e be a rigid nilpotent element of g.Then the index of g e is equal to ℓ.
Theorem 6.1 will complete the proof of Theorem 1.2 by Theorem 4.10.As explained in introduction, we can assume that g is simple.We consider two cases, according to g has classical type or exceptional type.
6.1.The classical case.We consider here the classical case.Theorem 6.2.Assume that g is simple of classical type and let e be a rigid nilpotent element.Then z(g e ) is generated by powers of e.In particular, the index of g e is equal to ℓ.
Proof.The second assertion results from Remark 5.17, Corollary 5.16(i) and Theorem 5.13.Furthermore, by Remark 5.17, we can assume that g has type B or D.
Set n := 2ℓ + 1 if g has type B ℓ and n := 2ℓ if g has type D ℓ .Denote by (n 1 , . . .,n k ) the partition of n corresponding to the nilpotent element e.By [Mo06c, Théorème 1.1.8],z(g e ) is not generated by powers of e if and only if n 1 and n 2 are both odd integers and n 3 < n 2 .On the other hand, as e is rigid, n k is equal to 1, n i ≤ n i+1 ≤ n i + 1 and all odd integers of the partition (n 1 , . . .,n k ) have a multiplicity different from 2; cf.[CMa93, Corollary 7.3.5].Hence, the preceding criterion is not satisfied for e; so z(g e ) is generated by powers of e. Remark 6.3.Yakimova's proof of Elashvili's conjecture in the classical case is shorter and more elementary [Y06a].The results of Section 5 will serve the exceptional case in a more relevant way.
6.2.The exceptional case.We let in this subsection g be simple of exceptional type and we assume that e is a nonzero rigid nilpotent element of g.The dimension of the center of centralizers of nilpotent elements has been recently described in [LT08, Theorem 4].On the other hand, we have explicit computations for the rigid nilpotent orbits in the exceptional types due to A.G. Elashvili.These computations are collected in [Sp82, Appendix of Chap.II] and a complete version was published later in [E85b].From all this, we observe that the center of g e has dimension 1 in most cases.In more details, we have: Proposition 6.4.Let e be nonzero rigid nilpotent element of g.
(i) Suppose that g has type G 2 , F 4 or E 6 .Then dim z(g e ) = 1.
(ii) Suppose that g has type E 7 .If g e has dimension 41, then dim z(g e ) = 2; otherwise dim z(g e ) = 1.
(iii) Suppose that g has type E 8 .If g e has dimension 112, 84, 76, or 46, then dim z(g e ) = 2, if g e has dimension 72, then dim z(g e ) = 3; otherwise dim z(g e ) = 1.By Corollary 5.16(ii), indg e = ℓ whenever dim z(g e ) = 1.So, as an immediate consequence of Proposition 6.4, we obtain: Corollary 6.5.Suppose that either g has type G 2 , F 4 , E 6 , or g has type E 7 and dim g e = 41, or g has type E 8 and dim g e ∈ {112, 84, 76, 72, 46}.Then dim z(g e ) = 1 and the index of g e is equal to ℓ.
According to Corollary 6.5, it remains 7 cases; there are indeed two rigid nilpotent orbits of dimension 46 in E 8 .We handle now these remaining cases.We process here in a different way; we will study technical conditions on g e under which indg e = ℓ.For the moment, we state general results about the index.
Let a be an algebraic Lie algebra.Recall that the stabilizer of ξ ∈ a * for the coadjoint representation is denoted by a ξ and that ξ is regular if dim a ξ = inda.Choose a commutative subalgebra t of a consisted of semisimple elements of a and denote by z a (t) the centralizer of t in a. Then a = z a (t) ⊕ [t, a].The dual z a (t) * of z a (t) identifies to the orthogonal complement of [t, a] in a * .Thus, ξ ∈ z a (t) * if and only if t is contained in a ξ .Lemma 6.6.Suppose that there exists ξ in z a (t) * such that dim (a ξ ∩ [t, a]) ≤ 2. Then inda ≤ indz a (t) + 1.
Proof.Let T be the closure in z a (t) * × Gr 3 ([t, a]) of the subset of elements (η, E) such that η is a regular element of z a (t) * and E is contained in a η .The image T 1 of T by the projection from z a (t) * × Gr 3 ([t, a]) to z a (t) * is closed in z a (t) * .By hypothesis, T 1 is not equal to z a (t) * since for all η in T 1 , dim (a η ∩ [t, a]) ≥ 3. Hence there exists a regular element ξ 0 in z a (t) * such that dim (a ξ 0 ∩ [t, a]) ≤ 2. Since t is contained in a ξ 0 , If [t, a] ∩ a ξ 0 = {0} then inda is at most indz a (t).Otherwise, a ξ 0 is not a commutative subalgebra since t is contained in a ξ 0 .Hence ξ 0 is not a regular element of a * , so ind a < dim a ξ 0 .Since dim a ξ 0 ≤ indz a (t) + 2, the lemma follows.
From now on, we assume that a = g e .As a rigid nilpotent element of g, e is a nondistinguished nilpotent element.So we can choose a nonzero commutative subalgebra t of g e consisted of semisimple elements.Denote by l the centralizer of t in g.As a Levi subalgebra of g, l is reductive Lie algebra whose rank is ℓ.Moreover its dimension is strictly smaller that dim g.In the preceding notations, we have z g e (t) = z g (t) e = l e .Let t 1 be a commutative subalgebra of l e containing t and consisted of semisimple elements of l.Then [t, g e ] is stable under the adjoint action of t 1 .For λ in t * 1 , denote by g e λ the λ-weight space of the adjoint action of t 1 in g e .Lemma 6.7.Let λ ∈ t * 1 be a nonzero weight of the adjoint action of t 1 in g e .Then −λ is also a weight for this action and λ and −λ have same multiplicity.Moreover, g e λ is contained in [t, g e ] if and only if the restriction of λ to t is not identically zero.
Proof.By definition, g e λ ∩ l e = {0} if and only if the restriction of λ to t is not identically zero.So g e λ is contained in [t, g e ] if and only if the restriction of λ to t is not equal to 0 since g e λ = (g e λ ∩ l e ) ⊕ (g e λ ∩ [t, g e ]).The subalgebra t 1 is contained in a reductive factor of g e .So we can choose h and f such that t 1 is contained in g e ∩ g f .As a consequence, any weight of the adjoint action of t 1 in g f is a weight of the adjoint action of t 1 in g e with the same multiplicity.Furthermore, the t 1 -module g f for the ajoint action is isomorphic to the t 1 -module (g e ) * for the coadjoint action.So −λ is a weight of the adjoint action of t 1 in g f with the same multiplicity as λ.Hence −λ is a weight of the adjoint action of t 1 in g e with the same multiplicity as λ, whence the lemma.
(ii) Suppose that g has type E 8 and that g e has dimension 84, 76, or 46.Then, for suitable choices of t and t 1 , Condition (2) of Proposition 6.4 is satisfied.
6.3.Proof of Theorem 1.2.We are now in position to complete the proof of Theorem 1.2: Proof of Theorem 1.2.We argue by induction on the dimension of g.If g has dimension 3, the statement is known.Assume now that indl e ′ = rkl for any reductive Lie algebras l of dimension at most dim g − 1 and any e ′ ∈ N(l).Let e ∈ N(g) be a nilpotent element of g.By Theorem 4.10 and Theorem 6.2, we can assume that e is rigid and that g is simple of exceptional type.Furthermore by Corollary 6.5, we can assume that dim z(g e ) > 1.Then we consider the different cases given by Proposition 6.9.
If, either g has type E 7 and dim g e = 41, or g has type E 8 and dim g equals 112, 72, or 46, then Condition (1) of Proposition 6.8 applies for suitable choices of t and t 1 by Proposition 6.9.Moreover, if l = z g (t), then l is a reductive Lie algebra of rank ℓ and strictly contained in g.So, from our induction hypothesis, we deduce that indg e = ℓ by Proposition 6.8.
If g has type E 8 and dim g e equals 84, 76, or 46, then Condition (2) of Proposition 6.8 applies for suitable choices of t and t 1 by Proposition 6.9.Arguing as above, we deduce that indg e = ℓ.In conclusion, Condition (1) of Proposition 6.8 is satisfied for t := kt and t 1 :=span(Bt1).
(2) E 8 , dim g e = 84: In this case, dim l e = 48 and dim t 1 = 3.The matrix A(7) has order 5 and it is singular of rank 4. The order of the other matrices is at most 2.
Theorem 2.8 ([CMa93],Theorem 7.1.1).Let O l be a nilpotent orbit of l.There exists a unique nilpotent orbit O g in g whose intersection with O l +p u is a dense open subset of O l +p u .Moreover, the intersection of O g and O l + p u consists of a single P -orbit and codim g (O g ) = codim l (O l ).The orbit O g only depends on l and not on the choice of a parabolic subalgebra p containing it [CMa93, Theorem 7.1.3].By definition, the orbit O g is called the induced orbit from O l ; it is denoted by Ind g l (O l ).If O l = 0, then we call O g a Richardson orbit.For example all even nilpotent orbits are Richardson [CMa93, Corollary 7.1.7].In turn, not all nilpotent orbits are induced from another one.A nilpotent orbit which is not induced in a proper way from another one is called rigid.
Proof.Remind that a sheet of g is by definition an irreducible component of a set {x ∈ g | dim G.x = m}, for some m ≥ 0. Dixmier sheets are those which contain semisimple elements.By [TY05, Theorem 39.4.8], the set of polarizable elements of g are the union of the Dixmier sheets of g.As e is Richardson, it is polarizable [TY05, Theorem 33.5.6].Hence, there is a Dixmier sheet S of g containing e. Let s be a semisimple elements of g belonging to S. Then C(s) is contained in S and by [TY05, Proposition 33.5.4],there is a nilpotent element e ′ in C(s) satisfying dim G.e ′ = dim G.s.Moreover, e ′ belongs to S and by [TY05, Proposition 35.3.5], the set of nilpotent elements belonging to S is precisely G.e ′ , whence G.e = G.e ′ .So, G.e is the subset of nilpotent elements of C(s) whose orbit has dimension dim G.e.