V Chernousov
email: chernous@math.ualberta.ca

V Egorov

P Gille
email: gille@ens.fr

A Pianzola
email: a.pianzola@math.ualberta.ca

A COHOMOLOGICAL PROOF OF PETERSON-KAC'S THEOREM ON CONJUGACY OF CARTAN SUBALGEBRAS FOR AFFINE KAC-MOODY LIE ALGEBRAS

Keywords: Affine Kac-Moody Lie algebra, Conjugacy, Reductive group scheme, Torsor, Laurent polynomials, Non-abelian cohomology. MSC 2000 17B67, 11E72, 14L30, 14E20

This paper deals with the problem of conjugacy of Cartan subalgebras for affine Kac-Moody Lie algebras. Unlike the methods used by Peterson and Kac, our approach is entirely cohomological and geometric. It is deeply rooted on the theory of reductive group schemes developed by Demazure and Grothendieck, and on the work of J. Tits on buildings.

Introduction

Chevalley's theorem on the conjugacy of split Cartan subalgebras is one of the cornerstones of the theory of simple finite dimensional Lie algebras over a field of characteristic 0. Indeed, this theorem affords the most elegant proof that the root system is an invariant of the Lie algebra.

The analogous result for symmetrizable Kac-Moody Lie algebras is the celebrated theorem of Peterson and Kac [PK] (see also [Kmr] and [MP] for detailed proofs). Beyond the finite dimensional case, by far the most important Kac-Moody Lie algebras are the affine ones. These algebras sit at the "border" of finite dimensional Lie theory and they can in fact be viewed as "finite dimensional" (not over the base field but over a Laurent polynomial ring) in the sense of [START_REF]Séminaire de Géométrie algébrique de l[END_REF]. This approach begs the question as to whether an SGA-inspired proof of conjugacy exists in the affine case. This paper, which builds in [CGP] and [GP], shows that the answer is yes. More precisely, in [P1] (the untwisted case) and [CGP] (general case) conjugacy is established for loop algebras by purely Galois cohomological methods. The step that is missing is extending this result to the "full" Kac-Moody Lie algebra. The central extension presents of course no difficulties, but the introduction of the derivation does. The present paper addresses this issue thus yielding a new cohomological proof of the conjugacy theorem of Peterson and Kac in the case of affine Kac-Moody Lie algebras.

V. Chernousov was partially supported by the Canada Research Chairs Program and an NSERC research grant.

A. Pianzola wishes to thank NSERC and CONICET for their continuous support.

Affine Kac-Moody Lie algebras

Split case. Let g be a split simple finite dimensional Lie algebra over an algebraically closed field k of characteristic 0 and let Aut(g) be its automorphism group. If x, y ∈ g we denote their product in g by [x, y]. We also let R = k[t ±1 ], and L(g) = g ⊗ k R. We still denote the Lie product in L(g) by [x, y] where x, y ∈ L(g).

The main object under consideration in this paper is the affine (split or twisted) Kac-Moody Lie algebra L corresponding to g. Any split affine Kac-Moody Lie algebra is of the form (see [Kac])

L = g ⊗ k R ⊕ k c ⊕ k d.
The element c is central and d is a degree derivation for a natural grading of L(g): if x ∈ g and p ∈ Z then [d, x ⊗ t p ] L = p x ⊗ t p .

If l 1 = x ⊗ t p , l 2 = y ⊗ t q ∈ L(g) are viewed as elements in L their Lie product is given by [x ⊗ t p , y ⊗ t q ] L = [x, y] ⊗ t p+q + p x, y δ 0,p+q • c where x, y is the Killing form on g and δ 0,p+q is Kronecker's delta.

Twisted case. Let m be a positive integer. Let S = k[t ±1 m ] be the ring of Laurent polynomials in the variable s = t 1 m with coefficients in k. Let L(g) S = L(g) ⊗ R S be the Lie algebra obtained from the R-Lie algebra L(g) by the base change R → S. Similarly we define Lie algebras L(g) S = L(g) S ⊕ kc and L(g) S = L(g) S ⊕ kc ⊕ kd. 1 Fix a primitive root of unity ζ ∈ k of degree m. The R-automorphism ζ × : S → S given by s → ζs generates the Galois group Γ = Gal(S/R) which we may identify with the abstract group Z/mZ by means of ζ × . Note that Γ acts naturally on Aut(g)(S) = Aut S-Lie (L(g) S ) and on L(g) S = L(g) ⊗ R S through the second factor.

Next, let σ be an automorphism of g of order m. This gives rise to an S-automorphism of L(g) S via x ⊗ s → σ(x) ⊗ s for x ∈ g, s ∈ S. It then easily follows that the assignment

1 → z 1 = σ -1 ∈ Aut S-Lie (L(g) S )
gives rise to a cocycle z = (z i ) ∈ Z 1 (Γ, Aut S-Lie (L(g) S )). This cocycle, in turn, gives rise to a twisted action of Γ on L(g) S . Applying Galois descent formalism we then obtain the Γ-invariant subalgebra L(g, σ) := (L(g) S ) Γ = (L(g) ⊗ R S) Γ . This is a "simple Lie algebra over R" in the sense of [START_REF]Séminaire de Géométrie algébrique de l[END_REF], which is a twisted form of the "split simple" R-Lie algebra L(g) = g ⊗ k R. Indeed S/R is an étale extension and from properties of Galois descent we have

L(g, σ) ⊗ R S ≃ L(g) S = (g ⊗ k R) ⊗ R S.
Note that L(g, id) = L(g).

For i ∈ Z/mZ, consider the eigenspace

g i = {x ∈ g : σ(x) = ζ i x}.
Simple computations show that

L(g, σ) = i∈Z g i ⊗ k[t ±1 ]s i . Let L(g, σ) := L(g, σ) ⊕ kc and L(g, σ) := L(g, σ) ⊕ kc ⊕ kd.
We give L(g, σ) a Lie algebra structure such that c is central element, d is the degree derivation, i.e. if x ∈ g i and p ∈ Z then

(2.0.1) [d, x ⊗ t p m ] := px ⊗ t p m and if y ⊗ t q m ∈ L(g, σ) we get [x ⊗ t p m , y ⊗ t q m ] L(g,σ) = [x, y] ⊗ t p+q m + p x, y δ 0,p+q • c,
where, as before, x, y is the Killing form on g and δ 0,p+q is Kronecker's delta.

2.1. Remark. Note that the Lie algebra structure on L(g, σ) is induced by that of on L(g) S if we view L(g, σ) as a subset of L(g) S .

2.2. Remark. Let σ be an automorphism of L(g) S such that σ| L(g

) S = σ, σ(c) = c, σ(d) = d. Then L(g, σ) = ( L(g) S ) σ .
Realization Theorem. (a) The Lie algebra L(g, σ) is an affine Kac-Moody Lie algebra, and every affine Kac-Moody Lie algebra is isomorphic to some L(g, σ).

(b) L(g, σ) ≃ L(g, σ ′
) where σ ′ is a diagram automorphism with respect to some Cartan subalgebra of g.

Proof.

See [START_REF] Kac | Infinite dimensional Lie algebras[END_REF]Theorems 7.4,8.3 and 8.5].

Let φ ∈ Aut k-Lie ( L(g) S ). Since L(g) S is the derived subalgebra of L(g) S the restriction φ| L(g) S induces a k-Lie automorphism of L(g) S . Furthermore, passing to the quotient L(g) S /kc ≃ L(g) S the automorphism φ| L(g) S induces an automorphism of L(g) S . This yields a well-defined morphism

Aut k-Lie ( L(g) S ) → Aut k-Lie (L(g) S ).
Similar considerations apply to Aut k-Lie ( L(g, σ)). The aim of the next few sections is to show that these two morphisms are surjective.

S-automorphisms of L(g) S

In this section we construct a "simple" system of generators of the automorphism group Aut(g)(S) = Aut S-Lie (L(g) S ) which can be easily extended to k-automorphisms of L(g) S . We produce our list of generators based on a well-known fact that the group in question is generated by S-points of the corresponding split simple adjoint algebraic group and automorphisms of the corresponding Dynkin diagram.

More precisely, let G be the split simple simply connected group over k corresponding to g and let G be the corresponding adjoint group. Choose a maximal split k-torus T ⊂ G and denote its image in G by T. The Lie algebra of T is a Cartan subalgebra h ⊂ g. We fix a Borel subgroup T ⊂ B ⊂ G.

Let Σ = Σ(G, T) be the root system of G relative to T. The Borel subgroup B determines an ordering of Σ, hence the system of simple roots Π = {α 1 , . . . , α n }. Fix a Chevalley basis [START_REF] Steinberg | Lectures on Chevalley Groups[END_REF] {H α 1 , . . . H αn , X α , α ∈ Σ} of g corresponding to the pair (T, B). This basis is unique up to signs and automorphisms of g which preserve B and T (see [START_REF] Steinberg | Lectures on Chevalley Groups[END_REF]§1,Remark 1]).

Since S is a Euclidean ring, by Steinberg [St62] the group G(S) is generated by the so-called root subgroups U α = x α (u) | u ∈ S , where α ∈ Σ and (3.0.1)

x α (u) = exp(uX α ) = ∞ n=0 u n X n α /n!
We recall also that by [START_REF] Steinberg | Lectures on Chevalley Groups[END_REF]§10,Cor. (b) after Theorem 29], every automorphism σ of the Dynkin diagram Dyn(G) of G can be extended to an automorphism of G (and hence of G) and g, still denoted by σ, which takes

x α (u) -→ x σ(α) (ε α u) and X α -→ ε α X σ(α) . Here ε α = ±1 and if α ∈ Π then ε α = 1. Thus we have a natural embedding Aut(Dyn(G)) ֒→ Aut S-Lie (L(g) S ).
The group G(S) acts by S-automorphisms on L(g) S through the adjoint representations ad : G → GL(L(g) S ) and hence we also have a canonical embedding G(S) ֒→ Aut S-Lie (L(g) S ). As we said before, it is well-known (see [P2] for example) that

Aut S-Lie (L(g) S ) = G (S) ⋊ Aut(Dyn(G)).
For later use we need one more fact.

3.1. Proposition. Let f : G → G be the canonical morphism. The group G(S) is generated by the root subgroups f (U α ), α ∈ Σ, and T(S).

Proof. Let Z ⊂ G be the center of G. The exact sequence 1 -→ Z -→ G -→ G -→ 1
gives rise to an exact sequence in Galois cohomology

f (G(S)) ֒→ G(S) -→ Ker [H 1 (S, Z) → H 1 (S, G)] -→ 1. Since H 1 (S, Z) → H 1 (S, G) factors through H 1 (S, Z) -→ H 1 (S, T) -→ H 1 (S, G)
and since H 1 (S, T) = 1 (because Pic S = 1) we obtain

(3.1.1) f (G(S)) ֒→ G(S) -→ H 1 (S, Z) -→ 1.
Similar considerations applied to

1 -→ Z -→ T -→ T -→ 1 show that (3.1.2) f (T(S)) ֒→ T(S) -→ H 1 (S, Z) -→ 1.
The result now follows from (3.1.1) and (3.1.2).

Corollary. One has

Aut S-Lie (L(g) S ) = Aut(Dyn(G)), U α , α ∈ Σ, T(S) .

k-automorphisms of L(g) S

We keep the above notation. Recall that for any algebra A over a field k the centroid of A is

Ctrd (A) = {χ ∈ End k (A) |χ(a • b) = a • χ(b) = χ(a) • b for all a, b ∈ A }.
It is easy to check that if χ 1 , χ 2 ∈ Ctrd(A) then both linear operators χ 1 •χ 2 and χ 1 + χ 2 are contained in Ctrd (A) as well. Thus, Ctrd (A) is a unital associative subalgebra of End k (A). It is also well-known that the centroid is commutative whenever A is perfect. Example. Consider the k-Lie algebra A = L(g) S . For any s ∈ S the linear k-operator χ s : L(g) S → L(g) S given by x → xs satisfy

χ s ([x, y]) = [x, χ s (y)] = [χ s (x), y],
hence χ s ∈ Ctrd (L(g) S ). Conversely, it is known (see [START_REF] Allison | Covering Algebras II: Loop algebras of Kac-Moody Lie algebras[END_REF]Lemma 4.2]) that every element in Ctrd (L(g) S ) is of the form χ s . Thus,

Ctrd (L(g) S ) = { χ s | s ∈ S } ≃ S. 4.1. Proposition. ([P2, Proposition 1]) One has Aut k-Lie (L(g) S ) ≃ Aut S-Lie (L(g) S ) ⋊ Aut k (Ctrd (L(g) S )) ≃ Aut S-Lie (L(g) S ) ⋊ Aut k (S).

Corollary. One has

Aut k-Lie (L(g) S ) = Aut k (S), Aut(Dyn(G)), U α , α ∈ Σ, T(S) .

Proof. This follows from Corollary 3.2 and Proposition 4.1.

Automorphisms of L(g) S

We remind the reader that the centre of L(g) S is the k-span of c and that L(g) S = L(g) S ⊕ kc. Since any automorphism φ of L(g) S takes the centre into itself we have a natural (projection) mapping µ : L(g) S → L(g) S /kc ≃ L(g) S which induces the mapping

λ : Aut k-Lie ( L(g) S ) → Aut k-Lie (L(g) S )
given by φ → φ ′ where φ ′ (x) = µ(φ(x)) for all x ∈ L(g) S . In the last formula we view x as an element of L(g) S through the embedding L(g) S ֒→ L(g) S .

5.1. Remark. It is straightforward to check that φ ′ is indeed an automorphism of L(g) S .

5.2. Proposition. The mapping λ is an isomorphism.

Proof. See [START_REF] Pianzola | Automorphisms of toroidal Lie algebras and their central quotients[END_REF]Proposition 4].

In what follows if φ ∈ Aut k-Lie (L(g) S ) we denote its (unique) lifting to Aut k-Lie ( L(g) S ) by φ. 5.3. Remark. For later use we need an explicit formula for lifts of automorphisms of L(g) S induced by some "special" points in T(S) (those which are not in the image of T(S) → T(S)). More precisely, choose the decomposition T ≃ G m,S × • • • × G m,S such that the canonical embedding G m,S → T into the i-th factor is the cocharacter of T dual to α i . As usual, we have the decomposition T(S) ≃ T(k) × Hom (G m , T). The second factor in the last decomposition is the cocharacter lattice of T and its elements correspond (under the adjoint action) to the subgroup in Aut S-Lie (L(g) S ) isomorphic to Hom(Q, Z) where Q is the corresponding root lattice: if φ ∈ Hom(Q, Z) it induces an S-automorphism of L(g) S (still denoted by φ) given by

X α → X α ⊗ s φ(α) , H α i → H α i .
It is straightforward to check the mapping φ : L(g) S → L(g) S given by

H α → H α + φ(α) X α , X -α • c, H α ⊗ s p → H α ⊗ s p if p = 0 and X α ⊗ s p → X α ⊗ s p+φ(α)
is an automorphism of L(g) S , hence it is the (unique) lift of φ.

Automorphisms of split affine Kac-Moody Lie algebras

Since L(g) S = [ L(g) S , L(g) S ] we have a natural (restriction) mapping τ : Aut k-Lie ( L(g) S ) → Aut k-Lie ( L(g) S ).

6.1. Proposition. The mapping τ is surjective.

Proof. By Proposition 5.2 and Corollary 4.2 the group Aut k-Lie ( L(g) S ) has the distinguished system of generators { φ } where

φ ∈ Aut(Dyn(G)), T(S), Aut k (S), U α .
We want to construct a mapping φ :

L(g) S → L(g) S which preserves the identity [d, x ⊗ t p m ] L = p x ⊗ t p m
for all x ∈ g and whose restriction to L(g) S coincides with φ. These two properties would imply that φ is an automorphism of L(g) S lifting φ. If φ ∈ U α is unipotent we define φ, as usual, through the exponential map

. If φ ∈ Aut(Dyn(G)) we put φ(d) = d. If φ is as in Remark 5.3 we extend it by d → d -X where X ∈ h is the unique element such that [X, X α ] = φ(α)X α for all roots α ∈ Σ. Note that automorphisms of L(g) S
given by points in T(k) are in the image of T(k) → T(k) and hence they are generated by unipotent elements. Lastly, if φ ∈ Aut k (S) is of the form s → as -1 where a ∈ k × (resp. s → as) we extend φ by φ(d) = -d (resp. φ(d) = d). We leave it to the reader to verify that in all cases φ preserves the above identity and hence φ is an automorphism of L(g) S .

6.2. Proposition. One has Ker τ ≃ V where V = Hom k (kd, kc).

Proof. We first embed V ֒→ Aut k-Lie ( L(g) S ). Let v ∈ V . Recall that any element x ∈ L(g) S can be written uniquely in the form x = x ′ + ad where x ′ ∈ L(g) S and a ∈ k. We define v : L(g) S → L(g) S by x → x + v(ad). One checks that v is an automorphism of L(g) S and thus the required embedding is given by v

→ v. Since v(x ′ ) = x ′ for all x ′ ∈ L we have v ∈ Ker τ . Conversely, let ψ ∈ Ker τ . Then ψ(x) = x for all x ∈ L(g) S . We need to show that ψ(d) = ac+d where a ∈ k. Let ψ(d) = x ′ + ac + bd where a, b ∈ k and x ′ ∈ L(g) S . Since [d, X α ] L(g) S = 0 we get [ψ(d), ψ(X α )] L(g) S = 0. Substituting ψ(d) = x ′ + ac + bd we obtain [x ′ + ac + bd, X α ] L(g) S = 0 or [x ′ , X α ] L(g) S = 0.
Since this is true for all roots α ∈ Σ, the element x ′ commutes with g and this can happen if and only if x ′ = 0.

It remains to show that b = 1. To see this we can argue similarly by considering the equality

[d, X α ⊗ t 1 m ] L(g) S = X α ⊗ t 1 m
and applying ψ.

6.3. Corollary. The sequence of groups

(6.3.1) 1 -→ V -→ Aut k-Lie ( L(g) S ) λ•τ -→ Aut k-Lie (L(g) S ) -→ 1 is exact.

Automorphism group of twisted affine Kac-Moody Lie algebras

We keep the notation introduced in § 2. In particular, we fix an integer m and a primitive root of unity

ζ = ζ m ∈ k of degree m. Consider the k-automorphism ζ × : S → S such that s → ζs which we view as a k- automorphism of L(g) S through the embedding Aut k (S) ֒→ Aut k-Lie (L(g) S ) ≃ Aut S-Lie (L(g) S ) ⋊ Aut k (S) (see Proposition 4.1). As it is explained in § 6 we then get the automorphism ζ × (resp. ζ × ) of L(g) S (resp. L(g) S ) given by x ⊗ s i + ac + bd -→ x ⊗ ζ i s i + ac + bd where a, b ∈ k and x ∈ g.
Consider now the abstract group Γ = Z/mZ (which can be identified with Gal (S/R) as already explained) and define its action on L(g) S (resp. L(g) S , L(g) S ) with the use of ζ × (resp. ζ × , ζ × ). More precisely, for every l ∈ L(g) S we let i(l) := ( ζ × ) i (l). Similarly, we define the action of Γ on Aut k-Lie ( L(g) S ) by

i : Aut k-Lie ( L(g) S ) -→ Aut k-Lie ( L(g) S ), x → ( ζ × ) i x( ζ × ) -i .
Therefore, Aut k-Lie ( L(g) S ) can be viewed as a Γ-set. Along the same lines one defines the action of Γ on Aut k-Lie (L(g) S ) and Aut S-Lie (L(g) S ) with the use of ζ × . It is easy to see that Γ acts trivially on the subgroup V ⊂ Aut k-Lie ( L(g) S ) introduced in Proposition 6.2. Thus, (6.3.1) can be viewed as an exact sequence of Γ-groups.

We next choose an element π ∈ Aut(Dyn(G)) ⊂ Aut k (g) of order m (clearly, m can take value 1, 2 or 3 only). Like before, we have the corresponding automorphism π of L(g) S given by

x ⊗ s i + ac + bd -→ π(x) ⊗ s i + ac + bd where a, b ∈ k and x ∈ g. Note that ζ × π = π ζ × . It then easily follows that the assignment 1 → z 1 = π-1 ∈ Aut k-Lie ( L(g) S )
gives rise to a cocycle z = (z

i ) ∈ Z 1 (Γ, Aut k-Lie ( L(g) S )
). This cocycle, in turn, gives rise to a (new) twisted action of Γ on L(g) S and Aut k-Lie ( L(g) S ). Analogous considerations (with the use of π) are applied to Aut k-Lie (L(g) S ) and L(g) S . For future reference note that π commutes with elements in V , hence the twisted action of Γ on V is still trivial. From now on we view (6.3.1) as an exact sequence of Γ-groups, the action of Γ being the twisted action.

7.1. Remark. As we noticed before the invariant subalgebra

L = L(g, π) = (L(g) S ) Γ = ((g ⊗ k R) ⊗ R S) Γ
is a simple Lie algebra over R, a twisted form of a split Lie algebra g ⊗ k R. The same cohomological formalism also yields that

(7.1.1) Aut R-Lie (L) ≃ (Aut S-Lie (L(g) S )) Γ .
7.2. Remark. It is worth mentioning that the canonical embedding

ι : (Aut k-Lie (L(g) S )) Γ ֒→ Aut k-Lie ((L(g) S ) Γ ) = Aut k-Lie (L) ≃ Aut R-Lie (L) ⋊ Aut k (R),
where the last isomorphism can be established in the same way as in Proposition 4.1, is not necessary surjective in general case. Indeed, one checks that if m = 3 then the k-automorphism of R given by t → t -1 and viewed as an Proof. Since Γ is cyclic of order m acting trivially on V ≃ k it follows that

element of Aut k-Lie (L) ≃ Aut R-Lie (L) ⋊ Aut k (R)
Z 1 (Γ, V ) = { x ∈ k | mx = 0 } = 0 as required.
The long exact cohomological sequence associated to (6.3.1) together with Lemma 7.4 imply the following. 7.5. Theorem. The following sequence

1 -→ V -→ (Aut k-Lie ( L(g) S )) Γ ν -→ (Aut k-Lie (L(g) S )) Γ -→ 1 is exact. In particular, the group Aut R-Lie (L) is in the image of the canon- ical mapping Aut k-Lie ( L) -→ Aut k-Lie (L) ≃ Aut R-Lie (L) ⋊ Aut k (R).
Proof. The first assertion is clear. As for the second one, note that as in Remark 7.2 we have the canonical embedding

(Aut k-Lie ( L(g) S )) Γ ֒→ Aut k-Lie (( L(g) S ) Γ ) = Aut k-Lie ( L)
and the commutative diagram

(Aut k-Lie ( L(g) S )) Γ ν ----→ (Aut k-Lie (L(g) S )) Γ     Aut k-Lie ( L) ----→ Aut k-Lie (L)
Then surjectivity of ν and Remark 7.2 yield the result.

Some properties of affine Kac-Moody Lie algebras

Henceforth we fix a simple finite dimensional Lie algebra g and a (diagram) automorphism σ of finite order m. For brevity, we will write L and ( L, L) for L(g, σ) and ( L(g, σ), L(g, σ)) respectively.

For all l 1 , l 2 ∈ L one has

(8.0.1) [l 1 , l 2 ] -[l 1 , l 2 ] L = ac
for some scalar a ∈ k. Using (2.0.1) it is also easy to see that for all y ∈ L one has

(8.0.2) [d, yt n ] L = mnyt n + [d, y] L t n 8.1. Remark. Recall that L has a natural R-module structure: If y = x ⊗ t p m ∈ L then yt := x ⊗ t p m +1 = x ⊗ t p+m m ∈ L. Therefore since [d, y] L is contained in L the expression [d, y] L t n is meaning- ful.
The infinite dimensional Lie algebra L admits a unique (up to non-zero scalar) invariant nondegenerate bilinear form (•, •). Its restriction to L ⊂ L is nondegenerate (see [START_REF] Kac | Infinite dimensional Lie algebras[END_REF]7.5.1 and 8.3.8]) and we have

(c, c) = (d, d) = 0, 0 = (c, d) = β ∈ k × and (c, l) = (d, l) = 0 for all l ∈ L.
8.2. Remark. It is known that a nondegenerate invariant bilinear form on L is unique up to nonzero scalar. We may view L as a subalgebra in the split Kac-Moody Lie algebra L(g) S . The last one also admits a nondegenerate invariant bilinear form and it is known that its restriction to L is nondegenerate. Hence this restriction is proportional to the form (•, •).

Let h 0 be a Cartan subalgebra of the Lie algebra g 0 .

8.3. Lemma. The centralizer of h 0 in g is a Cartan subalgebra h of g.

Proof. See [START_REF] Kac | Infinite dimensional Lie algebras[END_REF]Lemma 8.1].

The algebra H = h 0 ⊕ kc ⊕ kd plays the role of Cartan subalgebra for L. With respect to H our algebra L admits a root space decomposition. The roots are of two types: anisotropic (real) or isotropic (imaginary). This terminology comes from transferring the form to H * and computing the "length" of the roots.

The core L of L is the subalgebra generated by all the anisotropic roots. In our case we have L = L ⊕ kc. The correct way to recover L inside L is as its core modulo its centre. 2If m ⊂ L is an abelian subalgebra and α ∈ m * = Hom(m, k) we denote the corresponding eigenspace in L (with respect to the adjoint representation of L) by L α . Thus,

L α = { l ∈ L | [x, l] L = α(x)l for all x ∈ m }. The subalgebra m is called diagonalizable in L if L = α∈m * L α .
Every diagonalizable subalgebra of m ⊂ L is necessarily abelian. We say that m is a maximal (abelian) diagonalizable subalgebra (MAD) if it is not properly contained in a larger diagonalizable subalgebra of L.

8.4. Remark. Every MAD of L contains the center kc of L.

Example.

The subalgebra H is a MAD in L (see [START_REF] Kac | Infinite dimensional Lie algebras[END_REF]Theorem 8.5]).

Our aim is to show that an arbitrary maximal diagonalizable subalgebra m ⊂ L is conjugate to H under an element of Aut k ( L). For future reference we record the following facts: 8.6. Theorem. (a) Every diagonalizable subalgebra in L is contained in a MAD of L and all MADs of L are conjugate. More precisely, let G be the simple simply connected group scheme over R corresponding to L. Then for any MAD m of L there exists g ∈ G(R) such that Ad(g)(m) = h 0 .

(b) There exists a natural bijection between MADs of L and MADs of L.

Every diagonalizable subalgebra in L is contained in a MAD of L. All MADs of L are conjugate by elements in Ad(G(R)) ⊂ Aut k (L) ≃ Aut k ( L). (c) The image of the canonical map Aut k ( L) → Aut k ( L) ≃ Aut k (L) ob- tained by restriction to the derived subalgebra L contains Aut R-Lie (L).
Proof. (a) From the explicit realization of L one knows that h 0 is a MAD of L. Now (a) follows from [CGP]. In the next three sections we are going to prove some preliminary results related to a subalgebra A of the twisted affine Kac-Moody Lie algebra L which satisfies the following two conditions:

a) A is of the form A = A ⊕ kc ⊕ kd, where A is an R-subalgebra of L such that A ⊗ R K is a semisimple Lie algebra over K where K = k(t) is the fraction field of R.
b) The restriction to A of the non-degenerate invariant bilinear form (-, -) of L is non-degenerate.

In particular, all these results will be valid for A = L.

Weights of semisimple operators and their properties

Let x = x ′ + d ∈ A where x ′ ∈ A. It induces a k-linear operator ad(x) : A → A, y → ad(x)(y) = [x, y] A .
We say that x is a k-diagonalizable element of A if A has a k-basis consisting of eigenvectors of ad(x). Throughout we assume that x ′ = 0 and that x is k-diagonalizable.

For any scalar w ∈ k we let

A w = { y ∈ A | [x, y] A = wy }.
We say that w is a weight (= eigenvalue) of ad(x) if A w = 0. More generally, if O is a diagonalizable linear operator of a vector space V over k (of main interest to us are the vector spaces A, A = A ⊕ kc, A) and if w is its eigenvalue following standard practice we will denote by V w ⊂ V the corresponding eigenspace of O.

9.1. Lemma. (a) If w is a nonzero weight of ad(x) then A w ⊂ A. (b) A 0 = A 0 ⊕ x .
Proof. Clearly we have [ A, A] ⊂ A and this implies ad(x)( A) ⊂ A. It then follows that the linear operator ad

(x)| A is k-diagonalizable. Let A = ⊕ A w ′
where the sum is taken over all weights of ad(x)| A . Since x ∈ A 0 and since A = x ⊕ A we conclude that

A = x, A 0 ⊕ (⊕ w ′ =0 A w ′ ),
so that the result follows.

The operator ad(x)| A maps the center c = kc of A into itself, hence it induces a linear operator O x of A ≃ A/kc which is also k-diagonalizable. The last isomorphism is induced by a natural (projection) mapping λ : A → A. If w = 0 the restriction of λ to A w is injective (because A w does not contain kc). Since A = ⊕ w A w it then follows that λ| Aw : A w -→ A w is an isomorphism for w = 0. Thus the three linear operators ad(x), ad(x)| A and O x have the same nonzero weights. 9.2. Lemma. Let w = 0 be a weight of O x and let n ∈ Z. Then w + mn is also a weight of O x and A w+mn = t n A w .

Proof. Assume y ∈ A w ⊂ A, hence O x (y) = wy. Let us show that yt n ∈ A w+mn . We have (9.2.1)

O x (yt n ) = λ(ad(x)(yt n )) = λ([x, yt n ] A ). Substituting x = x ′ + d we get [x, yt n ] A = [x ′ , yt n ] A + [d, yt n ] A
Applying (8.0.1) and (8.0.2) we get that the right hand side is equal to

[x ′ , y] t n + ac + [d, y] A t n + mnyt n
where a ∈ k is some scalar. Substituting this into (9.2.1) we get

O x (yt n ) = λ([x ′ , y] t n + ac + [d, y] A t n + mnyt n ) = [x ′ , y] t n + λ([d, y] A t n ) + mnyt n By (8.0.1) there exists b ∈ k such that [x ′ , y] t n = ([x ′ , y] A + bc) t n .
Here we view [x ′ , y] t n as an element in A. Therefore

O x (yt n ) = mnyt n + λ(([x ′ , y] A + bc) t n + [d, y] A t n ) = mnyt n + λ(([x, y] A + bc) t n ).
We now note that by construction [x, y]

A + bc is contained in A ⊂ A. Hence λ(([x, y] A + bc)t n ) = λ([x, y] A + bc) t n = λ([x, y] A )) t n . Since λ([x, y] A ) = O x (y) = wy we finally get O x (yt n ) = mnyt n + wyt n = (w + mn)yt n .
Thus we have showed that A w t n ⊂ A w+nm . By symmetry A w+nm t -n ⊂ A w and we are done.

We now consider the case w = 0. 9.3. Lemma. Assume that dim A 0 > 1 and n ∈ Z. Then mn is a weight of ad(x).

Proof. Since dim A 0 > 1 there exists nonzero y ∈ A such that [x, y] A = 0. Then the same computations as above show that [x, yt n ] A = mnyt n .

Our next aim is to show that if w is a weight of ad(x) so is -w. We remind the reader that A is equipped with the nondegenerate invariant bilinear form (-, -). Hence for all y, z ∈ A one has

(9.3.1) ([x, y] A , z) = -(y, [x, z] A ).
9.4. Lemma. If w is a weight of ad(x) then so is -w.

Proof. If w = 0 there is nothing to prove. Assume w = 0. Consider the root space decomposition

A = w ′ A w ′ .
It suffices to show that for any two weights w 1 , w 2 of ad(x) such that w 1 + w 2 = 0 the subspaces A w 1 and A w 2 are orthogonal to each other. Indeed, the last implies that if -w were not a weight then every element in A w would be orthogonal to all elements in A, which is impossible. Let y ∈ A w 1 and z ∈ A w 2 . Applying (9.3.1) we have

w 1 (y, z) = ([x, y] A , z) = -(y, [x, z] A ) = -w 2 (y, z).
Since w 1 = -w 2 we conclude (y, z) = 0. Now we switch our interest to the operator O x and its weight subspaces. Since the nonzero weights of ad(x), ad(x)| A and O x are the same we obtain, by Lemmas 9.2 and 9.3, that for every weight w of O x all elements in the set

{ w + mn | n ∈ Z }
are also weights of O x . We call this set of weights by w-series. Recall that by Lemma 9.2 we have A w+mn = A w t n . 9.5. Lemma. Let w be a weight of O x and let A w R be the R-span of A w in A. Then the natural map ν : A w ⊗ k R → A w R given by l ⊗ t n → lt n is an isomorphism of k-vector spaces.

Proof. Clearly, the sum n A w+mn of vector subspaces A w+mn in A is a direct sum. Hence (9.5.1)

A w R = n A w t n = n A w+mn = n A w+mn Fix a k-basis {e i } of A w . Then {e i ⊗ t j } is a k-basis of A w ⊗ k R. Since ν(e i ⊗ t n ) = e i t n ∈ A w+mn
the injectivity of ν easily follows from (9.5.1). The surjectivity is also obvious.

Notation: We will denote the R-span A w R by A {w} . By our construction A {w} is an R-submodule of A and (9.5.2)

A = w A {w}
where the sum is taken over fixed representatives of weight series. 9.6. Corollary.

dim k A w < ∞.
Proof. Indeed, by the above lemma we have

dim k A w = rank R (A w ⊗ k R) = rank R A w R = rank R A {w} ≤ rank R A < ∞,
as required.

9.7.

Corollary. There are finitely many weight series.

Proof. This follows from the fact that A is a free R-module of finite rank. 9.8. Lemma. Let w 1 , w 2 be weights of

O x . Then [A w 1 , A w 2 ] ⊂ A w 1 +w 2 .
Proof. This is straightforward to check.

10. Weight zero subspace 10.1. Theorem. A 0 = 0.

Proof. Assume that A 0 = 0. Then, by Lemma 9.2, A mn = 0 for all n ∈ Z.

It follows that for any weight w, any integer n and all y ∈ A w , z ∈ A -w+mn we have [y, z] = 0. Indeed (10.1.1) [A w , A -w+mn ] ⊂ A w+(-w)+mn = A mn = 0.

For y ∈ A the operator ad(y) : A → A may be viewed as a k-operator or as an R-operator. When we deal with the Killing form -, -on the R-Lie algebra A we will view ad(y) as an R-operator of A. 10.2. Lemma. Let w 1 , w 2 be weights of ad(x) such that {w 1 } = {-w 2 }. Then for any integer n and all y ∈ A w 1 and z ∈ A w 2 +mn we have y, z = 0.

Proof. Let w be a weight of ad(x). By our condition we have {w} = {w + w 1 + w 2 }. Since (ad(y) • ad(z))(A {w} ) ⊂ A {w+w 1 +w 2 } , in any R-basis of A corresponding to the decomposition (9.5.2) the operator ad(y) • ad(z) has zeroes on the diagonal, hence Tr (ad(y) • ad(z)) = 0. 10.3. Lemma. Let w be a weight of ad(x), n be an integer and let y ∈ A w . Assume that ad(y) viewed as an R-operator of A is nilpotent. Then for every z ∈ A -w+mn we have y, z = 0.

Proof. Indeed, let l be such that (ad(y)) l = 0. Since by (10.1.1), ad(y) and ad(z) are commuting operators we have (ad(y) • ad(z)) l = (ad(y)) l • (ad(z)) l = 0.

Therefore ad(y) • ad(z) is nilpotent and this implies its trace is zero.

Since the Killing form is nondegenerate, it follows immediately from the above two lemmas that for every nonzero element y ∈ A w the operator ad(y) is not nilpotent. Recall that by Lemma 9.8 we have ad(y)(A w ′ ) ⊂ A w+w ′ . Hence taking into consideration Corollary 9.7 we conclude that there exits a weight w ′ and a positive integer l such that ad(y)(A {w ′ } ) = 0, (ad(y) • ad(y))(A {w ′ } ) = 0, . . . , (ad(y)) l (A {w ′ } ) = 0 and (ad(y) l (A {w ′ } ) ⊂ A {w ′ } . We may assume that l is the smallest positive integer satisfying these conditions. Then all consecutive scalars (10.3.1) w ′ , w ′ + w, w ′ + 2w, . . . , w ′ + lw are weights of ad(x), {w ′ + iw} = {w ′ + (i + 1)w} for i < l and {w ′ } = {w ′ +lw}. In particular, we automatically get that lw is an integer (divisible by m) which in turn implies that w is a rational number. Thus, under our assumption A 0 = 0 we have proved that all weights of ad(x) are rational numbers. We now choose (in a unique way) representatives w 1 , . . . , w s of all weight series such that 0 < w i < m and up to renumbering we may assume that

0 < w 1 < w 2 < • • • < w s < m.
10.4. Remark. Recall that for any weight w i , the scalar -w i is also a weight. Since 0 < -w i + m < m the representative of the weight series {-w i } is m -w i . Then the inequality m -w i ≥ w 1 implies m -w 1 ≥ w i . Hence out of necessity we have w s = m -w 1 .

We now apply the observation (10.3.1) to the weight w = w 1 . Let w ′ = w i be as in (10.3.1). Choose the integer j ≥ 0 such that w i + jw 1 , w i + (j + 1)w 1 are weights and w i + jw 1 < m, but w i + (j + 1)w 1 ≥ m. We note that since m is not a weight of ad(x) we automatically obtain w i + (j + 1)w 1 > m. Furthermore, we have where the sum is taken over linear mappings α ∈ m * = Hom (m, k). To find a contradiction we first make some simple observations about the structure of the corresponding eigenspace L 0 .

w i + jw 1 ≤ w s = m -w 1 (because w i + jw 1 is a weight of ad(x)). This implies m < w i + (j + 1)w 1 ≤ w s + w 1 = m -w 1 + w 1 = m -a contradiction
If L α = 0, it easily follows that α(c) = 0 (because c is in the center of L). Then α is determined uniquely by the value w = α(x) and so instead of L α we will write L w .

Recall that by Theorem 10.1, L 0 = 0. Our aim is first to show that L 0 contains a nonzero element y such that the adjoint operator ad(y) of L is k-diagonalizable. We will next see that y necessarily commutes with x viewed as an element in L and that it is k-diagonalizable in L as well. It then follows that the subspace in L spanned by c, x and y is a commutative k-diagonalizable subalgebra and this contradicts the fact that m is a MAD. 11.2. Lemma. Let y ∈ L be nonzero such that O x (y) = 0. Then [x, y] L = 0.

Proof. Assume that [x, y] L = bc = 0. Then (x, [x, y] L ) = (x, bc) = (x ′ + d, bc) = (d, bc) = βb = 0.
On the other hand, since the form is invariant we get

(x, [x, y] L ) = ([x, x] L , y) = (0, y) = 0
-a contradiction which completes the proof. 11.3. Lemma. Assume that y ∈ L 0 is nonzero and that the adjoint operator ad(y) of L is k-diagonalizable. Then ad(y) viewed as an operator of L is also k-diagonalizable.

Proof. Choose a k-basis { e i } of L consisting of eigenvectors of ad(y). Thus we have [y, e i ] = u i e i where u i ∈ k and hence

[y, e i ] L = u i e i + b i c where b i ∈ k. Case 1: Suppose first that u i = 0. Let ẽi = e i + b i u i • c ∈ L. Then we have [y, ẽi ] L = [y, e i ] L = u i e i + b i c = u i ẽi
and therefore ẽi is an eigenvector of the operator ad(y) : L → L.

Case 2: Let now u i = 0. Then [y, e i ] L = b i c and we claim that b i = 0. Indeed, we have

(x, [y, e i ] L ) = ([x, y] L , e i ) = (0, e i ) = 0
and on the other hand

(x, [y, e i ] A ) = (x, b i c) = (x ′ + d, b i c) = (d, b i c) = βb i .
It follows that b i = 0 and thus ẽi = e i is an eigenvector of ad(y).

Summarizing, replacing e i by ẽi we see that the set { ẽi } ∪ { c, x } is a k-basis of L consisting of eigenvectors of ad(y). 11.4. Proposition. The subalgebra L 0 contains an element y such that the operator ad(y) :

L → L is k-diagonalizable.
Proof. We split the proof in three steps.

Step 1: Assume first that there exists y ∈ L 0 which as an element in L K = L ⊗ R K is semisimple. We claim that our operator ad(y) is k-diagonalizable. Indeed, choose representatives w 1 = 0, w 2 , . . . , w l of the weight series of ad(x). The sets L w 1 , . . . , L w l are vector spaces over k of finite dimension, by Lemma 9.6, and they are stable with respect to ad(y) (because y ∈ L 0 ). In each k-vector space L w i choose a Jordan basis {e ij , j = 1, . . . , l i } of the operator ad(y)| Lw i . Then the set (11.4.1)

{ e ij , i = 1, • • • , l, j = 1, . . . , l i }
is an R-basis of L, by Lemma 9.5 and the decomposition given in (9.5.2). It follows that the matrix of the operator ad(y) viewed as a K-operator of L ⊗ R K is a block diagonal matrix whose blocks corresponds to the matrices of ad(y)| Lw i in the basis {e ij }. Hence (11.4.1) is a Jordan basis for ad(y) viewed as an operator on L ⊗ R K. Since y is a semisimple element of L ⊗ R K all matrices of ad(y)| Lw i are diagonal and this in turn implies that ad(y) is k-diagonalizable operator of L.

Step 2: We next consider the case when all elements in L 0 viewed as elements of the R-algebra L are nilpotent. Then L 0 , being finite dimensional, is a nilpotent Lie algebra over k. In particular its center is nontrivial since L 0 = 0. Let c ∈ L 0 be a nonzero central element of L 0 . For any z ∈ L 0 the operators ad(c) and ad(z) of L commute. Then ad(z) • ad(c) is nilpotent, hence c, z = 0. Furthermore, by Lemma 10.2 c, z = 0 for any z ∈ L w i , w i = 0. Thus c = 0 is in the radical of the Killing form of L -a contradiction.

Step 3: Assume now that L 0 contains an element y which as an element of L K has nontrivial semisimple part y s . Let us first show that y s ∈ L {0} ⊗ R K and then that y s ∈ L 0 . By Step 1, the last would complete the proof of the proposition.

By decomposition (9.5.2) applied to A = L we may write y s as a sum

y s = y 1 + y 2 + • • • + y l
where y i ∈ L {w i } ⊗ R K. In Step 1 we showed that in an appropriate R-basis (11.4.1) of L the matrix of ad(y) is block diagonal whose blocks correspond to the Jordan matrices of ad(y)| Lw i : L w i → L w i . It follows that the semisimple part of ad(y) is also a block diagonal matrix whose blocks are semisimple parts of ad(y)| Aw i . Since L K is a semisimple Lie algebra over a perfect field we get that ad(y s ) = ad(y) s . Hence for all weights w i we have (11.4.2) [y s , L w i ] ⊂ L w i .

On the other hand, for any u ∈ L w i we have

ad(y s )(u) = [y 1 , u] + [y 2 , u] + • • • + [y l , u]. Since [y j , u] ∈ L {w i +w j } ⊗ R K, it follows that ad(y s )(u) ∈ L {w i } if and only if [y 2 , u] = • • • = [y l , u] = 0.
Since this is true for all i and all u ∈ L w i and since the kernel of the adjoint representation of L K is trivial we obtain

y 2 = • • • = y l = 0. Therefore y s ∈ L {0} ⊗ R K.
It remains to show that y s ∈ L 0 . We may write y s in the form

y s = 1 g(t) (u 0 ⊗ 1 + u 1 ⊗ t + • • • + u m ⊗ t m )
where u 0 , • • • , u l ∈ L 0 and g(t) = g 0 + g 1 t + • • • + g n t n is a polynomial with coefficients g 0 , . . . , g n in k with g n = 0. The above equality can be rewritten in the form (11.4.3)

g 0 y s + g 1 y s ⊗ t + • • • + g n y s ⊗ t n = u 0 ⊗ 1 + • • • + u m ⊗ t m .
Consider an arbitrary index i and let u ∈ L w i . Recall that by (11.4.2) we have ad(y s )(L w i ) ⊂ L w i .

Applying both sides of (11.4.3) to u and comparing L w i +n -components we conclude that [g n y s , u] = [u n , u]. Since this is true for all u and all i and since the adjoint representation of L K has trivial kernel we obtain g n y s = u n . Since g n = 0 we get y s = u n /g n ∈ L 0 .

Now we can easily finish the proof of Theorem 11.1. Suppose the contrary. Then dim(m) < 3 and hence by Lemma 8.7 we have m = c, x ′ + d with x ′ ∈ L. Consider the operator O x on L. By Theorem 10.1 we have L 0 = 0. By Propositions 11.4 and 11.3 there exists a nonzero k-diagonalizable element y ∈ L 0 . Clearly, y is not contained in m. Furthermore, by Lemma 11.2, y viewed as an element of L commutes with m and by Lemma 11.3 it is kdiagonalizable in L. It follows that the subspace m 1 = m ⊕ y is an abelian k-diagonalizable subalgebra of L. But this contradicts maximality of m.

12. All MADs are conjugate 12.1. Theorem. Let G(R) be the preimage of {Ad(g) : g ∈ G(R)} under the canonical map Aut k ( L) → Aut k (L). Then all MADs of L are conjugate under G(R) to the subalgebra H in 8.5.

Proof. Let m be a M AD of L. By Lemma 8.7, m ⊂ L. Fix a vector x = x ′ + d ∈ m where x ′ ∈ L and let m ′ = m ∩ L. Thus we have m = x, c, m ′ . Note that m ′ = 0, by Theorem 11.1. Furthermore, since m ′ is k-diagonalizable in L, without loss of generality we may assume that m ′ ⊂ h 0 given that by Theorem 8.6(b) there exists g ∈ G(R) such that Ad(g)(m ′ ) ⊂ h 0 and that by Theorem 7.5 g has lifting to Aut k-Lie ( L).

Consider the weight space decomposition (12.1.1)

L = ⊕ i L α i
with respect to the k-diagonalizable subalgebra m ′ of L where α i ∈ (m ′ ) * and as usual

L α i = { z ∈ L | [t, z] = α i (t)z for all t ∈ m ′ }.
12.2. Lemma. L α i is invariant with respect to the operator O x .

Proof. The k-linear operator O x commutes with ad(t) for all t ∈ m ′ (because x and m ′ commute in L), so the result follows.

12.3. Lemma. We have x ′ ∈ L 0 .

Proof. By our construction m ′ is contained in h 0 , hence d commutes with the elements of m ′ . But x also commutes with the elements of m ′ and so does x ′ = x -d.

L 0 = C L (m ′ ), being the Lie algebra of the reductive group scheme C G (m ′ ) (see [CGP]), is of the form L 0 = z ⊕ A where z and A are the Lie algebras of the central torus of C G (m ′ ) and its semisimple part respectively. Our next goal is to show that A = 0. Suppose this is not true. To get a contradiction we will show that the subset A = A ⊕ kc ⊕ kd ⊂ L is a subalgebra satisfying conditions a) and b) stated at the end of § 8 and that it is stable with respect to ad(x). This, in turn, will allow us to construct an element y ∈ A which viewed as an element of L commutes with x and m ′ and is k-diagonalizable. The last, of course, contradicts the maximality of m.

Let H denote the simple simply connected Chevalley-Demazure algebraic k-group corresponding to g. Since G is split over S we have

H S = H × k S ≃ G S = G × R S.
Let C g (m ′ ) = t ⊕ r where t is the Lie algebra of the central torus of the reductive k-group C H (m ′ ) and r is the Lie algebra of its semisimple part. Since centralizers commute with base change, we obtain that

t S = t ⊗ k S = z ⊗ R S = z S , r S = r × k S = A ⊗ R S = A S .
12.4. Lemma. We have ad(d)(A) ⊂ A and in particular A is a subalgebra of L.

Proof. Since r consists of "constant" elements we have [d, r] L(g) S = 0, and this implies that [d, r S ] L(g) S ⊂ r S . Also, viewing L as a subalgebra of L(g) S we have [d, L] L ⊂ L. Furthermore, S/R is faithfully flat, hence A = A S ∩L = r S ∩ L. Since both subalgebras r S and L are stable with respect to ad(d), so is their intersection. 12.5. Lemma. The restriction of the nondegenerate invariant bilinear form (•, •) on L to L 0 is nondegenerate.

Proof. We mentioned before that the restriction of (•, •) to L is nondegenerate. Hence in view of decomposition (12.1.1) it suffices to show that for all a ∈ L 0 and b ∈ L α i with α i = 0 we have (a, b) = 0.

Let l ∈ m ′ be such that α i (l) = 0. Using the invariance of (•, •) we get

α i (l)(a, b) = (a, α i (l)b) = (a, [l, b]) = ([a, l], b) = 0.
Hence (a, b) = 0 as required.

12.6. Lemma. The restriction of (•, •) to A is nondegenerate.

Proof. By lemma(12.5) it is enough to show that z and A are orthogonal in L. Moreover, viewing z and A as subalgebras of the split affine Kac-Moody Lie algebra L(g) S and using Remark 8.2 we conclude that it suffices to verify that z S = t S and A S = r S are orthogonal in L(g) S .

Let a ∈ t and b ∈ r. We know that According to Lemma 12.3 we can write x ′ = x ′ 0 + x ′ 1 where x ′ 0 ∈ z and x ′ 1 ∈ A. 12.8. Lemma. We have O

x | A = O x ′ 1 +d | A . In particular, the operator O x ′ 1 +d | A of A is k-diagonalizable.
Proof. By Lemma 12.7, we have O x (A) ⊂ A. Since O x is k-diagonalizable (as an operator of L), so is the operator O x | A of A. Therefore the last assertion of the lemma follows from the first one.

Let now a ∈ A. Using the fact that x ′ 0 and a commute in L we have

[x ′ , a] L = [x ′ 0 , a] L + [x ′ 1 , a] L = [x ′ 1 , a] L + bc for some b ∈ k. Thus O x (a) = O x ′
1 +d (a). 12.9. Lemma. The operator ad( Note that x ′ ∈ C L (m ′ ), by Lemma 12.3, and that h 0 ⊂ C L (m ′ ) (because m ′ ⊂ h 0 , by construction). Since C L (m ′ ) is abelian and since x = x ′ +d it follows that ad(x)(h 0 ) = 0. Hence h 0 , x, c is a commutative k-diagonalizable subalgebra in L. But it contains our MAD m. Therefore m = h 0 , x, c . To finish the proof of Theorem 12.1 it now suffices to show that x ′ ∈ h 0 . For that, in turn, we may view x ′ as an element of L(g) S and it suffices to show that x ′ ∈ h because h ∩ L = h 0 . 12.12. Lemma. x ′ ∈ h.

x ′ 1 + d) : A → A is k-diagonalizable. Proof. Since by Lemma 12.8 O x ′ 1 +d | A : A → A is k-diagonalizable
Proof. Consider the root space decomposition of g with respect to the Cartan subalgebra h:

g = h ⊕ ( ⊕ α =0 g α ).
Every k-subspace g α has dimension 1. Choose a nonzero elements X α ∈ g α . It follows from m ′ = h 0 that C L(g) S (m ′ ) = h S . Thus x ′ ∈ h S . Then g α ⊗ k S is stable with respect to ad(x ′ ) and clearly it is stable with respect to ad(d).

Hence it is also stable with respect to O x . Arguing as in Lemma 9.2 one can easily see that the operator O x , viewed as an operator of L(g) S , is k-diagonalizable. Since g α ⊗ k S is stable with respect to O x , it is the direct sum of its weight subspaces. Hence g α ⊗ k S = ⊕ w (L(g) S ) {w} where {w} = {w + j/m | j ∈ Z} is the weight series corresponding to w. But g α ⊗ k S has rank 1 as an S-module. This implies that in the above decomposition we have only one weight series {w} for some weight w of O x .

We next note that automatically we have dim k (L(g) S ) w = 1. Any its nonzero vector which is a generator of the S-module g α ⊗ k S is of the form X α t j m . It follows from Lemma 9.2 that g α = X α is also a weight subspace of O x . Thus for every root α we have [x, X α ] L(g) S = [x ′ + d, X α ] L(g) S = [x ′ , X α ] = b α X α for some scalar b α ∈ k. Since x ′ ∈ h S this can happen if and only if x ′ ∈ h.

  is not in Im ι. However (7.1.1) implies that the group Aut R-Lie (L) is in the image of ι. 7.3. Remark. The k-Lie algebra L = ( L(g) S ) Γ is a twisted affine Kac-Moody Lie algebra. Conversely, by the Realization Theorem every twisted affine Kac-Moody Lie algebra can be obtained in such a way. 7.4. Lemma. One has H 1 (Γ, V ) = 1.

  (b) The correspondence follows from the fact that every MAD of L contains kc. A MAD m of L is necessarily of the form m ⊕ kc for some MAD m of L and conversely. The canonical map Aut k ( L) → Aut k (L) is an isomorphism by Proposition 5.2. (c) This was established in Theorem 7.5. 8.7. Lemma. If m ⊂ L is a MAD of L then m ⊂ L. Proof. Assume that m ⊂ L. By Theorem 8.6 (b), there exists a MAD m ′ of L containing m. Applying again Theorem 8.6 we may assume that up to conjugation by an element of Aut k ( L), in fact of G(R), we have m ⊂ m ′ = h 0 ⊕ kc. Then m is a proper subalgebra of the MAD H of L and this contradicts the maximality of m.

  that completes the proof of the theorem. 11. A lower bound of dimensions of MADs in L 11.1. Theorem. Let m ⊂ L be a MAD. Then dim m ≥ 3. By Lemma 8.7, m contains an element x of the form x = x ′ + d where x ′ ∈ L and it also contains c. Since x and c generate a subspace of m of dimension 2 the statement of the theorem is equivalent to x, c = m. Assume the contrary: x, c = m. Since m is k-diagonalizable we have the weight space decomposition L = α L α

  a, b δ i+j,0 where •, • is a Killing form of g. Since r is a semisimple algebra we have r = [r, r]. It follows that we can write b in the form b = [a i , b i ] for some a i , b i ∈ r. Using the facts that t and r commute and that the Killing form is invariant we havea, b = a, [a i , b i ] = [a, a i ], b i = 0, b i = 0. Lemma. The k-subspace A ⊂ L is invariant with respect to O x .Proof. Let a ∈ A. We need to verify that[x, a] L ∈ A ⊕ kc ⊂ L.But [d, A]L ⊂ A + kc by Lemma 12.4. We also have[x ′ , A] L ⊂ A ⊕ kc(because x ′ ∈ L 0 , by Lemma 12.3, and A viewed as a subalgebra in L 0 is an ideal). Since x = x ′ + d the result follows.

L

  we can apply the same arguments as in Lemma 11.3. Now we can produce the required element y. It follows from Lemma 12.6 that the Lie algebra A satisfies all the conditions stated at the end of Section 8. By Lemma 12.9, ad(x ′ 1 + d) is k-diagonalizable operator of A. Hence arguing as in Theorem 11.1 we see that there exists a nonzero y ∈ A such that [y, x ′ 1 + d] L = 0 and ad(y) is a k-diagonalizable operator on A. Then by Lemma 12.8 we have O x (y) = O x ′ 1 +d (y) = 0 and hence, by Lemma 11.2, x and y commute in L.According to our plan it remains to show that y is k-diagonalizable in L. To see this we need 12.10.Lemma. Let z ∈ m ′ . Then [z, y] L = 0. Proof. Since y ∈ A ⊂ C L (m ′ ) we have [z, y] L = 0. Then [z, y] L = bc for some b ∈ k. It follows 0 = (0, y) = ([x, z] L , y) = (x, [z, y] L ) = (x ′ + d, bc) = (d, bc) = βb.This yields b = 0 as desired.12.11. Proposition. The operator ad(y) : L → L is k-diagonalizable.Proof. According to Lemma 11.3, it suffices to prove that ad(y) : L → L is k-diagonalizable. Since y viewed as an element of A is semisimple it is still semisimple viewed as an element of L. In particular, the R-operator ad(y) : L → L is also semisimple.Recall that we have the decomposition of L into the direct sum of the weight spaces with respect to O x : {w i } .Since y and x commute in L, for all weights w we have ad(y)(L w ) ⊂ L w . If we choose any k-basis of L w it is still an R-basis of L {w} = L w ⊗ k Rand in this basis the R-operator ad(y)| L {w} and the k-operator ad(y)| Lw have the same matrices. Since the R-operator ad(y)| L {w} is semisimple, so is ad(y)| Lw , i.e. ad(y)| Lw is a k-diagonalizable operator. Thus ad(y) : L → L is k-diagonalizable. Summarizing, assuming A = 0 we have constructed the k-diagonalizable element y ∈ m = m ′ , x, c in L which commutes with m ′ and x in L. Then the subalgebra m, y in L is commutative and k-diagonalizable which is impossible since m is a MAD. Thus A is necessarily trivial and this implies C L (m ′ ) is the Lie algebra of the R-torus C G (m ′ ), in particular C L (m ′ ) is abelian.

Unlike L(g)S, these object exist over k but not over S.

In nullity one the core coincides with the derived algebra, but this is not necessarilty true in higher nullities.

By the previous lemma we have x

The proof of Theorem 12.1 is complete.