Conjugacy theorem for the strictly hyperbolic Kac-Moody algebras.

Yuly Billig Department of Mathematics, University of Alberta Edmonton, Alberta, Canada T6G 2G1

ABSTRACT. In this article we prove the conjugacy theorem for the (nonsymmetrizable) Kac-Moody algebras corresponding to strictly hyperbolic generalized Cartan matricies, i.e., without submatricies of finite type.

AMS Subject Classification numbers: 17B67, 22E65 Introduction. In [PK] Peterson and Kac proved the conjugacy theorem for the Cartan sub-

algebras of a symmetrizable Kac-Moody algebra. This theorem allows in particular to describe the group of the automorphisms of a Kac-Moody algebra.

The Peterson-Kac proof involves an interesting analysis of the action of a Kac-Moody group on the Kostant cone. To any element of the Kostant cone one associates a polyhedron which is a convex closure of the support of this element in the weight lattice. Peterson and Kac prove that

(i) every vertex of this polyhedron belongs to the W -orbit of the highest weight (ii) every edge is parallel to a real root.

This allows us to apply the procedure of “stripping off edges” which is the crucial part of the proof.

The detailed version of the Peterson-Kac proof is presented in the book [MP]. In order to deal with the sophisticated construction of the Kostant cone one

needs to use strong technical machinery, so the restriction to the symmetrizable Kac-Moody algebras seems to be unavoidable.

In this article we prove the conjugacy theorem for the strictly hyperbolic (non - symmetrizable) Kac-Moody algebras, i.e., corresponding to the generalized Cartan matricies with no rank 2 submatricies of finite type.

Our idea is to work directly with the convex closure of the support of an adlocally finite element in the root space. We prove that (i’) every vertex of this polyhedron is a real root (ii’) every edge is parallel to some (not necessarily real) root.

This allows us to apply the same procedure of “stripping off the edges”. Unfortunately, in the finite root system, edges may be not parallel to any root, so this proof doesn’t work in the finite type case.

The auther is grateful to Professors Robert Moody and Arturo Pianzola for fruitful discussions.

Definitions and notations.

Let A be a generalized Cartan matrix, i.e., a_{ij} ∈ ℤ; a_{ii} = 2; a_{ij} ≤ 0 if i ≠ j; a_{ij} = 0 ⟺ a_{ji} = 0; i, j = 1, … l.

Fix a base field, 𝔽 , of characteristic zero. A realization of A is a triple < 𝔥_{0}, Π, Π^{∨} >, where 𝔥_{0} is a vector space over 𝔽, dim 𝔥_{0} = l+corank(A), Π = {α_{1}, …, α_{l}} ⊂ 𝔥_{0}^{∗} and Π^{∨} = {h_{1}, …, h_{l}} ⊂ 𝔥_{0} , satisfying α_{j}(h_{i}) = a_{ij} .

Introduce first the Lie algebra 𝔤^{}˜ = 𝔤^{}˜ (A) over 𝔽 with the generators e_{i}, f_{i} and 𝔥_{0} , i = 1, …, l and the following defining relations:

[h, e_{i}] = α_{i}(h)e_{i}, [h, f_{i}] = −α_{i}(h)f_{i},
[e_{i}, f_{j}] = δ_{ij}h_{i}, [𝔥_{0}, 𝔥_{0}] = 0.

There exists the unique maximal ideal (radical) I in 𝔤^{}˜ that intersects with 𝔥_{0} trivially. The factor-algebra 𝔤(A) = 𝔤^{}˜ /I is called the Kac-Moody algebra.

The following relations hold in 𝔤(A) :

(ade_{i})^{−a_{ij}+1}e_{j} = (adf_{i})^{−a_{ij}+1}f_{j} = 0, i ≠ j.

Denote by 𝔤_{+} (resp. 𝔤_{−} ) the subalgebra in 𝔤, generated by e_{1}, …, e_{l} (resp. f_{1}, …, f_{l} ). Then we have a triangular decomposition for 𝔤:

𝔤 = 𝔤_{−} ⊕ 𝔥_{0} ⊕ 𝔤_{+}.

The Borel subalgebra 𝔟_{±} is defined as 𝔥_{0} ⊕ 𝔤_{±}.

We also have a root decomposition for 𝔤:

𝔤 = ⊕_{α∈𝔥_{0}^{∗}} 𝔤_{α}, where

𝔤_{α} = {x ∈ 𝔤| for all h ∈ 𝔥_{0} : [h, x] = α(h)x}.

The root system Δ = Δ(A) is the set of all α ∈ 𝔥_{0}^{∗} such that 𝔤_{α} ≠ 0. Obviously, Δ is a subset of the lattice ℤ^{l} generated by the fundamental roots {±α_{i}}, i = 1, …, l. It follows from the triangular decomposition for 𝔤 that Δ = Δ_{−} ∪ {0} ∪ Δ_{+}, where Δ_{+} = Δ ∩ ℕ^{l} and Δ_{−} = −Δ_{+} .

Let (V, ρ) be an infinite-dimensional 𝔤 -module. An element x ∈ 𝔤 is called ρ-locally finite if for any v ∈ V the linear span of {ρ(x)^{n}v}, n ∈ ℕ , is finitedimensional.

An element x ∈ 𝔤 is called ρ-locally nilpotent if for any v ∈ V there exists n ∈ ℕ such that ρ(x)^{n}v = 0.

We shall call x ∈ 𝔤 locally finite (resp. locally nilpotent) if it is ad-locally finite (resp. ad-locally nilpotent).

Let W denote the subgroup of Aut𝔽(𝔥_{0}^{∗}) generated by the “simple reflections” β ↦ β − ⟨β, α^{∨}⟩α for α ∈ Π. The set { wα | w ∈ W, α ∈ Π } of real roots will be denoted Δ^{re}. Define positive and negative real roots as usual by Δ_{±}^{re} = Δ^{re} ∩ Δ_{±}.

A 𝔤-module V, or (V, ρ) where ρ: 𝔤 → End𝔽(V), is said to be integrable if 𝔤_{α} is ρ-locally nilpotent for all α ∈ Δ^{re}. One associates a Kac-Moody group G(A) to 𝔤(A) as in [PK]; G(A) is the quotient G = G^{∗}/N of the free product G^{∗} of the root subgroups 𝔤_{α} (for real roots α) by the largest normal subgroup N of G^{∗} acting trivially on all integrable 𝔤-modules V (where 𝔤_{α} ⊂ G^{∗} acts on each such (V, ρ) by (e, v) ↦ exp(ρ(e))v). For real roots α, one has the canonical injection (also denoted exp) 𝔤_{α} → G^{∗} → G with image the one-parameter subgroup.

Highest weight representations for sl_{2} and for the Heisenberg algebra 𝔫.

Consider Lie algebra sl_{2}with the basis {e, h, f} and relations [h, e] = 2e , [h, f] = 2f , [e, f] = h.

Let (V, ρ) be an sl_{2}-module generated by a highest weight vector v : ρ(e)v = 0, ρ(h)v = cv.

The space V is a linear span of {ρ(f)^{k}v}, k ≥ 0. We can compute the action of ρ(e) on these vecors:

ρ(e)ρ(f)^{k}v = ∑_{i=0}^{k−1} ρ(f)^{k−1−i}ρ(h)ρ(f)^{i}v

= (∑_{i=0}^{k−1}(c − 2i))ρ(f)^{k−1}v

= k(c − (k − 1))ρ(f)^{k−1}v,

ρ(e)^{k}ρ(f)^{k}v = k! ∏_{i=0}^{k−1}(c − i)v.

This computation shows that if c is not a non-negative integer then ρ(e)^{k}v ≠ 0 for all k ∈ ℕ and V is an infinite-dimensional simple module.

If V is a finite-dimensional module then c ∈ ℕ ∪ {0} and ρ(f)^{c+1}v = 0 , so {ρ(f)^{k}v}, k = 0, …, c , is a basis for V.

We can rewrite the last formula as follows:

ρ(e)^{k}ρ(f)^{k}v = \frac{k!c!}{(c − k)!}v.

Denote v_{k} = \frac{1}{k!}ρ(f)^{k}v . Then

ρ(f)v_{k} = (k + 1)v_{k+1}, k < c,

ρ(e)v_{k} = (c − (k − 1))v_{k+1}, k > 0,
ρ(f)v_{c} = ρ(e)v_{0} = 0.

As the action of ρ(e) and ρ(f) on V is nilpotent, we can consider exponentials of these operators:

E(λ) = exp(λρ(e)) = ∑_{k=0}^{c} \frac{1}{k!}ρ(e)^{k},

F (λ) = exp(λρ(f)) = ∑_{k=0}^{c} \frac{1}{k!}ρ(f)^{k}.

The families of operators {E(λ)} and {F (λ)}, λ ∈ 𝔽, form abelian (additive) groups and together generate a group SL_{2}(𝔽) acting on V (if dim V > 1 ).

Consider now a 3-dimensional nilpotent Heisenberg algebra 𝔫 with the basis {x, y, z} , where z is a central element of 𝔫 and [x, y] = z.

Let (V, ρ) be a highest weight 𝔫 -module generated by vector v : ρ(x)v = 0, ρ(z)v = cv. As ρ(z) commutes with ρ(x) and ρ(y) we have ρ(z) = c Id_{V} . The space V is a linear span of {ρ(y)^{k}v}, k ≥ 0. Also,

ρ(x)ρ(y)^{k}v = ∑_{i=0}^{k−1} ρ(y)^{k−1−i}ρ(z)ρ(y)^{i}v =

= kcρ(y)^{k−1}v, and
ρ(x)^{k}ρ(y)^{k}v = k!c^{k}v,

so , if c ≠ 0, then ρ(y)^{k}v ≠ 0 and {ρ(y)^{k}v} are linearly independent. Hence V is an infinite-dimensional simple 𝔫-module.

Conjugacy Theorem.

Theorem 1. Let A be a generalized Cartan matrix such that a_{ij}a_{ji} ≥ 4 for all i, j and let 𝔤(A) be the corresponding Kac-Moody algebra over the field 𝔽of characteristic 0. Then all Cartan subalgebras in 𝔤(A) are conjugated under the action of a Kac-Moody group G(A).

For an element x ∈ 𝔤 consider a decomposition in homogeneous components:

x = ∑_{α∈Δ} x_{α}.

For a subset P of Δ denote

x_{P} = ∑_{α∈P} x_{α}.

Define the support of x in Δ as follows:

supp(x) = {α ∈ Δ|x_{α} ≠ 0},

and if 𝔥 is a subalgebra in 𝔤 then define

supp(𝔥) = ∪_{x∈𝔥} supp(x).

Let S(x) be a convex closure of supp(x) in ℤ^{l} . As supp(x) is finite it follows that S(x) is a polyhedron in ℤ^{l}. Note that every vertex of S(x) belongs to supp(x).

Proposition 1. Let x be a locally finite element in 𝔤 and let Γ be a generalized face of S(x).

(i) If Γ contains 0 then x_{Γ} is locally finite. (ii) If Γ doesn’t contain 0 then x_{Γ} is locally nilpotent.

Proof. Clearly, in order to prove local finitness (nilpotency) of an element, it is sufficient to consider its action on homogeneous elements. (i). Let π be a linear subspace spanned by Γ and let u ∈ 𝔤_{α}. Then

((adx)^{n}u)_{α+π} = (adx_{Γ})^{n}u.

As the linear span of {(adx)^{n}u}, n ∈ ℕ , is finite-dimensional then its projection on α + π is also finite-dimensional, hence x_{Γ} is locally finite.

(ii). In the same way

((adx)^{n}u)_{α+nΓ} = (adx_{Γ})^{n}u.

As Γ doesn’t contain 0 there exists n ∈ ℕ such that α+nΓ = {α+γ_{1}+…+γ_{n}|γ_{i} ∈ Γ} doesn’t intersect with the support of the span of {(adx)^{n}u} . Consequently, (adx_{Γ})^{n}u = 0.

Two following lemmas supply us with our main technical tools.

Lemma 1. Let V = ⊕_{k∈ℤ} V_{k} be a graded sl_{2}-module for which ρ(h)v_{k} = 2kv_{k} for all v_{k} ∈ V_{k}. If dimV_{0} < ∞ and dimV_{k} > 0 for all k > 0 then ρ(e) is not locally nilpotent.

Proof. Suppose, by contradiction, that ρ(e) is locally nilpotent. As dimV_{0} < ∞ then there exists m ∈ ℕ such that ρ(e)^{m}V_{0} = 0. Consider a non-zero element v_{n} ∈ V_{n} for some n > m. Again using local nilpotency of ρ(e) we can find s ≥ 0 such that ρ(e)^{s+1}v_{n} = 0 and u = ρ(e)^{s}v_{n} ≠ 0. Note that ρ(f)^{n+s}u ∈ V_{0}, hence ρ(e)^{n+s}ρ(f)^{n+s}u = 0 , but ρ(e)^{n+s}ρ(f)^{n+s}u = (2(n + s))!u ≠ 0 . This is a contradiction.

Lemma 2. Let V = ⊕_{k∈ℤ} V_{k} be a graded 𝔫-module for which ρ(x)V_{k} ⊂ V_{k+1}, ρ(y)V_{k} ⊂ V_{k−1} and for all v ∈ V , ρ(z)v = cv, c ≠ 0. If dimV_{0} < ∞ and dimV_{k} > 0 for all k > 0 then ρ(x) is not locally nilpotent.

Proof. The proof is essentially the same as for the previous lemma. Assume, by contradiction, that ρ(x) is locally nilpotent, then for some m > 0 ρ(e)^{m}V_{0} = 0. Let u be a non-zero element in V_{k} for some k > m such that ρ(x)u = 0. Then ρ(x)^{k}ρ(y)^{k}u = 0. On the other hand, ρ(x)^{k}ρ(y)^{k}u = k!c^{k}u ≠ 0 , a contradiction.

Lemma 3. Let α, β ∈ Δ^{re}, α ≠ ±β and let h_{α}, h_{β} be generators of [𝔤_{α}, 𝔤_{−α}] and [𝔤_{β}, 𝔤_{−β}] , respectively. If some non-zero linear combination c_{1}h_{α} + c_{2}h_{β} belongs to the center of 𝔤(A) then 𝔤(A) is an affine Lie algebra.

Proof. Considering the action of W on 𝔤 we may assume that α is a fundamental root: α = α_{i} ∈ Π, h_{α} = h_{i} and β ∈ Δ_{+}^{re}. From the action of W on the real co-roots we can deduce that in the decomposition c_{2}h_{β} = ∑_{j=1}^{l} b_{j}h_{j} all coefficients b_{j} have the same sign, so we may assume that they are all non-negative. The element c_{1}h_{i} + ∑_{j=1}^{l} b_{j}h_{j} belongs to Z(𝔤) if and only if the corresponding linear combination of rows of the generalized Cartan matrix A is zero.

If c_{1} +b_{i} < 0 then for the i-th column of A we have ∑_{{j=1 \atop j≠i}}^{l} b_{j}a_{ji} +2(c_{1} +b_{i}) < 0,

so this linear combination of rows of A is not a zero row.

If c_{1} + b_{i} ≥ 0 then this linear combination has all coefficients non-negative, so A is of affine type [K].

Lemma 4. Let 𝔤 be a non-affine Kac-Moody algebra and let β ∈ Δ_{+}^{re}, α_{i} ∈ Π , β − α_{i} ∉ Δ, α = β + nα_{i} ∈ Δ. If x ∈ 𝔤_{α}, x ≠ 0, then there exists y ∈ 𝔤_{−α} such that [x, y] ∉ Z(𝔤).

Proof. Consider the finite-dimensional < e_{i}, h_{i}, f_{i} > ≅ sl_{2}-module V = ⊕_{k≥0} 𝔤_{β+kα_{i}} . Without loss of generality we may assume that α(h_{i}) = −m ≤ 0. As V is completely reducible, x can be written as the following sum:

x = ∑_{k=0}^{n}(ade_{i})^{n−k}v_{k},

where v_{k} ∈ 𝔤_{β+kα_{i}}, [f_{i}, v_{k}] = 0.

Let u_{0} be a non-zero element of a real root space 𝔤_{−β} and let y = (adf_{i})^{n}u_{0}. We shall prove that [x, y] ∉ Z(𝔤).

We have

[(ade_{i})^{n−k}v_{k}, (adf_{i})^{n}u_{0}]

= ∑_{j=0}^{n−k}(−1)^{j}C_{n−k}^{j}(ade_{i})^{n−k−j}(adv_{k})(ade_{i})^{j}(adf_{i})^{n}u_{0}

= ∑_{j=0}^{n−k}(−1)^{j}C_{n−k}^{j} \frac{n!(m − j)!}{(n − j)!(m − n)!}(ade_{i})^{n−k−j}(adv_{k})(adf_{i})^{n−j}u_{0} =

= ∑_{j=0}^{n−k}(−1)^{j}C_{n−k}^{j} \frac{n!(m − j)!}{(n − j)!(m − n)!}(ade_{i})^{n−k−j}(adf_{i})^{n−j}[v_{k}, u_{0}].

As [v_{k}, u_{0}] ∈ 𝔤_{kα_{i}} , it follows [v_{k}, u_{0}] = 0 when k ≠ 0, 1. If k = 0 then [v_{0}, u_{0}] ∈ 𝔥_{0} , so (adf_{i})^{n−j}[v_{0}, u_{0}] = 0 when n − j > 1. If k = 1 then [v_{1}, u_{0}] ∈ 𝔤_{α_{i}} , so (adf_{i})^{n−j}[v_{1}, u_{0}] = 0 when n − j > 2. Hence,

[x, y] = (−1)^{n}(adv_{0})(ade_{i})^{n}(adf_{i})^{n}u_{0}+
+(−1)^{n−1}n(ade_{i})(adv_{0})(ade_{i})^{n−1}(adf_{i})^{n}u_{0}+
+(−1)^{n−1}(adv_{1})(ade_{i})^{n−1}(adf_{i})^{n}u_{0}

= (−1)^{n} \frac{n!m!}{(m − n)!}[v_{0}, u_{0}]+

+(−1)^{n−1}n\frac{n!(m − 1)!}{(m − n)!}(ade_{i})(adf_{i})[v_{0}, u_{0}]+

+(−1)^{n−1} \frac{n!(m − 1)!}{(m − n)!}(adf_{i})[v_{1}, u_{0}].

Note that [v_{0}, u_{0}] is a non-zero element of [𝔤_{β}, 𝔤_{−β}]. As [v_{0}, u_{0}] ∈ 𝔥_{0} and [v_{1}, u_{0}] ∈ 𝔤_{α_{i}} , we have [e_{i}[f_{i}[v_{0}, u_{0}]]], [f_{i}[v_{1}, u_{0}]] ∈ 𝔽h_{i}.

As 𝔤 is non-affine , applying Lemma 3 we conclude that [x, y] ∉ Z(𝔤).

Proposition 2. If x_{α} is a non-zero homogeneous locally-nilpotent element of 𝔤(A) and α ∈ Δ_{+} then α is a real root.

Proof. This statement is obvious for affine Kac-Moody algebras, so we may assume that 𝔤(A) is not affine. Suppose, by contradiction, that there exist imaginary roots in Δ_{+} with non-zero locally nilpotent elements in the corresponding root spaces. Let α be minimal with this property and let x_{α} ∈ 𝔤_{α} be locally nilpotent. Considering the action of the Weyl group we get that α ∈ −C^{∨}. Since 𝔤(A) is radical-free, [f_{i}, x_{α}] ≠ 0 for some i = 1, …, l.

Consider an automorphism F_{i}(1) of 𝔤(A):

F_{i}(1)x_{α} = ∑_{k=0}^{∞} \frac{(adf_{i})^{k}}{k!} x_{α}.

Let n be the maximal integer for which (adf_{i})^{n}x_{α} ≠ 0, n ≥ 1. According to Proposition 1(ii) (adf_{i})^{n}x_{a} is locally nilpotent. As α is the minimal positive imaginary root with non-zero locally nilpotent elements, α − nα_{i} should be a real root. Now, by Lemma 4 , there exists y_{−α} ∈ 𝔤_{−α} such that h_{α} = [x_{α}, y_{−α}] ∉ Z(𝔤).

Consider a 3-dimensional subalgebra < x_{α}, h_{α}, y_{−α} > . If this subalgebra is isomorphic to sl_{2}then taking a graded module ⊕_{k∈ℤ} 𝔤_{kα} and applying Lemma 1 we get a contradiction to local nilpotency of x_{α}.

If < x_{α}, h_{α}, y_{−α} > is isomorphic to the Heisenberg algebra then choose δ ∈ Δ_{+}^{im}∩−C^{∨} such that δ has the full support (i.e. all coefficients in the decomposition of δ as a linear combination of {α_{j}} are non-zero) and for all j = 1, …, l δ +α_{j} ∈ Δ_{+}^{im} ∪ −C^{∨} [K]. Since h_{α} ∉ Z(𝔤) then for at least one of these roots μ = δ + α_{i} we have μ(h_{α}) ≠ 0.

As μ, α ∈ Δ_{+}^{im} ∪−C^{∨} and μ has the full support it follows that μ+kα ∈ Δ_{+}^{im} for all k ∈ ℕ.

Now we can get a contradiction to the local nilpotency of x_{α} if we apply Lemma 2 to < x_{α}, h_{α}, y_{−α} > -module ⊕_{k∈ℤ} 𝔤_{μ+kα}. This completes the proof of Proposition 2.

Proposition 3. Let generalized Cartan matrix A satisfy a_{ij}a_{ji} ≥ 4 for all i, j. If x is a locally finite element of 𝔤(A) and Γ is an edge of S(x) then Γ is parallel to some root in Δ(A).

Proof. If Γ contains 0 then the proposition is obvious. If Γ doesn’t contain 0 then according to Proposition 1(ii) x_{Γ} is locally nilpotent. It follows from Proposition 2 that the vertices {β, γ} of edge Γ are real roots.

Consider now non-zero elements y_{−β} ∈ 𝔤_{−β} and y_{−γ} ∈ 𝔤_{−γ} . Let y = c_{1}y_{−β} + c_{2}y_{−γ}.

Suppose, by contradiction, that Γ is not parallel to any root. Then (Γ + (−Γ)) ∩ Δ = {0}, hence

[x_{Γ}, y] = c_{1}[x_{β}, y_{−β}] + c_{2}[x_{γ}, y_{−γ}] = c_{1}h_{β} + c_{2}h_{γ}.

As Γ doesn’t contain 0 then β and γ (resp. h_{β} and h_{γ} ) are linearly independent. It is possible to see that a subalgebra 𝔤_{1} in 𝔤(A) , generated by < x_{β}, x_{γ}, y_{−β}, y_{−γ} >, is isomorphic to a rank 2 Kac-Moody algebra, corresponding to a rank 2 root subsystem in Δ(A).

Since A doesn’t contain submatrices of rank 2 of finite type, Δ(A) doesn’t contain rank 2 root subsystems of finite type [HMR] . Hence 𝔤_{1} is infinite-dimensional.

If 𝔤_{1} is a Kac-Moody algebra of indefinite type then

det | \begin{matrix}β(h_{β}) & β(h_{γ}) \\ γ(h_{β}) & γ(h_{γ}) \end{matrix} | ≠ 0,

and there exist constants c_{1}, c_{2} such that β(c_{1}h_{β} + c_{2}h_{γ}) = γ(c_{1}h_{β} + c_{2}h_{γ}) = 2.

The subalgebra < x_{Γ}, c_{1}h_{β} + c_{2}h_{γ}, y > is isomorphic to sl_{2}and the module ⊕_{k∈ℤ} ( ⊕_{t∈ℚ} 𝔤_{kπ}) , where π is a line through Γ , satisfies the conditions of Lemma 1 which gives a contradiction.

If 𝔤_{1} is affine then

det | \begin{matrix}β(h_{β}) & β(h_{γ}) \\ γ(h_{β}) & γ(h_{γ}) \end{matrix} | = 0,

hence there exist constants c_{1}, c_{2} such that β(c_{1}h_{β} + c_{2}h_{γ}) = γ(c_{1}h_{β} + c_{2}h_{γ}) = 0.

If algebra 𝔤(A) itself is non-affine then by Lemma 3 c_{1}h_{β} + c_{2}h_{γ} ∉ Z(𝔤) and again there exists μ ∈ Δ_{+}^{im} ∩ −C^{∨} such that μ has the full support amd μ(c_{1}h_{β} + c_{2}h_{γ}) ≠ 0. If δ is a null-root of 𝔤_{1} then μ + kδ ∈ Δ_{+}^{im} for all k ≥ 0. Now the algebra < x_{Γ}, c_{1}h_{β} + c_{2}h_{γ}, y > is isomorphic to the Heisenberg algebra and the module ⊕_{k∈ℤ} 𝔤_{μ+kπ} satisfies the conditions of Lemma 2, which leads again to a contradiction.

Finally, suppose that both 𝔤_{1} and 𝔤(A) are affine. As a_{ij}a_{ji} ≥ 4 for all i, j we have that 𝔤(A) should have rank 2. In this case the required statement may be obtained by direct computation.

Lemma 5. Let α_{i} be a fundamental root and let the root string through a real root β_{1} along α_{i} contains four real roots β_{1} < β_{2} < β_{3} < β_{4}; β_{2} − β_{1} =

β_{4} − β_{3} = α_{i}; β_{3} − β_{2} = nα_{i}. If x = ∑_{j=0}^{n} x_{β_{2}+jα_{i}} ≠ 0 then x is not locally

nilpotent.

Proof. Roots ±α_{i}, ±β_{1} generate a rank 2 subsystem Δ_{1} in Δ(A).

Consider y = c_{1}y_{−β_{1}} + c_{2}y_{−β_{2}} + c_{3}y_{−β_{3}} + c_{4}y_{−β_{4}}. Then

[x, y] = c_{1}[x_{β_{2}}, y_{−β_{1}}] + c_{2}[x_{β_{2}}, y_{−β_{2}}] + c_{2}[x_{β_{2}+α_{i}}, y_{−β_{2}}]+
+c_{3}[x_{β_{3}−α_{i}}, y_{−β_{3}}] + c_{3}[x_{β_{3}}, y_{−β_{3}}] + c_{4}[x_{β_{3}}, y_{−β_{4}}].

If Δ_{1} is of indefinite type (resp. affine type) then there exist constants c_{2} and c_{3} such that h = c_{2}[x_{β_{2}}, y_{−β_{2}}] + c_{3}[x_{β_{3}}, y_{−β_{3}}] satisfies β_{2}(h) = β_{3}(h) = 2 (resp. β_{2}(h) = β_{3}(h) = 0 ).

Choose now constants c_{1} and c_{4} satisfying

c_{1}[x_{β_{2}}, y_{−β_{1}}] + c_{2}[x_{β_{2}+α_{i}}, y_{−β_{2}}] = 0,
c_{3}[x_{β_{3}−α_{i}}, y_{−β_{3}}] + c_{4}[x_{β_{3}}, y_{−β_{4}}] = 0.

If Δ_{1} is of indefinite type then < x, y, h >≅ sl_{2} and applying Lemma 1 to the module ⊕_{k∈ℤ} ( ⊕_{m∈ℤ} 𝔤_{kβ_{1}+mα_{i}}) we get that x is not locally nilpotent.

If Δ_{1} is of affine type and Δ(A) is non-affine then again choose μ ∈ Δ^{im}∩−C^{∨} with full support and satisfying μ(h) ≠ 0. Now applying Lemma 2 to the module ⊕_{k∈ℤ} ( ⊕_{m∈ℤ} 𝔤_{μ+kβ_{1}+mα_{i})} we get the required statement.

If, finally, both Δ_{1} and Δ(A) are affine then Δ_{1} is of type A_{2}^{(2)} and Δ(A) is of type A_{2k}^{(2)} . In this case the statement of the lemma may be obtained by direct computation.

Now we have all requisites to prove the theorem if the field 𝔽 is uncountable and algebraically closed. The idea could be the following: over an uncountable field every finite-dimensional abelian Cartan subalgebra 𝔥 contains a regular element x, such that C𝔤(x) = 𝔥, then applying the procedure of “ stripping off edges ” [PK] to S(x) we construct an automorphism g ∈ G(A) such that g(x) ∈ 𝔥_{0} , hence g(𝔥) = 𝔥_{0}. However, to be able to prove the theorem over an arbitrary field of characteristic 0 a little more work needs to be done.

Let V be a finite-dimensional sl_{2}-module with the set of weights P and the highest weight μ. Let v ∈ V, v = ∑_{γ∈P} v_{γ} ≠ 0. Consider a convex closure S(v) of the support of v in P, so S(v) is a segment or a point. Let 𝔽^{}¯ be an algebraic closure of 𝔽 and let V̄ = V ⊗𝔽 𝔽^{}¯ . Consider non-zero vectors in V with the following property:

For every g ∈ SL_{2}(𝔽^{}¯ ) vertices of S(gv) belong to the set {±μ}. (∗)

Lemma 6. Let V be a finite-dimensional < e_{α}, h_{α}, f_{α} >≅ sl_{2}-module over the field 𝔽 with the highest weight μ, 2μ = nα, n ≥ 2, dimV_{μ} = 1. Let v be a non-zero vector in V , satisfying the property (*). Then (i). There exists g_{0} ∈ SL_{2}(𝔽) such that g_{0}v ∈ V_{−μ}. (ii). If for vector w ∈ V w + cv satisfies (*) for all c ∈ 𝔽 then g_{0}w ∈ V_{−μ}.

Proof. (i) If μ is not a vertex of S(v) then −μ is the only vertex, hence v ∈ V_{−μ} and there is nothing to prove.

If μ is a vertex of S(v) , take a highest weight vector u_{0} and construct a basis for the irreducible submodule U generated by V_{μ}:

u_{k} = \frac{(adf_{α})^{k}}{k!} u_{0}.

Since V is completely reducible, there exists a canonical projection τ : V ↦ U. Let τ(v) = ∑_{k=0}^{n} a_{k}u_{k}, a_{0} ≠ 0. Note that for every w ∈ V, τ(w)_{±μ} = w_{±μ}.

Consider an automorphism F (λ) applied to τ(v) :

(F (λ)τ(v))_{−μ} = ∑_{k=0}^{n} C_{n}^{k}a_{k}λ^{n−k}.

As a_{0} ≠ 0 , there exists λ_{0} ∈ 𝔽^{}¯ such that (F (λ_{0})τ(v))_{−μ} = 0 . Consequently, (F (λ_{0})v) ∈ V̄_{μ} and τ(v) = v. Note that (F (λ_{0})v)_{μ−α} = a_{1} + λ_{0}a_{0} = 0, hence λ_{0} in fact belongs to 𝔽.

Finally, σ_{α}F (λ_{0})v ∈ V_{−μ} as was to be proven.

(ii). Now we may assume that v ∈ V_{−μ}. As in (i) we can prove that w ∈ U, w = ∑_{k=0}^{n} b_{k}u_{k}. As the vertices of S(cv + w) belong to {±μ} then w = b_{0}u_{0} + b_{n}u_{n}.

However, F (1)v = v and F (1)w = b_{0} ∑_{k=0}^{n} u_{k} + b_{n}u_{n} , hence b_{0} = 0 and w ∈ V_{−μ}. Consider now non-zero vectors in sl_{2}-module V with the following property:

For every g ∈ SL_{2}(𝔽^{}¯ ) vertices of S(gv) belong to the set

{±μ, ±(μ − α)}, but S(gv) is not a segment [α − μ, μ − α].  (∗∗)

The following lemma is an analogue of Lemma 6 adjusted for the situation of four real roots in a root string.

Lemma 7. Let V be a finite-dimensional < e_{α}, h_{α}, f_{α} >≅ sl_{2}-module over the field 𝔽 with the highest weight μ, 2μ = nα, n ≥ 3, dimV_{μ} = dimV_{μ−α} = 1. Let v be a non-zero vector in V , satisfying property (**). Then (i). There exists g_{0} ∈ SL_{2}(𝔽) such that g_{0}v ∈ V_{−μ} ⊕ V_{−μ+α}. (ii). If for vector w ∈ V w + cv satisfies property (**) for all c ∈ 𝔽 then g_{0}w ∈ V_{−μ} ⊕ V_{−μ+α}.

Proof. (i) Let τ(v) = ∑_{k=0}^{n} a_{k}u_{k}. If a_{0} = a_{1} = 0 then v ∈ V_{−μ} ⊕ V_{−μ+α}.

If a_{0} = 0, a_{1} ≠ 0 then as in Lemma 6 there exists λ_{0} ∈ 𝔽 such that (F (λ_{0})v)_{−μ} = 0 and hence F (λ_{0})v ∈ V_{μ−α} , so σ_{α}F (λ_{0})v ∈ V_{α−μ}.

If a_{0} ≠ 0 then there exists λ_{0} ∈ 𝔽^{}¯ such that (F (λ_{0})v)_{−μ} = 0. If (F (λ_{0})v)_{−μ+α} = 0 then F (λ_{0})v ∈ V̄_{μ} ⊕ V̄_{μ−α} and σ_{α}F (λ_{0})v ∈ V̄_{−μ} ⊕ V̄_{α−μ}.

If (F (λ_{0})v)_{−μ+α} ≠ 0 then there exists ϵ_{0} ∈ 𝔽^{}¯ such that (E(ϵ_{0})F (λ_{0})v)_{μ} = 0. Hence E(ϵ_{0})F (λ_{0})v ∈ V̄_{α−μ}.

We proved that there exists g_{1} ∈ SL_{2}(𝔽^{}¯ ) such that g_{1}v ∈ V̄_{−μ} ⊕ V̄_{α−μ} , let us prove now that the required element can be found in SL_{2}(𝔽) .

Using the Bruhat decomposition for SL_{2}(𝔽^{}¯ ) we get that g_{1}^{−1} may be written in the form σ_{α}F (κ)H(η) or in the form E(λ)F (κ)H(η) for some λ, κ, η ∈ 𝔽^{}¯ .

Note that F (κ)H(η)g_{1}(v) ∈ V̄_{−μ} ⊕ V̄_{α−μ}. If g_{1}^{−1} = σ_{α}F (κ)H(η) then σ_{α}^{−1}v ∈ V̄_{−μ} ⊕ V̄_{α−μ}. If g_{1}^{−1} = E(λ)F (κ)H(η) then v = g_{1}^{−1}g_{1}v = E(λ)(au_{n−1} + bu_{n}) , where

au_{n−1} + bu_{n} = F (κ)H(η)g_{1}(v), a, b ∈ 𝔽^{}¯ .

To prove that λ, a, b ∈ 𝔽 let us compute several components of v : v_{−μ} = (E(λ)(au_{n−1} + bu_{n}))_{−μ} = bu_{n} , hence b ∈ 𝔽 as v ∈ V. Also,

v_{−μ+α} = (a + λb)u_{n−1},
v_{−μ+2α} = (2λa + λ^{2}b)u_{n−2},
v_{−μ+3α} = (3λ^{2}a + λ^{3}b)u_{n−3}.

Consequently, c_{1} = a + λb; c_{2} = λ(2a + λb); c_{3} = λ^{2}(3a + λb) ∈ 𝔽.

Clearly, we may assume that b ≠ 0. We get:

λ(2c_{1} − λb) = c_{2}, λ^{2}(3c_{1} − 2λb) = c_{3}

and

λ^{2}c_{1} = 2λc_{2} − c_{3}.

If c_{1} = 0 then λ = \frac{c_{3}}{2c_{2}} ∈ 𝔽, or a = 0, λb = 0. If c_{1} ≠ 0 then λ^{2} = \frac{2c_{2}}{c_{1}} λ− \frac{c_{3}}{c_{1}} = −\frac{2a}{b}λ − \frac{c_{2}}{b} , hence a, λ ∈ 𝔽. Consequently, E(−λ)v ∈ V_{−μ} ⊕ V_{α−μ}, λ ∈ 𝔽 , as was to be proven.

The proof of part (ii) is analogous to the proof of (ii) of the previous lemma. Proof of the Theorem 1. It is sufficient to prove that every finite-dimensional

abelian subalgebra 𝔥 acting on 𝔤(A) semisimply can be conjugated into 𝔥_{0}. This will imply that every Cartan subalgebra in 𝔤(A) is finite-dimensional and hence is conjugated with 𝔥_{0}.

The arguments from [PK] show that it is enough to prove that 𝔥 can be conjugated into 𝔟_{−} .

Since 𝔥 is finite-dimensional, supp(𝔥) is finite and there exists x ∈ 𝔥 such that supp(x) = supp(𝔥). As x is locally finite then by Proposition 2 all non-zero vertices of S(x) are real roots.

Consider a natural partial ordering on ℤ^{l} ⊃ Δ. Let us prove that S(x) has a unique maximal vertex with respect to this partial ordering.

Indeed, let α be a maximal vertex of S(x). By Proposition 3 all edges going out of α are parallel to some roots, but Δ = Δ_{+} ∪ {0} ∪ Δ_{−} , Δ_{±} ⊂ ℤ_{±}^{l}. As α is maximal then no edges from α go in a positive direction, hence all edges from α go in a negative direction. This implies that α is the unique maximal vertex as S(x) is convex.

Applying the procedure of “stripping off edges” we shall show that we can always lower the maximal positive vertex α of supp(𝔥) .

Let σ_{i} be a fundamental reflection such that σ_{i}(α) < α. The root string Γ through α along α_{i} in Δ contains either 2 or 4 real roots. Due to Proposition 2 and Lemma 5, x_{Γ} satisfies the conditions of Lemma 6 or Lemma 7. By these lemmas there exists g_{0} ∈ SL_{2}(𝔽)_{i} such that supp(g_{0}(x)) has a maximal vertex in Γ which is lower than α. Moreower , by Lemmas 6,7 (ii) the maximal vertex of supp(g_{0}(𝔥)) is strictly less than α.

Continuing the process of lowering the maximal vertex of supp(𝔥) we shall eventually construct g ∈ G(A) such that g(𝔥) ⊂ 𝔟_{−} , as was to be proven.

References

[HMR]. J. Hurrelbrink, J. Morita and U. Rehmann, On the homological π_{0} of Kac-Moody groups over fields, Cont. Math., 126(1992) 71–77. [K]. V.G. Kac, Infinite-dimensional Lie algebras, Birkhauser, Boston, 1983. [MP]. R. V. Moody and A. Pianzola, Lie algebras with triangular decomposition, to appear. [PK]. D. H. Peterson and V. G. Kac, Infinite dimensional flag varieties and conjugacy theorems, Proc. Nat. Acad. Sci. U. S. A.,80 (1983)1778–1782.