Journal of Lie Theory Volume 4 (1994) 105–137 C ◯1994 Heldermann Verlag

Sums of Adjoint Orbits

Didier Arnal and Jean Ludwig

Communicated by K. H. Hofmann

Abstract. We show that the sum of two adjoint orbits in the Lie algebra of an exponential Lie group coincides with the Campbell-Baker-Hausdorff product of these two orbits.

Introduction

N. Wildberger and others have recently investigated the structure of the hypergroup of the adjoint orbits in relation with the class hypergroup of compact Lie groups. A generalization of the notion of this type of hypergroup to non-compact groups, for instance to nilpotent or exponential Lie groups, leads to the problem of determining a precise relation between the sum of adjoint orbits in the Lie algebra and the product of the corresponding conjugacy classes in the group (see [1], and [4]). In ([3]) Wildberger has shown that for nilpotent Lie groups G the exponential of the sum of two adjoint orbits Ω_{1} + Ω_{2} is equal to the product exp Ω_{1} ·exp Ω_{2} in G. In this paper we consider the same problem for exponential groups.

Let us recall that by the definition of exponential Lie groups, the mappings

exp: 𝔤 → G and log: G → 𝔤

are diffeomorphisms. We can transfer the group multiplication in G via exp to a group multiplication in the Lie algebra 𝔤 and we shall denote it by the symbol ∗. We obtain the so called Baker-Campbell-Hausdorff multiplication in 𝔤, which is given by

U ∗ V = U + V + \frac{1}{2}[U, V] + \frac{1}{12}[U, [U, V]] + \frac{1}{12}[V, [V, U]] + ⋯

for small U and V in 𝔤.

Let X and Y be two elements of the Lie algebra 𝔤 of the exponential group G. We denote by ^{A}X = Ad(A)X the adjoint action of the element A of G on X, and by

GX = {^{A}X | A ∈ G}

ISSN 0940–2268 / $2.50 ^{_{C} ◯} Heldermann Verlag the adjoint orbit of X. For h ∈ G, let

C(h) = {g · h · g^{−1} | g ∈ G}

be the conjugacy class of h.

We show in this note that exp(^{G}X+^{G}Y) is equal to C(exp X)·C(exp Y).

Theorem A. Let G be an exponential Lie group with Lie algebra 𝔤. For any elements X and Y of 𝔤 we have

GX +^{G}Y = ^{G}X ∗^{G}Y.

In fact in order to prove this identity, it suffices (see the end of the proof of Theorem A, after Lemma 11) to take two elements X and Y in 𝔤 and to show that there exist C, D, K, L in the subalgebra 𝔥 of 𝔤 generated by X and Y , such that

X ∗ Y = ^{C}X +^{D}Y, X + Y = ^{K}X ∗^{L}Y.

If we consider these identities on a purely formal level, they are almost trivial. Indeed, if 𝔥_{∞} is the free Lie algebra generated by X and Y , then we can form the formal CBH product U ∗ V as infinite power series in the brackets of X and Y and we obtain in this fashion a group structure on 𝔥_{∞}. It is easy to see (for instance [3]) that

(0.1)  ^{𝔥_{∞}}X ∗^{𝔥_{∞}}Y = X + Y + [𝔥_{∞}, 𝔥_{∞}] = ^{𝔥_{∞}}X +^{𝔥_{∞}}Y.

If 𝔥 is nilpotent then we get from this formal identity that

(0.2)  ^{𝔥}X ∗^{𝔥}Y = X + Y + [𝔥, 𝔥] = ^{𝔥}X +^{𝔥}Y.

In the exponential non-nilpotent case, (0.2) is no longer true (see the first example in the last section of this paper) and we are forced to use closures.

Theorem B. Let H be an exponential Lie group with Lie algebra 𝔥. If 𝔥 is generated by two elements X and Y , then

(^{H}X +^{H}Y)^{−} = X + Y + [𝔥, 𝔥] = (C(X) ∗ C(Y))^{−}.

(the symbol ‘^{−} ’ here means topological closure in 𝔥).

We see that in order to prove Theorem A we cannot use the result of Theorem B. In fact, Theorem A is much more delicate. Its proof requires a detailed analysis of the structure of a solvable Lie algebra generated by two elements.

The second example in the last section shows that in general solvable Lie groups the exponential of the sum of two adjoint orbits Ω_{1} and Ω_{2} may be much smaller than the product of exp Ω_{1} with exp Ω_{2} .

This example allows us finally to present in Theorem C a new characterisation of solvable exponential groups.

Proof of Theorem A

The proof of Theorem A needs some preparations.

Definition . Let 𝔥 be an exponential Lie algebra generated by two elements X and Y . Let 𝔟 be an ideal of 𝔥. We denote by S_{𝔟}^{∗}, resp. by S_{𝔟}^{+} , the set of all pairs (C, D), resp. (K, L) in G × G, such that

(1)  X ∗ Y = ^{C}X +^{D}Y mod 𝔟, resp. X + Y = ^{K}X ∗^{L}Y mod 𝔟.

If 𝔟 = {0}, then we abbreviate S_{β}^{∗}, resp. S_{β}^{+} to S^{∗}, resp. S^{+} .

Remark . If 𝔠 is another ideal of 𝔤 contained in 𝔟, then obviously

S_{𝔠}^{∗} ⊂ S_{𝔟}^{∗} resp. S_{𝔠}^{+} ⊂ S_{𝔟}^{+}.

We need the following well known formula. For any t ∈ ℝ define

f(t) = (−t)^{−1}(e^{−t} − 1).

Let us write for a linear operator Ψ on a finite dimensional vector space 𝔞

e(Ψ) = exp(Ψ) = ∑_{k=0}^{∞} \frac{1}{k!}Ψ^{k},

f(Ψ) = ∑_{k=1}^{∞} \frac{1}{k!}(−Ψ)^{k−1}
= ((−Ψ)^{−1}(e(−Ψ) − 1) if Ψ is invertible).

With these notations we have

Lemma 1. Let G be an exponential Lie group with Lie algebra 𝔤, let 𝔞 and 𝔗 be two abelian subalgebras of 𝔤 such that [𝔗, 𝔞] ⊂ 𝔞. Write Ψ = ad T_{|𝔞} for T ∈ 𝔗. Then for any T, T′ ∈ 𝔗, A, A′ ∈ 𝔞 we have

(4) (T + A) ∗ (T′ + A′) = T + T′ + f(Ψ + Ψ′)^{−1}(e(−Ψ′) · f(Ψ)A + f(Ψ′)A′).

In particular,

T ∗ f(Ψ)A = T + A = (e(Ψ) · f(Ψ)A) ∗ T.

Proof. Let 𝔠 = 𝔗 + 𝔞 and let us realize the exponential group C of 𝔠 as a semi-direct product of 𝔗 with 𝔞, i. e.,

C = 𝔗 × 𝔞 with multiplication (T, A) · (T′, A′) = (T + T′, exp(−Ψ′)A + A′) for any T, T′ ∈ 𝔗, A, A′ ∈ 𝔞. It is easy to see that the exponential mapping exp: 𝔠 → C is given by:

exp(T + A) = (T, f(Ψ)A).

Indeed, for any α and β ∈ ℝ we have

(αT, f(αΨ)αA) · (βT, f(βΨ)βA)
= ((α + β)T, exp(−βΨ) · f(αΨ)αA + f(βΨ)βA))
= ((α + β)T, f((α + β)Ψ)A).

Hence our mapping exp satisfies the functional equation:

exp(αX) · exp(βX) = exp(α + β)X.

Also,

\frac{d}{dt} exp(tX)_{|t=0} = X for any X ∈ 𝔠, α, β ∈ ℝ.

Hence exp must be the exponential mapping. The inverse mapping log is thus given by:

log(T, A) = T + f(Ψ)^{−1}A, T ∈ 𝔗, A ∈ 𝔞.

We can now compute the CBH product ∗ on C. Indeed

(T + A) ∗ (T′ + A′) = log(exp(T + A) · exp(T′ + A′))
= log((T, f(Ψ)A) · (T′, f(Ψ′)A′))
= log((T + T′, e(−Ψ′) · f(Ψ)A + f(Ψ′)A′))

= (T + T′) + f(Ψ + Ψ′)^{−1} · (e(−Ψ′) · f(Ψ)A + f(Ψ′)A′).

This finishes our proof.

Remark . If 𝔞_{ℂ} is the complexification of 𝔞 then we can extend Ψ, f(Ψ) and e(Ψ) ℂ-linearly to 𝔞_{ℂ}; we shall use the following relations for Z = X + iY ∈ 𝔞_{ℂ}:

(1.1) ℜ(Ψ(Z)) = Ψ(X), ℜ(f(Ψ)(Z)) = f(Ψ)(X), ℜ(e(Ψ)(Z)) = e(Ψ)(X).

The next lemma gives us special minimal ideals in 𝔥 which we shall use in the determination of the sets S^{∗} and S^{+} .

Lemma 2. Let 𝔥 be an exponential Lie algebra which is generated by two elements X and Y . Let 𝔪 be a noncentral ideal in 𝔥. Then 𝔪 contains an ideal 𝔟 of 𝔥 which is one of the following five types.

(i) 𝔟 = ℝU is one dimensional. There exists a nontrivial homomorphism Ψ: 𝔥 → ℝ such that

[A, U] = Ψ(A)U for any A ∈ 𝔥

and Ψ(X) ≠ 0 or Ψ(Y) ≠ 0.

(ii) 𝔟 = ℝU_{1} + ℝU_{2} is two-dimensional. There exists a complex nontrivial homomorphism Ψ of 𝔥 such that

[A, U_{1} + iU_{2}] = Ψ(A)(U_{1} + iU_{2}) for any A ∈ 𝔥

and Ψ(X) ≠ 0 or Ψ(Y) ≠ 0. (iii) 𝔟 = ℝU + ℝZ is two-dimensional and Z is contained in the center 𝔷

of 𝔥. There exists a nontrivial linear functional φ : 𝔥 → ℝ, which is a homomorphism on [𝔥, 𝔥], and a nontrivial homomorphism Ψ : 𝔥 → ℝ such that

[A, U] = Ψ(A)U + φ(A)Z, for all A ∈ 𝔥

and Ψ(X) · Ψ(Y) ≠ 0, φ(X) ≠ 0 or φ(Y) ≠ 0. (iv) 𝔟 = ℝU_{1} + ℝU_{2} + ℝZ_{1} + ℝZ_{2} is three or four dimensional and Z_{1}, Z_{2}

are contained in the center of 𝔥. We have [U_{1}, U_{2}] = 0 and there exist a nontrivial linear functional φ: 𝔥 → ℂ, which is a homomorphism on [𝔥, 𝔥], and a complex valued homomorphism Ψ: 𝔥 → ℂ such that

[A, U_{1} + iU_{2}] = Ψ(A)(U_{1} + iU_{2}) + φ(A)(Z_{1} + iZ_{2})

and Ψ(X) · Ψ(Y) ≠ 0, φ(X) ≠ 0 or φ(Y) ≠ 0.

(v) there exists an element U ≠ 0 in 𝔪 such that (0) ≠ [X, U], resp., 0 ≠ [Y, U] is contained in the center of 𝔥 and 𝔟 = ℝ[X, U], resp. 𝔟 = ℝ[Y, U].

Proof. Suppose first that there exists a minimal abelian ideal 𝔟 of 𝔥 contained in 𝔪 such that the intersection of 𝔟 with the center 𝔷 of 𝔥 is trivial. Since 𝔥 is solvable, 𝔟 must be of dimension 1 or 2. Furthermore since 𝔟 is not central we must have that [𝔥, 𝔟] ≠ (0). This gives us the cases (i) and (ii).

If no such ideal exists then 𝔷′ = 𝔪 ∩ 𝔷 ≠ (0), since now any minimal ideal of 𝔥 contained in 𝔪 is central. Let us choose a proper minimal ideal 𝔟̃ in 𝔪̃ = 𝔥/𝔷′. If 𝔟̃ is central in 𝔥̃ then 𝔟̃ is necessarily one dimensional since it is minimal, so we are in case (v). If 𝔟̃ is one dimensional and not central then we choose U′ in 𝔪 such that ℝ(U′ mod 𝔷′) = 𝔟̃. We have for any A ∈ 𝔥

[A, U′] = Ψ(A)U′ + Z_{A}

for some Ψ(A) in ℝ and some Z_{A} in 𝔷′. We can assume that the homomorphism Ψ of 𝔥 is not trivial, since otherwise 𝔟̃ would be central in 𝔥̃. Hence either Ψ(X) ≠ 0 or Ψ(Y) ≠ 0. If Ψ(X) = 0 and if Z_{X} = 0 then ℝ(U′ + Ψ(Y)^{−1} · Z_{Y}) is ad(X) and ad(Y) invariant and hence is a noncentral ideal of 𝔥 contained in 𝔪, which is impossible. If Ψ(X) = 0 but [X, U′] ≠ 0, then we are in case (v). If Ψ(X) ≠ 0, we replace U′ by U = U′ + Ψ(X)^{−1}Z_{X}. Whence [X, U] = Ψ(X)U and [Y, U] = Ψ(Y)U +Z for some Z in 𝔷′. The vector Z is not 0 since then ℝU would be a minimal noncentral ideal of 𝔥 contained in 𝔪. Now since Y and X generate 𝔥, we must have that [𝔥, U] ⊂ ℝU + ℝZ. An easy computation shows that φ is a homomorphism on ker Ψ ⊃ [𝔥, 𝔥]. This is case (iii).

Similarly, if 𝔟̃ is two-dimensional then we can find U′_{1} and U′_{2} in 𝔪 such that 𝔟̃ = span(U′_{1}, U′_{2}) mod 𝔷′ and such that for any A in 𝔥:

[A, U′_{1} + iU′_{2}] = Ψ(A)(U′_{1} + iU′_{2}) + Z_{A}

for some Ψ(A) in ℂ and some Z_{A} in (𝔷′)_{ℂ}. The homomorphism Ψ is not trivial since 𝔟̃ is two-dimensional. If Ψ(X) = 0 then we are either in the case ii) or in the case v). If Ψ(X) ≠ 0, we replace U′_{1} + iU′_{2} by U_{1} + iU_{2} = U′_{1} + iU′_{2} + Ψ(X)^{−1}Z_{X} and we get for Ξ = U_{1} + iU_{2} the relations

[X, Ξ] = Ψ(X)(Ξ), [Y, Ξ] = Ψ(Y)(Ξ) + Z

for some Z in (𝔷′)_{ℂ}. The vector Z cannot be 0 since otherwise span(U_{1}, U_{2}) would be a minimal noncentral ideal in 𝔥. Since [\overline{Ξ}, Ξ] ∈ ℂZ, necessarily, 0 = [X, [\overline{Ξ}, Ξ]]. On the other hand, [X, [\overline{Ξ}, Ξ]] = (\overline{Ψ(X)} + Ψ(X))[\overline{Ξ}, Ξ]. Since G is exponential, we have (\overline{Ψ(X)}+Ψ(X)) ≠ 0, hence [\overline{Ξ}, Ξ] = 0. This is case (iv).

Definition . We say that an ideal 𝔟 of 𝔥 is dangerous if it has the form (iii) or (iv) in Lemma 2.

Lemma 3. (a) Let 𝔥 be an exponential Lie algebra. Let U_{1} and U_{2} be two elements in 𝔥 such that [U_{1}, U_{2}] = 0 and such that there exists a nontrivial complex valued homomorphism Ψ of 𝔥 with

[A, U_{1} + iU_{2}] = Ψ(A)(U_{1} + iU_{2})

for any A in 𝔥.

Let X and Y be two elements in 𝔥 and suppose that Ψ(Y) ≠ 0. Let B be an element in 𝔟 = span(U_{1}, U_{2}). Then there exists for any (α_{1}, α_{2}) in ℝ^{2} an element (β_{1}, β_{2}) such that

(α_{1}U_{1}+α_{2}U_{2})X +^{(β_{1}U_{1}+β_{2}U_{2})}Y = X + Y + B.

Furthermore there exists for any (α_{1}, α_{2}) in ℝ^{2} another element (β_{1}, β_{2}) such that

(α_{1}U_{1}+α_{2}U_{2})X ∗^{(β_{1}U_{1}+β_{2}U_{2})}Y = X + Y + B.

(b) Let 𝔥 be an exponential Lie algebra. Let U be an element in 𝔥 such that there exists a nontrivial real valued homomorphism Ψ of 𝔥 such that

[A, U] = Ψ(A)U

for any A in 𝔥.

Let X and Y be two elements in 𝔥 and suppose that Ψ(Y) ≠ 0. Let B be an element in 𝔟 = span(U). There exists for any α in ℝ an element β in ℝ such that

αUX +^{βU}Y = X + Y + B.

Furthermore there exists for any α in ℝ another element β such that

αUX ∗^{βU}Y = X ∗ Y + B.

Proof. (a) We must make some precise computations involving the complexification 𝔥_{ℂ} of 𝔥. Any vector C = γ_{1}U_{1} + γ_{2}U_{2} of 𝔟 can be written as

C = ℜ(γ · Ξ),

where γ = γ_{1} − iγ_{2} ∈ ℂ and where Ξ = U_{1} + iU_{2} . Let us write

B = ℜ(ωΞ).

Now if we set α = −α_{1} + iα_{2}, β = −β_{1} + iβ_{2} , we get

(3.1)  \begin{matrix}^{(α_{1}U_{1}+α_{2}U_{2})}X +^{(β_{1}U_{1}+β_{2}U_{2})}Y \\ = & X + [α_{1}U_{1} + α_{2}U_{2}, X] + Y + [β_{1}U_{1} + β_{2}U_{2}, Y] \\ = & X + Y + ℜ((Ψ(X)α + Ψ(Y)β)Ξ). \end{matrix}

We see now that for every α in ℂ we find β in ℂ such that Ψ(X)α+Ψ(Y)β = ω, i.e. such that

(α_{1}U_{1}+α_{2}U_{2})X +^{(β_{1}U_{1}+β_{2}U_{2})}Y = X + Y + B.

In the same way we treat:

(α_{1}U_{1}+α_{2}U_{2})X ∗^{(β_{1}U_{1}+β_{2}U_{2})}Y = (X + [α_{1}U_{1} + α_{2}U_{2}, X]) ∗ (Y + [β_{1}U_{1} + β_{2}U_{2}, Y]) = (X + ℜ(Ψ(X)αΞ)) ∗ (Y + ℜ(Ψ(Y)βΞ)) = X ∗ (ℜ(f(Ψ(X))Ψ(X)αΞ)) ∗ (ℜ(e(Ψ(Y))f(Ψ(Y))Ψ(Y)βΞ) ∗ Y = X ∗ Y + ℜ(f(Ψ(X) + Ψ(Y))^{−1}{(e(−Ψ(Y))f(Ψ(X))Ψ(X)α + f(Ψ(Y))Ψ(Y)β)Ξ})

(by Lemma 1). Whence if we set

(3.2) f(Ψ(X) + Ψ(Y))^{−1}{e(−Ψ(Y))f(Ψ(X))Ψ(X)α + f(Ψ(Y))Ψ(Y)β} = ω,

then we get for every (α_{1}, α_{2}) in ℝ^{2} an element (β_{1}, β_{2}) in ℝ^{2} such that

(α_{1}U_{1}+α_{2}U_{2})X ∗^{(β_{1}U_{1}+β_{2}U_{2})}Y = X ∗ Y + B.

Part (b) is similar.

Lemma 4. (a) Let 𝔥 be an exponential Lie algebra. Let U_{1} and U_{2} be two elements of 𝔥 such that [U_{1}, U_{2}] = 0 and such that there exist a complex homomorphism Ψ, a complex linear functional φ of 𝔥, and a central vector Z_{1} + iZ_{2} = Z ≠ 0 in 𝔥_{ℂ}, such that

[A, Ξ] = Ψ(A)(Ξ) + φ(A)Z

for every A in 𝔥, where Ξ = U_{1} + iU_{2} .

Let X and Y be two elements in 𝔥 so that Ψ(X) · Ψ(Y) ≠ 0. Suppose furthermore that

det | \begin{matrix}Ψ(X) & Ψ(Y) \\ φ(X) & φ(Y) \end{matrix} | ≠ 0.

Let C = ℜ(γZ) and B = ℜ(ρΞ), for some γ, ρ ∈ ℂ. Then there exist (α_{1}, α_{2}) and (β_{1}, β_{2}) in ℝ^{2} such that

(α_{1}U_{1}+α_{2}U_{2})X +^{(β_{1}U_{1}+β_{2}U_{2})}Y = X + Y + B + C.

Furthermore there exist (ω_{1}, ω_{2}) and (τ_{1}, τ_{2}) in ℝ^{2} such that

(ω_{1}U_{1}+ω_{2}U_{2})X ∗^{(τ_{1}U_{1}+τ_{2}U_{2})}Y = X ∗ Y + B + C.

(b) Let 𝔥 be an exponential Lie algebra. Let U be an element of 𝔥 such that there exists a real homomorphism Ψ and a real linear functional φ of 𝔥 and a central vector Z ≠ 0 in 𝔥, with

[A, U] = Ψ(A)U + φ(A)Z for any A ∈ 𝔥.

Let X and Y be two elements in 𝔥 so that Ψ(X)·Ψ(Y) ≠ 0. Suppose furthermore that

det | \begin{matrix}Ψ(X) & Ψ(Y) \\ φ(X) & φ(Y) \end{matrix} | ≠ 0.

Let C = cZ and B = rU for some c, r ∈ ℝ.

Then there exist α and β in ℝ such that

αUX +^{βU}Y = X + Y + B + C.

Furthermore there exists for any ω in ℝ another element τ such that

ωUX ∗^{τU}Y = X ∗ Y + B + C.

Proof. (a) We set

α = −α_{1} + iα_{2}, resp. β = −β_{1} + iβ_{2},

and have

(α_{1}U_{1}+α_{2}U_{2})X +^{(β_{1}U_{1}+β_{2}U_{2})}Y
= X + [α_{1}U_{1} + α_{2}U_{2}, X] + Y + [β_{1}U_{1} + β_{2}U_{2}, Y]
= X + ℜ([X, α · Ξ]) + Y + ℜ([Y, β · Ξ])
= X + Y + ℜ((αΨ(X) + βΨ(Y)) · Ξ) + ℜ((αφ(X) + βφ(Y))Z).

Arnal and Ludwig 113

Since det | | ≠ 0, there exists a unique pair (α, β) in ℂ^{2} such that

(4.1)  αΨ(X) + βΨ(Y) = ρ and αφ(X) + βφ(Y) = γ.

This means of course that

X + Y + B + C = ^{(α_{1}U_{1}+α_{2}U_{2})}X +^{(β_{1}U_{1}+β_{2}U_{2})}Y.

Finally, for ω = −ω_{1} + iω_{2} and τ = −τ_{1} + iτ_{2} ∈ ℂ we have

E = ^{(ω_{1}U_{1}+ω_{2}U_{2})}X ∗^{(τ_{1}U_{1}+τ_{2}U_{2})}Y
= (X + ℜ([X, ω · Ξ]) ∗ (Y + ℜ([Y, τ · Ξ])
= (X + ℜ(Ψ(X)ω(Ξ + Ψ(X)^{−1} · φ(X)Z))∗
∗ (Y + ℜ(Ψ(Y)τ(Ξ + Ψ(Y)^{−1} · φ(Y)Z))
= X ∗ ℜ(ωf(Ψ(X))Ψ(X)(Ξ + Ψ(X)^{−1} · φ(X)Z))
∗ ℜ(τ · e(Ψ(Y))f(Ψ(Y))Ψ(Y)(Ξ + Ψ(Y)^{−1} · φ(Y)Z)) ∗ Y
= X ∗ ℜ((ωf(Ψ(X))Ψ(X) + τe(Ψ(Y))f(Ψ(Y))Ψ(Y))Ξ) ∗ Y
+ ℜ((ωf(Ψ(X))φ(X) + τe(Ψ(Y))f(Ψ(Y))φ(Y))Z).

by Lemma 1. Let us set

ωf(Ψ(X))Ψ(X) + τe(ΨY))f(Ψ(Y))Ψ(Y) = a

and

ωf(Ψ(X))φ(X) + τe(Ψ(Y))f(Ψ(Y))φ(Y) = b.

Then

E = X ∗ ℜ(a(Ξ + Ψ(Y)^{−1}φ(Y)Z)) ∗ Y + ℜ((−aΨ(Y)^{−1}φ(Y) + b)Z)

= X ∗ Y ∗ ℜ(e(−Ψ(Y))a(Ξ + Ψ(Y)^{−1}φ(Y)Z)) + ℜ((−aΨ(Y)^{−1}φ(Y) + b)Z) = X ∗ Y ∗ ℜ(e(−Ψ(Y))a)Ξ + ℜ((a(−Ψ(Y)^{−1}φ(Y)

+ e(−Ψ(Y))Ψ(Y)^{−1}φ(Y)) + b)Z) = X ∗ Y ∗ ℜ(e(−Ψ(Y))a(Ξ + Ψ(X ∗ Y)^{−1}φ(X ∗ Y)Z))

+ ℜ((a{−Ψ(Y)^{−1}φ(Y) − e(−Ψ(Y))Ψ(X ∗ Y)^{−1}φ(X ∗ Y)

+ e(−Ψ(Y))Ψ(Y)^{−1}φ(Y)} + b)Z)

= X ∗ Y + ℜ((f(Ψ(X ∗ Y))^{−1}e(−Ψ(Y))a)Ξ)

+ ℜ((a{f(Ψ(X ∗ Y))^{−1}e(−Ψ(Y))Ψ(X ∗ Y)^{−1}φ(X ∗ Y) − Ψ(Y)^{−1}φ(Y) − e(−Ψ(Y))Ψ(X ∗ Y)^{−1}φ(X ∗ Y) + e(−Ψ(Y))Ψ(Y)^{−1}φ(Y)} + b)Z).

Since det | | ≠ 0 we can choose ω and τ such that

(4.2) ωf(Ψ(X))Ψ(X) + τ · e(Ψ(Y))f(Ψ(Y))Ψ(Y) = ρe(Ψ(Y))f(Ψ(X ∗ Y))

and

ωf(Ψ(X))φ(X) + τe(Ψ(Y))f(Ψ(Y))φ(Y)
= γ − ρe(Ψ(Y))f(Ψ(X ∗ Y)){(f(Ψ(X ∗ Y))^{−1} − 1)
e(−Ψ(Y))Ψ(X ∗ Y)^{−1}φ(X ∗ Y) + (e(−Ψ(Y)) − 1)Ψ(Y)^{−1}φ(Y)},

and we get

(ω_{1}U_{1}+ω_{2}U_{2})X ∗^{(τ_{1}U_{1}+τ_{2}U_{2})}Y = X ∗ Y + B + C.

The proof of (b) is similar to the proof (a).

Lemma 5. (a) Let 𝔥 be an exponential Lie algebra. Let U_{1} and U_{2} be two elements in 𝔥, such that [U_{1}, U_{2}] = 0, let Z_{1}, Z_{2} be two central elements of 𝔥, such that

[A, U_{1} + iU_{2}] = Ψ(A)(U_{1} + iU_{2}) + φ(A)(Z_{1} + iZ_{2})

for any A in 𝔥. Let X and Y be two elements in 𝔥, such that Ψ(X)·Ψ(Y) ≠ 0, Ψ(X) + Ψ(Y) = 0, φ(X) = 0, φ(Y) = 1.

Let C be any element in the span of Z_{1}, Z_{2} . Then there exist α_{1}, α_{2} in ℝ, such that

α_{1}U_{1}+α_{2}U_{2}X +^{α_{1}U_{1}+α_{2}U_{2}}Y = X + Y + C.

Furthermore, there exist ω_{1}, ω_{2} in ℝ, such that ω_{1}U_{1}+ω_{2}U_{2}X ∗^{ω_{1}U_{1}+ω_{2}U_{2}}Y = X ∗ Y + C.

(b) Let 𝔥 be an exponential Lie algebra. Let U be an element in 𝔥, let Z be a central element of 𝔥, such that

[A, U_{1}] = Ψ(A)(U) + φ(A)(Z)

for any A in 𝔥. Let X and Y be two elements in 𝔥, such that Ψ(X)·Ψ(Y) ≠ 0, Ψ(X) + Ψ(Y) = 0, φ(X) = 0, and φ(Y) = 1.

Let C be any element in the span of Z. Then there exists an α in ℝ, such that

αUX +^{αU}Y = X + Y + C.

Furthermore, there exists ω in ℝ, such that ωUX ∗^{ωU}Y = X ∗ Y + C.

Proof. (a) Let us use the computations from the proof of Lemma 4 (a). For any (α_{1}, α_{2}) in ℝ^{2} we have

(α_{1}U_{1}+α_{2}U_{2})X +^{(α_{1}U_{1}+α_{2}U_{2})}Y
= X + ℜ([X, α · Ξ]) + Y + ℜ([Y, α · Ξ])
= X + Y + ℜ((αΨ(X) + αΨ(Y)) · Ξ) + ℜ((αφ(X) + αφ(Y))Z)
= X + Y + ℜ(αZ).

Hence it suffices to put α = γ, where γ is such that C = ℜ(γZ). Furthermore

(ω_{1}U_{1}+ω_{2}U_{2})X ∗^{(ω_{1}U_{1}+ω_{2}U_{2})}Y
= (X + ℜ([X, ω · Ξ])) ∗ (Y + ℜ([Y, ω · Ξ]))
= (X + ℜ(Ψ(X)ω(Ξ + Ψ(X)^{−1} · φ(X)Z)))
∗ (Y + ℜ(Ψ(Y)ω(Ξ + Ψ(Y)^{−1} · φ(Y)Z)))
= X ∗ (ℜ(ωf(Ψ(X))Ψ(X)(Ξ + Ψ(X)^{−1} · φ(X)Z)))
∗ (ℜ(ω · e(Ψ(Y))f(Ψ(Y))Ψ(Y)(Ξ + Ψ(Y)^{−1} · φ(Y)Z))) ∗ Y
= X ∗ (ℜ((ωf(Ψ(X))Ψ(X) + ω · e(Ψ(Y))f(Ψ(Y))Ψ(Y))Ξ)) ∗ Y
+ ℜ((ωe(Ψ(Y))f(Ψ(Y))φ(Y))Z).

Since ωf(Ψ(X))Ψ(X)+ω·e(Ψ(Y))f(Ψ(Y))Ψ(Y) = ω(e^{−Ψ(X)}−1+1−e^{Ψ(Y)}) = ω·0 = 0, it suffices to put ω = (e(Ψ(Y))f(Ψ(Y)))^{−1} ·γ, where γ is such that C = ℜ(γZ).

The proof of (b) is similar.

The structure of 𝔥

We shall now construct inductively a sequence of ideals

𝔥 ⊃ [𝔥, 𝔥] = 𝔫 ⊃ 𝔫_{r} ⊃ 𝔫_{r−1} ⋯ ⊃ 𝔫_{1} ⊃ 𝔫_{0} = (0)

such that 𝔫_{i}/𝔫_{i−1} = 𝔟_{i} is an ideal in 𝔥/𝔫_{i} of the form i) to v) in Lemma 2 (i = 1, 2, ⋯, r).

We shall use the root decomposition of 𝔥 relative to some regular element T ∈ 𝔥. Let us recall what a root Ψ of 𝔥 is. We choose any Jordan-Hölder sequence 𝔥 ⊃ 𝔥_{1} ⊃ ⋯ ⊃ 𝔥_{p} = {0} of 𝔥. The 𝔥-modules 𝔥̃_{j} = 𝔥_{j}/𝔥_{j+1} are irreducible, for j = 1 to p − 1, hence of dimension 1 or 2. We get in the dimension 1 case a real homomorphism Ψ_{j} of 𝔥. In the dimension 2 case, we have (𝔥̃_{j})_{ℂ} = ℂ Ξ + ℂ \overline{Ξ} and [A, Ξ] = Ψ_{j}(A)Ξ, [A, \overline{Ξ}] = \overline{Ψ_{j}(A)} \overline{Ξ}, for any A ∈ 𝔥 and Ψ_{j}, \overline{Ψ}_{j} are complex homomorphisms. The homomorphisms Ψ_{j} ’s and \overline{Ψ}_{j} ’s are called the roots of 𝔥. It is easy to see that the roots do not depend on a given Jordan-Hölder sequence and that for any T in 𝔥, the spectrum of ad T on 𝔥_{ℂ} is given by the numbers Ψ(T), Ψ =root of 𝔥.

Let first T be an element of 𝔥 which is in general position relatively to the roots of 𝔥, i. e. for any two distinct roots Ψ and Ψ′ of 𝔥 we have Ψ(T) ≠ Ψ′(T). We denote by

Ψ^{β}, β ∈ σ,

the corresponding root of 𝔥. Hence every root Ψ of 𝔥 is of the form Ψ^{β}, for some β ∈ σ. Furthermore, it is not difficult to see that a root Ψ, for which Ψ(T) = −β for some β ∈ σ, is equal to −Ψ^{β}.

Let 𝔥_{ℂ} = ∑_{β∈σ}(𝔥_{ℂ})_{β} be the decomposition of 𝔥_{ℂ} into the sum of the nilspaces of ad T, the summation being made over the spectrum σ of ad T. We have

[(𝔥_{ℂ})_{β}, (𝔥_{ℂ})_{β′}] ⊂ (𝔥_{ℂ})_{β+β′}, for any β, β′ in σ.

In particular (𝔥_{ℂ})_{0} = (𝔥_{0})_{ℂ} is a subalgebra of 𝔥_{ℂ} (which is in fact nilpotent) and

[(𝔥_{0})_{ℂ}, (𝔥_{ℂ})_{β}] ⊂ (𝔥_{ℂ})_{β}

for any β in σ. Furthermore for any S in 𝔥_{0} , ad S − Ψ^{β}(S) is nilpotent on (𝔥_{ℂ})_{β}. Let

𝔥_{0} = (𝔥_{0})_{ℂ} ∩ 𝔥, 𝔥_{β} = ((𝔥_{ℂ})_{β} + (𝔥_{ℂ})\overline{β})) ∩ 𝔥 = ((𝔥_{ℂ})_{β} + \overline{(𝔥_{ℂ})}_{β}) ∩ 𝔥.

Then 𝔥 = ∑_{β∈σ} 𝔥_{β} and 𝔥_{β} ⊂ 𝔫 for any β ≠ 0. If 𝔟 is any ideal of 𝔥 then

𝔟 = 𝔟 ∩ 𝔥_{0} + ∑_{β∈σ\0}(𝔥_{β}) ∩ 𝔟 =^{def} 𝔟_{0} + ∑_{β∈σ\0} 𝔟_{β}

and

𝔥/𝔟 = (𝔥/𝔟)_{0} + ∑_{β∈σ\0}(𝔥/𝔟)_{β} = (𝔥_{0}/𝔟) + ∑_{β∈σ\0}(𝔥_{β}/𝔟).

Furthermore, let Ψ be a root of 𝔥 and let β = Ψ(T). Suppose that −β is also an eigenvalue of ad_{𝔥_{ℂ}} T. Let Ψ′ be the root of 𝔥 corresponding to −β. If (𝔥_{ℂ})_{0} ⊃ [(𝔥_{ℂ})_{β}, (𝔥_{ℂ})_{−β}] ≠ {0}, then for any S in 𝔥_{0} , since ad S −(Ψ(S)+Ψ′(S)) is nilpotent on [(𝔥_{ℂ})_{β}, (𝔥_{ℂ})_{−β}], we must have that Ψ(S) + Ψ′(S) = 0. On the other hand every root of 𝔥 is trivial on ∑_{β∈σ\0} 𝔥_{β} and so Ψ′ = −Ψ.

We begin by choosing in [𝔫, 𝔫] an ideal 𝔟_{1} as in Lemma 2, if [𝔫, 𝔫] is not central. If [𝔫, 𝔫] is central, but 𝔫 is not, we choose the ideal 𝔟_{1} in 𝔫. If 𝔫 is central, but 𝔥 is not abelian, we choose 𝔟_{1} in 𝔥. If 𝔥 is abelian, we do nothing. Set 𝔫_{1} = 𝔟_{1} .

If 𝔟_{1} is dangerous, then we have Ξ_{1} in (𝔥_{ℂ})_{β} for some β ≠ 0 in σ and Z_{1} ≠ 0 in (𝔥_{0})_{ℂ} and we obtain the linear functional φ = φ_{1} of (iii) in Lemma 2 and the homomorphism Ψ_{1} , where

[A, Ξ_{1}] = Ψ_{1}(A)Ξ_{1} + φ_{1}(A)Z_{1}, A ∈ 𝔥,

and we let _{1}𝔫 = ker φ_{1} ∩ 𝔫 = {U ∈ 𝔫 | [U, Ξ_{1}] = 0}. Thus _{1}𝔫 is an ideal in 𝔥. If 𝔟_{1} is of the form (i), (ii) or (v) then we set _{1}𝔫 = 𝔫. Continuing in this fashion, we find inductively the ideals 𝔫_{2} ⊂ ⋯ ⊂ 𝔫_{j} of 𝔥 (contained in [𝔫, 𝔫] as long as [𝔫, 𝔫]/𝔫_{j−1} is not central in 𝔥/𝔫_{j−1} ), the ideals _{1}𝔫 ⊃ ⋯ ⊃ _{j}𝔫 ⊃ [𝔫, 𝔫], _{j}𝔫 ⊃ 𝔫_{j}. If _{j}𝔫/𝔫_{j} is not central in 𝔥/𝔫_{j}, then we again find an ideal 𝔟_{j+1} in _{j}𝔫/𝔫_{j}. In the case where 𝔟_{j+1} is of the form (iii) or (iv) we have Ξ_{j} in (𝔥_{ℂ})_{β} for some β_{j} = β ≠ 0 in σ and Z_{j} in (𝔥_{0})_{ℂ} and we obtain the linear functional φ = φ_{j} of iii) and the homomorphism Ψ_{j} in Lemma 2 where

(5.1)  [A, Ξ_{j}] = Ψ_{j}(A)Ξ_{j} + φ_{j}(A)Z_{j} mod (𝔫_{j})_{ℂ}, A ∈ 𝔥,

and we set _{j+1}𝔫 = _{j}𝔫 ∩ ker φ_{j} = {U ∈ _{j}𝔫 | [U, Ξ_{j}] = 0 mod (𝔫_{j})_{ℂ}}. Hence _{j+1}𝔫 is an ideal in 𝔥. If 𝔟_{j} is of the form i), ii), or v) then we set _{j+1}𝔫 = _{j}𝔫. Let 𝔫_{j+1} be the set of all elements x in 𝔥, such that x mod 𝔫_{j} ∈ 𝔟_{j+1} . This finishes step j.

We continue this process until we find some r in ℕ, for which _{r}𝔫/𝔫_{r} is central in 𝔥/𝔫_{r}. We set

𝔪 = _{r}𝔫.

We give now a precise description of 𝔥/𝔪.

Definition 1. Let J be the set of all the indices j in {1, ⋯, r} for which 𝔟_{j} is dangerous.

Thus

[𝔫, 𝔫] ⊂ 𝔪 = 𝔫 ∩ { ⋂_{j∈J} ker φ_{j}} ⊂ 𝔫

and 𝔪 is an ideal of 𝔥. Indeed 𝔫 ∩ ker φ_{j} = {u ∈ 𝔫 | [u, 𝔟_{j}] = {0} mod 𝔫_{j−1}} is an ideal of 𝔥 for any j in J. Furthermore we see from (5.1) that for every j in J, since [𝔥_{0}, (𝔥_{ℂ})_{β}] ⊂ (𝔥_{ℂ})_{β},

(5.2)  φ_{j}(𝔥_{0} ∩ 𝔫) = {0}.

Hence

𝔥_{0} ∩ 𝔫 ⊂ 𝔪.

Lemma 6. Let 𝔥̃ = 𝔥/𝔪. For any β ≠ 0 in the spectrum σ̃ of ad T on 𝔥̃ there exists j in J such that Ψ^{β} = −Ψ_{j}.

Proof. Suppose that β is not real. Choose a vector θ in (𝔥_{ℂ})_{β}, such that θ ∉ 𝔪_{ℂ} and such that

[T, θ] = βθ mod 𝔪_{ℂ} = Ψ^{β}(T)θ mod 𝔪_{ℂ}.

Since θ ∉ 𝔪_{ℂ}, there exists j in J such that φ_{j}(θ) ≠ 0. Choose Ξ_{j} and Z_{j} ≠ 0 in (𝔟_{j})_{ℂ} such that

[A, Ξ_{j}] = Ψ_{j}(A)Ξ_{j} + φ_{j}(A)Z_{j} mod (𝔫_{j−1})_{ℂ}

for any A in 𝔥. Hence

[[A, θ], Ξ_{j}] = [A, [θ, Ξ_{j}]] − [θ, [A, Ξ_{j}]] = 0 − [θ, Ψ_{j}(A)Ξ_{j}] mod 𝔫_{j−1}.

Hence [[A, θ]+Ψ_{j}(A)θ, Ξ_{j}] = 0 in (𝔥/𝔫_{j−1})_{ℂ} and so also φ_{j}([A, θ]+Ψ_{j}(A)θ) = 0, i.e.

[A, θ] = −Ψ_{j}(A)θ mod (ker φ_{j})_{ℂ} for all A ∈ 𝔥.

Since also

[T, θ] = βθ mod 𝔪_{ℂ} = βθ mod (ker φ_{j})_{ℂ}

we see that Ψ_{j}(T) = −β. Hence Ψ^{β} = −Ψ_{j} by 5.0.

In Lemma 7 and 8 we give a precise description of the elements X and Y mod𝔪 and in Lemma 9 we determine the structure of 𝔥/𝔫_{r}.

Lemma 7. Let 𝔥̃ = 𝔥/𝔪. For any β ≠ 0 in the spectrum σ̃ of ad T on 𝔥̃, ((𝔥̃)_{ℂ})_{β} is one dimensional and for T′ in 𝔥_{0} , θ in (𝔥_{ℂ})_{β}, we have [T′, θ] = Ψ^{β}(T′)θ mod 𝔪_{ℂ}. Furthermore 𝔥̃_{0} is one or two-dimensional.

Proof. Since [𝔥_{0}, 𝔥_{0}] ⊂ 𝔪 by (5.2) we have [𝔥̃_{0}, 𝔥̃_{0}] = {0}. Furthermore

[(𝔥_{ℂ})_{β}, (𝔥_{ℂ})_{γ}] ⊂ [𝔫, 𝔫] ⊂ 𝔪 for all β, γ ≠ 0 in σ

implies that

[(𝔥̃_{ℂ})_{β}, (𝔥̃_{ℂ})_{γ}] = {0}, for all β, γ ≠ 0 in σ.

Let us show that for any T′ in 𝔥_{0} , [T′, θ] − Ψ^{β}(T′)θ ∈ 𝔪_{ℂ}. Indeed, we have

[T′, θ] = Ψ^{β}(T′) · θ + θ_{1} for some θ_{1} in ((𝔥_{ℂ})_{β})

and θ is not a scalar multiple of θ_{1} , since ad(T′)−Ψ^{β}(T′) is nilpotent on ((𝔥_{ℂ})_{β}). The vector θ_{1} must be contained in 𝔪. Since otherwise we would have an index j in J, such that φ_{j}(θ_{1}) ≠ 0. Since θ_{1} is in ((𝔥_{ℂ})_{β}), [θ_{1}, Ξ_{j}] ∉ (𝔫_{j−1})_{ℂ} implies that β + Ψ_{j}(T) = 0 by 5.0 and so Ξ_{j} must be in (𝔥_{ℂ})_{−β}. Hence

[[T′, θ], Ξ_{j}] = [T′, [θ, Ξ_{j}]] − [θ, [T′, Ξ_{j}]] = 0 − Ψ_{j}(T′)φ_{j}(θ)Z_{j} mod (𝔫_{j−1})_{ℂ}.

On the other hand we also have that

[[T′, θ], Ξ_{j}] = [Ψ^{β}(T′)θ + θ_{1}, Ξ_{j}] = (Ψ^{β}(T′)φ_{j}(θ) + φ_{j}(θ_{1}))Z_{j} mod (𝔫_{j−1})_{ℂ}.

Thus φ_{j}(θ_{1}) = 0, a contradiction and θ_{1} must be an element of 𝔪_{ℂ}.

Suppose that (𝔥̃_{ℂ})_{β} is of dimension ≥ 2. Let V_{1} and V_{2} be two linearly independent vectors in ((𝔥̃)_{ℂ})_{β} and let 𝔅_{β} ⊂ ((𝔥̃)_{ℂ})_{β} be a supplementary subspace, i.e.:

((𝔥̃)_{ℂ})_{β} = ℂV_{1} ⊕ ℂV_{2} ⊕ 𝔅_{β}.

Let

𝔫′′ = 𝔅_{β} + ∑_{β′≠β,β′≠0}(𝔥̃_{ℂ})_{β′}, 𝔥′ = 𝔥̃_{ℂ}/𝔫′′, θ_{i} = V_{i} mod 𝔫′′, i = 1, 2.

Since X and Y generate 𝔥, the vectors X′ = X mod 𝔫′′ and Y′ = Y mod 𝔫′′ generate 𝔥′. Furthermore the subspace 𝔥′_{β} of 𝔥′ is spanned by θ_{i}, i = 1, 2 and 𝔥′ = 𝔥′_{0} + 𝔥′_{β}. We also have that [𝔥′_{β}, 𝔥′_{β}] and [𝔥̃_{0}, 𝔥̃_{0}] = (0). Let us write

X′ = T_{X} + α_{1}θ_{1} + α_{2}θ_{2}, Y′ = T_{Y} + β_{1}θ_{1} + β_{2}θ_{2},

where T_{X} and T_{Y} are the components of X′, resp. of Y′, in (𝔥′)_{0} . WHence

[X′, Y′] = Ψ^{β}(X)(β_{1}θ_{1} + β_{2}θ_{2}) − Ψ^{β}(Y)(α_{1}θ_{1} + α_{2}θ_{2}).

We see that 𝔥′_{1} = span(X′, Y′, Ψ^{β}(X)(β_{1}θ_{1} + β_{2}θ_{2}) − Ψ^{β}(Y)(α_{1}θ_{1} + α_{2}θ_{2})) is a subalgebra of 𝔥′ containing X′ and Y′ and so 𝔥′ = 𝔥′_{1} . But then (𝔥′)_{β} is of dimension 1, a contradiction. Obviously, since 𝔥_{0} ∩ 𝔫 ⊂ 𝔪, we have dim(𝔥̃)_{0} ≤ dim(𝔥/[𝔥, 𝔥]) ≤ 2.

Let now

σ′ = σ̃ \ {0}.

We have seen in Lemma 7 that for any β in σ′, ((𝔥)_{ℂ})_{β} mod 𝔪_{ℂ} is of dimension 1 over ℂ. We choose a vector θ′_{β} ≠ 0 in (𝔥_{ℂ})_{β} mod 𝔪_{ℂ} and we write:

\begin{aligned}X & = T_{X} + ∑_{β∈σ′} X_{β}θ′_{β} mod 𝔪_{ℂ}, \\ Y & = T_{Y} + ∑_{β∈σ′} Y_{β}θ′_{β} mod 𝔪_{ℂ}, \end{aligned}

where T_{X}, resp. T_{Y} ∈ 𝔥_{0} . We want to determine the X_{β} ’s, resp. Y_{β} ’s.

Lemma 8. We can assume that for any β in σ′, there exists a unique θ_{β} in (𝔥_{ℂ})_{β} with the following property: for every j in J we have φ_{j}(θ_{β}) = 1 or 0. Furthermore

X = T_{X} mod 𝔪, Y = T_{Y} + ℜ( ∑_{β∈σ′} θ_{β}) mod 𝔪.

For every j in J, there exists a unique β in σ′, such that φ_{j}(θ_{β}) = 1, φ_{j}(θ_{γ}) = 0 for all γ in σ′, γ ≠ β, and such that Ψ_{j} = −Ψ^{β} where Ψ^{β} is as in Lemma 7.

Proof. Let R = ℜ(∑_{β∈σ′} Ψ_{β}(X)^{−1} · X_{β}θ′_{β}) and let X′ = ^{R}X, Y′ = ^{R}Y . Then X′ = T_{X} mod 𝔪 and Y′ = T_{Y} + ∑_{β∈σ′} Y_{β}θ′_{β} mod 𝔪_{ℂ} for some new coefficients Y_{β}. We remark that 𝔥 is also generated by X′ and Y′, since R ∈ 𝔥 and so 𝔥 = ^{R}𝔥. We shall work from now on with X′ and Y′ and we shall show that there exist C, D, K, L in 𝔥 such that ^{C}X′ + ^{D}Y′ = X′ ∗ Y′ and KX′ ∗^{L}Y′ = X′ + Y′. But this means that

R^{−1}∗C∗RX +^{R^{−1}∗D∗R}Y = X ∗ Y and ^{R^{−1}∗K∗R}X ∗^{R^{−1}∗L∗R}Y = X + Y.

We can thus forget about X and Y and write X = X′, Y = Y′.

Let now j ∈ J. There exists some γ in σ and some Ξ_{j} in (𝔥_{ℂ})_{γ} ∩ 𝔫_{j} and Z_{j} in (𝔥_{ℂ})_{0} \ 𝔫_{j−1} such that (𝔫_{j})_{ℂ} mod (𝔫_{j−1})_{ℂ} = ℂΞ_{j} + ℂ(Ξ_{j})^{−} + ℂZ_{j} + ℂ(Z_{j})^{−} mod (𝔫_{j−1})_{ℂ} and such that

[A, Ξ_{j}] = Ψ_{j}(A) · Ξ_{j} + φ_{j}(A) · Z_{j} mod (𝔫_{j−1})_{ℂ}, for all A ∈ 𝔥.

Since φ_{j}(𝔫) ≠ 0, there exists some β in σ′ such that φ_{j}(θ′_{β}) ≠ 0 which implies that γ = −β, whence Ψ^{−β} = Ψ_{j}. Furthermore [θ′_{μ}, Ξ_{j}] ⊂ ((𝔥_{ℂ})_{μ−β} ∩ (𝔫_{j})_{ℂ}) ⊂ (𝔫_{j−1})_{ℂ} for all μ ≠ β. Hence φ_{j}(θ′_{μ}) = 0 for such a μ. Since X = T_{X} mod 𝔪 we have φ_{j}(X) = 0 and so φ_{j}(Y) must be ≠ 0 for every j. On the other hand, for any β in σ′, since θ′_{β} ∉ 𝔪_{ℂ}, there exists an index j in J, such that φ_{j}(θ′_{β}) ≠ 0. By rescaling Z_{j}, we may even assume that φ_{j}(Y) = 1 and so Y_{β} · φ_{j}(θ′_{β}) = 1 for any j in J associated with β. Replacing now θ′_{β} by Y_{β} · θ′_{β} = θ_{β} for all β in σ′, we finally get:

Y = T_{Y} + ∑_{β∈σ′} θ_{β} mod 𝔪_{ℂ}.

(8.1) Remark. We have just seen that for any j in J, the restriction of the linear functional φ_{j} to 𝔫 is of the form φ^{−β} for some unique β in σ′, where φ^{−β}(θ_{β}) = 1, φ^{−β}(θ_{β′}) = 0 for β′ ≠ β, φ^{−β}([𝔫, 𝔫]) = φ^{−β}(𝔪) = {0}. Furthermore Ψ^{−β} = Ψ_{j}.

Since _{r}𝔫/𝔫_{r} is central in 𝔥/𝔫_{r} we see from Lemma 8 that (𝔥)_{ℂ}/(𝔫_{r})_{ℂ} = (𝔥_{0})_{ℂ} = ∑_{β∈σ′} ℂθ_{β} mod (𝔫_{r})_{ℂ} and thus we can write

(8.2)  X = T_{X} mod 𝔫_{r}, Y = T_{Y} + ∑_{β∈σ′} θ_{β} mod (𝔫_{r})_{ℂ}

for some new elements T_{X}, T_{Y} in 𝔥_{0} . Let us set

(8.3)  S = [T_{X}, T_{Y}] ∈ [𝔥_{0}, 𝔥_{0}].

Lemma 9.

𝔥_{ℂ} = ℂT_{X} + ℂT_{Y} + ∑_{β∈σ′} ℂθ_{β} + ℂS + ∑_{β∈σ′} ℂ[θ_{β}, θ_{−β}] + (𝔫_{r})_{ℂ}.

Proof. First we observe that [𝔥, 𝔪] ⊂ 𝔫_{r}, since by the definition of 𝔫_{r}, 𝔪/𝔫_{r} is central in 𝔥/𝔫_{r}. This implies especially that for β ∈ σ′, ((𝔥_{ℂ})_{β}) ∩ 𝔪 is already contained in (𝔫_{r})_{ℂ}, since ad T_{X} is injective on ((𝔥_{ℂ})_{β}). Especially, [θ_{β}, θ′_{β}] ⊂ 𝔫_{r}, for β′ ≠ −β. Since for any β in σ′, [θ_{β}, θ_{−β}] ⊂ 𝔫_{ℂ} ∩ (𝔥_{0})_{ℂ} ⊂ 𝔪_{ℂ}, we thus have that [𝔥, [θ_{β}, θ_{−β}]] ⊂ (𝔫_{r})_{ℂ}. We have seen in Lemma 7 that for β ∈ σ′, T′ ∈ 𝔥_{0}, [T′, θ_{β}] = Ψ^{β}(T′)θ_{β} mod 𝔪_{ℂ}. Since [T′, θ_{β}] ∈ (𝔥_{β})_{ℂ}, we know now that [T′, θ_{β}] = Ψ^{β}(T′)θ_{β} mod (𝔫_{r})_{ℂ}.

Since by (5.2) [𝔥_{0}, 𝔥_{0}] ⊂ 𝔪, necessarily [[𝔥_{0}, 𝔥_{0}], 𝔥] ⊂ (𝔫_{r})_{ℂ}.

Now 𝔥′ = ℂT_{X} + ℂT_{Y} + ∑_{β∈σ′} ℂθ_{β} + ℂS + ∑_{β∈σ′} ℂ[θ_{β}, θ_{−β}] + (𝔫_{r})_{ℂ} is a subalgebra of 𝔥_{ℂ}. This subalgebra evidently contains X and Y . Hence 𝔥′ = 𝔥_{ℂ}.

(9.1) Remark. Lemma 9 tells us that

\begin{aligned}(𝔪)_{ℂ} & = [𝔥_{0}, 𝔥_{0}]_{ℂ} + ∑_{β∈σ′} ℂ[θ_{β}, θ_{−β}] + (𝔫_{r})_{ℂ} \\ & = ℂS + ∑_{β∈σ′} ℂ[θ_{β}, θ_{−β}] + (𝔫_{r})_{ℂ}. \end{aligned}

Let now for β ∈ σ′,

Z_{β} = [θ_{β}, θ_{−β}] ∈ (𝔥_{0})_{ℂ}.

Inductively we choose β_{1}, β_{2}, ⋯, β_{s} in σ′, such that

(9.2)  Z_{β_{j}} ∉ span_{ℂ}\{Z_{β_{i}}, Z_{\overline{β}_{i}} | i < j\} + (𝔫_{r})_{ℂ}

and such that

𝔪_{ℂ} = ℂS + span_{ℂ}\{Z_{β_{i}}, Z_{\overline{β}_{i}} | i = 1, ⋯, s\} + (𝔫_{r})_{ℂ}.

Let

(9.3)  σ^{+} = {β_{1}, ⋯, β_{s}} ⊂ σ′ ⊂ σ.

In particular, condition (9.2) implies that σ^{+} ∩ {−(σ^{+})} = Ø. Let

σ^{−} = −σ^{+}, σ1′ = σ′ \ \{σ^{−} ∪ (\overline{σ^{−}})\}, σ0′ = σ1′ \ \{σ^{+} ∪ \overline{σ^{+}}\}.

Let us choose a subset σ0 in σ0′ such that every real β in σ0′ is contained in σ0 and such that for any nonreal β in σ0′, \{β, \overline{β}\} ∩ σ0 contains one element and let

(9.4)  σ1 = σ0 ˙∪ σ^{+}.

Let

(_{1}𝔪)_{ℂ} = (𝔫_{r})_{ℂ} + span_{ℂ}\{θ_{β}, \overline{θ}_{β} | β ∈ σ^{−}\} + span_{ℂ}\{Z_{β}, \overline{Z}_{β} | β ∈ σ^{+}\},
(_{2}𝔪)_{ℂ} = (_{1}𝔪)_{ℂ} + ∑_{β∈σ1} ℂθ_{β}.

Hence

\begin{aligned}𝔥_{ℂ} & = ℂT_{X} + ℂT_{Y} + ∑_{β∈σ1′} ℂθ_{β} + ℂS + (_{1}𝔪)_{ℂ} \\ & = ℂT_{X} + ℂT_{Y} + ℂS + (_{2}𝔪)_{ℂ} \end{aligned}

and

(9.5) 𝔥 = ℝT_{X} + ℝT_{Y} + ∑_{β∈σ_{1}} ℜ(ℂθ_{β}) + ℝS + (_{1}𝔪) = ℝT_{X} + ℝT_{Y} + ℝS + (_{2}𝔪),

𝔪_{1} = (_{1}𝔪)_{ℂ} ∩ 𝔥 = (𝔫_{r}) + span{ℜ(ℂθ_{β}) | β ∈ σ^{−}} + span{ℜ(ℂZ_{β}) | β ∈ σ^{+}}.

We now begin the determination of some elements in S^{∗} and in S^{+} .

Lemma 10. S^{∗}_{𝔫_{r}}, resp. S^{+}_{𝔫_{r}}, contains an element (C, D) resp. (K, L) such that

0 ≠ det | \begin{matrix}Ψ_{j}(X) & Ψ_{j}(Y) \\ φ_{j}(^{C}X) & φ_{j}(^{D}Y) \end{matrix} |

resp. such that

0 ≠ det | \begin{matrix}Ψ_{j}(X) & Ψ_{j}(Y) \\ φ_{j}(^{K}X) & φ_{j}(^{L}Y) \end{matrix} |

for all j ∈ J.

Proof. The first step is to find a huge subset of S^{∗}_{_{2}𝔪} ⊃ S^{∗}_{_{1}𝔪} resp. of S^{+}_{_{2}𝔪} ⊃ S^{+}_{_{1}𝔪}. We want first to determine

(C, D) = (c_{0}T_{Y} , d_{0}T_{X}) in S^{∗}_{_{2}𝔪}

resp.

(K, L) = (k_{0}T_{y}, l_{0}T_{X}) in S^{+}_{_{2}𝔪}.

If S is an element of _{1}𝔪, then we can take any c_{0}, d_{0}, k_{0} and l_{0} in ℝ. If S ∉ _{1}𝔪, then given d_{0} , resp. k_{0} , in ℝ, we set c_{0} = d_{0} − \frac{1}{2} , resp. k_{0} = \frac{1}{2} + l_{0} .

Since _{2}𝔪 is an ideal, such that 𝔥/_{2}𝔪 is nilpotent of step ≤ 2, we have then that

CX +^{D}Y = X + Y + (d_{0} − c_{0})S = X + Y + \frac{1}{2}S = X ∗ Y mod_{2}𝔪.

Furthermore:

KX ∗^{L}Y = X + Y + (\frac{1}{2} + l_{0} − k_{0})S = X + Y mod_{2}𝔪.

We shall from now on fix c_{0}, d_{0}, k_{0} and l_{0} and we look for a large number of elements in S^{∗}_{_{1}𝔪} resp. S^{+}_{_{1}𝔪} of the form

(C, D) = ((ℜ( ∑_{β∈σ1} c_{β}θ_{β})) ∗ c_{0}T_{Y} , (ℜ( ∑_{β∈σ1} d_{β}θ_{β})) ∗ d_{0}T_{X})

resp.

(K, L) = (ℜ( ∑_{β∈σ1} k_{β}θ)) ∗ k_{0}T_{Y} , (ℜ( ∑_{β∈σ1} l_{β}θ)) ∗ l_{0}T_{X}).

Since

X = T_{X} mod _{1}𝔪, Y = T_{Y} + ∑_{β∈σ1} θ_{β} mod (_{1}𝔪)_{ℂ},

and since

X ∗ Y = T_{X} + T_{Y} + ρ_{0}S + ∑_{β∈σ1} ρ_{β}θ_{β} mod (_{1}𝔪)_{ℂ},

for some ρ_{0}, ρ_{β} ∈ ℂ, β ∈ σ1, we obtain the equations

(10.∗) \begin{matrix}^{C}X +^{D}Y \\ = T_{X} − ℜ( ∑_{β∈σ1} c_{β}Ψ^{β}(X)θ_{β}) + T_{Y} + ℜ( ∑_{β∈σ1}(e^{d_{0}Ψ^{β}(X)} − Ψ^{β}(Y)d_{β})θ_{β}) \\ = T_{X} + T_{Y} + ℜ( ∑_{β∈σ1} ρ_{β}θ_{β}) \\ = X ∗ Y mod (ℝS +_{1}𝔪), \end{matrix}

whence for any β in σ1:

(∗_{β})  c_{β}Ψ^{β}(X) + (−e^{d_{0}Ψ^{β}(X)} + Ψ^{β}(Y)d_{β}) = −(X ∗ Y)_{β} = −ρ_{β}.

We see that for any c_{β} in ℝ we can find a unique d_{β} such that (∗_{β}) is satisfied. In the same way

(10.+)  \begin{matrix}^{K}X ∗^{L}Y \\ = (T_{X} − ℜ( ∑_{β∈σ1} k_{β}Ψ^{β}(X)θ_{β})) \\ ∗ (T_{Y} + ℜ( ∑_{β∈σ1}(e^{l_{0}Ψ^{β}(X)} − Ψ^{β}(Y)l_{β})θ_{β}) \\ = X + Y mod (ℝS +_{1}𝔪) \end{matrix}

gives us for β in σ1 the equation

(+_{β}) e(−Ψ^{β}(Y))f(Ψ^{β}(X))Ψ^{β}(X)(−k_{β}) + f(Ψ^{β}(Y))(Ψ^{β}(Y)(−l_{β}) + e^{l_{o}Ψ^{β}(X)})

= f(Ψ^{β}(X) + Ψ^{β}(Y))(X + Y)_{β}.

(see Lemma 1). Again for any k_{β} in ℂ we find a unique l_{β} such that (+_{β}) is fulfilled.

In other words for any C = (ℜ(∑_{β∈σ1} c_{β}θ)) ∗ c_{0}T_{Y} , we find a unique D = (ℜ(∑_{β∈σ1} d_{β}θ_{β})) ∗ d_{0}T_{X}) such that (C, D) ∈ S^{∗}_{_{1}𝔪}, and the numbers d_{β} depend linearly on c_{β}, resp. for any

K = (ℜ( ∑_{β∈σ1} k_{β}θ)) ∗ k_{0}T_{Y}

there exists a unique L = (ℜ(∑_{β∈σ1} l_{β}θ_{β})) ∗ l_{0}T_{X} so that (K, L) ∈ S^{+}_{_{1}𝔪}, and the numbers l_{β} depend linearly on k_{β}.

We now proceed with to investigate S^{∗}_{𝔫_{r}}. As before we write σ^{+} = {β_{1}, ⋯, β_{s}}, and now

\begin{matrix}θ_{i} = θ_{β_{i}}, θ_{−i} = θ_{−β_{i}}, Z_{i} & = Z_{β_{i}}, c_{i} = c_{β_{i}}, d_{i} = d_{β_{i}}, \\ Ψ^{β_{i}} = Ψ^{i}, φ^{β_{i}} = φ^{i} & , Ψ^{−β_{i}} = Ψ^{−i}, φ^{−β_{i}} = φ^{−i}, \end{matrix}

and so on, i = 1 to s.

We recall that φ^{i}(θ_{−i}) = 1, φ^{i}(θ_{γ}) = 0 for γ ≠ −β_{i}, φ^{−i}(θ_{i}) = 1, φ^{−i}(θ_{γ}) = 0 for γ ≠ β_{i}. We shall use the following coordinates in the group H = exp 𝔥. Every element g in (exp 𝔫) · exp ℝT_{Y} can be written by (9.5) in a unique way as a product:

g = g_{r} ∗ (∏_{i=1}^{s} ℜ(c_{−i}θ_{−i}) ∗ ( ∑_{β∈σ1} re(c_{β}θ_{β})) ∗ c_{0}T_{Y} ,

where g_{r} ∈ N_{r} = exp 𝔫_{r} and where the c_{−i} and c_{β} are complex numbers. We will sometimes write

g_{θ_{−i}} = c_{−i}, resp. g_{θ_{β}} = c_{β}.

Our goal is now to construct rational functions c_{−i}, d_{−i}, i = 1 to s and d_{β}, β ∈ σ1, defined on Zariski open subsets in the variables c = {c_{β}, β ∈ σ1}, such that the pairs (C, D) with

\begin{aligned}C & = ∏_{i=1}^{s} ℜ(c_{−i}(c)θ_{−i}) ∗ ( ∑_{β∈σ1} ℜ(c_{β}θ_{β})) ∗ c_{0}T_{Y} , \\ D & = ∏_{i=1}^{s} ℜ(d_{−i}(c)θ_{−i}) ∗ ( ∑_{β∈σ1} ℜ(d_{β}(c)θ_{β})) ∗ d_{0}T_{X}) \end{aligned}

are in S^{∗}_{𝔫_{r}} for any c = (c_{β})_{β∈σ1} in the common domain of the functions c_{−i}, d_{β}, d_{−i} and that furthermore for any j in J, whenever Ψ_{j}(X) + Ψ_{j}(Y) ≠ 0, we have

0 ≠ det | \begin{matrix}Ψ_{j}(X) & Ψ_{j}(Y) \\ φ_{j}(^{C}X) & φ_{j}(^{D}Y) \end{matrix} | .

A similar result will of course be shown for S^{+}_{𝔫_{r}}.

Let for i = 1 to s,

(𝔪_{i})_{ℂ} = (𝔫_{r})_{ℂ} + ∑_{k=1}^{i}(ℂθ_{−k} + ℂ(θ_{−k})^{−} + ℂZ_{k} + ℂ(Z_{k})^{−}), (𝔪_{0})_{ℂ} = (𝔪)_{ℂ}.

In particular, (𝔪_{s})_{ℂ} = (_{1}𝔪)_{ℂ}.

The subspace 𝔪_{i} is an ideal of 𝔥 and 𝔪_{i}/𝔪_{i−1} ≅ ℂθ_{−i} + ℂ(θ_{−i})^{−} + ℂZ_{i} + ℂ(Z_{i})^{−} is dangerous. We determine now by induction on i the elements in S^{∗}_{𝔪_{i}}, resp. S^{+}_{𝔪_{i}}, for i = s to 0.

We have for any i in {1, ⋯, s}
[A, θ_{−i}] = Ψ^{−i}(A) · θ_{−i} + φ^{−i}(A)Z_{i} mod 𝔪_{i−1}, A ∈ 𝔥.

Taking (C, D) in S^{∗}_{𝔪_{i}} we try to find

(C′, D′) = (ℜ(c_{−i}θ_{−i}) ∗ C, ℜ(d_{−i}θ_{−i}) ∗ D) in S^{∗}_{𝔪_{i−1}},

using the formulas of Lemma 3, 4 and 5.

If Ψ^{i}(X) + Ψ^{i}(Y) = 0, we proceed in the following way. we have CX +^{D}Y = X ∗Y +B +Z′′ mod 𝔪_{i−1} for some B = ℜ(ρθ_{−i}) and Z′′ = ℜ(γZ_{i}). By Lemma 3 we can find a unique d′_{−i}, which depends linearly on ρ, such that

CX +^{d′_{−i}θ_{−i}∗D}Y = X ∗ Y + Z′ mod (𝔪_{i−1})

for some Z′ in span_{ℂ}(Z_{i}, \overline{Z}_{i}) ∩ 𝔥. Then we can apply Lemma 5 in order to determine c_{−i} and d_{−i}. By Lemma 5, we can find α in ℂ so that

ℜ(αθ_{−i})X ∗^{ℜ(αθ_{−i})}Y = X ∗ Y − Z′ mod 𝔪_{i−1}.

Whence

ℜ(αθ_{−i})CX +^{ℜ(αθ_{−i})d′_{−i}θ_{−i}∗D}Y
= ^{ℜ(αθ_{−i})}(^{C}X +^{d′_{−i}θ_{−i}∗D}Y)
= ^{ℜ(αθ_{−i})}(X ∗ Y + Z′ mod (𝔪_{i−1}))
= ^{ℜ(αθ_{−i})}X ∗^{ℜ(αθ_{−i})}Y + Z′ mod (𝔪_{i−1})
= X ∗ Y mod (𝔪_{i−1}).

Let us set c_{−i} = α, d_{−i} = α + d′_{−i}; c_{−i} and d_{−i} are rational functions of c_{i} and c_{β}, β ∈ σ0. If Ψ^{i}(X) + Ψ^{i}(Y) ≠ 0 consider the equations (given in 4.1)

(∗−i)  \begin{aligned}c_{−i}Ψ^{−i}(X) + d_{−i}Ψ^{−i}(Y) & = (X ∗ Y)_{−i} = ρ_{i} \\ c_{−i}φ^{−i}(^{C}X) + d_{−i}φ^{−i}(^{D}Y) & = (^{C}X +^{D}Y)_{Z_{i}} = γ_{i}, \end{aligned}

where γ_{i} depends on C and D. Since

\begin{aligned}C & = ℜ(c_{−(i+1)}θ_{−(i+1)}) ∗ ⋯ ∗ ℜ(c_{−s}θ_{−s}) ∗ ℜ( ∑_{β∈σ1} c_{β}θ_{β}) ∗ c_{0}T_{Y} resp. \\ D & = ℜ(d_{−(i+1)}θ_{−(i+1)}) ∗ ⋯ ∗ ℜ(d_{−s}θ_{−s}) ∗ ℜ( ∑_{β∈σ1} d_{β}θ_{β}) ∗ d_{0}T_{X}, \end{aligned}

we see that γ_{i} is rational in c_{β}, β ∈ σ0. In this way we obtain that the numbers c_{−i} and d_{−i} are rational functions in the c_{β} ’s, β ∈ σ1 for fixed d_{0} .

The condition

det | \begin{matrix}Ψ^{−i}(X) & Ψ^{−i}(Y) \\ φ^{−i}(^{C}X) & φ^{−i}(^{D}Y) \end{matrix} | ≠ 0

forces us to reject all the pairs (C, D) in S^{∗}_{𝔪_{i}} for which the corresponding determinant = 0. Since by (10.∗), resp. (10.+), we have for any β in σ′ that

(10.1)  ^{C}X = X − ℜ(c_{β}Ψ_{β}(X)θ_{β}) mod ker φ^{−β},

resp.

DY = Y + ℜ(e^{d_{0}·Ψ_{β}(X)} − d_{β}Ψ_{β}(Y)θ_{β}) mod ker φ^{−β}

we get φ^{−i}(^{C}X) = −c_{i}Ψ^{i}(X), φ^{−i}(^{D}Y) = e^{d_{0}·Ψ^{i}(X)} − d_{i}Ψ^{i}(Y) and so

0 = det | \begin{matrix}Ψ^{−i}(X) & Ψ^{−i}(Y) \\ φ^{−i}(^{C}X) & φ^{−i}(^{D}Y) \end{matrix} |

= Ψ^{−i}(X)(e^{d_{0}·Ψ^{i}(X)} − d_{i}Ψ^{i}(Y)) − Ψ^{−i}(Y)(−c_{i}Ψ^{i}(X)),

whence the condition det = 0 is equivalent to

(10.2)  c_{i} − d_{i} = −Ψ^{i}(Y)^{−1}e^{d_{0}·Ψ^{i}(X)},

since Ψ^{−i} + Ψ^{i} = 0.

Together with the condition (∗β) we obtain for fixed d_{0} a linear system in the variables c_{i} and d_{i}:

(∗i)  \begin{aligned}c_{i} − d_{i} & = −Ψ^{i}(Y)^{−1}e^{d_{0}·Ψ^{i}(X)} \\ c_{i}Ψ^{i}(X) + d_{i}Ψ^{i}(Y) & = e^{d_{0}Ψ^{i}(X)} − (X ∗ Y)_{i} \end{aligned}

with matrix

( \begin{matrix}1 & −1 \\ Ψ^{i}(X) & Ψ^{i}(Y) \end{matrix} )

whose determinant is Ψ^{i}(X) + Ψ^{i}(Y).

Since Ψ^{i}(X) + Ψ^{i}(Y) ≠ 0, there is a unique pair (c_{i}, d_{i}) which must be excluded, i.e. we must take out of S^{∗}_{𝔪_{i+1}} every pair (C, D) for which the coordinates c_{i}, resp. d_{i} satisfy (∗i). Finally when we arrive at i = 1 we have found elements C, D with (C, D) ∈ S^{∗}_{𝔫_{r}}, and we have just seen that we can write C, D in the form

\begin{aligned}C & = ℜ(c_{−1}θ_{−1}) ∗ ⋯ ∗ ℜ(c_{−s}θ_{−s}) ∗ ℜ( ∑_{β∈σ1} c_{β}θ_{β}) ∗ c_{0}T_{Y} , \\ D & = ℜ(d_{−1}θ_{−1}) ∗ ⋯ ∗ ℜ(d_{−s}θ_{−s}) ∗ re( ∑_{β∈σ1} d_{β}θ_{β}) ∗ d_{0}T_{Y}), \end{aligned}

where d_{β}, c_{−i} and d_{−i} are rational functions in the c_{β} ’s, β ∈ σ1 and where the c = (c_{β})_{β∈σ1} varies in a Zariski open set.

When we try to go from S^{∗}_{𝔫_{r}} to S^{∗} we shall encounter the dangerous ideals 𝔟_{j} with j ∈ J. If the corresponding root Ψ_{j} belongs to a root −β with β in σ1, i.e. if Ψ_{j} = Ψ^{−β} = −Ψ^{β}, φ_{j} = φ^{−β} for some β in σ1, and if Ψ_{j}(X) + Ψ_{J}(Y) ≠ 0, then we must again exclude all the solutions (C, D) in S^{∗}_{𝔫_{r}}, for which

0 = det | \begin{matrix}Ψ_{j}(X) & Ψ_{j}(Y) \\ φ_{j}(^{C}X) & φ_{j}(^{D}Y) \end{matrix} | .

But then the coordinates c_{β} and d_{β} are the unique solution of the system

(∗β)  \begin{aligned}c_{β} − d_{β} & = −Ψ^{β}(Y)^{−1}e^{d_{0}·Ψ^{β}(X)} \\ c_{β}Ψ^{β}(X) + d_{β}Ψ^{β}(Y) & = e^{d_{0}Ψ^{β}(X)} − (X ∗ Y)_{β} \end{aligned}

which gives us a Zariski closed subset of the β ’s, β ∈ σ1, which we must throw away. The case where Ψ_{j} belongs to a root β_{i} in σ^{+} is much more delicate. We recall that this means that Ψ_{j} = Ψ^{i}, φ_{j} = φ^{i} and so

[A, Ξ_{j}] = Ψ^{i}(A)Ξ_{j} + φ^{i}(A)Z_{j} mod (𝔫_{j−1})_{ℂ}, A ∈ 𝔥.

Now the coordinate c_{−i} of C had been obtained as solution of the system of equations:

(∗ − i)  \begin{aligned}c_{−i}Ψ^{−i}(X) + d_{−i}Ψ^{−i}(Y) & = (X ∗ Y)_{−i} \\ c_{−i}φ^{−i}(^{C}X) + d_{−i}φ^{−i}(^{D}Y) & = (^{C}X +^{D}Y)_{Z_{i}} = γ_{i}, \end{aligned}

where γ_{i} depends on the c_{β} ’s, β ∈ σ0. The condition

0 = det | \begin{matrix}Ψ_{j}(X) & Ψ_{j}(Y) \\ φ_{j}(^{C}X) & φ_{j}(^{D}Y) \end{matrix} | = Ψ_{j}(X)φ_{j}(^{D}Y) − Ψ_{j}(Y)φ_{j}(^{C}X)

⇔ c_{−i} − d_{−i} = Ψ^{i}(Y)^{−1}e^{−d_{0}·Ψ^{i}(X)},

(see 10.1) imposes another constraint on the solution, which, we recall, is a rational function of the c_{β}, β ∈ σ1.

Hence, if we determinec_{−i} and d_{−i} by the two equations (∗ − i) we get

c_{−i} = {Ψ^{−i}(X)φ^{−i}(^{D}Y) − Ψ^{−i}(Y)φ^{−i}(^{C}X)}^{−1}{(X ∗ Y)_{−i}φ^{−i}(^{D}Y) − γ_{i}Ψ^{−i}(Y)}

d_{−i} = {Ψ^{−i}(X)φ^{−i}(^{D}Y) − Ψ^{−i}(Y)φ^{−i}(^{C}X)}^{−1}{−(X ∗ Y)_{−i}φ^{−i}(^{C}X) + γ_{i}Ψ^{−i}(X)}.

The condition c_{−i} − d_{−i} = Ψ^{i}(Y)^{−1}e^{−d_{0}·Ψ^{−i}(X)} imposes another relation on c_{i}, namely:

\begin{matrix}Ψ^{i}(Y)^{−1}e^{−d_{0}·Ψ^{i}(X)}\{Ψ^{−i}(X)φ^{−i}(^{D}Y) − Ψ^{−i}(Y)φ^{−i}(^{C}X)\} \\ = (X ∗ Y)_{−i}φ^{−i}(^{D}Y) − γ_{i}Ψ^{−i}(Y) + (X ∗ Y)_{−i}(φ^{−i}(^{C}X)) − γ_{i}Ψ^{−i}(X). \end{matrix}

Since φ^{−i}(^{C}X + ^{D}Y) = φ^{−i}(X ∗ Y) = const, we obtain a nontrivial relation between c_{i} and the other variables c_{β}, β ∈ σ0:

{Ψ^{i}(Y)^{−1}e^{−d_{0}·Ψ^{i}(X)}(Ψ^{−i}(X) + Ψ^{−i}(Y))}φ^{−i}(^{C}X)
+ rational function in {c_{β}, β ∈ σ0} = 0.

Hence, using an appropriate induction hypothesis for j = r to 1, we can conclude that there exist rational functions c_{−i}, i = 1 to s, d_{β}, β ∈ σ1, defined on Zariski open subsets in the variables c_{β}, β ∈ σ1, such that the pairs

(C, D) =

(∏_{i=1}^{s} ℜ(c_{−i}θ_{−i}) ∗ ( ∑_{β∈σ1} ℜ(c_{β}θ_{β})) ∗ c_{0}T_{Y} ,

∏_{i=1}^{s} ℜ(d_{−i}θ_{−i}) ∗ ( ∑_{β∈σ1} ℜ(d_{β}θ_{β})) ∗ d_{0}T_{X})

are in S^{∗}_{𝔫_{r}} for any (c_{β})_{β∈σ1} in the common domain of the functions c_{−i}, d_{β}, d_{−i} and that furthermore for any j in J, whenever Ψ_{j}(X) + Ψ_{j}(Y) ≠ 0, we have

0 ≠ det | \begin{matrix}Ψ_{j}(X) & Ψ_{j}(Y) \\ φ_{j}(^{C}X) & φ_{j}(^{D}Y) \end{matrix} | .

We consider now S^{+}_{𝔫_{r}}. First we determine by induction on i the elements in S^{+}_{𝔪_{i}}. We have written

[A, θ_{−i}] = Ψ^{−i}(A) · θ_{−i} + φ^{−i}(A)Z_{i} mod (𝔪_{i−1})_{ℂ}, A ∈ 𝔥.

Taking now (K, L) in S^{+}_{𝔪_{i}}, we try to find

(K′ = ℜ(k_{−i}θ_{−i}) ∗ K, L′ = ℜ(l_{−i}θ_{−i}) ∗ L) in S^{+}_{𝔪_{i−1}}

using the formulas of Lemma 3, 4 and 5.

If Ψ^{i}(X) + Ψ^{i}(Y) = 0, we proceed in the following way. we have KX ∗^{L}Y = X +Y +B +Z′′ mod 𝔪_{i−1} for some B = ℜ(ρθ_{−i}) and Z′′ = ℜ(γZ_{i}). By Lemma 3 we can find a unique l′_{−i}, which depends linearly on ρ, such that

KX ∗^{l′_{−i}θ_{−i}∗L}Y = (X + Y) + Z′ mod (𝔪_{i−1})

for some Z′ in span_{ℂ}(Z_{i}, \overline{Z}_{i}) ∩ 𝔥. Then we can apply Lemma 5 in order to determine k_{−i} and l_{−i}. We can find α in ℂ so that

ℜ(αθ_{−i})X +^{ℜ(αθ_{−i})}Y = X + Y − Z′ mod 𝔪_{i−1}.

Whence

ℜ(αθ_{−i})KX ∗^{ℜ(αθ_{−i})l′_{−i}θ_{−i}∗L}Y
= ^{ℜ(αθ_{−i})}(^{K}X ∗^{l′_{−i}θ_{−i}∗L}Y)
= ^{ℜ(αθ_{−i})}(X + Y + Z′ mod (𝔪_{i−1}))
= ^{ℜ(αθ_{−i})}X +^{ℜ(αθ_{−i})}Y + Z′ mod (𝔪_{i−1})
= X + Y mod (𝔪_{i−1}).

Let us set k_{−i} = α, l_{−i} = α + l′_{−i}, k_{−i} and l_{−i} are rational functions in k_{β}, β ∈ σ0, and k_{i}. If Ψ^{i}(X) + Ψ^{i}(Y) ≠ 0, consider the equations (4.2) for X = ^{K}X, Y = ^{L}Y, Ψ = Ψ^{−i}, φ = φ^{−i}. We get

(+ − i)

k_{−i}f(Ψ^{−i}(X))Ψ^{−i}(X) + l_{−i}e(Ψ^{−i}(Y)) · f(Ψ^{−i}(Y))Ψ^{−i}(Y) = ρ′_{i}
(X + Y)_{Z_{i}} = k_{−i}f(Ψ^{−i}(X))φ^{−i}(^{K}X)
+ l_{−i}e(Ψ^{−i}(Y))f(Ψ^{−i}(Y))φ^{−i}(^{L}Y)
= γ_{i} − ⋯ = γ′_{i},

where γ_{i} depends rationally on k_{β}, β ∈ σ0 and where ρ′_{i} is a constant.

In this way we see that the numbers k_{−i} and l_{−i} are rational functions in the k_{β}, β ∈ σ0 and in k_{i}. The condition

det | \begin{matrix}Ψ^{−i}(X) & Ψ^{−i}(Y) \\ φ^{−i}(^{K}X) & φ^{−i}(^{L}Y) \end{matrix} | ≠ 0

forces us to reject all the pairs (K, L) in S^{+}_{𝔪_{i}} for which the corresponding determinant = 0. Since

(10.3)  ^{K}X = X − ℜ(k_{i}Ψ^{i}(X)θ_{i}) mod ker φ^{−i},

resp.

LY = T_{Y} + ℜ(e^{l_{o}·Ψ^{i}(X)} − l_{i}Ψ^{i}(Y)θ_{i}) mod ker φ^{−i},

(see 10.1) we get φ^{−i}(^{K}X) = −k_{i}Ψ^{i}(X), φ^{−i}(^{L}Y) = e^{l_{0}·Ψ^{i}(X)} − l_{i}Ψ^{i}(Y) and so

0 = det | \begin{matrix}Ψ^{−i}(X) & Ψ^{−i}(Y) \\ φ^{−i}(^{K}X) & φ^{−i}(^{L}Y) \end{matrix} |

is equivalent to

k_{i} − l_{i} = −Ψ^{i}(Y)^{−1}e^{l_{0}·Ψ^{i}(X)},

since Ψ^{−i} + Ψ^{i} = 0.

Together with the condition (^{K}X ∗^{L}Y)_{θ_{i}} = (X +Y)_{θ_{i}} we obtain a linear system in the variables k_{i} and l_{i}:

k_{i} − l_{i} = −Ψ^{i}(Y)^{−1}e^{l_{0}·Ψ^{i}(X)}f(Ψ^{i}(X) + Ψ^{i}(Y))^{−1}

{e(−Ψ^{i}(Y))f(Ψ^{i}(X))(−Ψ^{i}(X)k_{i}) + f(Ψ^{i}(Y))(e^{l_{0}Ψ^{i}(X)} − Ψ^{i}(Y)l_{i})}

= (X + Y)_{i}.

Replacing now k_{i} by l_{i} − Ψ^{i}(Y)^{−1}e^{l_{0}Ψ^{i}(X)} in the equation above, we get

f(Ψ^{i}(X) + Ψ^{i}(Y))(X + Y)_{i}
= (−1 + e(−Ψ^{i}(X) − Ψ^{i}(Y)))l_{i} + ⋯ independent of l_{i} and k_{i}.

Hence we find a unique l_{i} and k_{i} satisfying these equations. We must throw away all the pairs (K, L) for which the coordinates k_{i} and k_{i} satisfy these equations.

In this way we determine the elements (K, L) in S^{+}_{𝔫_{r}} and we have just seen that we can write them in the form:

\begin{matrix}(K, L) = ( & ℜ(k_{−1}θ_{−1}) ∗ ⋯ ∗ ℜ(k_{−s}θ_{−s}) ∗ ℜ( ∑_{β∈σ1} k_{β}θ_{β}) ∗ k_{o}T_{Y} , \\ ℜ(l_{−1}θ_{−1}) ∗ ⋯ ∗ ℜ(l_{−s}θ_{−s}) ∗ ℜ( ∑_{β∈σ1} l_{β}θ_{β}) ∗ l_{0}T_{Y} ) \end{matrix}

where the l_{β} ’s, k_{−i} ’s and l_{−i} ’s are rational functions in the variables k_{β} ’s, β ∈ σ1 defined on a Zariski open subset.

When we go from S^{+}_{𝔫_{r}} to S^{+} we shall have to deal again with the dangerous ideals 𝔟_{j} where j ∈ J. If the correspondig root Ψ_{j} belongs to a root −β contained in −σ1, i.e. Ψ_{j} = Ψ^{−β} and φ_{j} = φ^{−β} for some β in σ1 and if Ψ_{j}(X) + Ψ_{j}(Y) ≠ 0, we must again exclude all the solutions (K, L) in S^{+}_{𝔫_{r}}, for which

0 = det | \begin{matrix}Ψ_{j}(X) & Ψ_{j}(Y) \\ φ_{i}(^{K}X) & φ_{j}(^{L}Y) \end{matrix} | .

But then the coordinates k_{β} = (K)_{θ_{β}} and l_{β} = (L)_{θ_{β}} are given as the unique solution of the linear system of rank 2

k_{β} − l_{β} = −Ψ_{β}(Y)^{−1}e^{l_{0}·Ψ^{β}(X)}

and

(+_{β}) e(−Ψ^{β}(Y))f(Ψ^{β}(X))Ψ^{β}(X)(−k_{β}) + f(Ψ^{β}(Y))(Ψ^{β}(Y)(−l_{β}) + e^{l_{0}Ψ^{β}(X)})

= f(Ψ^{β}(X) + Ψ^{β}(Y))(X + Y)_{β}.

Hence it suffices to take out all the (K, L) which satisfy these two equations.

The case where Ψ_{j} belongs to a root β_{i} in σ^{+} is much more delicate. We recall that this means that

[A, Ξ_{j}] = Ψ^{i}(A)Ξ_{j} + φ^{i}(A)Z_{j}, A ∈ 𝔥.

Indeed the coordinate k_{−i} of K had been obtained as solution of the equation:

(+−_{i})

k_{−i}f(Ψ^{−i}(X))Ψ^{−i}(X) + l_{−i}e(Ψ^{−i}(Y)) · f(Ψ^{−i}(Y))Ψ^{−i}(Y)
= ρ_{i} · exp(Ψ^{−i}(Y))f(Ψ^{−i}(X ∗ Y))
k_{−i}f(Ψ^{−i}(X))φ^{−i}(^{K}X) + l_{−i}e(Ψ^{−i}(Y))f(Ψ^{−i}(Y))φ^{−i}(^{L}Y) = γ′_{i}.

The condition

0 = det | \begin{matrix}Ψ_{j}(X) & Ψ_{j}(Y) \\ φ_{j}(^{K}X) & φ_{j}(^{L}Y) \end{matrix} | = Ψ_{j}(X)φ_{j}(^{K}Y) − Ψ_{j}(Y)φ_{j}(^{L}X)

⇔ k_{−i} − l_{−i} = Ψ^{i}(Y)^{−1}e^{l_{0}·Ψ^{−i}(X)},

130 Arnal and Ludwig

imposes another constraint on the solution, which, we recall, is a rational function of the k_{β}, β ∈ σ1. This gives us three equations relating k_{i} = (K)_{θ_{i}}, l_{i} = (L)_{θ_{i}} with k_{−i} and l_{−i}. The first two equations tell us that

f(Ψ^{−i}(X)){Ψ^{−i}(X)φ_{−i}(^{L}Y) − Ψ^{−i}(Y)φ^{−i}(^{K}X)}k_{−i}
= ρ′_{i}φ^{−i}(^{L}Y) − γ′_{i} · Ψ^{−i}(Y)

f(Ψ^{−i}(Y))e(Ψ^{−i}(Y)){Ψ^{−i}(X)φ_{−i}(^{L}Y) − Ψ^{−i}(Y)φ^{−i}(^{K}X)}l_{−i}

= Ψ^{−i}(X)γ′_{i} − ρ′_{i}φ^{−i}(^{K}X)

where γ′_{i} = γ_{i} − ρ′_{i} · {(exp(−Ψ^{−i}(Y)) − 1)Ψ^{−i}(Y)^{−1} · φ^{−i}(^{L}Y) + ⋯} and where

ρ′_{i} = ρ_{i} · exp(Ψ^{−i}(Y))f(Ψ^{−i}(X ∗ Y)),

as in (4.2). If we introduce these values into the last of the three equations and if we use the identity

(10.4) φ_{−i}(^{K}X ∗^{L}Y) = e(−Ψ^{i}(Y))f(Ψ^{i}(X))φ_{−i}(^{K}X) + f(Ψ^{i}(Y))φ_{−i}(^{L}X)

we get, since γ′_{i} = γ_{i} − ⋯ = −ρ′_{i}(exp(−Ψ^{−i}(Y)) − 1)Ψ^{−i}(Y)^{−1} · φ^{−i}(^{L}Y) + a function in the variables k_{β}, β ∈ σ0:

Ψ^{i}(Y)^{−1}e^{l_{0}·Ψ^{−i}(X)} · {Ψ^{−i}(X)φ^{−i}(^{L}Y) − Ψ^{−i}(Y)φ^{−i}(^{K}X)} = f(Ψ^{−i}(X))^{−1} · (ρ′_{i}φ^{−i}(^{L}Y) − γ′_{i}Ψ^{−i}(Y)) − (f(Ψ^{−i}(Y))e(Ψ^{−i}(Y))^{−1}(Ψ^{−i}(X)γ′_{i} − ρ′_{i}φ^{−i}(^{K}X)) = ρ′_{i}f(Ψ^{−i}(X))^{−1}φ^{−i}(^{L}Y)(1 + (exp(−Ψ^{−i}(Y)) − 1)Ψ^{−i}(Y)^{−1}Ψ^{−i}(Y)) − ρ′_{i}f(−Ψ^{−i}(Y))^{−1}{(−Ψ^{−i}(X)(exp(−Ψ^{−i}(Y)) − 1)Ψ^{−i}(Y)^{−1}φ^{−i}(^{L}Y)) − φ^{−i}(^{K}X)} + a function in the other variables = ρ′_{i}f(Ψ^{−i}(X))^{−1}f(Ψ^{i}(Y))^{−1}{φ^{−i}(^{L}Y)((f(Ψ^{i}(Y))(1 + (exp(Ψ^{i}(Y)) − 1) + f(Ψ^{−i}(X))(−Ψ^{−i}(X))(exp(Ψ^{i}(Y)) − 1)Ψ^{−i}(Y)^{−1}) + f(Ψ^{−i}(X)) · φ^{−i}(^{K}X)} + a function in the other variables = ρ′_{i}f(Ψ^{−i}(X))^{−1}f(Ψ^{i}(Y))^{−1}.

{φ^{−i}(^{L}Y)(f(Ψ^{i}(Y)) exp(Ψ^{i}(Y)) + (exp(Ψ^{i}(X)) − 1)f(Ψ^{−i}(Y)))

+ φ^{−i}(^{K}X)f(Ψ^{−i}(X))}

+ a function in the other variables = ρ′_{i}f(Ψ^{−i}(X))^{−1}f(Ψ^{i}(Y))^{−1}.

{φ^{−i}(^{L}Y)f(Ψ^{−i}(Y))e(Ψ^{i}(X)) + f(Ψ^{−i}(X))φ^{−i}(^{K}X)}

+ a function in the other variables = ρ′_{i}(f(Ψ^{−i}(X))f(Ψ^{i}(Y)))^{−1} · (e^{Ψ^{i}(Y)+Ψ^{i}(X)}).

{e(−Ψ^{i}(Y))f(Ψ^{i}(X))φ_{−i}(^{K}X) + f(Ψ^{i}(Y))φ_{−i}(^{L}Y)} + a function in the other variables

= ρ′_{i}(f(Ψ^{−i}(X))f(Ψ^{i}(Y)))^{−1} · (e^{Ψ^{i}(Y)+Ψ^{i}(X)}) · φ^{−i}(^{K}X ∗^{L}Y) + a function in the other variables = a function in the other variables ,

by (10.4) and since φ^{−i}(^{K}X ∗^{L}Y) = φ^{−i}(X + Y) + constant.

On the other hand , using again (10.4)

Ψ^{i}(Y)^{−1}e^{l_{0}·Ψ^{−i}(X)} · {Ψ^{−i}(X)φ^{−i}(^{L}Y) − Ψ^{−i}(Y)φ^{−i}(^{K}X)}

= ⋯ = Ψ^{i}(Y)^{−1}e^{l_{0}·Ψ^{−i}(X)}f(Ψ^{−i}(Y))^{−1}{e^{−Ψ^{i}(X)} − e^{Ψ^{i}(Y)}}φ^{−i}(^{K}X) + a function in the other variables

Hence we get an identity of the form:

φ^{−i}(^{K}X) + a rational function of the other variables = 0.

This gives us a nontrivial rational condition on the k_{β} ’s, β ∈ σ0 and k_{i}. Hence, using an appropriate induction hypothesis for j = r to 1, we can conclude that for any l_{0} ∈ ℝ, there exist rational functions l_{i}, i = 1 to s, l_{β}, β ∈ σ1, k_{−}i, i = 1 to s in the variables k_{β} ∈ ℝ, β ∈ σ1, defined on Zariski open subsets, such that the pairs

\begin{matrix}(K, L) = \\ (∏_{i=1}^{s} ℜ(k_{−i}θ_{−i}) ∗ ( ∑_{β∈σ1} ℜ(k_{β}θ_{β})) ∗ k_{0}T_{Y} , ∏_{i=1}^{s} ℜ(l_{−i}θ_{−i}) ∗ ( ∑_{β∈σ1} ℜ(l_{β}θ_{β})) ∗ l_{0}T_{X}) \end{matrix}

are in S^{+}_{𝔫_{r}} for any (k_{β})_{β∈σ1} in the common domain of the functions k_{−i}, l_{β}, l_{−i} and such the determinants are ≠ 0.

Lemma 11. Let 𝔥 be a subalgebra generated by two elements X and Y of the exponential Lie algebra 𝔤. Then S^{∗} and S^{+} are not empty.

Proof. Let us take any element (C, D) in S^{∗}_{𝔫_{r}} with the property of Lemma 10. It is now easy to see that for some appropriate elements N, M in 𝔫_{r}, we have (N ∗ C, M ∗ D) ∈ S^{∗}. We proceed by a backwards induction on j = r to 1. Having found (C_{j}, D_{j}) in S_{𝔫_{|}}^{∗}, such that (C_{j} mod 𝔫_{r}, D_{j} mod 𝔫_{r}) = (C mod 𝔫_{r}, D mod 𝔫_{r}) we look for elements N_{j−1}, M_{j−1} in 𝔥_{j−1} , such that (N_{j−1}∗C_{j}, M_{j−1}∗D_{j}) ∈ S^{∗}_{𝔫_{j−1}} . It suffices to consider the corresponding Lemma 3, 4 or 5. Since for any N in 𝔪 and U in 𝔥 we have φ_{j}(N ∗ U) = φ_{j}(U) we see that in the dangerous cases

det | \begin{matrix}Ψ_{j}(X) & Ψ_{j}(Y) \\ φ_{j}(^{C_{j−1}}X) & φ_{j}(^{D_{j−1}}Y) \end{matrix} | = det | \begin{matrix}Ψ_{j}(X) & Ψ_{j}(Y) \\ φ_{j}(^{C}X) & φ_{j}(^{D}Y) \end{matrix} | ≠ 0

and so we can solve the given equations. This shows us that S^{∗} ≠ Ø. We proceed in the same way to show that S^{+} is not empty.

End of the proof of Theorem A: Let A = ^{C}X +^{D}Y be an element of ^{G}X +^{G}Y . Let us write ^{C}X = X′ and ^{D}Y = Y′. Let 𝔥′ be the subalgebra generated by X′ and Y′. By Lemma 11, there exist K′ and L′ in 𝔥′, such that

X′ + Y′ = (^{K′}X′ ∗^{L′}Y′).

Hence A = (^{K′∗C}X) ∗ (^{L′∗D}Y) ∈ (^{G}X) ∗ (^{G}Y). If B = ^{K}X ∗^{L}Y ∈ ^{G}X ∗^{G}Y , then writing ^{K}X = X′, ^{L}Y = Y′, we can find by Lemma 11 elements C′ and D′ in the subalgebra generated by X′ and Y′, such that X′ ∗Y′ = ^{C′}X′ +^{D′}Y′. Hence

B = ^{K}X ∗^{L}Y = X′ ∗ Y′ = ^{C′∗K}X +^{D′∗L}Y ∈ ^{G}X +^{G}Y.

Proof of Theorem B

It is easy to see that ^{H}X +^{H}Y ⊂ X + Y + [𝔥, 𝔥], hence also
(^{H}X +^{H}Y)^{−} ⊂ X + Y + [𝔥, 𝔥].

In order to prove that X + Y + [𝔥, 𝔥] ⊂ (^{H}X +^{H}Y)^{−} we proceed by induction on the dimension of 𝔥. If 𝔥 is one dimensional then there is nothing to prove.

We may suppose that [𝔥, 𝔥] = 𝔫 is not central, since otherwise 𝔥 is nilpotent and we know then by Wildberger’s result that X + Y + [𝔥, 𝔥] = HX +^{H}Y . We look at minimal noncentral ideals 𝔟 contained in 𝔫. If 𝔟∩𝔷 = {0}, then we consider 𝔭 : 𝔥 → 𝔥/𝔟 = 𝔥̃. We have cases (i) and (ii) of Lemma 2. We shall treat only the case (ii) and leave the other case to the reader.

There exists a basis U_{1}, U_{2} of 𝔟 such that for any A ∈ 𝔥,

[A, U_{1} + iU_{2}] = Ψ(A)(U_{1} + iU_{2}),

where A → Ψ(A) is a nontrivial linear functional which satisfies Ψ([𝔥, 𝔥]) = {0}. Let us suppose that Ψ(Y) ≠ 0 (otherwise we replace Y by X). We shall use now the induction hypothesis for 𝔥̃ = 𝔥/𝔟. Let X̃ = X mod 𝔟, Ỹ = Y mod 𝔟 etc. We get:

(_{H}X̃ + _{H}Ỹ)^{−} = X̃ + Ỹ + [𝔥̃, 𝔥̃].

Let now p ∈ [𝔥, 𝔥]. By the induction hypothesis, there exists for ϵ > 0 an element O(ϵ) in 𝔥 of length < ϵ, an element B in 𝔟, C, D in 𝔥 such that

X + Y + P = ^{C}X +^{D}Y + B + O(ϵ).

We have seen in Lemma 3 that there exists β_{1}, β_{2} in ℝ such that

CX +^{(β_{1}U_{1}+β_{2}U_{2})∗D}Y = X + Y + P − O(ϵ).

We continue now with case (iv) of Lemma 2. We write as before:

[X, θ] = Ψ(X)θ, [Y, θ] = Ψ(Y)θ + Z,

where Z ∈ 𝔷 + i𝔷 and where Ψ(X) and Ψ(Y) are the two nonreal complex numbers, which are not purely imaginary (since 𝔥 is exponential). We divide through 𝔷′ = span_{ℂ}(Z, Z^{−}) ∩ 𝔥 and we apply the induction hypothesis.

For any P ∈ [𝔥, 𝔥] and ϵ > 0 there exist C, D, O(\frac{ϵ}{2}) in 𝔥, ∥O(\frac{ϵ}{2})∥ < \frac{ϵ}{2}, W in 𝔷′ such that

X + Y + P = ^{C}X +^{D}Y + W + O(\frac{ϵ}{2}).

We look at the complex matrices

M = ( \begin{matrix}Ψ(X) & Ψ(Y) \\ φ(^{C}X) & φ(^{D}Y) \end{matrix} ) .

If the rank of M is not 2, we take

R = (−Ψ(X))^{−1}[X, Y]

and we set for any δ ≠ 0 in ℝ:

C′ = (δR) ∗ C.

The corresponding matrix

M′ = ( \begin{matrix}Ψ(X) & Ψ(Y) \\ φ(^{C′}X) & φ(^{D}Y) \end{matrix} )

is then of rank 2 for all δ ≠ 0. Indeed we have

[R, θ] = (−Ψ(X))^{−1}[[X, Y], θ] = (−Ψ(X))^{−1}([X, [Y, θ]] − [Y, [X, θ]])
= (−Ψ(X))^{−1}(Ψ(X)Ψ(Y)θ − Ψ(X)Ψ(Y)θ − Ψ(X)Z) = Z.

Hence

\begin{aligned}[^{C′}X, θ] & = ^{δR}[^{C}X,^{−δR}θ] = ^{δR}[^{C}X, θ] = ^{δR}(Ψ(X)θ + φ(^{C}X)Z) \\ & = Ψ(X)θ + (φ(^{C}X) + δΨ(X))Z. \end{aligned}

Thus

φ(^{C′}X) = φ(^{C}X) + Ψ(X)δ

and

detM′ = −Ψ(X)Ψ(Y)δ ≠ 0.

For δ very small the element

O(δ) = ^{C}X −^{C′}X

of [𝔥, 𝔥] is of length < \frac{ϵ}{2} and so we can write O(δ) + O(\frac{ϵ}{2}) = O(ϵ) and also

X + Y + P = ^{C′}X +^{D}Y + W + O(ϵ)

and if we now write C instead of C′ we can assume that rank M = 2. By Lemma 4 we can choose α_{1}, α_{2}, β_{1}, β_{2} in ℝ such that

(α_{1}U_{1}+α_{2}U_{2})∗CX +^{(β_{1}U_{1}+β_{2}U_{2})∗D}Y = ^{C}X +^{D}Y + W.

This means of course that

X + Y + P = ^{(α_{1}U_{1}+α_{2}U_{2})∗C}X +^{(β_{1}U_{1}+β_{2}U_{2})∗D}Y + O(ϵ).

The case (v) of Lemma 2 is very easy and is left to the reader.

It has been shown by Wildberger in [3] that if H is nilpotent then we do not need closures. In fact the following slightly stronger result holds.

Proposition 1. Let 𝔥 be a nilpotent Lie algebra generated by two elements X and Y as an ideal. Then:

HX +^{H}Y = X + Y + [𝔥, 𝔥] = ^{H}X ∗^{H}Y.

Proof. Indeed, if 𝔥 is abelian, then the result is clear. If not, let us proceed by induction on the dimension of 𝔥. There exists a noncentral element U in 𝔥, such that [U, 𝔥] ⊂ 𝔷. If now

[X, U] = 0 and [Y, U] = 0,

then X and Y are contained in the centralizer 𝔷(U) of U. But 𝔷(U) is an ideal in 𝔥. This implies that 𝔷(U) = 𝔥 and so U is central in 𝔥. Hence we may suppose that [X, U] = Z ≠ 0. We divide through ℝZ and we use the induction hypothesis for 𝔥/ℝZ and so on.

Two examples

First Example: ^{H}X +^{H}Y ≠ X + Y + [𝔥, 𝔥].

We give now an example of an exponential Lie algebra𝔤 generated by two elements X, Y in 𝔤 such that

GX +^{G}Y = ^{G}X ∗^{G}Y ≠ X + Y + [𝔤, 𝔤].

Let 𝔤 be a Lie algebra spanned by the vectors T, U, V, Z and equipped with the following nontrivial brackets:

[T, U] = −U, [T, V] = V, [U, V] = Z.

The Lie algebra𝔤 is an extension by T of the two step nilpotent algebra 𝔫 = [𝔤, 𝔤] = span(U, V, Z) and the center of 𝔤 is given by the span of Z. Let now

X = T + U, Y = −T + V.

X and Y generate 𝔤: indeed let 𝔤_{0} be the subalgebra of 𝔤 generated by X and Y . Then X + Y = U + V ∈ 𝔤_{0}, [X, U + V] = −U + V ∈ 𝔤_{0} mod ℝZ and thus [U, V] = Z ∈ 𝔤_{0} . Finally 𝔤_{0} contains T, U, V and Z and so 𝔤 = 𝔤_{0} .

We shall realize the group G associated with 𝔤 as a semidirect product of ℝ with 𝔫 = [𝔤, 𝔤], i.e. G = ℝ × 𝔫 and the multiplication in G is given by:

\begin{matrix}(t, & uU + vV + zZ) · (t′, u′U + v′V + z′Z) \\ = (t + t′, (e^{t′}u + u′)U + (e^{−t′}v + v′)V + (z + z′ + \frac{1}{2}(e^{t′}uv′ − e^{−t′}v′u)Z). \end{matrix}

Let us show now that ^{G}X + ^{G}Y ≠ X + Y + [𝔤, 𝔤]. We remark first that G = [G, G] · exp(ℝX) = [G, G] · exp(ℝY), hence

GX = ^{[G,G]}X, resp. ^{G}Y = ^{[G,G]}Y

and so

\begin{aligned}^{G}X & = \{^{exp uU∗exp vV} (T + U) | u, v ∈ ℝ\} \\ & = \{T + (1 + u)U + (−v)V + (−v − uv)Z | u, v ∈ ℝ\}. \end{aligned}

And similarly

\begin{aligned}^{G}Y & = \{^{exp u′U∗exp v′V} (−T + V) | u′, v′ ∈ ℝ\} \\ & = \{−T + (−u′)U + (1 + v′)V + (u′(1 + v′)Z) | u′, v′ ∈ ℝ\}. \end{aligned}

Hence

\begin{matrix}^{G}X +^{G}Y = \\ = \{(1 + u − u′)U + (1 − v + v′)V + (v(−1 − u) + u′(1 + v′)Z | u, v, u′, v′ ∈ ℝ\}. \end{matrix}

Let A = αU + βV + δZ be any element in [𝔤, 𝔤]. We try to solve the equation:

A ∈ ^{G}X +^{G}Y,

i.e.

α = 1 + u − u′, β = −v + (1 + v′), δ = v(−1 − u) + u′(1 + v′),

for some u, u′, v, v′ ∈ ℝ. If now 1 + u − u′ = α = 0, then δ = −u′v + u′(1 + v′) = u′(−v + (1 + v′)) = u′β. Thus if β = 0, δ must also be 0 and no element

A = δZ of [𝔤, 𝔤], δ ≠ 0,

is contained in ^{G}X +^{G}Y . We also see that ^{G}X +^{G}Y contains every element B = αU + βV + δZ, with α^{2} + β^{2} ≠ 0 and finally

GX +^{G}Y = [𝔤, 𝔤] \ ℝ^{∗}Z.

Second Example: exp(^{H}X +^{H}Y) ≠ C(exp X) · C(exp Y).

Let us show by a last example that for solvable nonexponential groups we do no longer have that

exp(^{G}X +^{G}Y) = C(exp X) · C(exp Y).

Let 𝔤 = 𝔢(2) = ℝT + ℂ be the three dimensional Lie algebra with the brackets:

[T, Ξ] = i Ξ, Ξ ∈ ℂ.

The Lie group associated with 𝔤 can be described as G = E(2) = ℝ × ℂ with group law:

(t, Ξ) · (t′, Ξ′) = (t + t′, e^{−it′} · Ξ + Ξ′).

In Lemma 1 we have seen that

exp(tT + Ξ) = (t, \frac{(e^{−it} − 1)}{−it} · Ξ).

Let now X = sT, Y = tT + 1, with s · t ≠ 0. Then X and Y generate 𝔤 and we have

GX = ^{ℂ}X = sT + ℂ,^{G}Y = tT + ℂ and ^{G}X +^{G}Y = (s + t)T + ℂ

and also

C(exp X) = (s, ℂ) and C(exp Y) = (t, ℂ).

Thus

C(exp X) · C(exp Y) = ((s + t), ℂ).

But

exp(^{G}X +^{G}Y) = exp((s + t)T + ℂ) = ((s + t), f(s + t)ℂ).

Hence, if s + t = 2kπ ≠ 0, then

exp(^{G}X +^{G}Y) = exp(2kπ + ℂ) = (2kπ, {0 · ℂ})
= (2kπ, {0}) ≠ exp(^{G}X) · exp(^{G}Y) = (2kπ, ℂ).

Theorem C. Let G = exp 𝔤 be a simply connected, connected solvable Lie group. Then G is exponential if and only if for every X and Y in 𝔤, exp X · exp Y ∈ exp(^{G}X +^{G}Y).

Proof. If G is exponential, then the condition is satisfied by Theorem A. If G is not exponential, then the exponential mapping is not surjective. However, for any solvable Lie groupS with Lie algebra 𝔰, for any subspace 𝔴 of 𝔰 such that 𝔰 = 𝔴 + [𝔰, 𝔰], we have

S = exp 𝔴 · exp[𝔰, 𝔰] = exp 𝔴 · exp 𝔰.

Hence if we choose g in G, such that g ∉ exp 𝔤, then we can take X, Y in 𝔤, such that

g = exp X · exp Y.

Hence g ∈ exp^{G}X · exp^{G}Y , but g ∉ exp(^{G}X +^{G}Y).

Final Question: Would it be possible to obtain our result directly by using a special expression for the Baker-Campbell-Hausdorff-formula (see for instance [2] in a different context)?

References

[1] Dooley, A. H., Repka. J, and N. J. Wildberger, Sums of adjoint orbits, Lin. Multilin. Alg. 36 (1993), 79–101.

[2] Rouvière, F., Espaces symétriques et méthode de Kashiwara-Vergne, Ann. scient. Ec. Norm. Sup. 4^{e} série, t.19, 1986, 553–581.

[3] Wildberger, N. J., On a relationship between adjoint orbits and conjugacy classes of a Lie group, Canad. Math. Bull. Vol. 33, (1990), 297–304.

Arnal and Ludwig 137

[4] —, Hypergroups and harmonic analysis, Proc. Center Mathem. Anal. (ANU), 29 (1992), 238–253.

Dép. de Mathématiques et d’ Informatique Université de Metz Ile du Saulcy F-57045 Metz Cédex 01 France

Received September 19, 1994 and in final form November 25, 1994