Seminar Sophus Lie 2 (1992) 41–55

Compact elements in solvable real Lie algebras

Karl H. Hofmann

1. Observations on solvable algebras

We consider a real solvable Lie algebra 𝔤.

The intersection of the descending central series will be called C^{∞}(𝔤) or C^{∞} for short. If 𝔥 is a Cartan subalgebra then 𝔤 = 𝔥 + C^{∞}. Since the characteristic ideal C^{∞} is contained in the commutator algebra 𝔤′, it is nilpotent.

All Cartan algebras in the solvable algebra 𝔤 are conjugate.

1.1. Definition. Fixing a Cartan algebra 𝔥 we shall write

(1)  𝔢 =^{def} 𝔥 ∩ C^{∞}.

1.2. Lemma. All roots vanish on 𝔢.

Proof. We assume the contrary and find an element X ∈ 𝔢 = 𝔥∩C^{∞} such that there is a root λ on 𝔥_{ℂ} with λ =^{def} λ(X) ≠ 0. By the definition of the root λ, there is a smallest natural number n such that (ad X − λ·1)^{n}(𝔤_{ℂ}) = {0}. Hence there is a Y_{0} ∈ 𝔤_{ℂ}^{λ} such that Y =^{def} (ad X − λ·1)^{n−1}Y_{0} ≠ 0. Hence (ad X)Y = λ·Y and (ad X)^{n}Y = λ^{n}·Y ≠ 0 for all m ∈ ℕ. But X ∈ C^{∞} and thus ad X is nilpotent, i.e., there is a natural number m with (ad X)^{m} = 0. This contradiction proves the claim.

1.3. Lemma. Under the circumstances of Lemma 1.2, 𝔥 = 𝔥_{1} ⊕ 𝔢 with a suitable vector space complement 𝔥_{1} for 𝔢 ∈ 𝔥. Every element X ∈ 𝔥 then decomposes uniquely as X = X_{1} + Y with X_{1} ∈ 𝔥_{1} and Y ∈ 𝔢. Also, X is regular if and only if X_{1} is regular. In particular, the regular elements of 𝔥_{1} are dense in 𝔥_{1} .

Proof. By Lemma 1.2 we have λ(X) = λ(X_{1}) for every root λ. Since X ∈ 𝔥 is regular iff λ(X) ≠ 0 for all roots λ, the assertion follows.

1.4. Lemma. (i) If X ∈ 𝔤 is such that ad X is nilpotent as well as semisimple, then X is central.

(ii) If 𝔫 is the nilradical of 𝔤 and S ⊆ 𝔤 is the set of all elements X with ad X semisimple, then S ∩ 𝔫 = 𝔷, where 𝔷 is the center of 𝔤.

Proof. (i) If ad X is semisimple and nilpotent, then ad X = 0 and thus X is central.

(ii) Since X ∈ 𝔫 implies that ad X is nilpotent, by (i) we conclude that S ∩ 𝔫 ⊆ 𝔷. Conversely, X ∈ 𝔷 implies ad X = 0. In particular, X ∈ S. But 𝔷 ⊆ 𝔫 and thus 𝔷 ⊆ S ∩ 𝔫.

Note C^{∞} ⊆ 𝔫.

2. Compact elements in solvable algebras

2.1. Definition. (i) Set

(2)  comp(𝔤) = {X ∈ 𝔤 : Spec(ad X) ⊆ iℝ and ad X is semisimple}.

If no confusion is possible we shall briefly write comp for this set.

(ii) We fix a Cartan subalgebra and set

(3)  𝔱 = 𝔥 ∩ comp .

The elements of comp are precisely the elements X ∈ 𝔤 for which e^{ℝ· ad X} is relatively compact in Aut(𝔤) ⊆ Gl(|𝔤|). If 𝔤 is solvable and G is the simply connected Lie group associated with 𝔤 then exp ℝ·X is isomorphic to ℝ for all X ∈ 𝔤, in particular for those in comp. The adjoint group Ad(G) agrees with Inn(𝔤) ⊆ Aut(𝔤). Our arguments will remain entirely within 𝔤 and its group Inn(𝔤) of inner automorphisms.

Let Γ = ⟨e^{ad C^{∞}}⟩ = e^{ad C^{∞}} ⊆ Inn(𝔤).

2.2. Lemma. (i) 𝔱 is a central subalgebra of 𝔥.

(ii) 𝔱 contains the center 𝔷 of 𝔤.

(iii) comp = Γ·𝔱.

Proof. (i) Every compact element X in 𝔥 generates a semisimple ad X with purely imaginary spectrum. In particular, ad X|𝔥 is nilpotent and semisimple and thus X is central in 𝔥 by 1.4(i). It follows that the set 𝔱 is central in 𝔥. Clearly 𝔱 is closed under scalar multiplication. In order to show that 𝔱 is closed under addition, take X_{j} ∈ 𝔱, j = 1, 2. Then [X_{1}, X_{2}] = 0 and thus [ad X_{1}, ad X_{2}] = ad[X_{1}, X_{2}] = 0. Hence

\overline{e^{ℝ· ad(X_{1}+X_{2})}} ⊆ \overline{e^{ℝ· ad X_{1}}}·\overline{e^{ℝ· ad X_{2}}}

is compact. Thus X_{1} + X_{2} ∈ comp ∩𝔥 = 𝔱.

(ii) If X ∈ 𝔷, then ad X = 0 whence X ∈ comp. Also, 𝔷 ⊆ 𝔥 since the center is contained in every Cartan algebra. Thus 𝔷 ⊆ 𝔱.

(iii) The Cartan algebras of 𝔤 are conjugate under Γ. Every compact element element X of 𝔤 is semisimple, and thus is contained in a Cartan algebra of 𝔤 (namely, in each Cartan algebra of its centralizer 𝔷(X, 𝔤)). Thus there is a γ ∈ Γ such that γ^{−1}X ∈ 𝔥. Then X ∈ γ·𝔥 ∩ comp = γ(𝔥 ∩ comp) = γ·𝔱. Thus comp ⊆ Γ·𝔱. The converse is clear.

The linear span of comp 𝔤 is invariant under automorphisms and is therefore a fully characteristic ideal of 𝔤 (i.e, it is invariant under all derivations). We want to investigate the structure of this ideal and shall therefore assume until further notice that

(4)  𝔤 = span(comp 𝔤).

Let A denote the set of nonzero weights of 𝔤_{ℂ} with respect to 𝔱_{ℂ} and 𝔤_{ℂ}^{α}, α ∈ A the respective weight spaces of 𝔤_{ℂ}. Now set 𝔱^{+} = 𝔤 ∩ ⨁_{α∈A} 𝔤_{ℂ}^{α}. This definition yields immediately the following remark:

2.3. Remark. The vector space 𝔱^{+} is the same as the Fitting one component of ad X for any regular X ∈ 𝔱. For the action of the torus Ω = \overline{e^{ad 𝔱}} on 𝔤, the vector space 𝔱^{+} is precisely the span of all simple nonzero nontrivial modules.

Write 𝔷(𝔱, 𝔤) for the centralizer of 𝔱 in 𝔤.

2.4. Lemma. (i) 𝔤 = 𝔱 + C^{∞}(𝔤) and C^{∞} = 𝔤′.

(ii) 𝔷(𝔱, 𝔤) = 𝔥 = 𝔱 + 𝔢.

(iii) 𝔤 = 𝔥 ⊕ 𝔱^{+} .

(iv) C^{∞} = 𝔢 ⊕ 𝔱^{+} .

(v) ⟨𝔱^{+}⟩ is an ideal of 𝔤.

Proof. (i) By Lemma 2.2(iii) we have comp(𝔤) = Γ·𝔱, and Γ·𝔱 = e^{ad C^{∞}}𝔱 ⊆ 𝔱 + C^{∞} since every summand except the first of the exponential series contains a factor from C^{∞}, and since C^{∞} is an ideal. Thus 𝔤 = span comp ⊆ 𝔱 + C^{∞} ⊆ 𝔤 and the first assertion follows. In order to see the second we recall that C^{∞} ⊆ 𝔤′ always and that 𝔤/C^{∞} is isomorphic to a subalgebra of 𝔱 by what we just saw. Thus 𝔤/C^{∞} is abelian and hence C^{∞} ⊇ 𝔤′.

(ii) By Lemma 1.3 there is a regular element X of 𝔤 contained in 𝔱. Thus 𝔥 is the nilspace of ad X. But ad X is semisimple, thus 𝔥 = ker ad X = 𝔷(X, 𝔤) ⊇ 𝔷(𝔱, 𝔤). Now 𝔥 ⊆ 𝔷(𝔱, 𝔤) by Lemma 2.2(i). This proves the first equality. By (i) We have 𝔤 = 𝔱 + C^{∞} and thus 𝔥 = 𝔥 ∩ (𝔱 + C^{∞}) = 𝔱 + (𝔥 ∩ C^{∞}) = 𝔱 + 𝔢 by the modular law and the definition of 𝔢 in (1).

(iii) is the Fitting decomposition of ad X for a regular X inside 𝔱. (iv) By (iii) we have 𝔤 = 𝔥 + 𝔱^{+} . If X is regular and contained in

𝔱, then ad X|𝔱^{+} is an automorphism. Thus 𝔱^{+} is contained in every term of the descending central series and thus in C^{∞}. Hence C^{∞} = (𝔥 + 𝔱^{+}) ∩ C^{∞} = (𝔥 ∩ C^{∞}) + 𝔱^{+} = 𝔢 + 𝔱^{+} by the modular law and the definition of 𝔢 again.

(v) We note that [𝔥, 𝔱^{+}] ⊆ 𝔱^{+} . Therefore [𝔥, ⟨𝔱^{+}⟩] ⊆ ⟨𝔱^{+}⟩. Hence ⟨𝔱^{+}⟩ is an 𝔥 + 𝔱^{+} -module and thus a 𝔤-module in view of (iii). Hence ⟨𝔱^{+}⟩ is an ideal.

We shall set L[Γ] =^{def} {X ∈ 𝔤 : e^{ℝ· ad X} ⊆ Γ}.

2.5. Lemma. (i) If A is a subgroup of a Lie group B, then A = ⟨exp 𝔞⟩ holds for a Lie algebra 𝔞 ⊆ 𝔟 = L(B) if and only if A is analytic and 𝔞 = L(A).

(ii) If 𝔳 ⊆ 𝔟 is a vector space then A = ⟨exp 𝔳⟩ if an only if ⟨𝔳⟩ = 𝔞. (iii) Γ = ⟨e^{ad 𝔱^{+}}⟩. (iv) C^{∞} = ⟨𝔱^{+}⟩.

(v) L[Γ] = 𝔷 + ⟨𝔱^{+}⟩.

Proof. (i) The “if” portion is clear. Suppose now that A = ⟨exp 𝔞⟩. Then A is pathwise connected and hence is analytic by the Theorem of Yamabe and Gôto. Hence there is a unique intrinsic Lie group topology on A making A into a Lie group A_{ℓ} with exponential function exp |L(A): L(A) → A_{ℓ}. Here X ∈ L(A) iff exp ℝ·X ⊆ A. This implies 𝔞 ⊆ L(A). Assume for the moment that 𝔞 ≠ L(A). Then the analytic subgroup generated by 𝔞 in the Lie group A_{ℓ} is proper in A_{ℓ}, since proper subalgebras generate proper analytic subgroups. On the other hand this subgroup is ⟨exp 𝔞⟩ = A. This is a contradiction since A = A_{ℓ} as sets. Hence 𝔞 = L(A) follows and the claim is proved.

(ii) Firstly, suppose A = ⟨exp 𝔳⟩. Then X ∈ 𝔳 implies exp ℝ·X ⊆ ⟨exp 𝔳⟩ = A and this implies X ∈ L(A) = 𝔞. It follows that ⟨𝔳⟩ ⊆ 𝔞. Then A = ⟨exp 𝔳⟩ ⊆ ⟨exp⟨𝔳⟩⟩ ⊆ ⟨exp 𝔞⟩ = A. Thus A = ⟨exp⟨𝔳⟩⟩ and hence, by (i) we conclude ⟨𝔳⟩ = 𝔞 as asserted.

Secondly suppose that ⟨𝔳⟩ = 𝔞. Then ⟨exp 𝔳⟩ ⊆ ⟨exp⟨𝔳⟩⟩ = ⟨exp 𝔞⟩ = A. On the other hand, 𝔳 ⊆ L(⟨exp 𝔳⟩) by the definition of the right hand side, and thus 𝔞 = ⟨𝔳⟩ ⊆ L(⟨exp 𝔳⟩) and therefore A = ⟨exp 𝔞⟩ ⊆ ⟨exp L⟨exp 𝔳⟩)⟩ = ⟨exp 𝔳⟩ by (i).

(iii) Clearly Γ_{1} =^{def} ⟨e^{ad 𝔱^{+}}⟩ is contained in Γ and is normal by 2.4(v). If X ∈ 𝔱, then the centralizer of X contains 𝔢 as [𝔱, 𝔢] = {0} by 2.2(i). Thus for the isotropy group Γ_{X} at X we have 𝔢 ⊆ L[Γ_{X}]. Hence Γ = Γ_{1}Γ_{X}. If γ ∈ Γ, then Γ = γΓγ^{−1} = Γ_{1}Γ_{γX}. Thus Γ = Γ_{1}Γ_{X} for all X ∈ comp by 2.2(iii). Hence Γ·X = Γ_{1}.X for all X ∈ comp. But e^{ad⟨𝔱^{+}⟩}X ⊆ X + ⟨𝔱^{+}⟩ by 2.4(v). Thus comp ⊆ 𝔱 + ⟨𝔱^{+}⟩ and hence 𝔤 = span comp ⊆ 𝔱 + ⟨𝔱^{+}⟩. Thus 𝔤/⟨𝔱^{+}⟩ is an abelian algebra isomorphic to a subalgebra of 𝔱 and thus C^{∞} = 𝔤′ ⊆ ⟨𝔱^{+}⟩ in view of 2.4(i). We conclude C^{∞} = ⟨𝔱^{+}⟩ and thus Γ = e^{ad⟨𝔱^{+}⟩}. The Lie algebra generated by ad 𝔱^{+} in Der 𝔤 is ad⟨𝔱^{+}⟩. By (ii) we obtain Γ_{1} = ⟨e^{ad 𝔱^{+}}⟩ = exp^{⟨ad 𝔱^{+}⟩} = Γ.

(iv) was proved in the process of proving (iii).

(v) The analytic subgroup Γ of Aut 𝔤 has the Lie algebra ad⟨𝔱^{+}⟩ by (iii) and (i) above. Hence the Lie algebra L[Γ] is exactly ⟨𝔱^{+}⟩ + 𝔷.

2.6. Lemma. Let 𝔧 denote the largest ideal of 𝔤 contained in 𝔢. Then

(5)  𝔧 = 𝔷 ∩ 𝔢 = 𝔷 ∩ 𝔤′ = 𝔱 ∩ 𝔢 = 𝔱 ∩ 𝔤′.

Proof. Firstly, as 𝔢 = 𝔥 ∩ 𝔤′ we have 𝔷 ∩ 𝔢 = 𝔷 ∩ 𝔥 ∩ 𝔤′ = 𝔷 ∩ 𝔤′ and t ∩ 𝔢 = 𝔱 ∩ 𝔥 ∩ 𝔤′ = 𝔱 ∩ 𝔤′.

Secondly, since 𝔧 ⊆ 𝔢 ⊆ 𝔥 and 𝔱^{+} is an 𝔥-module, we have [𝔧, 𝔱^{+}] ⊆ 𝔧 ∩ 𝔱^{+} ⊆ 𝔢 ∩ 𝔱^{+} = {0}. It follows that [𝔧, ⟨𝔱^{+}⟩] = [𝔱, 𝔤′] = {0}. Since 𝔱 is central in 𝔥 we also know [𝔧, 𝔱] = {0}. Then [𝔧, 𝔤] = [𝔧, 𝔱 + 𝔤′] = {0}, i.e., 𝔧 ⊆ 𝔷. Thus 𝔧 ⊆ 𝔷∩𝔤′, and since 𝔷∩𝔤′ is an ideal in 𝔤, the equality 𝔧 = 𝔷∩𝔤′ follows. Thirdly, 𝔷 ⊆ 𝔱. Hence 𝔷 ∩ 𝔤′ ⊆ 𝔱 ∩ 𝔤′. On the other hand, X ∈ 𝔱 ∩ 𝔤′ implies that ad X is both nilpotent and semisimple. Hence X ∈ 𝔷 by Lemma 1.4. Thus 𝔱 ∩ 𝔤′ ⊆ 𝔷. Hence 𝔷 ∩ 𝔤′ = 𝔱 ∩ 𝔢.

We also need the following lemmas for the purpose of induction with respect to dimension.

2.7. Lemma. Let 𝔞 denote an ideal of 𝔤 and write 𝔤_{1} = 𝔤/𝔞. Then 𝔤_{1} is spanned by comp(𝔤_{1}).

Proof. We know that 𝔥_{1} is a Cartan algebra of 𝔤_{1} . Furthermore, all elements of (𝔱 + 𝔞)/𝔞 are semisimple and have purely imaginary spectrum, hence are contained in 𝔱_{1} = comp 𝔥_{1} . Since 𝔤 is spanned by comp = Γ·𝔱 and then 𝔤_{1} is spanned by comp(𝔤_{1}) = Γ_{1}·𝔱_{1} .

2.8. Lemma. Let 𝔞 be a nonzero minimal ideal (which exists as 𝔤 is solvable).

(i) Then there are two mutually exclusive cases:

(a) [𝔱, 𝔞] = {0}. Then 𝔞 is central and dim 𝔞 = 1.

(b) [𝔱, 𝔞] ≠ {0}. Then 𝔞 ⊆ 𝔱^{+} and dim 𝔞 = 2. in both cases, [𝔤′, 𝔞] = {0}.

(ii) 𝔷 ∩ 𝔤′ = {0} if and only if all nonzero minimal ideals are of type (b). In this case, 𝔥 is abelian.

Proof. We prove (i): Case (a): We know 𝔞 ⊆ 𝔷(𝔱, 𝔤) = 𝔥. Also [𝔞, 𝔱^{+}] ⊆ 𝔞 ∩ 𝔱^{+} ⊆ 𝔥 ∩ 𝔱^{+} = {0}. Then also [𝔞, 𝔤′] = [𝔞, ⟨𝔱^{+}⟩] = {0} whence {0} = [𝔞, 𝔱 + 𝔤′] = [𝔞, 𝔤], i.e., 𝔞 ⊆ 𝔷.

Case (b): We briefly consider the compact connected abelian group Ω =^{def} \overline{e^{ad 𝔱}}. Then 𝔤 is an Ω-module. Its unique fixed point module is 𝔥 = 𝔷(𝔱, 𝔤). Its unique effective submodule (spanned by all nontrivial simple submodules) is 𝔱^{+} . Now 𝔞 is an Ω-module. Hence 𝔞 is of the form 𝔞 = (𝔞 ∩ 𝔥) ⊕ (𝔞 ⊕ 𝔱^{+}). The assumption [𝔱, 𝔞] ≠ {0} implies 𝔞 ⊕ 𝔱^{+} ≠ {0}. Since Ω is connected, any nonzero submodule of the effective module 𝔱^{+} is at least 2-dimensional. But 𝔞, as a minimal ideal of 𝔤, is at most two dimensional. Hence 𝔞 ∩ 𝔱^{+} = 𝔞 and thus 𝔞 ⊆ 𝔱^{+} ⊆ 𝔤′. In this case, dim 𝔞 = 2.

In order to show [𝔤′, 𝔞] = {0} consider the representation π: 𝔤 → 𝔤𝔩(𝔞) ≅ 𝔤𝔩(2, ℝ), π(X)(Y) = [X, Y]. It image π(𝔤) contains π(𝔱) ≅ 𝔰𝔬(2) ⊆ 𝔤𝔩(2, ℝ)

and is solvable. But ℝ·E_{2} ⊕ 𝔰𝔬(2) = \{( \begin{matrix}r & s \\ −s & r \end{matrix} ) : r, s ∈ ℝ\}, the unique Cartan

algebra containing 𝔰𝔬(2) is also a unique Borel subalgebra (i.e., maximal solvable subalgebra) containing 𝔰𝔬(2). Thus π(𝔤) is contained in an abelian algebra and thus π(𝔤′) = 0. This means [𝔤′, 𝔞] = {0}.

(ii) If 𝔧 = 𝔷 ∩ 𝔤′ ≠ {0} then every one-dimensional vector subspace 𝔞 of 𝔧 is a nonzero minimal ideal of type (a). Conversely, if those minimal ideals exist, then 𝔧 ≠ {0} by (i).

Now we assume 𝔧 = {0} and show [𝔥, 𝔥] = {0}. We prove the claim by induction. The situation is known through simple inspection up to dimension 3. We assume that the claim is proven for all dimensions below n > 3 and consider 𝔤 with dim 𝔤 = n. Let 𝔞 denote a nonzero minimal ideal. Since 𝔧 = {0} then 𝔞 is of type (b). In particular, 𝔞 ⊆ 𝔱^{+} ⊆ 𝔤′. From Lemma 2.7 we know that 𝔤_{1} =^{def} 𝔤/𝔞 satisfies our general hypothesis to be spanned by its compact elements. Suppose that 𝔧_{1} is the maximal ideal of 𝔤_{1} contained in 𝔥_{1} ∩ 𝔤′_{1} = \frac{𝔥+𝔞}{𝔞} ∩ \frac{𝔤′}{𝔞} = \frac{(𝔥+𝔞)∩𝔤′}{𝔞} = \frac{(𝔥∩𝔤′)+𝔞}{𝔞} = \frac{𝔧⊕𝔞}{𝔞} (by the modular law). Now 𝔧 was assumed to be {0}. Hence 𝔧_{1} = {0}. Thus the induction hypothesis applies to 𝔤/𝔞 and shows that the Cartan algebra \frac{𝔥+𝔞}{𝔞} is abelian. Thus [𝔥, 𝔥] ⊆ 𝔞. But then [𝔥, 𝔥] ⊆ 𝔥 ∩ 𝔞 = {0}. The induction is complete.

2.9. Lemma. The Cartan algebra 𝔥 satisfies [𝔥, 𝔥] ⊆ 𝔷 ∩ 𝔤′. In particular, 𝔥 is nilpotent of class ≤ 2.

Proof. Recall 𝔧 = 𝔷 ∩ 𝔤′. Consider 𝔤_{1} = 𝔤/𝔧. Let 𝔧^{∗} contain 𝔧 and be such that 𝔧^{∗}/𝔧 = 𝔷(𝔤_{1}) ∩ 𝔤′_{1} ⊆ 𝔥_{1} ∩ 𝔤′_{1} = \frac{𝔥}{𝔧} ∩ \frac{𝔤′}{𝔧} = \frac{𝔥∩𝔤′}{𝔧} = 𝔢/𝔧. Thus 𝔧^{∗} is an ideal of 𝔤 contained in 𝔢. Then 𝔧^{∗} ⊆ 𝔧 by 2.6 and thus 𝔧^{∗} = 𝔧. Thus 𝔤_{1} satisfies the hypotheses of 2.8(ii) and thus [𝔥_{1}, 𝔥_{1}] = {0}. Hence [𝔥, 𝔥] ⊆ 𝔧.

We collect some of the information which was accumulated so far:

2.10. Proposition. Let 𝔤 be a real solvable Lie algebra which is spanned by its compact elements. Fix a Cartan algebra 𝔥 and set 𝔱 = 𝔥 ∩ comp(𝔤). Then we have

(i) [𝔱, 𝔥] = {0}, (ii) 𝔤′ = C^{∞}, (iii) 𝔷 ∩ 𝔤′ is the largest ideal contained in 𝔥 ∩ 𝔤′, and (iii) [𝔥, 𝔥] ⊆ 𝔷 ∩ 𝔤′. In particular, 𝔥 is nilpotent of class ≤ 2.

M. Wüstner has constructed a 13-dimensional solvable Lie algebra spanned by its compact elements in which 𝔥 ≅ ℝ^{2} ⊕ H_{3} with the threedimensional Heisenberg algebra H_{3} so that Part (iii) of Lemma 2.9 cannot be improved in that 𝔥 need not be abelian.

The structure of a solvable algebra 𝔤 spanned by comp therefore may be roughly summarized in the following diagram:

𝔤
𝔤′ = C^{∞} = ⟨𝔱^{+}⟩
𝔥  𝔱^{+}
𝔢 = 𝔥 ∩ 𝔤′
𝔱
𝔷
𝔧 = 𝔷 ∩ 𝔤′
𝔱_{1}  𝔱_{2}  𝔧  𝔢_{1}  𝔱^{+}

The following remarks will be used for purposes of induction:

2.11. Lemma. Let 𝔞 denote an ideal of 𝔤. Set

Γ_{1} = ⟨e^{ad 𝔤′_{1}}⟩ with 𝔤′_{1} = (𝔤′ + 𝔞)/𝔞.

(i) There is a surjective homomorphism φ: Γ → Γ_{1} with φ(γ)(X + 𝔞) = γ(X) + 𝔞. and ker φ = {γ ∈ Γ: (∀X ∈ 𝔤)γ(X) − X ∈ 𝔞}. In particular, ker φ ⊇ \overline{⟨e^{ad 𝔞}⟩} with the closure taken in Γ.

(ii) The orbit Γ_{1}·(X + 𝔞) equals (Γ·X + 𝔞)/𝔞.

Proof. (i) The assignment γ ↦ (X+𝔞 ↦ γ(X)+𝔞): Inn(𝔤) → Inn(𝔤_{1}) is a well defined homomorphism whose kernel consists of all those γ with γ(X) − X ∈ 𝔞 for all X ∈ 𝔤.

Now let γ = e^{ad X} with X ∈ 𝔤′. Let γ_{1} = e^{ad(X+𝔞)} ∈ Γ_{1} . Take Y ∈ 𝔤 and notice φ(γ(Y +𝔞)) = γ(Y)+𝔞 = e^{ad X}Y +𝔞 = Y +[X, Y]+\frac{1}{2}[X, [X, Y]]+⋯+𝔞 on the one hand and γ_{1}(Y + 𝔞) = e^{ad(X+𝔞)}(Y + 𝔞) = Y + 𝔞 + [X + 𝔞, Y + 𝔞] + \frac{1}{2}[X + 𝔞, [X + 𝔞, Y + 𝔞]] + ⋯ = Y + [X, Y] + \frac{1}{2}[X, [X, Y]] + ⋯ + 𝔞 on the other. Hence γ_{1} = φ(γ). Thus φ(Γ) ⊆ Γ_{1} . Since Γ_{1} is generated by e^{ad 𝔤_{1}} = φ(e^{ad 𝔤′}), we have φ(Γ) = Γ_{1} .

If X ∈ 𝔞, then e^{ad X}Y − Y = [X, Y] + ⋯ ∈ 𝔞.

(ii) Finally, for γ_{1} ∈ Γ_{1} we find γ ∈ Γ with γ_{1} = φ(γ). Then γ_{1}·(X + 𝔞) = φ(γ)·(X + 𝔞) = γ·X + 𝔞. Hence Γ_{1}·(X + 𝔞) = (Γ·X + 𝔞)/𝔞.

We consider the semidirect product ℝ^{2}⋊D of a plane with a compact connected group D acting irreducibly on ℝ^{2} . Let 𝔤 = ℝ^{2}⋊𝔡 denote the Lie algebra and X = (0, x) a vector acting nontrivially on ℝ^{2} . Let Y = (v, 0). Then in the subalgebra ℝ^{2}⋊ℝ·x the subspaces ℝ·X = {0} × ℝ·x and ℝ·(X − Y) = ℝ.(−v, x) are Cartan subalgebras and hence are conjugate under e^{ad ℝ^{2}×{0}}. In particular, there is a g ∈ ℝ × {1} ⊆ G with Ad(g)ℝ.(X − Y) = ℝ·X.

2.12. Lemma. Let X ∈ comp 𝔤 be regular and 𝔞 a minimal abelian ideal such that [X, 𝔞] ≠ {0}. Then X − Y ∈ comp(𝔤) for all Y ∈ 𝔞.

Proof. The element ad X|𝔞 is semisimple and has nonzero purely imaginary spectrum. Thus the subalgebra ℝ·X ⊕ 𝔞 is invariant under the action of the group e^{ℝ· ad X} which transitively permutes the the one-dimensional subspaces of 𝔞. Hence it permutes transitively the one parameter subgroups of the analytic subgroup A =^{def} e^{ad 𝔞} under conjugation. This subgroup therefore is either closed or relatively compact. In the second case, its closure would be a compact normal abelian subgroup of \overline{Inn(𝔤)} and would therefore be central, which is not the case on account of the nontrivial action of X on 𝔞. Thus A is closed in \overline{Inn 𝔤} and isomorphic to ℝ^{2} . The group D =^{def} \overline{e^{ℝ· ad X}} ⊆ int(𝔤) is compact and contained in the group M =^{def} DA ⊆ \overline{⟨e^{ℝ·X⊕𝔞}⟩} ⊆ int(𝔤). The product DA is semidirect. Now let Y ∈ 𝔞. Consider the subgroup D_{1} =^{def} \overline{e^{ℝ· ad(X−Y)}} . Since ℝ·(X − Y) is conjugate to ℝ·X under the action of A by the remarks preceding this lemma, we know that D_{1} is conjugate to D in Inn(𝔤). Thus D_{1} is compact. Hence X − Y ∈ comp(𝔤) as asserted.

3. On some group actions

We let 𝔤 denote a real Lie algebra and consider a mixture of actions by translation and inner automorphisms. If Γ is a closed subgroup of Inn(𝔤), as before, we let L[Γ] denote the unique Lie algebra within 𝔤 such that X ∈ L[Γ] means e^{ℝ·X} ⊆ Γ. Then L[Γ]/𝔷 ≅ L(Γ) is the Lie algebra of Γ. We identify the Lie algebra of Γ with L[Γ]/𝔷.

3.1. Lemma. Let Γ ⊆ Inn(𝔤) be a subgroup and 𝔞 a subalgebra invariant under Γ. We let 𝔞⋊Γ denote the semidirect product Lie group with multiplication (Z_{1}, γ_{1})(Z_{2}, γ_{2}) = (Z_{1} + γ_{1}(Z_{2}), Z_{1} + Z_{2}) and we set

(Z, γ)·X = Z + γ(X) for γ ∈ Γ, Z ∈ 𝔞, X ∈ 𝔤.

Then ((Z, γ) X) ↦ (Z, γ)·X : (𝔞⋊Γ) × 𝔤 → 𝔤 is an action.

Proof. If (0, η) is the identity of 𝔞⋊Γ then (0, η)·X = 0 + η(X) = X and (Z_{1}, γ_{1})·(Z_{2}, γ_{2})·X) = Z_{1} + γ_{1}(Z_{2} + γ_{2}(X)) = Z_{1} + γ_{1}(Z_{2}) + γ_{1}γ_{2}(X) = ((Z_{1}, γ_{1})(Z_{2}, γ_{2}))·X.

3.2. Definition. We call the action described in the preceding lemma the affine action of 𝔞⋊Γ.

3.3. Lemma. (i) Let A and B two subgroups of a group C. Then A × B acts on C on the left via (a, b)·c = acb^{−1} .

The function a ↦ (a, c^{−1}ac) : A ∩ cBc^{−1} → A × B implements an isomorphism onto the isotropy group (A × B)_{c}.

The function (a, b) ↦ acb: A × B → G induces a bijection

(A × B)/{(a, c^{−1}ac): a ∈ A ∩ cBc^{−1}} → AcB.

(ii) If A, B, and C are Lie groups, then AcB is a submersed submanifold of dimension dim L(A) + dim L(B) − dim(L(A) ∩ Ad(c)L(B))

(iii) If, under the hypotheses of (ii),

dim(L(A) ∩ Ad(c)L(B)) = dim L(A) + dim L(B) − dim L(C) for all c ∈ C,

then all AcB, c ∈ C are open closed submanifolds of C. If C is connected, then AcB = G for all c ∈ C.

(iv) Suppose that C is a connected Lie group acting differentiably on a manifold M. Let B denote the isotropy subgroup of C at m. Then cBc^{−1} is the isotropy subgroup at c·m. If A is a Lie subgroup such that

(†)  dim(L(A) ∩ Ad(c)L(B)) = const for all c ∈ C,

then A·m = C·m.

Proof. (i) The action is clear. We have (a, b) ∈ (A × B)_{c} iff acb^{−1} = c iff ac = cb iff b = c^{−1}ac. If this is satisfied, then a = cbc^{−1} ∈ cBc^{−1} , and conversely, if a satisfies this condition then (a, c^{−1}ac) ∈ (A × B)_{c}.

(ii) The dimension of the submersed manifold AcB equals the dimension of the homogeneous space (A × B)/{(a, c^{−1}ac): a ∈ A ∩ cBc^{−1}} and its dimension is that of (L(A) × L(B))/{(X, Ad(c)^{−1}X): X ∈ L(A) ∩ Ad(c)L(B)}. The assertion follows.

(iii) Under the present hypotheses, by (ii) all sets AcB are open submanifolds of C. Since they form the orbits decomposition of a group action, each of the sets AcB is also closed. The remainder is then clear.

(iv) There are bijective immersions AB/B → A·m and C/B → C·m and AB/B ⊆ C/B. The claim then follows if we establish AB/B = C/B. By (iii) we have AcB = C for all c ∈ C, in particular AB = C. The assertion follows.

For the following discussion it is useful to recall some general facts on the smooth action of Lie groups on manifolds.

Let (g, m) ↦ g·m : G × M → M denote a smooth action of a Lie group on a manifold M. Let X ↦ X̃: 𝔤 → V (M) denote the map which associates with an X ∈ 𝔤 the G-invariant vector field X̃: M → T (M) on M which is given by X̃(x) = \frac{d}{dt} |_{t=0} ( exp(t·X)·x) for x ∈ M. Then the integral manifolds of X̃ are exactly the orbits of the action (r, x) ↦ (exp r·X)·x : ℝ × M → M. Thus m is a fixed point for this action iff X̃(m) = 0. As a consequence, we obtain

3.4. Lemma. Let H = G_{m} denote the isotropy group of G at m. Then X ∈ 𝔥 = L(H) if and only if \frac{d}{dt} |_{t=0} ( exp(t·X)·m) = 0.

3.5. Lemma. Suppose that Γ is an analytic subgroup of the group Inn(𝔤) of inner automorphisms of 𝔤 and 𝔞 a subalgebra invariant under Γ and let 𝔞⋊Γ act on 𝔤 by the affine action. (See Definition 3.2.)

(i) Let (𝔞⋊Γ)_{X} denote the isotropy group of the affine action at X. Then

L((𝔞⋊Γ)_{X}) = {(Z, Y + 𝔷) ∈ 𝔞⋊L(Γ) = 𝔞⋊(L[Γ]/𝔷) : [X, Y] = Z}.

(ii) The orbit (𝔞⋊Γ)·X is a submersed submanifold whose tangent space at X is 𝔞 + [L[Γ], X] when the tangent space T_{X}(𝔤) of 𝔤 at X is elementarily identified with 𝔤.

Proof. (i) First we note that L(𝔞) = 𝔞 and L(𝔞⋊Γ) = 𝔞⋊L(Γ) = 𝔞⋊(L[Γ]/𝔷). For Z ∈ 𝔞, Y ∈ L[Γ], and sufficiently small t ∈ ℝ we have (t·Z, 0)∗(0, t·Y +𝔷) = t·(Z, Y +𝔷)+(\frac{t^{2}}{2} ·[Z, Y], 0)+O(t^{3}) and exp((t·Z, 0)∗(0, t·Z +𝔷)) = (t·Z, e^{t· ad Y}). Then

exp(t·(Z, Y + 𝔷))·X + O(t^{2}) = exp(t·(Z, Y + 𝔷) + O(t^{2}))·X
= t·Z + e^{t· ad Y} X = X + t·(Z + [Y, X]) + O(t^{2}).

Hence the assertion is a consequence of Lemma 3.4.

(ii) We recall exp(Z, Y + 𝔷)·X − X = t·([Y, X] + Z) + O(t^{2}) and this implies the assertion.

3.6. Lemma. Let 𝔫 denote a subalgebra of the nilradical of a Lie algebra 𝔤 and suppose that 𝔫 contains the center 𝔷. Set Γ = ⟨e^{ad 𝔫}⟩ and let and 𝔞 be any ideal of 𝔤. We consider the Lie group 𝔞⋊Γ, the isotropy group (𝔞⋊Γ)_{X} of the affine action of 𝔞⋊Γ at an element X. Then

(i) L({0} × Γ) ∩ L((Y, α)^{−1}(𝔞⋊Γ)_{X}(Y, α)) ≅ (𝔷(X − Y, 𝔤) ∩ 𝔫)/𝔷 for all (Y, α) ∈ 𝔞⋊Γ.

(ii) Suppose now that, in addition, 𝔫 is an ideal, X a regular compact element of 𝔤, and 𝔞 a minimal ideal with [X, 𝔞] ≠ {0}. Then for any Cartan algebra 𝔥 we have

L({0} × Γ) ∩ L((Y, α)^{−1}(𝔞⋊Γ)_{X}(Y, α)) ≅ (𝔥 ∩ 𝔫)/𝔷 for all (Y, α) ∈ 𝔞⋊Γ.

Proof. (i) We identify the Lie algebra of Γ with L[𝔫]/𝔷.

For (Y, α) ∈ 𝔞⋊Γ we have (Y, α) = (Y, 1)(0, α). Hence

L({0} × Γ) ∩ L((Y, α)^{−1}(𝔞⋊Γ)_{X}(Y, α)) =
L({0} × Γ) ∩ L((0, α)^{−1}(Y, 1)^{−1}(𝔞⋊Γ)_{X}(Y, 1)(0, α)) =
L({0} × Γ) ∩ Ad(0, α)^{−1}L((Y, 1)^{−1}(𝔞⋊Γ)_{X}(Y, 1)) ≅
Ad(0, α)L({0} × Γ) ∩ L((Y, 1)^{−1}(𝔞⋊Γ)_{X}(Y, 1)) =
L((0, α)({0} × Γ)(0, α^{−1}) ∩ L((−Y, 1)(𝔞⋊Γ)_{X}(Y, 1)) =
L({0} × Γ) ∩ L((−Y, 1)(𝔞⋊Γ)_{X}(Y, 1)).

In order to determine the right portion of the intersection, we note first that (Z, γ) ∈ (𝔞⋊Γ)_{X} iff Z + γ(X) = X iff Z = X − γ(X) ∈ 𝔞. Further, (−Y, 1)(𝔞⋊Γ)_{X}(Y, 1) = (𝔞⋊Γ)_{(−Y,1)·X} and (−Y, 1)·X = X − Y . Thus

(−Y, 1)(𝔞⋊Γ)(Y, 1) = {((X − Y) − γ(X − Y), γ) : γ ∈ Γ and γ(X − Y) ∈ X − Y + 𝔞}.

In order to compute the Lie algebra

𝔟 =^{def} L((−Y, 1)(𝔞⋊Γ)_{X}(Y, 1)) = L(𝔞⋊Γ)_{X−Y}

we want to apply Lemma 3.5 and recall that 𝔫 is nilpotent and thus Γ = e^{ad 𝔫} and L(Γ) = L[Γ]/𝔷 = 𝔫/𝔷. Then

\begin{aligned}𝔟 & = \{(Z_{1}, Z_{2} + 𝔷) ∈ 𝔞⋊L(Γ) : [X − Y, Z_{2}] = Z_{1}\} \\ & = \{([X − Y, Z], Z + 𝔷) : Z ∈ 𝔫 and [X − Y], Z] ∈ 𝔞\}. \end{aligned}

Thus

L({0} × Γ) ∩ L((−Y, 1)(𝔞⋊Γ)_{X}(Y, 1))

= {0} × L(Γ) ∩ {(ad(X − Y)(Z), Z + 𝔷): Z ∈ 𝔫, ad(X − Y)(Z) ∈ 𝔞} = {(0, Z + 𝔷): Z ∈ 𝔫, ad(X − Y)(Z) = 0} = {0} × 𝔷(X − Y, 𝔫)/𝔷 = {0} × (𝔷(X − Y, 𝔤) ∩ 𝔫)/𝔷 ≅ (𝔷(X − Y, 𝔤) ∩ 𝔫)/𝔷.

This proves (i).

(ii) Under the additional hypotheses, by Lemma 2.12 we know that X − Y ∈ comp(𝔤) for all Y ∈ 𝔞. Hence X − Y is semisimple. Now X is regular and Y ∈ 𝔞 and 𝔞 as a minimal ideal is abelian hence contained in the nilradical. Thus Y is in the nilradical. Therefore X − Y is regular. Therefore 𝔷(X − Y, 𝔤) is the Cartan algebra 𝔥(X − Y) generated by X − Y . Since 𝔤 is solvable there is an inner automorphism β with 𝔥(X − Y) = β(𝔥). As 𝔫 is an ideal, we have 𝔷(X − Y, 𝔤) + 𝔫 = β(𝔥) + β(𝔫) = β(𝔥 + 𝔫). The assertion now follows from (i).

3.7. Lemma. Let 𝔤 denote a solvable Lie algebra which is spanned by comp and let 𝔱 = 𝔥 ∩ comp as in (3) for a Cartan algebra 𝔥. Let X ∈ 𝔱 be a regular element and 𝔞 a minimal ideal with [X, 𝔞] ≠ {0}. If Γ = e^{ad 𝔤′}, then 𝔞 + Γ·X = Γ·X.

Proof. We apply the preceding Lemma 3.6 with 𝔫 = 𝔤′ + 𝔷. Thus

L({0} × Γ) ∩ Ad((Y, α))(𝔞 × Γ)_{X} ≅ (𝔥 ∩ (𝔷 + 𝔤′))/𝔷 for all (Y, α) ∈ 𝔞 × Γ.

Now we apply Lemma 3.3 with C = 𝔞⋊Γ, A = {0} × Γ, M = 𝔤 and m = X, the affine action of the group 𝔞⋊Γ, and B = (𝔞⋊Γ)_{X}. Condition (†) of Lemma 3.3(iv) is satisfied. We conclude 𝔞 + Γ·X = C·m = A·m = Γ·X.

We are ready for the main result. From Γ = e^{ad 𝔤′} we know that Γ·X ⊆ X + 𝔤′ for all X. In the four-dimensional oscillator algebra, for regular elements X ∈ 𝔱, the orbits ΓX are three dimensional quadratic hyperboloids. The next theorem, however, shows that the such orbits are almost affine under rather general circumstances; they need just the subalgebra 𝔢 = 𝔥 ∩ 𝔤′ as filling material in order to fill X + 𝔤′.

3.8. Theorem. Suppose that 𝔤 is a solvable Lie algebra which is the span of its compact elements. We set Γ = e^{ad 𝔤′}. Let 𝔥 be a Cartan algebra and set 𝔱 = 𝔥 ∩ comp. Then 𝔥 ∩ 𝔤′ is a nilpotent subalgebra of class at most two with commutator algebra contained in 𝔷 ∩ 𝔤′, and for every regular X ∈ 𝔱 we have

(†)  (𝔥 ∩ 𝔤′) + Γ·X = X + 𝔤′.

Proof. The first assertions were summarized in Proposition 2.10. It remains to show (†). We prove the claim by induction w.r.t. dim 𝔤.

The assertion is true for dim 𝔤 = 0, 1, 2, 3 as a direct inspection shows. Now suppose that 𝔤 is a counterexample with smallest possible dimension.

Claim (i): 𝔷 = 𝔧.

For the center 𝔷 we know 𝔷 ∈ 𝔱 and we write 𝔷 = 𝔧 ⊕ 𝔷_{1} . Then 𝔱 = 𝔷_{1} ⊕ 𝔱_{1} and 𝔤 = 𝔷_{1} ⊕ (𝔱_{1} + 𝔤′). If we set 𝔤_{1} = 𝔱_{1} + 𝔤′, then 𝔤′_{1} = 𝔤′

and 𝔢_{1} =^{def} 𝔥_{1} ∩ 𝔤′_{1} = (𝔱_{1} + 𝔢) ∩ 𝔤′ = 𝔢. 𝔷(𝔤_{1}) = 𝔷 ∩ 𝔤_{1} ⊆ 𝔤′. If 𝔷_{1} ≠ 0, then

dim 𝔤_{1} < dim 𝔤 and so 𝔤_{1} is not a counterexample. As X is a regular element of 𝔱, we have X = Z + X_{1} with Z ∈ 𝔷_{1} and X_{1} ∈ 𝔱_{1} with X_{1} regular. Then 𝔢+Γ·X = 𝔢+Γ(Z +X_{1}) = 𝔢+Z +Γ·X_{1} = Z +𝔢_{1} +Γ·X_{1} = Z +X_{1} +𝔤′_{1} = X +𝔤′.

So 𝔤 cannot be a counterexample. Hence 𝔷_{1} = {0} and thus 𝔷 ⊆ 𝔧. On the other hand 𝔧 ⊆ 𝔷 by Lemma 2.6.

Step (ii): 𝔷 = {0} and 𝔥 is abelian. In particular, 𝔢 is abelian. Suppose not. By (i) above we know 𝔷 ⊆ 𝔤′. Then we consider 𝔤_{1} = 𝔤/𝔷. By Lemma 2.7, 𝔤_{1} is spanned by its compact elements. We write 𝔷_{2} for the full inverse image of 𝔷_{1} = 𝔷(𝔤_{1}) in 𝔤. Since 𝔥 is the full inverse image of 𝔥_{1} = 𝔥/𝔷 we have 𝔷_{2} ⊆ 𝔥. Then 𝔷_{2} ∩ 𝔤′ is an ideal in 𝔢 = 𝔥 ∩ 𝔤′ and is therefore contained in 𝔧 = 𝔷 by Lemma 2.6. Then 𝔧_{1} = 𝔷_{1} ∩ 𝔤′_{1} = \frac{𝔷_{2}}{𝔷} ∩ \frac{𝔤′}{𝔷} = \frac{𝔷_{2}∩𝔤′}{𝔷} ⊆ \frac{𝔷∩𝔤′}{𝔷} = \{0\}. Further,

𝔢_{1} = 𝔥_{1} ∩ 𝔤_{1} = \frac{𝔥}{𝔷} ∩ \frac{𝔤′}{𝔷} = \frac{𝔢}{𝔷}.

Let Γ_{1} = e^{ad 𝔤′_{1}} ⊆ Aut 𝔤_{1} . We have dim 𝔤_{1} < dim 𝔤 by assumption. Then 𝔤_{1} cannot be a counterexample to the theorem. For a regular X ∈ 𝔱 also X + 𝔷 is regular in 𝔤_{1} and we obtain

𝔢_{1} + Γ_{1}·(X + 𝔷) = 𝔷_{1} + Γ_{1}·X = X + 𝔤_{1}.

In view of Lemma 2.11 we pass to inverse images and find

𝔢 + Γ·X = 𝔢 + Γ·X + 𝔷 = X + 𝔷 + 𝔤′ = X + 𝔤′

as 𝔷 is contained in 𝔤′ by (i). But this contradicts the fact that 𝔤 is a counterexample. Thus 𝔷 = {0}. Then 𝔥 is abelian by 2.8(ii).

Step (iii): Let 𝔞 be a nonzero minimal ideal of 𝔤. By Lemma 2.6(b) we know 𝔞 ⊆ 𝔱^{+} ⊆ 𝔤′. We consider the factor algebra 𝔤_{1} = 𝔤/𝔞. It is spanned by its compact elements in view of Lemma 2.7 and satisfies dim 𝔤_{1} < dim 𝔤. Hence 𝔤_{1} not a counterexample. We let Γ_{1} = ⟨e^{ad(𝔤′/𝔞)}⟩ ⊆ Aut 𝔤_{1} . Then for X regular in 𝔱, the element X + 𝔞 ∈ \frac{𝔱+𝔞}{𝔞} is compact and regular. Thus we have

(∗)  𝔢_{1} + Γ_{1}·(X + 𝔞) = (X + 𝔞) + 𝔤′_{1}

with 𝔢_{1} = 𝔥_{1} ∩ 𝔤′_{1} .

Now we let 𝔢^{∗} denote the full inverse image of 𝔢_{1} in 𝔤. Since 𝔢_{1} = 𝔥_{1} ∩𝔤′_{1} where 𝔥_{1} = \frac{𝔥+a}{𝔞} we know 𝔢^{∗}/𝔞 = 𝔢_{1} = \frac{𝔥+𝔞}{𝔞} ∩ \frac{𝔤′}{𝔞} = \frac{(𝔥+𝔞)∩𝔤′}{𝔞} . Because of the modular law this equals \frac{(𝔥∩𝔤′)+𝔞}{𝔞} = \frac{𝔢+𝔞}{𝔞}. Hence we have 𝔢^{∗} = 𝔢+𝔞. (Note that 𝔢 is abelian by Step (ii). By 2.8(i) we know [𝔤′, 𝔞] = {0}. Thus [𝔢, 𝔞] ⊆ [𝔤′, 𝔞] = {0}. Hence 𝔢 + 𝔞 abelian.)

In view of Lemma 2.11, passing to the inverse images in (∗) we obtain

(∗∗)  𝔢 + 𝔞 + Γ·X = (𝔢 + 𝔞) + (Γ·X + 𝔞) = X + 𝔞 + 𝔤′ = X + 𝔤′.

Now the hypotheses of Lemma 3.7(iii) are satisfied whence we know that 𝔞 + Γ·X = Γ·X. Thus 𝔢 + Γ·X = X + 𝔤′ and this again shows that 𝔤 is not a counterexample. Thus case (b) cannot occur either—a contradiction which finishes the proof.

We shall identify the tangent space T_{X}(Γ·X) at X of the orbit Γ·X ⊆ 𝔤 with a vector subspace of 𝔤 via translation.

3.9. Proposition. Under the hypotheses of Theorem 3.8, the tangent space of the orbit Γ·X at the point X is T_{X}(Γ·X) = 𝔱^{+} .

Proof. From Lemma 3.5(ii) we know that T_{X}(Γ·X) = [L[Γ], X]. But L[Γ] = L[e^{ad 𝔤′}] = 𝔷 + 𝔤 by Lemmas 2.4(i) and 2.5(iv,v). Now 𝔤′ = 𝔢 ⊕ 𝔱^{+} by Lemma 2.4(i,iv). Thus [L[Γ], X] = [𝔢 ⊕ 𝔱^{+}, X] = [𝔢, X] + [𝔱^{+}, X]. But [𝔢, X] = {0} by Lemma 2.2(i), and [𝔱^{+}, X] = ad(X)(𝔱^{+}) = 𝔱^{+} by Remark 2.3. This proves the proposition.

This proposition shows that we cannot expect to reduce the size of the filling space 𝔢 = 𝔥 ∩ 𝔤′ in Theorem 3.8, because by the preceding proposition we have exactly (𝔥 ∩ 𝔤′) + T_{X}(Γ·X) = 𝔢 + 𝔱^{+} = 𝔤′.

We obtain the following result as an immediate consequence of Theorem 3.8:

3.10. Theorem. Let 𝔤 be a solvable Lie algebra which is spanned by its compact elements. Let 𝔢_{1} denote any vector space complement of 𝔷 ∩ 𝔤′ in 𝔢 = 𝔥 ∩ 𝔤′. Then 𝔤 = 𝔢_{1} + \overline{comp(𝔤)}.

Proof. Let R denote the set of regular elements in 𝔱. Now, by Theorem 3.8 and Lemma 2.2(iii) we conclude R+𝔤′ = 𝔢+Γ·R ⊆ 𝔢+comp(𝔤) = (𝔢_{1} +(𝔷∩𝔥))+ comp(𝔤) = 𝔢_{1}+comp(𝔤) since 𝔷+comp = comp. Hence R+𝔤′ ⊆ 𝔢+\overline{comp(𝔤)} ⊆ 𝔤. But R is dense in 𝔱 and R + 𝔤′ is dense in 𝔤 by Lemma 2.4(i). The assertion follows.

In particular we note that 𝔢_{1} = {0} means 𝔱 = 𝔥, i.e., 𝔥 ⊆ comp, and, equivalently, 𝔥 ∩ 𝔤′ = 𝔷 ∩ 𝔤′, and that, in this case, the set comp of compact elements is dense in 𝔤.

We now abandon the general hypothesis that 𝔤 is spanned by comp and summarize that which our results yield for real solvable Lie algebras in general. Thus let 𝔤 denote a solvable Lie algebra. Then span(comp(𝔤)) is a characteristic ideal 𝔤_{comp} . Let 𝔥 be a Cartan algebra of 𝔤 and set 𝔱 = 𝔥 ∩ comp(𝔤). Define 𝔱^{+} to be the Fitting one component of ad X for a regular X ∈ 𝔱. Now 𝔥_{comp} = 𝔷(𝔱, 𝔤_{comp}) = 𝔥 ∩ 𝔤_{comp} is a Cartan algebra of 𝔤_{comp} containing 𝔱. We set 𝔢 = 𝔥_{comp} ∩ C^{∞}(𝔤_{comp}) and let 𝔢_{1} denote any complement of 𝔷 ∩ 𝔢 in 𝔢.

3.11. Corollary. In a solvable Lie algebra 𝔤 we have

𝔤_{comp} = 𝔱 + ⟨𝔱^{+}⟩ = 𝔱 + (𝔢 ⊕ 𝔱^{+})

The subalgebra 𝔢 is nilpotent of class at most 2 and centralizes 𝔱. It does not, in general, centralize 𝔱^{+} even though all roots of 𝔤_{comp} with respect to 𝔥_{comp} vanish on 𝔢.

Further, 𝔢 ⊕ 𝔱^{+} = ⟨𝔱^{+}⟩ = 𝔤′_{comp} = C^{∞}(𝔤_{comp}) and 𝔷(𝔤_{comp}) = 𝔱 ∩ 𝔢 = 𝔱 ∩ 𝔤′_{comp} .

Finally,

𝔤_{comp} = 𝔢_{1} + \overline{comp}.

Proof. This is just a summary of what has been shown in the process.

4. Examples

In order to see that the structural information on solvable Lie algebras which are spanned by their compact elements cannot be much improved we consider a convenient class of examples.

Let A denote a finite dimensional commutative algebra over ℝ with identity and an augmentation morphism α: A → ℝ whose kernel A_{0} is nilpotent. If A_{0}^{n} = {0} and A_{0}^{n−1} ≠ {0}, then n is called the nilpotent class of A_{0} . We identify ℝ with the multiples of 1.

Let 𝔨 = 𝔲 ⊕ 𝔳 denote a Lie algebra such that 𝔲 = ℝ·U + ℝ·[IX, X] and 𝔳 = ℝ·X ⊕ ℝ·IX with [U, X] = −IX and [U, IX] = X. Then dim 𝔨 = 3 or 4 according as [IX, X] = or ≠ 0. In other words, 𝔨 is one of the standard low dimensional test algebras 𝔰𝔬(3), 𝔰𝔩(2, ℝ), 𝔪_{3}, 𝔬_{4} where 𝔪_{3} is the motion algebra of the euclidean plane and 𝔬_{4} the oscillator algebra with compactly embedded Cartan algebra 𝔲 and real root space 𝔳.

4.1. Example. The Lie algebra A⊗𝔨 with bracket [a⊗Y, a′⊗Y′] = aa′⊗[Y, Y′] is of mixed type with Levi complement 1⊗𝔨 if 𝔨 = 𝔰𝔬(3), 𝔰𝔩(2, ℝ) and is solvable if 𝔨 is solvable. The subalgebra

𝔤 = (ℝ ⊗ 𝔲) ⊕ (A_{0} ⊗ 𝔨)

is a maximal solvable subalgebra. The following statements hold

(i) The nilradical 𝔫 equals A_{0} ⊗ 𝔨. The nilpotent class of 𝔫 is that of A_{0} , and dim 𝔫 = (dim A_{0})(dim 𝔨). Further 𝔤′ = C^{∞} = (A_{0}^{2} ⊗ 𝔲) ⊕ (A_{0} ⊗ 𝔳).

(ii) 𝔥 =^{def} A ⊗ 𝔲 is a Cartan algebra of A ⊗ 𝔨 and of 𝔤. We have [𝔥, 𝔥] = 0, 𝔱 =^{def} comp(𝔥) = 1 ⊗ 𝔲, 𝔢 =^{def} 𝔥 ∩ C^{∞} = A_{0} ⊗ 𝔲, and 𝔱^{+} = A_{0} ⊗ 𝔳. In particular,

dim 𝔥 = (dim A)(dim 𝔲), dim 𝔱 = dim 𝔲, dim 𝔱^{+} = 2(dim A_{0}).

(iii) 𝔤_{comp} = 𝔱 + ⟨𝔱^{+}⟩ = (1 ⊗ 𝔲) ⊕ (A_{0}^{2} ⊗ 𝔲) ⊕ (A_{0} ⊗ 𝔳). Accordingly we have 𝔢_{comp} = A_{0}^{2} ⊗ 𝔲, and 𝔷_{comp} = B ⊗ 𝔲, where B = {a ∈ A : Aa = {0}}. Further 𝔤′_{comp} = 𝔤′ and 𝔢_{comp} = 𝔢.

References

[1] Wüstner, M., An example of a real solvable Lie algebra spanned by its compact elements, Seminar Sophus Lie 2 (1992), this volume.

Karl H. Hofmann Fachbereich Mathematik Technische Hochschule Darmstadt Schloßgartenstraße 7 D-6100 Darmstadt, Germany

Received February 3, 1992