Journal of Lie Theory Volume 12 (2002) 205–216 C ◯2002 Heldermann Verlag

The Approximative Centre of a Lie Algebra

Grant Cairns

Communicated by M. Cowling

Abstract. This paper examines the approximative centre of a Lie algebra; this is the set of elements which are not sent uniformly to infinity by the adjoint action of the underlying Lie group.

The centre Z(𝔤) of a Lie algebra 𝔤 is the set of elements that are fixed by the adjoint action of the underlying Lie group. A natural generalization would be to consider the set of elements with bounded orbit under the adjoint action. It turns out that another useful notion is the set of elements which are not sent uniformly to infinity. More generally, one has:

Definition 1. If 𝔤 is a finite dimensional real Lie algebra, and μ ∈ ℝ, let ℭ_{μ}(𝔤), or simply ℭ_{μ}, denote the complement of the set of elements X ∈ 𝔤 for which there exists a neighbourhood U ∋ X and an element Y ∈ 𝔤 such that the map

(exp tμ. tr(ad Y)) exp ad(t.Y)|_{U}: U → 𝔤

tends uniformly to infinity as t goes to positive infinity.

When 𝔤 is unimodular, ℭ_{μ} is independent of μ and is called the approximative centre of 𝔤; this notion is due to Etienne Ghys and was introduced in [4]. Some of the ideas presented below were developed in [3], but were not published. A further application for μ = 1 is given in [5]. The present paper is devoted to the following:

Theorem. Let 𝔤 be a finite dimensional real Lie algebra. Then:

(a) ℭ_{μ} = 𝔰 ⊕ 𝔫, where 𝔰 is a compact semisimple direct summand of 𝔤, and

𝔫 is a nilpotent characteristic ideal of 𝔤. Moreover:

(i) 𝔫 is metabelian, (ii) Z(𝔫)(= Z(ℭ_{μ})) is an ideal of 𝔤 and 𝔫 = Z(𝔫) + [𝔫, 𝔤], (iii) ℭ_{μ} ⊂ Z(𝔤) + [𝔤, 𝔤].

(b) If 𝔤 is unimodular or μ = 0, then 𝔰 is the maximal compact semisimple direct summand of 𝔤, and Z(𝔤) ⊂ 𝔫.

(c) If 𝔤 is not unimodular and μ ≠ 0, then 𝔰 = 0 and 𝔫 is abelian.

ISSN 0949–5932 / $2.50 ^{_{C} ◯} Heldermann Verlag Corollary. With the above notation, one has:

(a) ℭ_{μ} is unimodular,

(b) If 𝔤 is nilpotent, ℭ_{μ} is abelian,

(c) If 𝔤 is solvable, ℭ_{μ} is metabelian,

(d) ℭ_{μ} = 𝔤 if and only if 𝔤 is the Lie algebra of a compact Lie group.

Following the proofs of the theorem and its corollary, the paper concludes with 3 examples.

Notation. In the following, if 𝔨 is a subalgebra of 𝔤, then

(a) R(𝔨) denotes the radical of 𝔨, and we write R(𝔤) = 𝔯,

(b) Z(𝔨) is the centre of 𝔨, and Z_{𝔤}(𝔨) is the centralizer of 𝔨 in 𝔤. Proof of the theorem. For convenience, throughout this proof, we will use the l_{1} norm ∥(z_{1}, …, z_{n})∥ = ∑_{i=1}^{n} |z_{i}| on ℂ^{n}, rather than the Euclidean norm. We begin by studying the automorphism exp(t.C) of ℂ^{n}, where C is the following Jordan form:

C = ( \begin{matrix}λ & & & & 0 \\ 1 & λ \\ & ... & ... \\ & & 1 & λ \\ 0 & & & 1 & λ \end{matrix} ) .

First, note that

(*)  exp t.C = exp t.λ ( \begin{matrix}1 & & & & 0 \\ t & 1 \\ t^{2}/2! & t & 1 \\ ⋮ & ... & ... & ... \\ t^{n−1}/(n − 1)! & … & t^{2}/2! & t & 1 \end{matrix} ) .

Lemma 1. Let T ∈ ℝ.

(a) If the real part of λ+T is positive (or negative), then for all x ∈ ℂ^{n}\{0}, there exists a neighbourhood U of x in ℂ^{n} such that

(exp t.T) exp t.C|_{U}: U → ℂ^{n}

tends uniformly to infinity as t → +∞ (or t → −∞).

(b) If λ + T is imaginary, then for all x = (x_{1}, …, x_{n}) ∈ ℂ^{n}, the following

conditions are equivalent:

(i) there exists a neighbourhood U of x in ℂ^{n} such that

(exp t.T) exp t.C|_{U}: U → ℂ^{n}

tends uniformly to infinity when t tends to infinity, (ii) there exists 1 ≤ l < \frac{n+1}{2} such that x_{l} ≠ 0.

Proof. (a) Suppose λ + T > 0. Setting A_{t} = (exp t.T) exp t.C, one has from (*):

(A_{t})^{−1} = (exp −t.(λ + T)) ( \begin{matrix}1 & & & & 0 \\ −t & 1 \\ t^{2}/2! & −t & 1 \\ ⋮ & ... & ... & ... \\ (−t)^{n−1}/(n − 1)! & … & t^{2}/2! & −t & 1 \end{matrix} ) .

Let x ∈ ℂ^{n}\{0}, and 0 < K < 1. Let U be a relatively compact neighbourhood of x in ℂ^{n} which doesn’t contain zero in its closure.

Let B > 0. Clearly, there exists m ∈ ℕ such that ∥z∥/K^{m} > B for all z ∈ U. One has

∥z∥ = ∥A_{t}^{−1} ∘ A_{t}(z)∥ ≤ ∥A_{t}^{−1}∥.∥A_{t}(z)∥

and so ∥A_{t}(z)∥ ≥ ∥z∥/∥A_{t}^{−1}∥. There exists t_{0} > 0 such that for all t ≥ t_{0} the operator norm ∥A_{t}^{−1}∥ of the automorphism A_{t}^{−1} is less than K. Hence, for all t > t_{0} ,

∥A_{mt}(z)∥ ≥ ∥z∥/∥A_{mt}^{−1}∥ = ∥z∥/∥A_{t}^{−m}∥ ≥ ∥z∥/∥A_{t}^{−1}∥^{m} > ∥z∥/K^{m} > B.

Thus, ∥A_{s}(z)∥ > B for all z ∈ U and s > m.t_{0} . The case where Re(λ + T) < 0 is analogous.

(b) First suppose that x_{l} = 0 for all 1 ≤ l < (n + 1)/2. We will show that there does not exist any neighbourhood U of x for which the map (exp t.T) exp t.C|_{U}: U → ℂ^{n} tends uniformly to infinity when t tends to infinity. Let U be a relatively compact neighbourhood of x. It suffices to show that there exists N > 0 such that, for all sufficiently large t > 0, there exists z_{t} ∈ U such that ∥(exp t.T) exp t.C(z_{t})∥ < N.

Let k be the smallest natural number such that x_{k} ≠ 0. By hypothesis, 2k ≥ n + 1. For all sufficiently large t > 0 we will construct an element z_{t} of U of the form z_{t} = (ϵ_{1}(t), …, ϵ_{k−1}(t), x_{k}, x_{k+1}, …, x_{n}). First note that if z = (ϵ_{1}, …, ϵ_{k−1}, x_{k}, …, x_{n}) ∈ ℂ^{n}, then, by (∗), one has

∥(exp t.T) exp tC(z)∥ = |ϵ_{1}| + |ϵ_{1}t + ϵ_{2}| + ⋯ + |ϵ_{1} \frac{t^{k−1}}{(k − 1)!} + ⋯ + ϵ_{k−1}t + x_{k}|

+|ϵ_{1} \frac{t^{k}}{k!} + ⋯ + ϵ_{k−1} \frac{t^{2}}{2} + x_{k}t + x_{k+1}| + ⋯ + |ϵ_{1} \frac{t^{n−1}}{(n − 1)!} + ⋯ + x_{n}|.

We choose the (k − 1) numbers ϵ_{j} such that the last (n − k) terms of the preceding expression are zero. To do this, consider the following equation:

(1) ( ..._{\frac{t^{n−1}}{(n−1)!}  \frac{t^{n+1−k}}{(n+1−k)!}}^{\frac{t^{k}}{k!}  \frac{t^{2}}{2}} ) ( ϵ_{1}(t) ; ⋮ ; ϵ_{k−1}(t) ) = ( −x_{k}t − x_{k+1} ; ⋮ ; −x_{k} \frac{t^{n−k}}{(n−k)!} − ⋯ − x_{n} )

We have (k−1) variables (the ϵ_{j}(t)) and (n−k) equations. Because 2k ≥ n+1, one has n − k ≤ k − 1. In order to solve equation (1), set ϵ_{j}(t) = 0 for all n − k < j ≤ k − 1, and solve the following equation:

(2) ( ..._{\frac{t^{n−1}}{(n−1)!}  \frac{t^{k}}{k!}}^{\frac{t^{k}}{k!}  \frac{t^{2k−n+1}}{(2k−n+1)!}} ) ( ϵ_{1}(t) ; ⋮ ; ϵ_{n−k}(t) ) = ( −x_{k}t − x_{k+1} ; ⋮ ; −x_{k} \frac{t^{n−k}}{(n−k)!} − ⋯ − x_{n} )

This equation possesses a unique solution (ϵ_{1}(t), …, ϵ_{n−k}(t)) since the coefficient matrix is invertible for all 1 ≤ 2k − n + 1 ≤ k ≤ n and t > 0. Let F denote the coefficient matrix of the left hand side of (2), and let F_{j} denote the matrix obtained by replacing the j^{th} column of F by the right hand side of (2). By Cramer’s method, ϵ_{j}(t) = det F_{j}/ det F, for all 1 ≤ j ≤ n − k. Notice that det F = at^{k(n−k)} for some non-zero real a, while det F_{j} is a polynomial in t of degree k(n − k) − k + j which is divisible by t^{k(n−k)−n+j}. Hence the solution (ϵ_{1}(t), …, ϵ_{n−k}(t)) of equation (2) has the following form; for all 1 ≤ j ≤ k − 1

(3)  ϵ_{j}(t) = \frac{a_{j1}}{t^{k−j}} + \frac{a_{j2}}{t^{k−j+1}} + ⋯ + \frac{a_{j(n−k+1)}}{t^{n−j}}

where the coefficients a_{jl} are constants depending only on x_{k}, …, x_{n}. Note that as 2k ≥ n + 1, one has k − j ≥ 1. Let

(4)  z_{t} = (ϵ_{1}(t), …, ϵ_{k−1}(t), x_{k}, …, x_{n}) ∈ ℂ^{n}.

By (3), it is clear that z_{t} ∈ U for all sufficiently large t. By construction, one has

∥(exp t.T) exp tC(z_{t})∥ = |ϵ_{1}(t)| + |tϵ_{1}(t) + ϵ_{2}|

+ ⋯ + |ϵ_{1}(t)\frac{t^{k−1}}{(k − 1)!} + ⋯ + ϵ_{n−k}(t)t^{2k−n} + x_{k}|.

But, by (3), each of the terms of the preceding expression is bounded as t → ∞. In other words, there exists N > 0 such that ∥(exp t.T) exp tC(z_{t})∥ < N for all sufficiently large t.

Conversely, we will show that if there exists 1 ≤ l < (n + 1)/2 such that x_{l} ≠ 0, then there exists a neighbourhood U of x such that the map (exp t.T) exp t.C|_{U}: U → ℂ^{n} tends uniformly to infinity as t → ∞. Let k again denote the smallest natural number such that x_{k} ≠ 0. By hypothesis, 2k < n + 1. Let U be a relatively compact neighbourhood of x in ℂ^{n} such that for all z = (z_{1}, …, z_{n}) ∈ U one has z_{k} ≠ 0. One has, for all t > 0,

∥(exp t.T) exp t.C(z)∥ = ∥z_{1}∥ + ∥z_{1}t + z_{2}∥ + ⋯ + ∥\frac{t^{n−1}}{(n − 1)!}z_{1} + ⋯ + z_{n}∥.

Our strategy is consider only the k last terms, and to eliminate the terms z_{j} for j < k. Set T_{j}(z_{1}, …, z_{j}) = ∥\frac{t^{j−1}}{(j−1)!}z_{1} + \frac{t^{j−2}}{(j−2)!}z_{2} + ⋯ + z_{j}∥. Note that

T_{j}(z_{1}, …, z_{j}) + T_{j+1}(z_{1}, …, z_{j+1}) = \frac{t^{j−1}}{(j − 1)!} (∥z_{1} + a∥ + \frac{t}{j} ∥z_{1} + b∥) ,

where

a = \frac{(j − 1)!}{t^{j−1}} (\frac{t^{j−2}}{(j − 2)!}z_{2} + ⋯ + z_{j}) ,
b = \frac{j!}{t^{j}} (\frac{t^{j−1}}{(j − 1)!}z_{2} + ⋯ + z_{j+1}) .

For t > j one has ∥z_{1} + a∥ + \frac{t}{j} ∥z_{1} + b∥ ≥ ∥b − a∥ and so

T_{j}(z_{1}, …, z_{j}) + T_{j+1}(z_{1}, …, z_{j+1}) ≥ \frac{t^{j−1}}{(j − 1)!}∥b − a∥

= ∥\frac{t^{j−2}}{(j − 1)!}z_{2} + \frac{t^{j−3}}{(j − 2)!}2z_{3} + ⋯ + \frac{t^{j−k}}{(j − k + 1)!}(k − 1)z_{k} + ⋯ + \frac{j}{t} z_{j+1}∥.

Thus,

(5)  T_{j}(z_{1}, …, z_{j}) + T_{j+1}(z_{1}, …, z_{j+1}) ≥ \frac{1}{t}T_{j}(z_{2}, 2z_{3}, …, jz_{j+1}).

From above,

∥(exp t.T) exp t.C(z)∥ = ∑_{j=1}^{n} T_{j}(z_{1}, …, z_{j}) ≥ ∑_{j=n−k+1}^{n} T_{j}(z_{1}, …, z_{j}).

Hence, by (5), for all t > n,

∥( exp t.T) exp t.C(z)∥ ≥ \frac{1}{2t} ∑_{j=n−k+1}^{n−1} T_{j}(z_{2}, 2z_{3}, …, jz_{j+1})

≥ \frac{1}{2^{2}t^{2}} ∑_{j=n−k+1}^{n−2} T_{j}(2z_{3}, 3!z_{4}, …, j(j + 1)z_{j+2})

⋮

≥ \frac{1}{2^{k−1}t^{k−1}} T_{n−k+1}((k − 1)!z_{k}, \frac{k!}{1!}z_{k+1}, …, \frac{(n − 1)!}{(n − k)!}z_{n})

= \frac{1}{2^{k−1}} ∥\frac{t^{n−2k+1}(k − 1)!}{(n − k)!}z_{k} + \frac{t^{n−2k}k!}{(n − k − 1)!1!}z_{k+1} + ⋯ + \frac{(n − 1)!}{(n − k)!t^{k−1}} z_{n}∥ .

So, as n − 2k + 1 ≥ 1 and z_{k} ≠ 0 for all z ∈ U, it is clear that for all M > 0 there exists N > 0 such that ∥(exp t.T) exp t.C(z)∥ > M for all t > N and z ∈ U. This establishes Lemma 1.

Let y be an element of 𝔤. Let M_{y} denote the Jordan form of the linear endomorphism of ℂ^{n} determined by ad(y), and denote the Jordan blocks of M_{y} by C_{1}(y), …, C_{r}(y), and the corresponding eigenvalues λ_{1}, …, λ_{r}. If x ∈ 𝔤, we write x_{y} = x^{1} ⊕ ⋯ ⊕ x^{r} ∈ ℂ^{n} where x^{i} = (x_{1}^{i}, …, x_{n_{i}}^{i}) are the components of x with respect to the decomposition of ℂ^{n} determined by the Jordan form M_{y}; one has M_{y}(x_{y}) = C_{1}(y)(x^{1}) ⊕ ⋯ ⊕ C_{r}(y)(x^{r}).

Lemma 2. Let x, y ∈ 𝔤. The following two conditions are equivalent: (a) there exists a neighbourhood U of x in 𝔤 such that

(exp tμ. tr(ad y)) exp ad(t.y)|_{U}: U → 𝔤

tends uniformly to infinity as t → ∞,

(b) there exists 1 ≤ i ≤ r and a neighbourhood U_{i} of x^{i} in ℂ^{n_{i}} such that

(exp tμ. tr(ad y)) exp(t.C_{i}(y))|_{U_{i}}: U_{i} → ℂ^{n_{i}}

tends uniformly to infinity as t → ∞.

Proof. (a) follows immediately from (b) because

∥ exp ad(t.y)(z)∥ = ∑_{i=1}^{r} ∥ exp ad(t.C_{i}(y))(z^{i})∥

for all z ∈ ℂ^{n}. To see the converse, notice that if

(exp tμ. tr(ad y)) exp(t.C_{i}(y))|_{U_{i}}: U_{i} → ℂ^{n_{i}}

does not tend uniformly to infinity as t → ∞, then the point z_{t} ∈ ℂ^{n_{i}} given by (4), in the proof of Lemma 1, has real coordinates; the required result follows easily.

Set

ℭ_{μ}(y, i) = {x^{i} ∈ ℂ^{n_{i}} | x^{i} ≡ 0, if Re(μ tr(ad y) + λ_{i}) ≠ 0, and
x_{l}^{i} = 0 for all 1 ≤ l < (n_{i} + 1)/2, if Re(μ tr(ad y) + λ_{i}) = 0}.

From Lemmas 1 and 2, one has:

Lemma 3. With the notation introduced above, the approximative centre of 𝔤 is

ℭ_{μ} = ⋂_{y∈𝔤}{x ∈ 𝔤 | x_{y} ∈ ⨁_{i=1}^{r} ℭ_{μ}(y, i)}.

In particular, ℭ_{μ} is a vector space. In fact, we have:

Lemma 4. ℭ_{μ} is a characteristic ideal of 𝔤.

Proof. Let ϕ: 𝔤 → 𝔤 be a derivation and consider the automorphism Φ_{s} = e^{sϕ} of 𝔤. Note that ℭ_{μ} is Φ_{s} -invariant. Indeed, if x and y belong to 𝔤, then

Φ_{s}(exp(t. ad y)(x)) = exp(t. ad(Φ_{s}y))(Φ_{s}x)

and hence

Φ_{s}(exp μ tr(ad(t.y)) exp(t. ad y)(x)) = (exp μ tr(ad(t.Φ_{s}y)) exp(t. ad(Φ_{s}y))(Φ_{s}x).

It follows that if x is an element of ℭ_{μ}, then Φ_{s}x is too, for all s ∈ ℝ. Differentiating with respect to s, one has that ℭ_{μ} is invariant under ϕ.

Remark 1. Let 𝔤_{ℂ} = 𝔤 ⊗_{ℝ} ℂ denote the complexification of 𝔤. One can define the approximative centre ℭ_{μ}(𝔤_{ℂ}) of 𝔤_{ℂ} is an analogous manner to that of 𝔤. It is clear from the above that ℭ_{μ}(𝔤_{ℂ}) is an ideal of 𝔤_{ℂ}. However, it is not true in general that ℭ_{μ}(𝔤_{ℂ}) = ℭ_{μ} ⊗_{ℝ} ℂ. Nevertheless, ℭ_{μ} ⊗_{ℝ} ℂ is clearly an ideal of 𝔤_{ℂ}. Thus, if y ∈ ℭ_{μ} and z ∈ 𝔤_{ℂ}, then ad(y)(z) ∈ ℭ_{μ} ⊗_{ℝ} ℂ and in particular, C_{i}(y)(z) ∈ ℭ_{μ}(y, i), for all i.

If y ∈ ℭ_{μ}, let ad_{ℭ_{μ}}(y): ℭ_{μ} → ℭ_{μ} denote the restriction of ad(y) to ℭ_{μ}.

Lemma 5. If y ∈ ℭ_{μ}, then:

(a) the eigenvalues λ_{1}, …, λ_{r} of ad(y): 𝔤 → 𝔤 are all imaginary, whence tr(ad y) = 0,

(b) for all i, one has:

(i) if λ_{i} = 0, then the corresponding Jordan block has size at most 3; that is, n_{i} ≤ 3,

(ii) if λ_{i} ≠ 0, then the corresponding Jordan block has size 1; that

is, n_{i} = 1,

(c) ℭ_{μ} = ker(ad_{ℭ_{μ}}^{2}(y)) ⊕ im(ad_{ℭ_{μ}}^{2}(y)),

Proof. First note that y is an eigenvector of ad(y) with eigenvalue 0, and so Re(μ tr(ad y)) = 0, by Lemma 3. Hence μ tr(ad y) = 0. Fix i and let {Z_{j}^{i} | 1 ≤ j ≤ n_{i}} be the canonical basis of ℂ^{n_{i}}. Using the notation from above, one has

(6)  C_{i}(y)(Z_{j}^{i}) = \{ \begin{matrix}λ_{i}Z_{j}^{i} + Z_{j+1}^{i} & ; if j < n_{i}, \\ λ_{i}Z_{j}^{i} & ; if j = n_{i}. \end{matrix}

(a). If Re(λ_{i}) ≠ 0, then Re(λ_{i}+μ tr(ad y)) ≠ 0, since μ tr(ad y) = 0. By Lemmas 1 and 2, one has Z_{n_{i}}^{i} ∉ ℭ_{μ}(y, i), but by (6) and the previous remark, one has Z_{n_{i}}^{i} = \frac{1}{λ_{i}} C_{i}(y)(Z_{n_{i}}^{i}) ∈ ℭ_{μ}(y, i), which is a contradiction. Thus Re(λ_{i}) = 0, for all i.

(b)(i). If λ_{i} = 0, then by the previous remark, Z_{2}^{i} = C_{i}(y)(Z_{1}^{i}) ∈ ℭ_{μ}(y, i). Hence, by Lemma 1(b), 2 ≥ \frac{n_{i}+1}{2} ; that is, n_{i} ≤ 3.

(b)(ii). If λ_{i} ≠ 0, then by the previous remark,

Z_{n_{i}}^{i} = \frac{1}{λ_{i}} C_{i}(y)(Z_{n_{i}}^{i}) ∈ ℭ_{μ}(y, i)

and Z_{j}^{i} = \frac{1}{λ_{i}} (C_{i}(y)(Z_{j}^{i}) − Z_{j+1}^{i}) for all 1 ≤ j ≤ n_{i} − 1. So by induction, Z_{j}^{i} ∈ ℭ_{μ}(y, i) for all 1 ≤ j ≤ n_{i}. In particular, Z_{1}^{i} ∈ ℭ_{μ}(y, i), and hence, by Lemma 1(b), 1 ≥ \frac{n_{i}+1}{2} ; that is, n_{i} = 1.

(c). Now consider ad^{2}(y). Note that:

(i) if n_{i} = 1, ℭ_{μ}(y, i) = ⟨Z_{1}^{i}⟩, and C_{i}^{2}(y)(Z_{1}^{i}) = 0 if λ_{i} = 0, and Z_{1}^{i} = \frac{1}{λ_{i}^{2}} C_{i}^{2}(Z_{1}^{i}) if λ_{i} ≠ 0.

(ii) if n_{i} = 2, ℭ_{μ}(y, i) = ⟨Z_{2}^{i}⟩, and C_{i}^{2}(y)(Z_{2}^{i}) = 0.

(iii) if n_{i} = 3, ℭ_{μ}(y, i) = ⟨Z_{2}^{i}, Z_{3}^{i}⟩, C_{i}^{2}(y)(Z_{2}^{i}) = 0 and C_{i}^{2}(y)(Z_{3}^{i}) = 0.

Hence each of the basis elements of ℭ_{μ}(y, i) belongs to either the kernel or the image of C_{i}^{2}(y). Since ℭ_{μ} ⊂ {x ∈ 𝔤 | x_{y} ∈ ⨁_{i=1}^{r} ℭ_{μ}(y, i)}, it follows that ℭ_{μ} is the vector subspace sum ker(ad_{ℭ_{μ}}^{2}(y)) + im(ad_{ℭ_{μ}}^{2}(y)). So for dimension reasons, ℭ_{μ} = ker(ad_{ℭ_{μ}}^{2}(y)) ⊕ im(ad_{ℭ_{μ}}^{2}(y)).

Lemma 6.

(a) If y ∈ 𝔤, one has ℭ_{μ} ⊂ ker(ad(y)) + im(ad(y)),

(b) ℭ_{μ} ⊂ Z(𝔤) + [𝔤, 𝔤].

Proof. (a). Suppose that y ∈ 𝔤. With the above notation,

ℭ_{μ}(y, i) = \{ \begin{matrix}0 & ; if Re(μ tr(ad y) + λ_{i}) ≠ 0, \\ ⟨Z_{l}^{i} | l ≥ (n_{i} + 1)/2⟩ & ; if Re(μ tr(ad y) + λ_{i}) = 0. \end{matrix}

Suppose that Re(μ tr(ad y) + λ_{i}) = 0. If λ_{i} = 0, then:

(i) if n_{i} = 1, then C_{i}(y)(Z_{1}^{i}) = 0.

(ii) if n_{i} ≥ 2, then Z_{l}^{i} = C_{i}(y)(Z_{l−1}^{i}), for all l ≥ (n_{i} + 1)/2.

If λ_{i} ≠ 0, then Z_{n_{i}}^{i} = C_{i}(y)(\frac{1}{λ_{i}} Z_{n_{i}}^{i}) and Z_{l}^{i} = C_{i}(y)(\frac{1}{λ_{i}} Z_{l}^{i}) − \frac{1}{λ_{i}} Z_{l+1}^{i}, for all 1 ≤ l ≤ n_{i} − 1. In particular, Z_{l}^{i} ∈ im C_{i}(y), for all l ≥ (n_{i} + 1)/2.

Hence each of the basis elements of ℭ_{μ}(y, i) belongs to either the kernel or the image of C_{i}(y). Since ℭ_{μ} ⊂ {x ∈ 𝔤 | x_{y} ∈ ⨁_{i=1}^{r} ℭ_{μ}(y, i)}, it follows that ℭ_{μ} is a subset of the vector subspace sum ker(ad(y)) + im(ad(y)).

(b). From (a), one has:

ℭ_{μ} ⊂ ⋂_{y∈𝔤} ker(ad(y)) + ⨁_{y∈𝔤} im(ad(y)).

That is, ℭ_{μ} ⊂ Z(𝔤) + [𝔤, 𝔤].

Let K denote the Killing–Cartan form of ℭ_{μ}; this is the map K: ℭ_{μ} × ℭ_{μ} → ℝ defined by K(x, y) = tr(ad_{ℭ_{μ}}(x) ∘ ad_{ℭ_{μ}}(y): ℭ_{μ} → ℭ_{μ}).

Remark 2. If y ∈ ℭ_{μ} and {λ_{i} | i = 1, …, r} are the eigenvalues of ad_{ℭ_{μ}}(y), one has K(y, y) = ∑_{i=1}^{r} n_{i}λ_{i}^{2} . By Lemma 5, one has K(y, y) ≤ 0 and K(y, y) = 0 if and only if the map ad_{ℭ_{μ}}^{2}(y) is identically zero.

Let 𝔫 denote the maximal nilpotent ideal of ℭ_{μ}. Consider the radical R(ℭ_{μ}) of ℭ_{μ}. Recall that R(ℭ_{μ}) is the K -orthogonal complement of the derived algebra [ℭ_{μ}, ℭ_{μ}]. Let ℭ_{μ}^{⊥} denote the K -orthogonal complement of ℭ_{μ}. Recall that

(7)  R(ℭ_{μ}) ⊃ ℭ_{μ}^{⊥} ⊃ 𝔫.

Definition 2. For convenience, we introduce four sets:

(a) A = {x ∈ ℭ_{μ} | ad_{ℭ_{μ}}^{2}(x) ≡ 0},

(b) B = {x ∈ ℭ_{μ} | K(x, x) = 0}.
(c) D = ⊕_{x∈ℭ_{μ}} im ad_{ℭ_{μ}}^{2}(x),
(d) E = ∩_{x∈ℭ_{μ}} ker ad_{ℭ_{μ}}^{2}(x),

Lemma 7. R(ℭ_{μ}) = 𝔫, and 𝔫 is metabelian.

Proof. First notice that ℭ_{μ}^{⊥} ⊂ B ⊂ A ⊂ 𝔫 ⊂ ℭ_{μ}^{⊥}. Indeed,

ℭ_{μ}^{⊥} = {x ∈ ℭ_{μ} | K(x, y) = 0 for all y ∈ ℭ_{μ}}

and so ℭ_{μ}^{⊥} ⊂ B. Remark 2 gives B ⊂ A. Recall that 𝔫 is the set of elements x for which ad_{ℭ_{μ}}(x): ℭ_{μ} → ℭ_{μ} is nilpotent [1]. So A ⊂ 𝔫. Equation (7) gives 𝔫 ⊂ ℭ_{μ}^{⊥}. So B = A = 𝔫 = ℭ_{μ}^{⊥}. In particular, 𝔫 = A and so 𝔫 is metabelian.

Now note that E ⊂ A. Indeed, if x ∈ E, then for all y ∈ ℭ_{μ}, x ∈ ker ad_{ℭ_{μ}}^{2}(y) and x ∈ ker ad_{ℭ_{μ}}^{2}(x + y). So

0 = [x + y, [x + y, x]] = [x + y, [y, x]] = [x, [y, x]] + [y, [y, x]] = [x, [y, x]].

Thus ad_{ℭ_{μ}}^{2}(x)(y) = 0 for all y ∈ ℭ_{μ}. So x ∈ A.

From Lemma 5, for all x ∈ ℭ_{μ}, ker ad^{2}(x) ∩ im ad^{2}(x) = {0}. Hence ℭ_{μ} = D ⊕ E. As E ⊂ A = 𝔫, we have E ⊂ R(ℭ_{μ}), from (7). Thus

(8)  R(ℭ_{μ}) = (R(ℭ_{μ}) ∩ D) ⊕ E.

Note also that R(ℭ_{μ}) ∩ D ⊂ A. Indeed, it suffices to show that if x, y ∈ ℭ_{μ}, and z = ad_{ℭ_{μ}}^{2}(x)(y) ∈ R(ℭ_{μ}), then z ∈ B. But

K(z, z) = K([x, [x, y]], z) = K([x, y], [x, z]).

Now [x, z] ∈ [ℭ_{μ}, R(ℭ_{μ})] ⊂ ℭ_{μ}^{⊥}. So K(z, z) = 0.

Since E ⊂ A and R(ℭ_{μ}) ∩ D ⊂ A, (8) gives R(ℭ_{μ}) ⊂ A = 𝔫, and hence by (7), R(ℭ_{μ}) = 𝔫.

Let 𝔰 be a Levi subalgebra of ℭ_{μ}.

Lemma 8. 𝔫 is a characteristic ideal of 𝔤 and 𝔰 is a direct summand of 𝔤.

Proof. As 𝔯 is a characteristic ideal [1], and by Lemma 4, ℭ_{μ} is a characteristic ideal, so 𝔯 ∩ ℭ_{μ} is one too. But 𝔫 = 𝔯 ∩ ℭ_{μ} since by [6, Theorem 3.8.1], 𝔯 ∩ ℭ_{μ} = R(ℭ_{μ}), and by the previous Lemma R(ℭ_{μ}) = 𝔫.

For all y ∈ 𝔯, Lemma 6 and the fact that 𝔯 is an ideal gives

𝔰 ⊂ ℭ_{μ} ⊂ ker ad(y) + im ad(y) ⊂ ker ad(y) + 𝔯.

Hence 𝔰 ⊂ Z_{𝔤}(𝔯) + 𝔯. Thus, as 𝔰 is semisimple,

𝔰 = [𝔰, 𝔰] ⊂ [Z_{𝔤}(𝔯) + 𝔯, Z_{𝔤}(𝔯) + 𝔯] ⊂ [Z_{𝔤}(𝔯), Z_{𝔤}(𝔯)] + [𝔯, 𝔯] ⊂ Z_{𝔤}(𝔯) + [𝔯, 𝔯].

Taking repeated brackets of 𝔰 with itself, and using the fact that 𝔯 is solvable, one obtains 𝔰 ⊂ Z_{𝔤}(𝔯).

By the Malcev–Harish-Chandra Theorem, 𝔤 has a Levi subalgebra 𝔖 such that 𝔰 ⊂ 𝔖 (See [6, Cor. 3.14.3]). Note that 𝔰 = 𝔖 ∩ ℭ_{μ}. Indeed, clearly 𝔰 ⊂ 𝔖 ∩ ℭ_{μ}. On the other hand, since 𝔖 is a subalgebra and ℭ_{μ} is an ideal of 𝔤, 𝔖 ∩ ℭ_{μ} is an ideal of 𝔖. So 𝔖 ∩ ℭ_{μ} is a semisimple subalgebra. Thus, as 𝔰 is a maximal semisimple subalgebra of ℭ_{μ}, 𝔰 = 𝔖 ∩ ℭ_{μ}. In particular, 𝔰 is an ideal of 𝔖. As 𝔖 is semisimple, 𝔰 is a direct summand of 𝔖; that is, there is an ideal 𝔰′ of 𝔖 such that 𝔖 is an internal direct sum of ideals 𝔖 = 𝔰 ⊕ 𝔰′. Consider the vector space direct sum 𝔱 = 𝔰′ ⊕ 𝔯. By construction 𝔱 is an ideal of 𝔤. From above, 𝔰 ⊂ Z_{𝔤}(𝔯). So we have an internal direct sum of ideals: 𝔤 = 𝔰 ⊕ 𝔱.

Remark 3. By Remark 2, K(x, x) ≤ 0 for all x ∈ 𝔰, where K is the Killing– Cartan form of ℭ_{μ}. So, by the previous Lemma, the Killing–Cartan form of 𝔰 is negative semi-definite; that is, 𝔰 is compact.

Returning to the statement of the theorem, note that with the exception of (a)(ii), part (a) follows from Lemma 8, Remark 3, and Lemmas 7 and 6(b). From Lemma 6(a), one has:

ℭ_{μ} ⊂ ⋂_{z∈𝔫} ker(ad(z)) + ⨁_{z∈𝔫} im(ad(z)).

That is, ℭ_{μ} ⊂ Z_{g}(𝔫) + [𝔫, 𝔤]. Hence 𝔫 ⊂ Z_{g}(𝔫) + [𝔫, 𝔤]. It follows that as 𝔫 is an ideal, by Lemma 8,

𝔫 ⊂ 𝔫 ∩ Z_{g}(𝔫) + [𝔫, 𝔤] = Z(𝔫) + [𝔫, 𝔤].

Clearly Z(𝔫) + [𝔫, 𝔤] ⊂ 𝔫. To see that Z(𝔫) is an ideal of 𝔤, notice that as 𝔫 is an ideal of 𝔤, the Jacobi identity gives

[𝔫, [Z(𝔫), 𝔤]] = [𝔤, [Z(𝔫), 𝔫]] + [Z(𝔫), [𝔫, 𝔤]] = 0,

and so [Z(𝔫), 𝔤] ⊂ Z(𝔫). So we have established (a)(ii).

Suppose that 𝔤 is unimodular or that μ = 0. One sees directly from Lemma 3 that Z(𝔤) ⊂ ℭ_{μ}, and hence Z(𝔤) ⊂ 𝔫. Notice that if 𝔤_{1} is a compact semisimple Lie algebra, then 𝔤_{1} is the Lie algebra of a compact Lie group and so the orbits of the adjoint action are bounded. Hence ℭ_{μ}(𝔤_{1}) = 𝔤_{1} . As 𝔤 is unimodular or μ = 0, Lemma 3 implies that if 𝔤 is a direct sum of ideals, 𝔤 = 𝔤_{1} ⊕ 𝔤_{2} , then ℭ_{μ} = ℭ_{μ}(𝔤_{1}) ⊕ ℭ_{μ}(𝔤_{2}). In particular, if 𝔤_{1} is compact semisimple, 𝔤_{1} ⊂ ℭ_{μ}, from which it follows that 𝔤_{1} ⊂ 𝔰. Thus 𝔰 is the maximal compact semisimple direct summand of 𝔤.

Finally, suppose that 𝔤 is not unimodular and μ ≠ 0. Let x ∈ 𝔤 with τ := − tr(ad(x)) ≠ 0. Consider the Jordan form of the induced derivation ad_{ℭ_{μ}}(x): ℭ_{μ} → ℭ_{μ}. Suppose that λ_{1}, λ_{2} are (not necessarily distinct) eigenvalues of ad_{ℭ_{μ}}(x) and for each i = 1, 2 let {Z_{j}^{i} | 1 ≤ j ≤ n_{i}} be linearly independent vectors in ℭ_{μ} ⊗_{ℝ} ℂ with ad(x)(Z_{j}^{i}) = λ_{i}Z_{j}^{i} + Z_{j−1}^{i} for all j, where by definition Z_{0}^{i} = 0. We will show by induction on p = j + k that [Z_{j}^{1}, Z_{k}^{2}] = 0 for all j ≤ n_{1}, k ≤ n_{2} . The claim is obviously true for p = 0. Suppose that it holds for p = l. Then for p = l + 1, the inductive hypothesis gives:

ad(x)[Z_{j}^{1}, Z_{k}^{2}] = [ad(x)Z_{j}^{1}, Z_{k}^{2}] + [Z_{j}^{1}, ad(x)Z_{k}^{2}]
= [λ_{1}Z_{j}^{1}, Z_{k}^{2}] + [Z_{j−1}^{1}, Z_{k}^{2}] + [Z_{j}^{1}, λ_{2}Z_{k}^{2}] + [Z_{j}^{1}, Z_{k−1}^{2}]

= (λ_{1} + λ_{2})[Z_{j}^{1}, Z_{k}^{2}]. (9)

As Z_{j}^{1}, Z_{k}^{2} ∈ ℭ_{μ} ⊗_{ℝ} ℂ, one has [Z_{j}^{1}, Z_{k}^{2}] ∈ ℭ_{μ} ⊗_{ℝ} ℂ. By Lemma 3, the eigenvalues of ad_{ℭ_{μ}}(x) all have real part equal to μτ. So Re(λ_{1}) = Re(λ_{2}) = μτ and Re(λ_{1} + λ_{2}) = 2μτ ≠ μτ. So λ_{1} + λ_{2} is not an eigenvalue of ad_{ℭ_{μ}}(x), and thus (9) gives [Z_{j}^{1}, Z_{k}^{2}] = 0. This completes the induction. Thus ℭ_{μ} is abelian. So 𝔰 = 0 and 𝔫 is abelian, as required. This completes the proof of the theorem.

Proof of the corollary. Part (a) follows immediately from Lemma 5(a), while (c) follows immediately from the theorem.

(b) If 𝔤 is nilpotent, then ℭ_{μ} = 𝔫 and by part (a)(ii) of the theorem,

ℭ_{μ} = Z(ℭ_{μ}) + [ℭ_{μ}, 𝔤] = Z(ℭ_{μ}) + [Z(ℭ_{μ}) + [ℭ_{μ}, 𝔤], 𝔤]

= Z(ℭ_{μ}) + [Z(ℭ_{μ}), 𝔤] + [[ℭ_{μ}, 𝔤], 𝔤]. (10)

Now by part (a)(ii) of the theorem, Z(ℭ_{μ}) is an ideal of 𝔤, and so [Z(ℭ_{μ}), 𝔤] ⊂ Z(ℭ_{μ}) and (10) gives ℭ_{μ} = Z(ℭ_{μ}) + [[ℭ_{μ}, 𝔤], 𝔤]. Repeating this argument, one has ℭ_{μ} = Z(ℭ_{μ}) + 𝔤^{k}(ℭ_{μ}) for all k ≥ 1, where 𝔤^{i}(ℭ_{μ}) = [𝔤, 𝔤^{i−1}(ℭ_{μ})] and 𝔤^{1}(ℭ_{μ}) = [𝔤, ℭ_{μ}]. Thus, if 𝔤 is nilpotent, ℭ_{μ} = Z(ℭ_{μ}); that is, ℭ_{μ} is abelian.

(d) If 𝔤 is the Lie algebra of a compact Lie group, the orbits of the adjoint action of 𝔤 are bounded and thus ℭ_{μ} = 𝔤. Conversely, if ℭ_{μ} = 𝔤, then by the theorem, 𝔤 = 𝔰 ⊕ 𝔫, where 𝔰 is compact semisimple and 𝔫 is metabelian. In fact, by part (a)(ii) of the theorem, 𝔫 = Z(𝔫) + [𝔫, 𝔤] = Z(𝔫), and so 𝔫 is abelian. Hence 𝔤 is the Lie algebra of a compact Lie group.

Example 1. The approximative centre of the following (solvable unimodular) Lie algebra is nilpotent non-abelian:

𝔤 = ⟨x, y, z, w | [x, y] = z, [x, z] = −y, [y, z] = w⟩.

Indeed, it is easy to see from Lemma 3 that ℭ_{μ} = ⟨y, z, w⟩.

Example 2. Consider the standard filiform nilpotent Lie algebra (see [2]):

𝔤 = ⟨x, y_{1}, …, y_{n} | [x, y_{i}] = y_{i+1}, ∀i < n⟩.

The approximative centre of 𝔤 is abelian and strictly greater than the centre; indeed, it is easy to see from Lemma 3 that ℭ_{μ} = ⟨y_{i} | i ≥ (n + 1)/2⟩.

Example 3. Consider the Lie algebra 𝔤 = ℝ^{3} ⋊ 𝔰𝔬(3, ℝ), where the action of 𝔰𝔬(3, ℝ) on ℝ^{3} is the standard linear one. Here 𝔤 has a Levi subalgebra which is compact and simple, but the approximative centre has no simple factor (ℭ_{μ} = ℝ^{3} ).

Remark 4. If a Lie algebra 𝔤 has an ideal 𝔞 which is the Lie algebra of a compact Lie group, it doesn’t necessarily follow that 𝔞 ⊂ ℭ_{μ}, or that ℭ_{μ} ⊂ 𝔞. In Example 2, the ideal ⟨y_{1}, …, y_{n}⟩ is abelian but it is not contained in ℭ_{μ}. In Example 1, ℭ_{μ} is not contained in the (maximal) abelian ideal ⟨z, w⟩.

Thanks. The author is very grateful to the referee, whose careful reading and sensible suggestions improved the presentation of this paper.

References

[1] N. Bourbaki, “Lie groups and Lie algebras. Chapters 1–3,” Springer-Verlag, Berlin, 1998.

[2] D. Burde, Affine cohomology classes for filiform Lie algebras, in “Crystallographic groups and their generalizations,” Amer. Math. Soc., Providence, RI, Contemp. Math. 262 (2000), 159–170.

[3]  G. Cairns,“Feuilletages géodésibles,” Thesis, Montpellier, 1987.

[4] G. Cairns and E. Ghys, Totally geodesic foliations on 4-manifolds, J. Diff. Geom. 23 (1986), 241–254.

[5] G. Cairns and P. Molino, Weakly involutive totally geodesic distributions of constant rank, preprint.

[6] V.S. Varadarajan, “Lie groups, Lie Algebras, and their Representations,” Springer-Verlag, 1984.

G. Cairns Department of Mathematics La Trobe University Melbourne, Australia 3083 G.Cairns@latrobe.edu.au

Received November 17, 2000 and in final form May 15, 2001