Proc. Summer School on Diff. Geometry Dep. de Matemática, Universidade de Coimbra September 1999, 91-98.

Degeneracy of Some Lie Algebras^{1}

Inês Cruz

Departamento de Matemática Aplicada Centro de Matemática Aplicada da Universidade do Porto Rua das Taipas, 135, 4050 Porto, Portugal e-mail: imcruz@fc.up.pt

Abstract. We consider real Lie algebras with only zero and two-dimensional coadjoint orbits and having a nontrivial center. We show that, apart from the already known cases (studied for example by Dufour and Weinstein), such Lie algebras are degenerate in both the smooth and analytic category.

1. Introduction

A Poisson structure {, } on a manifold M is a Lie algebra structure on C^{∞}(M) satisfying the Leibniz identity:

{fg, h} = f{g, h} + {f, h}g, ∀f, g, h ∈ C^{∞}(M).

Alternatively it can be given by a contravariant skew-symmetric 2-tensor P such that [P, P] = 0, where [,] stands for the Schouten bracket. In local coordinates the Poisson tensor P can be written in the form:

P = ∑_{1≤i<j≤n}\{x_{i}, x_{j}\}\frac{∂}{∂x_{i}} ∧ \frac{∂}{∂x_{j}} .

Using Weinstein’s splitting theorem ([7]), the local study of P can be reduced to zero rank points, which translates into the following local expression for P:

P = ∑_{1≤i<j≤n} ∑_{k=1}^{n} C_{ij}^{k}x_{k} \frac{∂}{∂x_{i}} ∧ \frac{∂}{∂x_{j}} + higher order terms.

1This is an original research article and no version has been submitted for publication elsewhere.

The numbers C_{ij}^{k} are the structure constants of a Lie algebra and they sometimes determine the possibility of bringing P to a (local) linear form:

∑_{1≤i<j≤n} ∑_{k=1}^{n} C_{ij}^{k}y_{k} \frac{∂}{∂y_{i}} ∧ \frac{∂}{∂y_{j}} ,

through a smooth or analytic change of coordinates. P is then said to be linearizable. In this context a Lie algebra is said to be (smoothly or analytically) nondegenerate if every Poisson tensor associated with it in the above way, is (smoothly or analytically) linearizable. For example, semisimple Lie algebras are analytically nondegenerate (see [7] and [2]).

In this result we consider real Lie algebras of dimension at least four, whose connected and 1-connected Lie group has only zero and two-dimensional coadjoint orbits (such Lie algebras will be called nice and an exhaustive list of them can be found in [1]). We restrict to the case where the Lie algebra has a nontrivial center although the result is valid without this assumption (see [3]). We conclude that, apart from the already known cases (those to which the results of Weinstein in [7] and of Dufour in [5] apply), all these Lie algebras are degenerate in both the smooth and analytic category. Our proof is constructive, i.e., we associate a nonlinearisable Poisson structure to every Lie algebra 𝔤 being studied. This is done by perturbing the Lie-Poisson tensor in 𝔤^{∗} with second order terms in such a way that higher dimensional symplectic leaves appear around the singular point. This technique was used by A. Weinstein in [8] to prove that noncompact semisimple Lie algebras of real rank at least two are smoothly degenerate.

Notation. We follow the notation in [1] for the nice Lie algebras. These are, up to a direct sum with a central ideal:

1. type (i) – 𝔰o(3) or 𝔰l(2, ℝ);

2. type (ii) – ℝT +_{E} 𝔞, where 𝔞 is an abelian ideal and the action of T on 𝔞 is by an endomorphism E of 𝔞;

3. type (iii) – ℝT + 𝔥, where 𝔥 is the three-dimensional Heisenberg algebra spanned by X, Y, Z with [X, Y] = Z and either:

[T, X] = Y, [T, Y] = −X, [T, Z] = 0

or:

[T, X] = X, [T, Y] = −Y, [T, Z] = 0;

4. type (iv) – 𝔤 is six-dimensional with basis X_{i}, Y_{i}, 1 ≤ i ≤ 3 and the nonvanishing

brackets are:

[X_{1}, X_{2}] = Y_{3}, [X_{2}, X_{3}] = Y_{1}, [X_{3}, X_{1}] = Y_{2};

5. type (v) – 𝔤 is five-dimensional with basis X_{i}, 1 ≤ i ≤ 3, Y_{j}, 1 ≤ j ≤ 2 and the multiplicative law reads:

[X_{1}, X_{2}] = X_{3}, [X_{1}, X_{3}] = Y_{1}, [X_{2}, X_{3}] = Y_{2}.

The Lie-Poisson tensors in the dual of any of these Lie algebras will be denoted by Lie-Poisson tensor of type (i)–(v).

The main result is the following:

Theorem 1. Let 𝔤 be a nice Lie algebra (with dim 𝔤 ≥ 4) with a nontrivial center. Then 𝔤 is smoothly and analytically degenerate except if 𝔤 = 𝔰o(3) ⊕ ℝ or 𝔤 = 𝔰l(2, ℝ) ⊕ ℝ.

Remark 1. The condition on the center of 𝔤 means that Lie algebras of type (ii) will not be under consideration. Nevertheless, Theorem 1 holds without this assumption. The proof in that case envolves the use of Jordan normal form for the endomorphism E of 𝔞 (see [3] for more details).

Remark 2. The result is no longer valid if we restrict to the Poisson-Lie group case: Poisson-Lie groups associated with nice Lie algebras are linearizable (Mohammed Sbai, personal communication).

The proof of this theorem can be found in Section 4.

2. Raising the rank of Poisson structures

Definition 1. A Lie algebra is said to be nice if the coadjoint orbits of its connected and 1-connected Lie group have dimension zero or two.

Definition 2. Let P be a Poisson tensor on a manifold M. Then P is said to be nice if its symplectic leaves have dimension zero or two and not nice at a singular point if it has symplectic leaves of dimension at least four, in some set whose closure contains the singular point.

We start with a nice Lie algebra (𝔤, [ ,]), or equivalently with a nice Lie-Poisson tensor P on V = 𝔤^{∗}. We want to perturb P with second order terms so that symplectic leaves of higher dimension appear in any neighbourhood of the origin. Let (x_{1}, …, x_{n}) (with n ≥ 4) be linear coordinates on V and P be a linear Poisson tensor on V. Then the expression of P in the basis:

\{\frac{∂}{∂x_{i}} ∧ \frac{∂}{∂x_{j}} \}_{1≤i<j≤n}

is linear. Let Q be an alternating contravariant 2-tensor whose expression in the above basis is quadratic. Then P′ = P + Q is said to be a quadratic perturbation of P. Such P′ will be a Poisson tensor if and only if:

[P + Q, P + Q] = 0,

where [,] stands for the Schouten bracket. Equivalently:

[P, Q] = 0 and [Q, Q] = 0.  (1)

The last equation means that Q itself is a Poisson tensor. Our goal is then to find a quadratic Poisson tensor Q such that [P, Q] = 0 and P + Q is not nice at the origin.

3. Nice tensors admitting Casimir functions

Let be a nice Lie algebra with a nontrivial center. This means that the Lie-Poisson tensor P on its dual space admits a linear coordinate function as a Casimir function. This is the case of the tensors of type (iii)–(v). Assume that x_{n} is such a Casimir function for P. We will consider quadratic perturbations P + Q of P by taking Q to be of the form x_{n}L, with L a Lie-Poisson tensor.

Lemma 1. Let P and L be nice Lie-Poisson tensors on V and x_{n} a Casimir function for P. Then P + x_{n}L is a Poisson tensor if and only if [P, L] = 0.

Proof. Using properties of the Schouten bracket we can write:

[P + x_{n}L, P + x_{n}L] = [P, P] + 2[P, x_{n}L] + [x_{n}L, x_{n}L].

Now [P, P] = 0 since P is a Poisson tensor and:

[x_{n}L, x_{n}L] = x_{n}^{2}[L, L] − 2x_{n}L^{♯}(dx_{n}) ∧ L.

Again [L, L] = 0 and the fact that L is nice implies that L^{♯}(dx_{n}) ∧ L = 0, so that:

[P + x_{n}L, P + x_{n}L] = 2[P, x_{n}L]
= 2x_{n}[P, L] − 2P^{♯}(dx_{n}) ∧ L.

The conclusion follows using the fact that x_{n} is a Casimir function for P. □

We consider now the problem of raising the rank of P.

Lemma 2. Let P and L be as in the previous lemma and suppose that:

1. [P, L] = 0;

2. there is a subset U of V, whose closure contains the origin, where the following conditions hold:

(a) im(P^{♯}) ∩ im(L^{♯}) = {0};

(b) ker(P^{♯}) ≠ ker(L^{♯}). Then the tensor P + x_{n}L is not nice at the origin.

Proof. First we remark that, if L is any nontrivial Lie-Poisson tensor, then the set:

M_{0}(L) = {p ∈ V : rank(L)_{p} = 0}

is an hyperplane of codimension at least one. Since both P and L are in these conditions, and furthermore they are nice, then this implies that in any neighbourhood of the origin there is a point p such that:

rank(P)_{p} = 2 and rank(L)_{p} = 2.

Furthermore we can choose p such that x_{n}(p) ≠ 0. The hypothesis on the image of P and L then implies that:

ker(P^{♯} + x_{n}L^{♯})_{p} = ker P_{p}^{♯} ∩ ker L_{p}^{♯}.

Since both ker P_{p}^{♯} and ker L_{p}^{♯} have codimension two, then the hypothesis on the kernel of P^{♯} and L^{♯} implies that ker(P^{♯} + x_{n}L^{♯})_{p} has codimension four, which concludes the lemma. □

Our goal is now to find L such that conditions 1, 2a and 2b of Lemma 2 hold.

3.1. Choice of L

We will choose L from the Lie-Poisson tensors of type (ii), as this will give us some freedom to choose (by choosing the endomorphism E). We will use the notation L_{E} for the Lie-Poisson tensor on the dual of the Lie algebra 𝔤_{E} = ℝT +_{E} 𝔞.

We first write L_{E} in a coordinate free way. Let V be the given vector space (base space for the Poisson tensor P). Then a Lie algebra 𝔤_{E} of type (ii) on V^{∗} is determined by α ∈ V, z ∈ V^{∗} (with z(α) ≠ 0) and E a nonzero endomorphism of ker(α), in the following sense:

𝔤_{E} = ℝz + ker(α),  (2)

where the action of z on ker(α) is by the endomorphism E. Our goal is to show that there exist α, z and E such that the Lie-Poisson tensor L_{E} satisfies conditions 1, 2a and 2b. Let β = {X_{1}, …, X_{n}} be a basis for V and (x_{1}, …, x_{n}) be coordinates in that basis. We write α as α_{1}X_{1} + ⋯ + α_{n}X_{n} and assume that α_{1} ≠ 0. Assuming furthermore that z(α) = 1, the expression of L_{E} in x-coordinates is given by:

L_{E}(dx_{1}, dx_{i}) = E(α_{1}x_{i} − α_{i}x_{1})

and

L_{E}(dx_{i}, dx_{j}) = \frac{α_{i}}{α_{1}} E(α_{1}x_{j} − α_{j}x_{1}) − \frac{α_{j}}{α_{1}} E(α_{1}x_{i} − α_{i}x_{1}),

where j > i > 1. Now let u and v be two generators for the image of P^{♯}. Condition 2a is equivalent to saying that u and v form a free system together with the vector fields:

u′ = α_{1} \frac{∂}{∂x_{1}} + ⋯ + α_{n} \frac{∂}{∂x_{n}} and v′ = E(x_{1} − α_{1}z)\frac{∂}{∂x_{1}} + ⋯ + E(x_{n} − α_{n}z)\frac{∂}{∂x_{n}}

(see Lemma 3 in Appendix for the details).

We now remark that for fixed α and z it is easy to find an endomorphism E such that equation [P, L] = 0 holds. Conditions 2a and 2b merely restrict the field of those solutions. We choose α = X_{1} and z = x_{1} so that L_{E}(dx_{i}, dx_{j}) will be zero for all j > i > 1 and u′ and v′ will be given by:

u′ = \frac{∂}{∂x_{1}} and v′ = E(x_{2})\frac{∂}{∂x_{2}} + ⋯ + E(x_{n})\frac{∂}{∂x_{n}} .

Using these simplifications the problem of finding L_{E} is easily solved.

In Table 1 we present (up to an isomorphism of coordinates in the base space V) the generators for the image and kernel of the nice Lie-Poisson tensors of types (ii)–(v). For the tensor of type (ii) we are using the just described choice of α and z and we denote by E_{i} the function E(x_{i}). We have also assumed that E_{2} ≠ 0. This can always be achieved by permuting the coordinates (x_{2}, …, x_{n}), unless E = 0. From this table it is easy to see that, taking L_{E} to be of type (ii), condition 2b holds automatically. The choice of E that forces L_{E} to satisfy conditions 1 and 2a can be found in Table 2. In that table we present the functions E_{2}, …, E_{n} which determine the endomorphism E, where in (iv) one of E_{4}, E_{5} or E_{6} is nonzero and in (v) one of E_{4} or E_{5} is nonzero. This endomorphism, together with the just described choice of α and z, determines the Lie algebra 𝔤_{E}, and therefore the tensor L_{E}. We can then conclude that:

Type of  Generators for
Tensor  Image  Kernel
(ii)  \frac{∂}{∂x_{1}} , E_{2} \frac{∂}{∂x_{2}} + ⋯ + E_{n} \frac{∂}{∂x_{n}}  E_{2}dx_{i} − E_{i}dx_{2}, i = 3, …, n
(iii) x_{2} \frac{∂}{∂x_{1}} − x_{4} \frac{∂}{∂x_{2}} , x_{3} \frac{∂}{∂x_{1}} − x_{4} \frac{∂}{∂x_{3}} dx_{4}, x_{4}dx_{1} + x_{2}dx_{2} + x_{3}dx_{3}
x_{3} \frac{∂}{∂x_{1}} − x_{4} \frac{∂}{∂x_{2}} , x_{2} \frac{∂}{∂x_{1}} − x_{4} \frac{∂}{∂x_{3}} dx_{4}, x_{4}dx_{1} + x_{3}dx_{2} + x_{2}dx_{3}
(iv)  x_{6} \frac{∂}{∂x_{1}} − x_{4} \frac{∂}{∂x_{3}} , x_{6} \frac{∂}{∂x_{2}} − x_{5} \frac{∂}{∂x_{3}}  dx_{4}, dx_{5}, dx_{6},
x_{4}dx_{1} + x_{5}dx_{2} + x_{6}dx_{3}
(v)  x_{4} \frac{∂}{∂x_{1}} + x_{5} \frac{∂}{∂x_{2}} , x_{3} \frac{∂}{∂x_{1}} − x_{5} \frac{∂}{∂x_{3}}  dx_{4}, dx_{5},
x_{5}dx_{1} − x_{4}dx_{2} + x_{3}dx_{3}

Table 1: Generators for the image and kernel of some nice Lie-Poisson tensors

Theorem 2. Let P be one of the tensors of type (iii), (iv) or (v) as denoted in [1]. Then there is a nice tensor L_{E} of type (ii) such that P + x_{n}L_{E} is not nice at the origin.

Type of P Endomorphism E determining L_{E}
(iii)  E_{2} = a_{2}x_{2} + a_{3}x_{3} + a_{4}x_{4}

E_{3} = b_{2}x_{2} + b_{3}x_{3} + b_{4}x_{4} E_{4} = (a_{2} + b_{3})x_{4} ≠ 0

(iv)  E_{2} = a_{2}x_{2} + ⋯ + a_{6}x_{6}

E_{3} = b_{2}x_{2} + ⋯ + b_{6}x_{6} E_{4} = (a_{2} + b_{3})x_{4} E_{5} = c_{4}x_{4} + c_{5}x_{5} + c_{6}x_{6} E_{6} = d_{4}x_{4} + d_{5}x_{5} + d_{6}x_{6}

(v)  E_{2} = a_{2}x_{2} + ⋯ + a_{5}x_{5}

E_{3} = b_{2}x_{2} + ⋯ + b_{5}x_{5} E_{4} = c_{4}x_{4} + c_{5}x_{5} E_{5} = (a_{2} + b_{3})x_{5}

Table 2: Tensor L_{E} of type (ii) to be associated with P

4. Proof of the main theorem

As proved in Theorem 2, the Lie algebras of type (iii), (iv) and (v) are degenerate in both the smooth and analytic category. In fact that theorem shows that it is possible to perturb the Lie-Poisson tensor on the dual of these Lie algebras with second order terms, in such a way that the perturbed tensor is no longer nice.

Because abelian Lie algebras of dimension at least two are degenerate in any category, then taking their direct sum with central ideals will always produce a degenerate Lie algebra. This leaves us to classify:

𝔰o(3) ⊕ ℝ and 𝔰l(2, ℝ) ⊕ ℝ.

The Lie algebra 𝔰o(3) ⊕ ℝ is smoothly and analytically nondegenerate as a consequence of the theorem in [5]. Using the same result one concludes that 𝔰l(2, ℝ) ⊕ ℝ is analytically nondegenerate. It is, however, degenerate in the smooth sense since the Lie algebra 𝔰l(2, ℝ) is smoothly degenerate (see [7]). This concludes the proof of Theorem 1. □

Appendix: Coordinate changes in the base space

Let ϕ : M ⟶ N be a diffeomorphism between manifolds M and N, and suppose that P is a Poisson tensor on M. Then there exists a unique Poisson tensor Q on N making ϕ into a Poisson diffeomorphism. Such tensor Q is given by ϕ_{∗}P, the pushforward of P by ϕ. Furthermore the image of Q^{♯} is just the pushforward by ϕ of the image of P^{♯}. In the case we are interested in, ϕ is an automorphism of a vector space V. If ϕ is represented by the matrix A and the tensor matrix for P is M, then the tensor matrix for Q is just N = AMA^{T}. Furthermore im(Q^{♯}) = A(im(P^{♯})).

Lemma 3. Let V be a real vector space and let L_{E} denote the Lie-Poisson tensor on the dual of the Lie algebra:

𝔤_{E} = ℝz +_{E} ker(α).

Then there exist coordinates x = (x_{1}, …, x_{n}) in V such that:

L_{E}(dx_{1}, dx_{j}) = E(α_{1}x_{j} − α_{j}x_{1}), for j > 1

and

L_{E}(dx_{i}, dx_{j}) = \frac{α_{i}}{α_{1}} E(α_{1}x_{j} − α_{j}x_{1}) − \frac{α_{j}}{α_{1}} E(α_{1}x_{i} − α_{i}x_{1}) for j > i > 1.

Furthermore the image of L♯_{E} is spanned by the vector fields:

u = α_{1} \frac{∂}{∂x_{1}} + ⋯ + α_{n} \frac{∂}{∂x_{n}} and v = E(x_{1} − α_{1}z)\frac{∂}{∂x_{1}} + ⋯ + E(x_{n} − α_{n}z)\frac{∂}{∂x_{n}} .

Proof. Let (x_{1}, …, x_{n}) be coordinates in V such that x_{1}(α) ≠ 0. Then the following is a basis for ker(α):

{y_{2}, …, y_{n}} = {α_{1}x_{2} − α_{2}x_{1}, …, α_{1}x_{n} − α_{n}x_{1}},

where α_{i} stands for x_{i}(α). We complete this basis with y_{1} = z_{1}x_{1} + … + z_{n}x_{n} to get a basis for 𝔤_{E}. In y-coordinates L is represented by the matrix:

M = ( \begin{matrix}0 & −E(y_{2}) & ⋯ & −E(y_{n}) \\ E(y_{2}) & 0 & ⋯ & 0 \\ ⋮ & ⋮ & & ⋮ \\ E(y_{n}) & 0 & ⋯ & 0 \end{matrix} ) .

The change of coordinates that takes us back to x coordinates is given by A the inverse matrix of:

( \begin{matrix}z_{1} & z_{2} & z_{3} & ⋯ & z_{n} \\ −α_{2} & α_{1} & 0 & ⋯ & 0 \\ −α_{3} & 0 & α_{1} & ⋯ & 0 \\ ⋮ & ⋮ & ⋮ & & ⋮ \\ −α_{n} & 0 & 0 & ⋯ & α_{1} \end{matrix} ) .

This is just:

A = ( \begin{matrix}α_{1} & −z_{2} & −z_{3} & ⋯ & −z_{n} \\ α_{2} & \frac{1−α_{2}z_{2}}{α_{1}} & \frac{−α_{2}z_{3}}{α_{1}} & ⋯ & \frac{−α_{2}z_{n}}{α_{1}} \\ α_{3} & \frac{−α_{3}z_{2}}{α_{1}} & \frac{1−α_{3}z_{3}}{α_{1}} & ⋯ & \frac{−α_{3}z_{n}}{α_{1}} \\ ⋮ & ⋮ & ⋮ & & ⋮ \\ α_{n} & \frac{−α_{n}z_{2}}{α_{1}} & \frac{−α_{n}z_{3}}{α_{1}} & ⋯ & \frac{1−α_{n}z_{n}}{α_{1}} \end{matrix} )

and a tedious but straight forward calculation will complete the proof of the lemma. □

Acknowledgements. I. Cruz would like to thank Prof. John Rawnsley for all the help during the Ph.D. at Warwick; also the conversations with Prof. Michel Cahen in Belgium were very much appreciated.

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