On Riemann-Poisson Lie groups

TL;DR

This paper studies Riemann-Poisson Lie groups, characterizing their Lie algebras, providing construction methods, and listing all such Lie algebras up to dimension five.

Contribution

It offers a comprehensive characterization and classification of Riemann-Poisson Lie groups and their Lie algebras, including explicit constructions and classifications up to dimension five.

Findings

Characterization of Lie algebras of Riemann-Poisson Lie groups

Construction methods for these Lie algebras

Complete list of such Lie algebras up to dimension five

Abstract

A Riemann-Poisson Lie group is a Lie group endowed with a left invariant Riemannian metric and a left invariant Poisson tensor which are compatible in the sense introduced in C.R. Acad. Sci. Paris s\'er. {\bf I 333} (2001) 763-768. We study these Lie groups and we give a characterization of their Lie algebras. We give also a way of building these Lie algebras and we give the list of such Lie algebras up to dimension 5.

Equations

$$[\alpha ,\beta {]}_{\pi }={\mathcal{L}}_{{\#}_{\pi }(\alpha )}\beta -{\mathcal{L}}_{{\#}_{\pi }(\beta )}\alpha -d\pi (\alpha ,\beta ),\alpha ,\beta \in {\Omega }^1(M).$$

$$2⟨{\mathcal{D}}_{\alpha }\beta ,\gamma {⟩}^{\ast }$$

$${\mathcal{D}}_{\alpha }\beta -{\mathcal{D}}_{\beta }\alpha =[\alpha ,\beta {]}_{\pi }\text{mbox}and{\#}_{\pi }(\alpha ).⟨\beta ,\gamma {⟩}^{\ast }=⟨{\mathcal{D}}_{\alpha }\beta ,\gamma {⟩}^{\ast }+⟨\beta ,{\mathcal{D}}_{\alpha }\gamma {⟩}^{\ast }.$$

$$\mathcal{D}\pi (\alpha ,\beta ,\gamma ):={\#}_{\pi }(\alpha ).\pi (\beta ,\gamma )-\pi ({\mathcal{D}}_{\alpha }\beta ,\gamma )+\pi (\beta ,{\mathcal{D}}_{\alpha }\gamma )=0.$$

$$\pi =\sum _{i,j}{X}_i\wedge X_j.$$

$$\omega (u,v)=-{\omega }_J\left(\sqrt{-A^2}u,v\right),u,v\in TM.$$

$$\pi (\alpha ,\beta )(a)=r({\mathrm{L}}_a^{\ast }\alpha ,{\mathrm{L}}_a^{\ast }\beta ),$$

$$[r,r]=0,$$

$$[r,r](\alpha ,\beta ,\gamma ):=\prec \alpha ,[r_{\#}(\beta ),r_{\#}(\gamma )]\succ +\prec \beta ,[r_{\#}(\gamma ),r_{\#}(\alpha )]\succ +\prec \gamma ,[r_{\#}(\alpha ),r_{\#}(\beta )]\succ ,\alpha ,\beta ,\gamma \in {\mathfrak{g}}^{\ast },$$

$$[\alpha ,\beta {]}_r={\mathrm{ad}}_{r_{\#}(\alpha )}^{\ast }\beta -{\mathrm{ad}}_{r_{\#}(\beta )}^{\ast }\alpha ,\alpha ,\beta \in {\mathfrak{g}}^{\ast },$$

$$r_{\#}([\alpha ,\beta {]}_r)=[r_{\#}(\alpha ),r_{\#}(\beta )],\alpha ,\beta \in {\mathfrak{g}}^{\ast }.$$

$$⟨u,v⟩(a)=\rho (T_a{\mathrm{L}}_{a^{-1}}u,T_a{\mathrm{L}}_{a^{-1}}v),$$

$$2\varrho (A_{u}v,w)=\varrho ([u,v],w)+\varrho ([w,u],v)+\varrho ([w,v],u),u,v,w\in \mathfrak{g}.$$

$$A_{u}v-A_{v}u=[u,v]\text{mbox}and\varrho (A_{u}v,w)+\varrho (v,A_{u}w)=0,$$

$$\omega (A_{u}v,w)+\omega (u,A_{u}v)=0,$$

$$u^{\ell }(a)=T_e{\mathrm{L}}_a(u)\text{mbox}and{\alpha }^{\ell }(a)=T_a^{\ast }{\mathrm{L}}_{a^{-1}}(\alpha ),a\in G,{\mathrm{L}}_a(b)=ab.$$

$$\{\begin{array}{l}[\pi ,\pi {]}_S({\alpha }^{\ell },{\beta }^{\ell },{\gamma }^{\ell })=[r,r](\alpha ,\beta ,\gamma ),{\#}_{\pi }({\alpha }^{\ell })=(r_{\#}(\alpha ){)}^{\ell },{\mathcal{L}}_{{\#}_{\pi }({\alpha }^{\ell })}{\beta }^{\ell }={\left({\mathrm{ad}}_{r_{\#}(\alpha )}^{\ast }\beta \right)}^{\ell },\\ [{\alpha }^{\ell },{\beta }^{\ell }{]}_{\pi }={\left([\alpha ,\beta {]}_r\right)}^{\ell },{\mathcal{D}}_{{\alpha }^{\ell }}{\beta }^{\ell }={\left(A_{\alpha }\beta \right)}^{\ell }.\end{array}$$

$$[\pi ,\pi {]}_S({\alpha }^{\ell },{\beta }^{\ell },{\gamma }^{\ell })=0\text{mbox}and\mathcal{D}\pi ({\alpha }^{\ell },{\beta }^{\ell },{\gamma }^{\ell })=0.\text{qed}$$

$$\delta {\omega }_r(u,v,w):={\omega }_r(u,[v,w])+{\omega }_r(v,[w,u])+{\omega }_r(w,[u,v])=0$$

$$\prec \gamma ,r_{\#}([\alpha ,\beta {]}_r)-[r_{\#}(\alpha ),r_{\#}(\beta )]\succ =-[r,r](\alpha ,\beta ,\gamma ),\alpha ,\beta ,\gamma \in {\mathfrak{g}}^{\ast }$$

$$[r,r](\alpha ,\beta ,\gamma )=-\delta {\omega }_r(r_{\#}(\alpha ),r_{\#}(\beta ),r_{\#}(\gamma )).\text{qed}$$

$$r(A_{\alpha }\beta ,\gamma )+r(\beta ,A_{\alpha }\gamma )=0.$$

$$\{\begin{array}{l}[r,r]=0,\\ r(A_{\alpha }\beta ,\gamma )=0,\alpha \in \mathcal{I},\beta \in \mathcal{I},\gamma \in {\mathcal{I}}^{\perp },\\ r(A_{\alpha }\beta ,\gamma )+r(\beta ,A_{\alpha }\gamma )=0,\alpha \in \mathcal{I},\beta \in {\mathcal{I}}^{\perp },\gamma \in {\mathcal{I}}^{\perp },\\ r(A_{\alpha }\beta ,\gamma )=0,\alpha \in {\mathcal{I}}^{\perp },\beta \in \mathcal{I},\gamma \in {\mathcal{I}}^{\perp },\\ r(A_{\alpha }\beta ,\gamma )+r(\beta ,A_{\alpha }\gamma )=0,\alpha \in {\mathcal{I}}^{\perp },\beta \in {\mathcal{I}}^{\perp },\gamma \in {\mathcal{I}}^{\perp }.\end{array}$$

$$r_{\#}([\alpha ,\beta {]}_r)=r_{\#}(A_{\alpha }\beta )-r_{\#}(A_{\beta }\alpha )\stackrel{\text{eqref}mo}{=}[r_{\#}(\alpha ),r_{\#}(\beta )]=0.$$

$$\prec \gamma ,r_{\#}(A_{\alpha }\beta )\succ$$

$$\varrho ({\mathrm{ad}}_{u}v,w)+\varrho (v,{\mathrm{ad}}_{u}w)=0,u,v,w\in \mathfrak{g}.$$

$$0$$

$${\mathfrak{g}}^{\ast }=\mathcal{I}\oplus {\mathcal{I}}^{\perp }\text{mbox}and\mathfrak{g}=S\oplus S^{\perp },$$

$$\varrho (\#(\alpha ),r_{\#}(\beta ))=\prec \alpha ,r_{\#}(\beta )\succ =r(\beta ,\alpha ),$$

$${\omega }_r(u,v)=\rho (Ju,v),u,v\in S.$$

Full text

1 Introduction

In this paper, we study Lie groups endowed with a left invariant Riemannian metric and a left invariant Poisson tensor satisfying a compatibility condition to be defined below. They constitute a subclass of the class of Riemann-Poisson manifolds introduced and studied by the second author (see [2, 3, 4, 5]).

Let $(M,\pi,\langle\;,\;\rangle)$ be smooth manifold endowed with a Poisson tensor $\pi$ and a Riemannian metric $\langle\;,\;\rangle$. We denote by $\langle\;,\;\rangle^{*}$ the Euclidean product on $T^{*}M$ naturally associated to $\langle\;,\;\rangle$. The Poisson tensor defines a Lie algebroid structure on $T^{*}M$ where the anchor map is the contraction $\#_{\pi}:T^{*}M\longrightarrow TM$ given by $\prec\beta,\#_{\pi}(\alpha)\succ=\pi(\alpha,\beta)$ and the Lie bracket on $\Omega^{1}(M)$ is the Koszul bracket given by

[TABLE]

This Lie algebroid structure and the metric $\langle\;,\;\rangle^{*}$ define a contravariant connection ${\cal D}:\Omega^{1}(M)\times\Omega^{1}(M)\longrightarrow\Omega^{1}(M)$ by Koszul formula

[TABLE]

This is the unique torsionless contravariant connection which is metric, i.e., for any $\alpha,\beta,\gamma\in\Omega^{1}(M)$,

[TABLE]

The notion of contravariant connection was introduced by Vaisman in [13] and studied in more details by Fernandes in the context of Lie algebroids [8]. The connection ${\cal D}$ defined above is called contravariant Levi-Civita connection associated to the couple $(\pi,\langle\;,\;\rangle)$ and it appeared first in [2].

The triple $(M,\pi,\langle\;,\;\rangle)$ is called a Riemannian-Poisson manifold if ${\cal D}\pi=0$, i.e., for any $\alpha,\beta,\gamma\in\Omega^{1}(M),$

[TABLE]

This notion was introduced by the second author in [2]. Riemann-Poisson manifolds turned out to have interesting geometric properties (see[2, 3, 4, 5]). Let’s mention some of them.

The condition of compatibility (3) is weaker than the condition $\nabla\pi=0$ where $\nabla$ is the Levi-Civita connection of $\langle\;,\;\rangle$. Indeed, the condition (3) allows the Poisson tensor to have a variable rank. For instance, linear Poisson structures which are Riemann-Poisson exist and were characterized in [5]. Furthermore, let $(M,\langle\;,\;\rangle)$ be a Riemannian manifold and $(X_{1},\ldots,X_{r})$ a family of commuting Killing vector fields. Put

[TABLE]

Then $(M,\pi,\langle\;,\;\rangle)$ is a Riemann-Poisson manifold. This example illustrates also the weakness of the condition (3) and, more importantly, it is the local model of the geometry of noncommutative deformations studied by Hawkins (see [9, Theorem 6.6]). 2. 2.

Riemann-Poisson manifolds can be thought of as a generalization of Kähler manifolds. Indeed, let $(M,\pi,\langle\;,\;\rangle)$ be a Poisson manifold endowed with a Riemannian metric such that $\pi$ is invertible. Denote by $\omega$ the symplectic form inverse of $\pi$. Then $(M,\pi,\langle\;,\;\rangle)$ is Riemann-Poisson manifold if and only if $\nabla\omega=0$ where $\nabla$ is the Levi-Civita connection of $\langle\;,\;\rangle$. In this case, if we define $A:TM\longrightarrow TM$ by $\omega(u,v)=\langle Au,v\rangle$ then $-A^{2}$ is symmetric definite positive and hence there exists a unique $Q:TM\longrightarrow TM$ symmetric definite positive such that $Q^{2}=-A^{2}$. It follows that $J=AQ^{-1}$ satisfies $J^{2}=-\mathrm{Id}_{TM}$, skew-symmetric with respect $\langle\;,\;\rangle$ and $\nabla J=0$. Hence $(M,J,\langle\;,\;\rangle)$ is a Kähler manifold and its Kähler form $\omega_{J}(u,v)=\langle Ju,v\rangle$ is related to $\omega$ by the following formula:

[TABLE]

Having this construction in mind, we will call in this paper a Kähler manifold a triple $(M,\langle\;,\;\rangle,\omega)$ where $\langle\;,\;\rangle$ is a Riemannian metric and $\omega$ is a nondegenerate 2-form $\omega$ such that $\nabla\omega=0$ where $\nabla$ is the Levi-Civita connection of $\langle\;,\;\rangle$. 3. 3.

The symplectic foliation of a Riemann-Poisson manifold when $\pi$ has a constant rank has an important property namely it is both a Riemannian foliation and a Kähler foliation.

Recall that a Riemannian foliation is a foliated manifold $(M,\mathcal{F})$ with a Riemannian metric $\langle\;,\;\rangle$ such that the orthogonal distribution $T^{\perp}\mathcal{F}$ is totally geodesic.

Kähler foliations are a generalization of Kähler manifolds (see [6]) and, as for the notion of Kähler manifold, we call in this paper a Kähler foliation a foliated manifold $(M,\mathcal{F})$ endowed with a leafwise metric $\langle\;,\;\rangle_{\mathcal{F}}\in\Gamma(\otimes^{2}T^{*}\mathcal{F})$ and a nondegenerate leafwise differential 2-form $\omega_{\mathcal{F}}\in\Gamma(\otimes^{2}T^{*}\mathcal{F})$ such any leaf with the restrictions of $\langle\;,\;\rangle_{\mathcal{F}}$ and $\omega_{\mathcal{F}}$ is a Kähler manifold.

Theorem 1.1** ([4]).**

Let $(M,\langle\;,\;\rangle,\pi)$ be a Riemann-Poisson manifold with $\pi$ of constant rank. Then its symplectic foliation is both a Riemannian and a Kähler foliation.

Having in mind these properties particularly Theorem 1.1, it will be interesting to find large classes of examples of Riemann-Poisson manifolds. This paper will describe the rich collection of examples which are obtained by providing an arbitrary Lie group $G$ with a Riemannian metric $\langle\;,\;\rangle$ and a Poisson tensor $\pi$ invariant under left translations and such that $(G,\langle\;,\;\rangle,\pi)$ is Riemann-Poisson. We call $(G,\langle\;,\;\rangle,\pi)$ a Riemann-Poisson Lie group. This class of examples can be enlarged substantially, with no extra work, as follows. If $(G,\langle\;,\;\rangle,\pi)$ is a Riemann-Poisson Lie group and $\Gamma$ is any discrete subgroup of $G$ then $\Gamma\backslash G$ carries naturally a structure of Riemann-Poisson manifold.

The paper is organized as follows. In Section 2, we give the material needed in the paper and we describe the infinitesimal counterpart of Riemann-Poisson Lie groups, namely, Riemann-Poisson Lie algebras. In Section 3, we prove our main result which gives an useful description of Riemann-Poisson Lie algebras (see Theorem 3.2). We use this theorem in Section 4 to derive a method for building Riemann-Poisson Lie algebras. We explicit this method by giving the list of Riemann-Poisson Lie algebras up to dimension 5.

2 Riemann-Poisson Lie groups and their infinitesimal characterization

Let $G$ be a Lie group and $(\mathfrak{g}=T_{e}G,[\;,\;])$ its Lie algebra.

A left invariant Poisson tensor $\pi$ on $G$ is entirely determined by

[TABLE]

where $a\in G,\alpha,\beta\in T_{a}^{*}G$, $\mathrm{L}_{a}$ is the left multiplication by $a$ and $r\in\wedge^{2}\mathfrak{g}$ satisfies the classical Yang-Baxter equation

[TABLE]

where $[r,r]\in\wedge^{3}\mathfrak{g}$ is given by

[TABLE]

and $r_{\#}:\mathfrak{g}^{*}\longrightarrow\mathfrak{g}$ is the contraction associated to $r$. In this case, the Koszul bracket (1) when restricted to left invariant differential 1-forms induces a Lie bracket on $\mathfrak{g}^{*}$ given by

[TABLE]

where $\prec{\mathrm{ad}}^{*}_{u}\alpha,v\succ=-\prec\alpha,[u,v]\succ$. Moreover, $r_{\#}$ becomes a morphism of Lie algebras, i.e.,

[TABLE] 2. 2.

A let invariant Riemannian metric $\langle\;,\;\rangle$ on $G$ is entirely determined by

[TABLE]

where $a\in G,u,v\in T_{a}G$ and $\rho$ is a scalar product on $\mathfrak{g}$. The Levi-Civita connection of $\langle\;,\;\rangle$ is left invariant and induces a product $A:\mathfrak{g}\times\mathfrak{g}\longrightarrow\mathfrak{g}$ given by

[TABLE]

It is the unique product on $\mathfrak{g}$ satisfying

[TABLE]

for any $u,v,w\in\mathfrak{g}$. We call $A$ the Levi-Civita product associated to $(\mathfrak{g},[\;,\;],\rho)$. 3. 3.

Let $(G,\langle\;,\;\rangle,\Omega)$ be a Lie group endowed with a left invariant Riemannian metric and a nondegenerate left invariant 2-form. Then $(G,\langle\;,\;\rangle,\Omega)$ is a Kähler manifold if and only if, for any $u,v,w\in\mathfrak{g}$,

[TABLE]

where $\omega=\Omega(e)$, $\rho=\langle\;,\;\rangle(e)$ and $A$ is the Levi-Civita product of $(\mathfrak{g},[\;,\;],\rho)$. In this case we call $(\mathfrak{g},[\;,\;],\rho,\omega)$ a Kähler Lie algebra.

As all the left invariant structures on Lie groups, Riemann-Poison Lie groups can be characterized at the level of their Lie algebras.

Proposition 2.1**.**

Let $(G,\pi,\langle\;,\;\rangle)$ be a Lie group endowed with a left invariant bivector field and a left invariant metric and $(\mathfrak{g},[\;,\;])$ its Lie algebra. Put $r=\pi(e)\in\wedge^{2}\mathfrak{g}$, ${\varrho}=\langle\;,\;\rangle_{e}$ and ${\varrho}^{*}$ the associated Euclidean product on $\mathfrak{g}^{*}$. Then $(G,\pi,\langle\;,\;\rangle)$ is a Riemann-Poisson Lie group if and only if

1. $(i)$

$[r,r]=0$, 2. $(ii)$

for any $\alpha,\beta,\gamma\in\mathfrak{g}^{*}$, $r(A_{\alpha}\beta,\gamma)+r(\beta,A_{\alpha}\gamma)=0$,

where $A$ is the Levi-Civita product associated to $(\mathfrak{g}^{*},[\;,\;]_{r},{\varrho}^{*})$.

Proof.

For any $u\in\mathfrak{g}$ and $\alpha\in\mathfrak{g}^{*}$, we denote by $u^{\ell}$ and $\alpha^{\ell}$, respectively, the left invariant vector field and the left invariant differential 1-form on $G$ given by

[TABLE]

Since $\pi$ and $\langle\;,\;\rangle$ are left invariant, one can see easily from (1) and (1) that we have, for any $\alpha,\beta,\gamma\in\mathfrak{g}^{*}$,

[TABLE]

The proposition follows from these formulas, (3) and the fact that $(G,\pi,\langle\;,\;\rangle)$ is a Riemann-Poisson Lie group if and only if, for any $\alpha,\beta,\gamma\in\mathfrak{g}^{*}$,

[TABLE]

Conversely, given a triple $(\mathfrak{g},r,{\varrho})$ where $\mathfrak{g}$ is a real Lie algebra, $r\in\wedge^{2}\mathfrak{g}$ and ${\varrho}$ a Euclidean product on $\mathfrak{g}$ satisfying the conditions $(i)$ and $(ii)$ in Proposition 2.1 then, for any Lie group $G$ whose Lie algebra is $\mathfrak{g}$, if $\pi$ and $\langle\;,\;\rangle$ are the left invariant bivector field and the left invariant metric associated to $(r,{\varrho})$ then $(G,\pi,\langle\;,\;\rangle)$ is a Riemann-Poisson Lie group.

Definition 2.1**.**

A Riemann-Poisson Lie algebra is a triple $(\mathfrak{g},r,{\varrho})$ where $\mathfrak{g}$ is a real Lie algebra, $r\in\wedge^{2}\mathfrak{g}$ and ${\varrho}$ a Euclidean product on $\mathfrak{g}$ satisfying the conditions $(i)$ and $(ii)$ in Proposition 2.1.

To end this section, we give another characterization of the solutions of the classical Yang-Baxter equation (5) which will be useful later.

We observe that $r\in\wedge^{2}{\mathfrak{g}}$ is equivalent to the data of a vector subspace $S\subset\mathfrak{g}$ and a nondegenerate 2-form $\omega_{r}\in\wedge^{2}S^{*}$.

Indeed, for $r\in\wedge^{2}{\mathfrak{g}}$, we put $S=\mathrm{Im}r_{\#}$ and $\omega_{r}(u,v)=r(r_{\#}^{-1}(u),r_{\#}^{-1}(v))$ where $u,v\in S$ and $r_{\#}^{-1}(u)$ is any antecedent of $u$ by $r_{\#}$.

Conversely, let $(S,\omega)$ be a vector subspace of $\mathfrak{g}$ with a non-degenerate 2-form. The 2-form $\omega$ defines an isomorphism $\omega^{b}:S\longrightarrow S^{*}$ by $\omega^{b}(u)=\omega(u,.)$, we denote by $\#:S^{*}\longrightarrow S$ its inverse and we put $r_{\#}=\#\circ i^{*}$ where $i^{*}:{\mathfrak{g}}^{*}\longrightarrow S^{*}$ is the dual of the inclusion $i:S\hookrightarrow\mathfrak{g}$.

With this observation in mind, the following proposition gives another description of the solutions of the Yang-Baxter equation.

Proposition 2.2**.**

Let $r\in\wedge^{2}{\mathfrak{g}}$ and $(S,\omega_{r})$ its associated vector subspace. The following assertions are equivalent:

$[r,r]=0.$** 2. 2.

$S$* is a subalgebra of $\mathfrak{g}$ and*

[TABLE]

for any $u,v,w\in S$.

Proof.

The proposition follows from the following formulas:

[TABLE]

and, if $S$ is a subalgebra,

[TABLE]

This proposition shows that there is a correspondence between the set of solutions of the Yang-Baxter equation the set of symplectic subalgebras of $\mathfrak{g}$. We recall that a symplectic algebra is a Lie algebra $S$ endowed with a non-degenerate 2-form $\omega$ such that $\delta\omega=0.$

3 A characterization of Riemann-Poisson Lie algebras

In this section, we combine Propositions 2.1 and 2.2 to establish a characterization of Riemann-Poisson Lie algebras which will be used later to build such Lie algebras. We establish first an intermediary result.

Proposition 3.1**.**

Let $(\mathfrak{g},r,{\varrho})$ be a Lie algebra endowed with $r\in\wedge^{2}\mathfrak{g}$ and a Euclidean product ${\varrho}$. Denote by $\mathcal{I}=\ker r_{\#}$, $\mathcal{I}^{\perp}$ its orthogonal with respect to ${\varrho}^{*}$ and $A$ the Levi-Civita product associated to $(\mathfrak{g}^{*},[\;,\;]_{r},{\varrho}^{*})$. Then $(\mathfrak{g},r,{\varrho})$ is a Riemann-Poisson Lie algebra if and only if:

1. $(c_{1})$

$[r,r]=0$. 2. $(c_{2})$

For all $\alpha\in\mathcal{I},A_{\alpha}=0.$ 3. $(c_{3})$

For all $\alpha,\beta,\gamma\in\mathcal{I}^{\perp},$ $A_{\alpha}\beta\in\mathcal{I}^{\perp}$ and

[TABLE]

Proof.

By using the splitting $\mathfrak{g}^{*}=\mathcal{I}\oplus\mathcal{I}^{\perp}$, on can see that the conditions $(i)$ and $(ii)$ in Proposition 2.1 are equivalent to

[TABLE]

Suppose that the conditions $(c_{1})$-$(c_{3})$ hold. Then for any $\alpha\in\mathcal{I}$ and $\beta\in\mathcal{I}^{\perp}$, $A_{\beta}\alpha=[\beta,\alpha]_{r}$ and hence $r_{\#}(A_{\beta}\alpha)=[r_{\#}(\beta),r_{\#}(\alpha)]=0$ and hence the equations in (11) holds.

Conversely, suppose that (11) holds. Then $(c_{1})$ holds obviously.

For any $\alpha,\beta\in\mathcal{I}$, the second equation in (11) is equivalent to $A_{\alpha}\beta\in\mathcal{I}$ and we have from (7) and (9) $[\alpha,\beta]_{r}=0$ and $A_{\alpha}\beta\in\mathcal{I}^{\perp}$. Thus $A_{\alpha}\beta=0$.

Take now $\alpha\in\mathcal{I}$ and $\beta\in\mathcal{I}^{\perp}$. For any $\gamma\in\mathcal{I}$, ${\varrho}^{*}(A_{\alpha}\beta,\gamma)=-{\varrho}^{*}(\beta,A_{\alpha}\gamma)=0$ and hence $A_{\alpha}\beta\in\mathcal{I}^{\perp}$. On the other hand,

[TABLE]

So, for any $\gamma\in\mathcal{I}^{\perp}$,

[TABLE]

This shows that $A_{\alpha}\beta\in\mathcal{I}$ and hence $A_{\alpha}\beta=0$. Finally, $(c_{2})$ is true. Now, for any $\alpha\in\mathcal{I}^{\perp}$, the fourth equation in (11) implies that $A_{\alpha}$ leaves invariant $\mathcal{I}$ and since it is skew-symmetric it leaves invariant $\mathcal{I}^{\perp}$ and $(c_{3})$ follows. This completes the proof. ∎

Proposition 3.2**.**

Let $(\mathfrak{g},{\varrho},r)$ be a Lie algebra endowed with a solution of classical Yang-Baxter equation and a bi-invariant Euclidean product, i.e.,

[TABLE]

Then $(\mathfrak{g},{\varrho},r)$ is Riemann-Poisson Lie algebra if and only if $\mathrm{Im}r_{\#}$ is an abelian subalgebra.

Proof.

Since ${\varrho}$ is bi-invariant, one can see easily that for any $u\in\mathfrak{g}$, ${\mathrm{ad}}_{u}^{*}$ is skew-symmetric with respect to ${\varrho}^{*}$ and hence the Levi-Civita product $A$ associated to $(\mathfrak{g}^{*},[\;,\;]_{r},{\varrho}^{*})$ is given by $A_{\alpha}\beta={\mathrm{ad}}_{r_{\#}(\alpha)}^{*}\beta$. So, $(\mathfrak{g},{\varrho},r)$ is Riemann-Poisson Lie algebra if and only if, for any $\alpha,\beta,\gamma\in\mathfrak{g}^{*}$,

[TABLE]

and the result follows. ∎

Let $(\mathfrak{g},[\;,\;])$ be a Lie algebra, $r\in\wedge^{2}\mathfrak{g}$ and ${\varrho}$ a Euclidean product on $\mathfrak{g}$. Denote by $(S,\omega_{r})$ the symplectic vector subspace associated to $r$ and by $\#:\mathfrak{g}^{*}\longrightarrow\mathfrak{g}$ the isomorphism given by ${\varrho}$. Note that the Euclidean product on $\mathfrak{g}^{*}$ is given by ${\varrho}^{*}(\alpha,\beta)={\varrho}(\#(\alpha),\#(\beta))$. We have

[TABLE]

where $\mathcal{I}=\ker r_{\#}$. Moreover, $r_{\#}:\mathcal{I}^{\perp}\longrightarrow S$ is an isomorphism, we denote by $\tau:S\longrightarrow\mathcal{I}^{\perp}$ its inverse. From the relation

[TABLE]

we deduce that $\#:\mathcal{I}\longrightarrow S^{\perp}$ is an isomorphism and hence $\#:\mathcal{I}^{\perp}\longrightarrow S$ is also an isomorphism.

Consider the isomorphism $J:S\longrightarrow S$ linking $\omega_{r}$ to ${\varrho}_{|S}$, i.e.,

[TABLE]

On can see easily that $J=-\#\circ\tau$.

Theorem 3.2**.**

With the notations above, $(\mathfrak{g},r,{\varrho})$ is a Riemann-Poisson Lie algebra if and only if the following conditions hold:

$(S,{\varrho}_{|S},\omega_{r})$* is a Kähler Lie subalgebra, i.e., for all $s_{1},s_{2},s_{3}\in S$,*

[TABLE]

where $\nabla$ is the Levi-Civita product associated to $(S,[\;,\;],{\varrho}_{|S})$. 2. 2.

for all $s\in S$ and all $u,v\in S^{\perp},$

[TABLE]

where $\phi_{S}:S\longrightarrow\mathrm{End}(S^{\perp})$, $u\mapsto\mathrm{pr}_{S^{\perp}}\circ{\mathrm{ad}}_{u}$ and $\mathrm{pr}_{S^{\perp}}:\mathfrak{g}\longrightarrow S^{\perp}$ is the orthogonal projection. 3. 3.

For all $s_{1},s_{2}\in S$ and all $u\in S^{\perp},$

[TABLE]

where $\phi_{S^{\perp}}:S^{\perp}\longrightarrow\mathrm{End}(S)$, $u\mapsto\mathrm{pr}_{S}\circ{\mathrm{ad}}_{u}$ and $\mathrm{pr}_{S}:\mathfrak{g}\longrightarrow S$ is the orthogonal projection.

Proof.

Suppose first that $(\mathfrak{g},r,{\varrho})$ is a Riemann-Poisson Lie algebra. According to Propositions 3.1 and 2.2, this is equivalent to

[TABLE]

where $A$ is the Levi-Civita product of $(\mathfrak{g}^{*},[\;,\;]_{r},{\varrho}^{*})$.

For $\alpha,\beta\in\mathcal{I}$ and $\gamma\in\mathcal{I}^{\perp}$,

[TABLE]

Since $\#:\mathcal{I}\longrightarrow S^{\perp}$ and $r_{\#}:\mathcal{I}^{\perp}\longrightarrow S$ are isomorphisms, we deduce from (16) that $A_{\alpha}\beta=0$ for any $\alpha,\beta\in\mathcal{I}$ is equivalent to (13).

For $\alpha\in\mathcal{I}$ and $\beta,\gamma\in\mathcal{I}^{\perp}$,

[TABLE]

Now, $\#(\beta),\#(\gamma)\in S$ and $r_{\#}(\beta),r_{\#}(\gamma)\in S$ and since $S$ is a subalgebra we deduce that $[r_{\#}(\beta),\#(\gamma)],[r_{\#}(\gamma),\#(\beta)]\in S$ and hence

[TABLE]

We have also $\#:\mathcal{I}\longrightarrow S^{\perp}$ and $r_{\#}:\mathcal{I}^{\perp}\longrightarrow S$ are isomorphisms so that, by virtue of (17), $A_{\alpha}\beta=0$ for any $\alpha\in\mathcal{I}$ and $\beta\in\mathcal{I}^{\perp}$ is equivalent to (14).

On the other hand, for any $\alpha,\beta,\gamma\in\mathcal{I}^{\perp}$, since $\#=-J\circ r_{\#}$, the relation

[TABLE]

can be written

[TABLE]

But $r_{\#}([\alpha,\beta]_{r})=[r_{\#}(\alpha),r_{\#}(\beta)]$ and hence

[TABLE]

where $\langle u,v\rangle_{J}={\varrho}(Ju,Jv)$. This shows that $r_{\#}(A_{\alpha}\beta)=\nabla_{r_{\#}(\alpha)}r_{\#}(\beta)$ where $\nabla$ is the Levi-Civita product of $(S,[\;,\;],\langle\;,\;\rangle_{J})$ and the third relation in (15) is equivalent to

[TABLE]

This is equivalent to $\nabla_{u}Jv=J\nabla_{u}v$. Let us show that $\nabla$ is actually the Levi-Civita product of $(S,[\;,\;],{\varrho})$. Indeed, for any $u,v,w\in S$, $\nabla_{u}v-\nabla_{v}u=[u,v]$ and

[TABLE]

So we have shown the direct part of the theorem. The converse can be deduced easily from the relations we established in the proof of the direct part. ∎

Example 1**.**

Let $G$ be a compact connected Lie group, $\mathfrak{g}$ its Lie algebra and $T$ an even dimensional torus of $G$. Choose a bi-invariant Riemannian metric $\langle\;,\;\rangle$ on $G$, a nondegenerate $\omega\in\wedge^{2}S^{*}$ where $S$ is the Lie algebra of $T$ and put ${\varrho}=\langle\;,\;\rangle(e)$. Let $r\in\wedge^{2}\mathfrak{g}$ be the solution of the classical Yang-Baxter associated to $(S,\omega)$. By using either Proposition 3.2 or Theorem 3.2, one can see easily that $(\mathfrak{g},{\varrho},r)$ is a Riemann-Poisson Lie algebra and hence $(G,\langle\;,\;\rangle,\pi)$ is a Riemann-Poisson Lie group where $\pi$ is the left invariant Poisson tensor associated to $r$. According to Theorem 1.1, the orbits of the right action of $T$ on $G$ defines a Riemannian and Kähler foliation. For instance, $G=\mathrm{SO}(2n)$, $T=\operatorname{Diagonal}(D_{1},\ldots,D_{n})$ where $D_{i}=\left(\begin{matrix}\cos(\theta_{i})&\sin(\theta_{i})\\ -\sin(\theta_{i})&\cos(\theta_{i})\end{matrix}\right)$ and $\langle\;,\;\rangle=-K$ where $K$ is the Killing form.

4 Construction of Riemann-Poisson Lie algebras

In this section, we give a general method for building Riemann-Poisson Lie algebras and we use it to give all Riemann-Poisson Lie algebras up to dimension 5.

According to Theorem 3.2, to build Riemann-Poisson Lie algebras one needs to solve the following problem.

Problem 1**.**

We look for:

A Kähler Lie algebra $({\mathfrak{h}},[\;,\;]_{\mathfrak{h}},{\varrho}_{\mathfrak{h}},\omega)$, 2. 2.

a Euclidean vector space $({\mathfrak{p}},{\varrho}_{\mathfrak{p}})$, 3. 3.

a bilinear skew-symmetric map $[\;,\;]_{\mathfrak{p}}:{\mathfrak{p}}\times{\mathfrak{p}}\longrightarrow{\mathfrak{p}}$, 4. 4.

a bilinear skew-symmetric map ${\mu}:{\mathfrak{p}}\times{\mathfrak{p}}\longrightarrow{\mathfrak{h}}$, 5. 5.

two linear maps $\phi_{\mathfrak{p}}:{\mathfrak{p}}\longrightarrow\mathrm{sp}({\mathfrak{h}},\omega)$ and $\phi_{\mathfrak{h}}:{\mathfrak{h}}\longrightarrow\mathrm{so}({\mathfrak{p}})$ where $\mathrm{sp}({\mathfrak{h}},\omega)=\left\{J:{\mathfrak{h}}\longrightarrow{\mathfrak{h}},J^{\omega}+J=0\right\}$ and $\mathrm{so}({\mathfrak{p}})=\left\{A:{\mathfrak{p}}\longrightarrow{\mathfrak{p}},A^{*}+A=0\right\}$, $J^{\omega}$ is the adjoint with respect to $\omega$ and $A^{*}$ is the adjoint with respect to ${\varrho}_{\mathfrak{p}}$,

such that the bracket $[\;,\;]$ on $\mathfrak{g}={\mathfrak{h}}\oplus{\mathfrak{p}}$ given, for any $a,b\in{\mathfrak{p}}$ and $u,v\in{\mathfrak{h}}$, by

[TABLE]

is a Lie bracket.

In this case, $(\mathfrak{g},[\;,\;])$ endowed with $r\in\wedge^{2}\mathfrak{g}$ associated to $({\mathfrak{h}},\omega)$ and the Euclidean product ${\varrho}={\varrho}_{\mathfrak{h}}\oplus{\varrho}_{\mathfrak{p}}$ becomes, by virtue of Theorem 3.2, a Riemann-Poisson Lie algebra.

Proposition 4.1**.**

With the data and notations of Problem 1, the bracket given by (18) is a Lie bracket if and only if, for any $u,v\in{\mathfrak{h}}$ and $a,b,c\in{\mathfrak{p}}$,

[TABLE]

where $\oint$ stands for the circular permutation.

Proof.

The equations follow from the Jacobi identity applied to $(a,u,v)$, $(a,b,u)$ and $(a,b,c)$. ∎

We tackle now the task of determining the list of all Riemann-Poisson Lie algebras up to dimension 5. For this purpose, we need to solve Problem 1 in the following four cases: $(a)$ $\dim{\mathfrak{p}}=1$, $(b)$ $\dim{\mathfrak{h}}=2$ and ${\mathfrak{h}}$ non abelian, $(c)$ $\dim{\mathfrak{h}}=\dim{\mathfrak{p}}=2$ and ${\mathfrak{h}}$ abelian, $(d)$ $\dim{\mathfrak{h}}=2$, $\dim{\mathfrak{p}}=3$ and ${\mathfrak{h}}$ abelian.

It is easy to find the solutions of Problem 1 when $\dim{\mathfrak{p}}=1$ since in this case $\mathrm{so}({\mathfrak{p}})=0$ and the three last equations in (19) hold obviously.

Proposition 4.2**.**

If $\dim{\mathfrak{p}}=1$ then the solutions of Problem 1 are a Kähler Lie algebra $({\mathfrak{h}},{\varrho},\omega)$, $\phi_{\mathfrak{h}}=0$, $[\;,\;]_{\mathfrak{p}}=0$, $\mu=0$ and $\phi_{\mathfrak{p}}(a)\in\mathrm{sp}({\mathfrak{h}},\omega)\cap\mathrm{Der}({\mathfrak{h}})$ where $a$ is a generator of ${\mathfrak{p}}$ and $\mathrm{Der}({\mathfrak{h}})$ the Lie algebra of derivations of ${\mathfrak{h}}$.

Let us solve Problem 1 when ${\mathfrak{h}}$ is 2-dimensional non abelian.

Proposition 4.3**.**

Let $(({\mathfrak{h}},\omega,{\varrho}_{\mathfrak{h}}),({\mathfrak{p}},[\;,\;]_{\mathfrak{p}},{\varrho}_{\mathfrak{p}}),\mu,\phi_{\mathfrak{h}},\phi_{\mathfrak{p}})$ be a solution of Problem 1 with ${\mathfrak{h}}$ is 2-dimensional non abelian. Then there exists an orthonormal basis $\hbox{\bb B}=(e_{1},e_{2})$ of ${\mathfrak{h}}$, $b_{0}\in{\mathfrak{p}}$ and two constants $\alpha\not=0$ and $\beta\not=0$ such that:

1. $(i)$

$[e_{1},e_{2}]_{\mathfrak{h}}=\alpha e_{1}$, $\omega=\beta e_{1}^{*}\wedge e_{2}^{*}$, 2. $(ii)$

$({\mathfrak{p}},[\;,\;]_{\mathfrak{p}},{\varrho}_{\mathfrak{p}})$* is a Euclidean Lie algebra,* 3. $(iii)$

$\phi_{\mathfrak{h}}(e_{1})=0$, $\phi_{\mathfrak{h}}(e_{2})\in\mathrm{Der}({\mathfrak{p}})\cap\mathrm{so}({\mathfrak{p}})$ and, for any $a\in{\mathfrak{p}}$, $M(\phi_{\mathfrak{p}}(a),\hbox{\bb B})=\left(\begin{matrix}0&{\varrho}_{\mathfrak{p}}(a,b_{0})\\ 0&0\end{matrix}\right)$, 4. $(iv)$

for any $a,b\in{\mathfrak{p}}$, $\mu(a,b)=\mu_{0}(a,b)e_{1}$ with $\mu_{0}$ is a 2-cocycle of $({\mathfrak{p}},[\;,\;]_{\mathfrak{p}})$ satisfying

[TABLE]

Proof.

Note first that from the third relation in (19) we get that $\phi_{\mathfrak{h}}({\mathfrak{h}})$ is a solvable subalgebra of $\mathrm{so}({\mathfrak{p}})$ and hence must be abelian. Since ${\mathfrak{h}}$ is 2-dimensional non abelian then $\dim\phi_{\mathfrak{h}}({\mathfrak{h}})=1$ and $[{\mathfrak{h}},{\mathfrak{h}}]\subset\ker\phi_{\mathfrak{h}}$. So there exists an orthonormal basis $(e_{1},e_{2})$ of ${\mathfrak{h}}$ such that $[e_{1},e_{2}]_{\mathfrak{h}}=\alpha e_{1}$, $\phi_{\mathfrak{h}}(e_{1})=0$ and $\omega=\beta e_{1}^{*}\wedge e_{2}^{*}$. If we identify the endomorphisms of ${\mathfrak{h}}$ with their matrices in the basis $(e_{1},e_{2})$, we get that $\mathrm{sp}({\mathfrak{h}},\omega)=\mathrm{sl}(2,\hbox{\bb R})$ and there exists $a_{0},b_{0},c_{0}\in{\mathfrak{p}}$ such that, for any $a\in{\mathfrak{p}}$,

[TABLE]

The first equation in (19) is equivalent to

[TABLE]

for any $a\in{\mathfrak{p}}$. Since $\phi_{\mathfrak{h}}(e_{2})$ is sekw-symmetric, this is equivalent to

[TABLE]

This implies that $a_{0}=c_{0}=0$. The second equation in (19) implies that $\phi_{\mathfrak{h}}(e_{2})$ is a derivation of $[\;,\;]_{\mathfrak{p}}$. If we take $u=e_{1}$ in the forth equation in (19), we get that $[e_{1},\mu(a,b)]=0$, for any $a,b\in{\mathfrak{p}}$ and hence $\mu(a,b)=\mu_{0}(a,b)e_{1}$. If we take $u=e_{2}$ in the forth equation in (19) we get (20). The two last equations are equivalent to $[\;,\;]_{\mathfrak{p}}$ is a Lie bracket and $\mu_{0}$ is 2-cocycle of $({\mathfrak{p}},[\;,\;]_{\mathfrak{p}})$. ∎

The following proposition gives the solutions of Problem 1 when ${\mathfrak{h}}$ is 2-dimensional abelian and $\dim{\mathfrak{p}}=2$.

Proposition 4.4**.**

Let $(({\mathfrak{h}},\omega,{\varrho}_{\mathfrak{h}}),({\mathfrak{p}},[\;,\;]_{\mathfrak{p}},{\varrho}_{\mathfrak{p}}),\mu,\phi_{\mathfrak{h}},\phi_{\mathfrak{p}})$ be a solution of Problem 1 with ${\mathfrak{h}}$ is 2-dimensional abelian and $\dim{\mathfrak{p}}=2$. Then one of the following situations occurs:

$\phi_{\mathfrak{h}}=0$, $({\mathfrak{p}},[\;,\;]_{\mathfrak{p}},{\varrho}_{\mathfrak{p}})$ is a 2-dimensional Euclidean Lie algebra, there exists $a_{0}\in{\mathfrak{p}}$ and $D\in\mathrm{sp}({\mathfrak{h}},\omega)$ such that, for any $a\in{\mathfrak{p}}$, $\phi_{\mathfrak{p}}(a)={\varrho}_{\mathfrak{p}}(a_{0},a)D$ and there is no restriction on $\mu$. Moreover, $a_{0}\in[{\mathfrak{p}},{\mathfrak{p}}]_{\mathfrak{p}}^{\perp}$ if $D\not=0$. 2. 2.

$\phi_{\mathfrak{h}}=0$, $({\mathfrak{p}},[\;,\;]_{\mathfrak{p}},{\varrho}_{\mathfrak{p}})$ is a 2-dimensional non abelian Euclidean Lie algebra, $\phi_{\mathfrak{p}}$ identifies ${\mathfrak{p}}$ to a two dimensional subalgebra of $\mathrm{sp}({\mathfrak{h}},\omega)$ and there is no restriction on $\mu$. 3. 3.

$({\mathfrak{p}},[\;,\;]_{\mathfrak{p}},{\varrho}_{\mathfrak{p}})$* is a Euclidean abelian Lie algebra and there exists an orthonormal basis $\hbox{\bb B}=(e_{1},e_{2})$ of ${\mathfrak{h}}$ and $b_{0}\in{\mathfrak{p}}$ such that $\omega=\alpha e_{1}^{*}\wedge e_{2}^{*}$, $\phi_{\mathfrak{h}}(e_{1})=0$, $\phi_{\mathfrak{h}}(e_{2})\not=0$ and, for any $a\in{\mathfrak{p}}$, $M(\phi_{\mathfrak{p}}(a),\hbox{\bb B})=\left(\begin{matrix}0&{\varrho}_{\mathfrak{p}}(b_{0},a)\\ 0&0\end{matrix}\right)$ and there is no restriction on $\mu$.*

Proof.

Note first that since $\dim{\mathfrak{p}}=2$ the last two equations in (19) hold obviously and $({\mathfrak{p}},[\;,\;]_{\mathfrak{p}})$ is a Lie algebra. We distinguish two cases:

1. $(i)$

$\phi_{\mathfrak{h}}=0$. Then (19) is equivalent to $\phi_{\mathfrak{p}}$ is a representation of ${\mathfrak{p}}$ in $\mathrm{sp}({\mathfrak{h}},\omega)\simeq\mathrm{sl}(2,\hbox{\bb R})$. Since $\mathrm{sl}(2,\hbox{\bb R})$ doesn’t contain any abelian two dimensional subalgebra, if ${\mathfrak{p}}$ is an abelian Lie algebra then $\dim\phi_{\mathfrak{p}}({\mathfrak{p}})\leq 1$ and the first situation occurs. If ${\mathfrak{p}}$ is not abelian then the first or the second situation occurs depending on $\dim\phi_{\mathfrak{p}}({\mathfrak{p}})$. 2. $(ii)$

$\phi_{\mathfrak{h}}\not=0$. Since $\dim\mathrm{so}({\mathfrak{p}})=1$ there exists an orthonormal basis $\hbox{\bb B}=(e_{1},e_{2})$ of ${\mathfrak{h}}$ such that $\phi_{\mathfrak{h}}(e_{1})=0$ and $\phi_{\mathfrak{h}}(e_{2})\not=0$. We have $\mathrm{sp}({\mathfrak{h}},\omega)=\mathrm{sl}(2,\hbox{\bb R})$ and hence, for any $a\in{\mathfrak{p}}$, $M(\phi_{\mathfrak{p}}(a),\hbox{\bb B})=\left(\begin{matrix}{\varrho}_{\mathfrak{p}}(a_{0},a)&{\varrho}_{\mathfrak{p}}(b_{0},a)\\ {\varrho}_{\mathfrak{p}}(c_{0},a)&-{\varrho}_{\mathfrak{p}}(a_{0},a)\end{matrix}\right)$. Choose an orthonormal basis $(a_{1},a_{2})$ of ${\mathfrak{p}}$. Then there exists $\lambda\not=0$ such that $\phi_{\mathfrak{h}}(e_{2})(a_{1})=\lambda a_{2}$ and $\phi_{\mathfrak{h}}(e_{2})(a_{2})=-\lambda a_{1}$.

The first equation in (19) is equivalent to

[TABLE]

This is equivalent to

[TABLE]

Then $a_{0}=c_{0}=0$ and hence $\phi_{\mathfrak{p}}(a)=\left(\begin{matrix}0&{\varrho}_{\mathfrak{p}}(b_{0},a)\\ 0&0\end{matrix}\right)$. The second equation in (19) gives

[TABLE]

and hence $\phi_{\mathfrak{h}}(e_{2})([a_{1},a_{2}]_{\mathfrak{p}})=0$. Thus $[a_{1},a_{2}]_{\mathfrak{p}}=0$. All the other equations in (19) hold obviously.∎

To tackle the last case, we need the determination of 2-dimensional subalgebras of $\mathrm{sl}(2,\hbox{\bb R})$.

Proposition 4.5**.**

The 2-dimensional subalgebras of $\mathrm{sl}(2,\hbox{\bb R})$ are

[TABLE]

where $x\in\hbox{\bb R}\setminus\{0\}$. Moreover, $\mathfrak{g}_{x}=\mathfrak{g}_{y}$ if and only if $x=y$.

Proof.

Let $\mathfrak{g}$ be a 2-dimensional subalgebra of $\mathrm{sl}(2,\hbox{\bb R})$. We consider the basis $\hbox{\bb B}=(h,e,f)$ of $\mathrm{sl}(2,\hbox{\bb R})$ given by

[TABLE]

Then

[TABLE]

If $h\in\mathfrak{g}$ then ${\mathrm{ad}}_{h}$ leaves $\mathfrak{g}$ invariant. But ${\mathrm{ad}}_{h}$ has three eigenvalues $(0,2,-2)$ with the associated eigenvectors $(h,e,f)$ and hence it restriction to $\mathfrak{g}$ has $(0,2)$ or $(0,-2)$ as eigenvalues. Thus $\mathfrak{g}=\mathfrak{g}_{1}$ or $\mathfrak{g}=\mathfrak{g}_{2}$.

Suppose now that $h\notin\mathfrak{g}$. By using the fact that $\mathrm{sl}(2,\hbox{\bb R})$ is unimodular, i.e., for any $w\in\mathrm{sl}(2,\hbox{\bb R})$ ${\mathrm{tr}}({\mathrm{ad}}_{w})=0$, we can choose a basis $(u,v)$ of $\mathfrak{g}$ such that $(u,v,h)$ is a basis of $\mathrm{sl}(2,\hbox{\bb R})$ and

[TABLE]

If $(x_{1},x_{2},x_{3})$ and $(y_{1},y_{2},y_{3})$ are the coordinates of $u$ and $v$ in 𝔹, the brackets above gives

[TABLE]

Note first that if $x_{1}=0$ then $(x_{2},x_{3})=(0,0)$ which impossible so we must have $x_{1}\not=0$ and hence $d=4-a^{2}$. If we replace in the third equation in the second system and the last equation, we get $x_{3}=\frac{1}{4}$ and $y_{3}=-\frac{a}{4}$. The third equation in the first system gives $x_{2}=-\frac{1}{16x_{1}}$ and hence $y_{1}=(2-a)x_{1}$ and $y_{2}=\frac{(a+2)}{16x_{1}}$. Thus

[TABLE]

But

[TABLE]

and hence

[TABLE]

One can check easily that $\mathfrak{g}_{x}=\mathfrak{g}_{y}$ if and only if $x=y$. This completes the proof. ∎

The following two propositions give the solutions of Problem 1 when ${\mathfrak{h}}$ is 2-dimensional abelian and $\dim{\mathfrak{p}}=3$.

Proposition 4.6**.**

Let $(({\mathfrak{h}},\omega,{\varrho}_{\mathfrak{h}}),({\mathfrak{p}},[\;,\;]_{\mathfrak{p}},{\varrho}_{\mathfrak{p}}),\mu,\phi_{\mathfrak{h}},\phi_{\mathfrak{p}})$ be a solution of Problem 1 with ${\mathfrak{h}}$ is 2-dimensional abelian and $\dim{\mathfrak{p}}=3$ and $\phi_{\mathfrak{h}}=0$. Then one of the following situations occurs:

1. $(i)$

$({\mathfrak{p}},[\;,\;]_{\mathfrak{p}},{\varrho}_{\mathfrak{p}})$* is 3-dimensional Euclidean Lie algebra, $\phi_{\mathfrak{p}}=0$ and $\mu$ is 2-cocycle for the trivial representation.* 2. $(ii)$

$\phi_{\mathfrak{p}}$* is an isomorphism of Lie algebras between $({\mathfrak{p}},[\;,\;]_{\mathfrak{p}})$ and $\mathrm{sl}(2,\hbox{\bb R})$ and there exists an endomorphism $L:{\mathfrak{p}}\longrightarrow{\mathfrak{h}}$ such that for any $a,b\in{\mathfrak{p}}$,*

[TABLE] 3. $(iii)$

There exists a basis $\hbox{\bb B}_{\mathfrak{p}}=(a_{1},a_{2},a_{3})$ of ${\mathfrak{p}}$, $\alpha\not=0$, $\beta\not=0$, $\gamma,\tau\in\hbox{\bb R}$ such that $[\;,\;]_{\mathfrak{p}}$ has one of the two following forms

[TABLE]

In both cases, there exists an orthonormal basis $\hbox{\bb B}_{\mathfrak{h}}=(e_{1},e_{2})$ of ${\mathfrak{h}}$, $x\not=0$, $u\not=0$ and $v\in\hbox{\bb R}$ such that $\phi_{\mathfrak{p}}$ has one of the following forms

[TABLE]

Moreover, $\mu$ is a 2-cocycle for $({\mathfrak{p}},[\;,\;]_{\mathfrak{p}},\phi_{\mathfrak{p}})$. 4. $(iv)$

There exists an orthonormal basis $\hbox{\bb B}=(a_{1},a_{2},a_{3})$ of ${\mathfrak{p}}$ such that $\phi_{\mathfrak{p}}(a_{1})=\phi_{\mathfrak{p}}(a_{2})=0$, $\phi_{\mathfrak{p}}(a_{3})$ is a non zero element of $\mathrm{sp}({\mathfrak{h}},\omega)$ and

[TABLE]

Moreover, $\mu$ is a 2-cocycle for $({\mathfrak{p}},[\;,\;]_{\mathfrak{p}},\phi_{\mathfrak{p}})$.

Proof.

In this case, (19) is equivalent to $({\mathfrak{p}},[\;,\;]_{\mathfrak{p}})$ is a Lie algebra and $\phi_{\mathfrak{p}}$ is a representation and $\mu$ is a 2-cocycle of $({\mathfrak{p}},[\;,\;]_{\mathfrak{p}},\phi_{\mathfrak{p}})$.

We distinguish four cases:

$\phi_{\mathfrak{p}}=0$ and the case $(i)$ occurs. 2. 2.

$\dim\phi_{\mathfrak{p}}({\mathfrak{p}})=3$ and hence ${\mathfrak{p}}$ is isomorphic to $\mathrm{sp}({\mathfrak{h}},\omega)\simeq\mathrm{sl}(2,\hbox{\bb R})$ and hence $\mu$ is a coboundary. Thus $(ii)$ occurs. 3. 3.

$\dim\phi_{\mathfrak{p}}({\mathfrak{p}})=2$ then $\ker\phi_{\mathfrak{p}}$ is a one dimensional ideal of ${\mathfrak{p}}$. But $\phi_{\mathfrak{p}}({\mathfrak{p}})$ is a 2-dimensional subalgebra of $\mathrm{sp}({\mathfrak{h}},\omega)\simeq\mathrm{sl}(2,\hbox{\bb R})$, therefore it is non abelian so ${\mathfrak{p}}/\ker{\mathfrak{p}}$ is non abelian.

If $\ker{\mathfrak{p}}\subset[{\mathfrak{p}},{\mathfrak{p}}]_{\mathfrak{p}}$ then $\dim[{\mathfrak{p}},{\mathfrak{p}}]_{\mathfrak{p}}=2$ so there exists an orthonormal basis $(a_{1},a_{2},a_{3})$ of ${\mathfrak{p}}$ such that $a_{1}\in\ker{\mathfrak{p}}$ and

[TABLE]

and we must have $\xi=0$ in order to have the Jacobi identity.

If $\ker{\mathfrak{p}}\nsubset[{\mathfrak{p}},{\mathfrak{p}}]$ then $\ker{\mathfrak{p}}\subset Z({\mathfrak{p}})$ and $\dim[{\mathfrak{p}},{\mathfrak{p}}]=1$. Then there exits a basis $(a_{1},a_{2},a_{3})$ of ${\mathfrak{p}}$ such that $a_{1}\in\ker{\mathfrak{p}}$, $a_{2}\in[{\mathfrak{p}},{\mathfrak{p}}]$, $a_{3}\in\{a_{1},a_{2}\}^{\perp}$ and

[TABLE]

The matrix of ${\varrho}_{\mathfrak{p}}$ in $(a_{1},a_{2},a_{3})$ is given by

[TABLE]

We choose an orthonormal basis $(e_{1},e_{2})$ of ${\mathfrak{h}}$ and identify $\mathrm{sp}({\mathfrak{h}},\omega)$ to $\mathrm{sl}(2,\hbox{\bb R})$. Now $\phi_{\mathfrak{p}}({\mathfrak{p}})=\{\phi_{p}(a_{2}),\phi_{p}(a_{3})\}$ is a subalgebra of $\mathrm{sl}(2,\hbox{\bb R})$ and, according to Proposition 4.5, $\phi_{\mathfrak{p}}({\mathfrak{p}})=\mathfrak{g}_{1}$, $\mathfrak{g}_{2}$ or $\mathfrak{g}_{x}$. But

[TABLE]

So in order for $\phi_{\mathfrak{p}}$ to be a representation we must have

[TABLE]

[TABLE]

or

[TABLE] 4. 4.

$\dim\phi_{\mathfrak{p}}({\mathfrak{p}})=1$ then $\ker\phi_{\mathfrak{p}}$ is a two dimensional ideal of ${\mathfrak{p}}$. Then there exists an orthonormal basis $(a_{1},a_{2},a_{3})$ of ${\mathfrak{p}}$ such that

[TABLE]

The Jacobi identity gives $\alpha=0$ or $(p,r)=(0,0)$. We take $\phi_{\mathfrak{p}}(a_{1})=\phi_{\mathfrak{p}}(a_{2})=0$ and $\phi_{\mathfrak{p}}(a_{3})\in\mathrm{sl}(2,\hbox{\bb R})$.∎

Proposition 4.7**.**

Let $(({\mathfrak{h}},\omega,{\varrho}_{\mathfrak{h}}),({\mathfrak{p}},[\;,\;]_{\mathfrak{p}},{\varrho}_{\mathfrak{p}}),\mu,\phi_{\mathfrak{h}},\phi_{\mathfrak{p}})$ be a solution of Problem 1 with ${\mathfrak{h}}$ is 2-dimensional abelian, $\dim{\mathfrak{p}}=3$ and $\phi_{\mathfrak{h}}\not=0$. Then there exists an orthonormal basis $(e_{1},e_{2})$ of ${\mathfrak{h}}$, an orthonormal basis $(a_{1},a_{2},a_{3})$ of ${\mathfrak{p}}$, $\lambda>0$, $\alpha,p,q,\mu_{1},\mu_{2},\mu_{3}\in\hbox{\bb R}$ such that

[TABLE]

[TABLE]

and one of the following situations occurs:

1. $(i)$

$p\not=0$, $\alpha=0$ and

[TABLE] 2. $(ii)$

$p=0$, $\mu_{3}\not=0$, $\alpha=0$ and

[TABLE] 3. $(iii)$

$p=0$, $\mu_{3}=0$ and

[TABLE]

Proof.

Since $\phi_{\mathfrak{h}}\not=0$ then $\phi_{\mathfrak{h}}({\mathfrak{h}})$ is a non trivial abelian subalgebra of $\mathrm{so}({\mathfrak{p}})$ and hence it must be one dimensional. Then there exists an orthonormal basis $(e_{1},e_{2})$ of ${\mathfrak{h}}$ and an orthonormal basis $(a_{1},a_{2},a_{3})$ of ${\mathfrak{p}}$ and $\lambda>0$ such that $\phi_{\mathfrak{h}}(e_{1})=0$ and

[TABLE]

The first equation in (19) is equivalent to

[TABLE]

This is equivalent to

[TABLE]

Thus $\phi_{\mathfrak{p}}(a_{i})=\left(\begin{matrix}0&\mu_{i}\\ 0&0\end{matrix}\right)$ for $i=1,2$ and $\phi_{\mathfrak{p}}(a_{3})=\left(\begin{matrix}u&v\\ w&-u\end{matrix}\right)$. Consider now the second equation in (19)

[TABLE]

This equation is obviously true when $u=e_{1}$ and $(a,b)=(a_{1},a_{2})$. For $u=e_{1}$ and $(a,b)=(a_{1},a_{3})$, we get

[TABLE]

and hence $w=0$.

For $u=e_{2}$ and $(a,b)=(a_{1},a_{2})$, we get $\phi_{\mathfrak{h}}(e_{2})([a_{1},a_{2}]_{\mathfrak{p}})=0$ and hence $[a_{1},a_{2}]_{\mathfrak{p}}=\alpha a_{3}$.

For $u=e_{2}$ and $(a,b)=(a_{1},a_{3})$ or $(a,b)=(a_{2},a_{3})$ , we get

[TABLE]

This implies that $[a_{1},a_{3}]_{\mathfrak{p}},[a_{2},a_{3}]_{\mathfrak{p}}\in\mathrm{span}\{a_{1},a_{2}\}$ and hence

[TABLE]

So

[TABLE]

This is equivalent to

[TABLE]

To summarize, we get

[TABLE]

Let consider now the fourth equation in (19)

[TABLE]

This equation is obviously true for $u=e_{1}$.

For $u=e_{2}$ and $(a,b)=(a_{1},a_{2})$, $(a,b)=(a_{1},a_{3})$ or $(a,b)=(a_{2},a_{3})$, we get

[TABLE]

The last two equations are equivalent to

[TABLE]

$\bullet$ $p\not=0$ then

[TABLE]

$\bullet$ $p=0$ and $\mu_{3}\not=0$ then $\alpha=0$ and

[TABLE]

$\bullet$ $p=0$ and $\mu_{3}=0$ then

[TABLE]

∎

By using Propositions 4.2-4.7, we can give all the Riemann-Poisson Lie algebras of dimension 3, 4 or 5.

Let $(\mathfrak{g},[\;,\;],{\varrho},r)$ be a Riemann-Poisson Lie algebra of dimension less or equal to 5. According to what above then $\mathfrak{g}={\mathfrak{h}}\oplus{\mathfrak{p}}$ and the Lie bracket on $\mathfrak{g}$ is given by (18) and $(({\mathfrak{h}},\omega,{\varrho}_{\mathfrak{h}}),({\mathfrak{p}},[\;,\;]_{\mathfrak{p}},{\varrho}_{\mathfrak{p}}),\mu,\phi_{\mathfrak{h}},\phi_{\mathfrak{p}})$ are solutions of Problem 1.

$\bullet$ $\dim\mathfrak{g}=3$. In this case $\dim{\mathfrak{h}}=2$ and $\dim{\mathfrak{p}}=1$ and, by applying Proposition 4.2, the Lie bracket of $\mathfrak{g}$, ${\varrho}$ and $r$ are given in Table 1, where $e^{12}=e_{1}\wedge e_{2}$.

[TABLE]

$\bullet$ $\dim\mathfrak{g}=4$. We have three cases:

1. $(c41)$

$\dim{\mathfrak{h}}=2$, $\dim{\mathfrak{p}}=2$ and ${\mathfrak{h}}$ is non abelian and we can apply Proposition 4.3 to get the Lie brackets on $\mathfrak{g}$, ${\varrho}$ and $r$. They are described in rows 1 and 2 in Table 2. 2. $(c42)$

$\dim{\mathfrak{h}}=2$, $\dim{\mathfrak{p}}=2$ and ${\mathfrak{h}}$ is abelian and we can apply Propositions 4.4 and 4.5 to get the Lie brackets on $\mathfrak{g}$, ${\varrho}$ and $r$. They are described in rows 3 and 8 in Table 2. 3. $(c43)$

$\dim{\mathfrak{h}}=4$. In this case $\mathfrak{g}$ is a Kähler Lie algebra. We have used [11] to derive all four dimensional Kähler Lie algebra together with their symplectic derivations. The results are given in Table 3. The notation $\mathrm{Der}^{s}({\mathfrak{h}})$ stands for the vector spaces of derivations which are skew-symmetric with respect the symplectic form. The vector space $\mathrm{Der}^{s}({\mathfrak{h}})$ is described by a family of generators and $E_{ij}$ is the matrix with 1 in the $i$ row and $j$ column and 0 elsewhere.

[TABLE]

[TABLE]

$\bullet$ $\dim\mathfrak{g}=5$. We have:

1. $(c51)$

$\dim{\mathfrak{h}}=4$ and ${\mathfrak{h}}$ abelian and hence a symplectic vector space. We can apply Proposition 4.2 and $\mathfrak{g}$ is semi-direct product. 2. $(c52)$

$\dim{\mathfrak{h}}=4$ and ${\mathfrak{h}}$ non abelian. We can apply Proposition 4.2 and Table 3 to get the Lie brackets on $\mathfrak{g}$, ${\varrho}$ and $r$. The result is summarized in Table 4. 3. $(c53)$

$\dim{\mathfrak{h}}=2$ and ${\mathfrak{h}}$ non abelian. We apply Proposition 4.3. In this case $({\mathfrak{p}},[\;,\;]_{\mathfrak{p}},{\varrho}_{\mathfrak{p}})$ is a 3-dimensional Euclidean Lie algebra and one must compute $\mathrm{Der}({\mathfrak{p}})\cap\mathrm{so}({\mathfrak{p}})$ and solve (20). Three dimensional Euclidean Lie algebras were classified in [10]. For each of them we have computed $\mathrm{Der}({\mathfrak{p}})\cap\mathrm{so}({\mathfrak{p}})$ and solved (20) by using Maple. The result is summarized in Table 5 when ${\mathfrak{p}}$ is unimodular and Table 6 when ${\mathfrak{p}}$ is nonunimodular. 4. $(c54)$

$\dim{\mathfrak{h}}=2$ and ${\mathfrak{h}}$ abelian and $\phi_{\mathfrak{h}}=0$. We apply Proposition 4.6 and we perform all the needed computations. We use the classification of 3-dimensional Euclidean Lie algebras given in [10]. The results are given in Tables 7-8. 5. $(c55)$

$\dim{\mathfrak{h}}=2$ and ${\mathfrak{h}}$ abelian and $\phi_{\mathfrak{h}}\not=0$. We apply Proposition 4.7 and we perform all the needed computations. The results are given in Table 9.

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

This theorem unknown to our knowledge can be used to build examples of Riemann-Poisson Lie algebras.

Theorem 4.3**.**

Let $(G,\langle\;,\;\rangle)$ be an even dimensional flat Riemannian Lie group. Then there exists a left invariant differential $\Omega$ on $G$ such that $(G,\langle\;,\;\rangle,\Omega)$ is a Kähler Lie group.

Proof.

Let $\mathfrak{g}$ be the Lie algebra of $G$ and ${\varrho}=\langle\;,\;\rangle(e)$. According to Milnor’s Theorem [12, Theorem 1.5] and its improved version [1, Theorem 3.1] the flatness of the metric on $G$ is equivalent to $[\mathfrak{g},\mathfrak{g}]$ is even dimensional abelian, $[\mathfrak{g},\mathfrak{g}]^{\perp}=\{u\in\mathfrak{g},{\mathrm{ad}}_{u}+{\mathrm{ad}}_{u}^{*}=0\}$ is also even dimensional abelian and $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]\oplus[\mathfrak{g},\mathfrak{g}]^{\perp}$. Moreover, the Levi-Civita product is given by

[TABLE]

and there exists a basis $(e_{1},f_{1},\ldots,e_{r},f_{r})$ of $[\mathfrak{g},\mathfrak{g}]$ and $\lambda_{1},\ldots,\lambda_{r}\in[\mathfrak{g},\mathfrak{g}]^{\perp}\setminus\{0\}$ such that for any $a\in[\mathfrak{g},\mathfrak{g}]^{\perp}$,

[TABLE]

We consider a nondegenerate skew-symmetric 2-form $\omega_{0}$ on $[\mathfrak{g},\mathfrak{g}]^{\perp}$ and $\omega_{1}$ the nondegenerate skew-symmetric 2-form on $[\mathfrak{g},\mathfrak{g}]^{\perp}$ given by $\omega_{1}=\sum_{i=1}^{r}e_{i}^{*}\wedge f_{i}^{*}$. One can sees easily that $\omega=\omega_{0}\oplus\omega_{1}$ is a Kähler form on $\mathfrak{g}$. ∎