#  On the Existence and Properties of Left Invariant $k$-Symplectic Structures
on Lie Groups with Bi-Invariant Pseudo-Riemannian Metric

Ait Brik Ilham, Mohamed Boucetta  Université Hassan II
Faculté des Sciences Ain Chock
e-mail<EMAIL_ADDRESS>Université Cadi-Ayyad
Faculté des sciences et techniques
BP 549 Marrakech Maroc
e-mail<EMAIL_ADDRESS>
###### Abstract

$k$-symplectic manifolds are a convenient framework to study classical field
theories and they are a generalization of polarized symplectic manifolds. This
paper focus on the existence and the properties of left invariant
$k$-symplectic structures on Lie groups having a bi-invariant pseudo-
Riemannian metric. We show that compact semi-simple Lie groups and a large
class of Lie groups having a bi-invariant pseudo-Riemannian metric does not
carry any left invariant $k$-symplectic structure. This class contains the
oscillator Lie groups which are the only solvable non abelian Lie groups
having a bi-invariant Lorentzian metric. However, we built a natural left
invariant $n$-symplectic structure on $\mathrm{SL}(n,\hbox{\bb R})$. Moreover,
up to dimension 6, only three connected and simply connected Lie groups have a
bi-invariant indecomposable pseudo-Riemannian metric and a left invariant
k-symplectic structure, namely, the universal covering of
$\mathrm{SL}(2,\hbox{\bb R})$ with a 2-symplectic structure, the universal
covering of the Lorentz group $\mathrm{SO}(3,1)$ with a 2-symplectic
structure, and a 2-step nilpotent 6-dimensional connected and simply connected
Lie group with both a 1-symplectic structure and a 2-symplectic structure.

###### keywords:

$k$-symplectic structure, semi-simple Lie algebras , bi-invariant metric

###### MSC:

17B60 ,

###### MSC:

17B99

##  1 Introduction

Symplectic geometry is the natural arena to develop classical mechanics. The
$k$-symplectic geometry is a generalization of the polarized symplectic
geometry which was developed by A. Awane [3], Awane and Goze [4] and C.
Günther in [10] as an attempt to develop a convenient geometric framework to
study classical field theories (see [12]). This geometry was also an attempt
to formalize Nambu mechanic which is a generalization of Hamiltonian mechanics
involving multiple Hamiltonians.

A $k$-symplectic manifold is a smooth manifold $M$ of dimension $(k+1)n$
endowed with an involutive vector subbundle $E\subset TM$ and a family
$(\omega_{1},\ldots,\omega_{k})$ of closed differential 2-forms such that:
$\mathrm{rank}(E)=nk$, the family $(\omega_{1},\ldots,\omega_{k})$ is
nondegenerate, i.e., $\displaystyle\cap_{i=1}^{k}\ker\omega_{i}=\\{0\\}$ and
$E$ is isotropic with respect all the $\omega_{i}$. A 1-symplectic manifold is
a symplectic manifold endowed with a Lagrangian foliation. There are few known
examples of $k$-symplectic manifolds and one way to build new one is to
consider the class of left invariant $k$-symplectic structures on a connected
Lie group $G$ and their quotients by lattices. In this case, a left invariant
$k$-symplectic structure on a connected Lie group $G$ of dimension $(k+1)n$ is
equivalent to the data of a Lie subalgebra ${\mathfrak{h}}$ of dimension $nk$
of the Lie algebra $\mathfrak{g}$ of $G$ and a family
$\\{\theta_{i}\in\wedge^{2}\mathfrak{g}^{*},i=1,\ldots,k\\}$ of nondegenerate
2-cocycle of $\mathfrak{g}$ such that
$\theta_{i}({\mathfrak{h}},{\mathfrak{h}})=0$ for $i=1,\ldots,n$. We call
$(\mathfrak{g},{\mathfrak{h}},\theta_{1},\ldots,\theta_{k})$ a $k$-symplectic
Lie algebra. The determination of Lie groups having a left invariant
$k$-symplectic structure is an open problem and their study is at its debut
[2, 1].

In this paper, we study left invariant $k$-symplectic structures on Lie groups
having a bi-invariant pseudo-Riemannian metric. This class of Lie group is
very large and contains all semi-simple Lie groups. They became relevant some
years ago when they were useful in the formulation of some physical problems,
for instance in the so known Adler-Kostant-Symes scheme. More recently they
appeared in conformal field theory [8]. One can see [13], for a survey on the
Lie algebras of Lie groups having a bi-invariant metric.

In Section 2, we show that compact semi-simple Lie groups and a large class of
Lie groups having a bi-invariant pseudo-Riemannian metric does not carry any
left invariant $k$-symplectic structure. This class contains the oscillator
Lie groups which are the only solvable non abelian Lie groups having a bi-
invariant Lorentzian metric. We end the section, by building a natural left
invariant $n$-symplectic structure on $\mathrm{SL}(n,\hbox{\bb R})$. Moreover,
In Section 3, we show that, up to dimension 6, only three connected and simply
connected Lie groups have a bi-invariant indecomposable pseudo-Riemannian
metric and a left invariant k-symplectic structure, namely, the universal
covering of $\mathrm{SL}(2,\hbox{\bb R})$ with a 2-symplectic structure, the
universal covering of the Lorentz group $\mathrm{SO}(3,1)$ with a 2-symplectic
structure, and a 2-step nilpotent 6-dimensional connected and simply connected
Lie group with both a 1-symplectic structure and a 2-symplectic structure.

##  2 Left invariant $k$-symplectic structures on quadratic Lie groups

A $k$-symplectic manifold is a smooth manifold $M$ of dimension $(k+1)n$
endowed with a vector subbundle $E\subset TM$ and a family
$(\omega_{1},\ldots,\omega_{k})$ of differential 2-forms such that:

  1. $(i)$

$E$ is of rank $nk$ and is involutive, i.e, for any $X,Y\in\Gamma(E)$,
$[X,Y]\in\Gamma(E)$ hence $E$ defines a foliation on $M$,

  2. $(ii)$

for each $i\in\\{1,\ldots,k\\}$, $d\omega_{i}=0$ and the family
$(\omega_{1},\ldots,\omega_{k})$ is nondegenerate, i.e.,
$\displaystyle\cap_{i=1}^{k}\ker\omega_{i}=\\{0\\}$,

  3. $(iv)$

$E$ is isotropic with respect all the $\omega_{i}$, i.e., $\omega_{i}(E,E)=0$.

A 1-symplectic manifold is a symplectic manifold endowed with a Lagrangian
foliation.

A left invariant $k$-symplectic structure on a Lie group $G$ is a
$k$-symplectic structure $(E,\omega_{1},\ldots,\omega_{k})$ on $G$ such that
$E$ and the $\omega_{i}$ are invariant by all the left multiplications of $G$.
In particular, ${\mathfrak{h}}=E_{e}$ is a Lie subalgebra of the Lie algebra
$\mathfrak{g}$ of $G$ and $\omega_{1}(e),\ldots,\omega_{k}(e)$ are 2-cocycles
of $\mathfrak{g}$.

A $k$-symplectic Lie algebra is a real Lie algebra $\mathfrak{g}$ of dimension
$nk+n$ with a subalgebra ${\mathfrak{h}}$ of dimension $nk$ and a family
$(\theta_{1},\ldots,\theta_{k})$ of 2-forms satisfying:

  1. $(i)$

The family $(\theta_{1},\ldots,\theta_{k})$ is nondegenerate, i.e.,
$\bigcap_{i=1}^{k}\ker\theta_{i}=\\{0\\}$,

  2. $(ii)$

for $i=1,\ldots,k$, $\theta_{i}$ is a 2-cocycle, i.e.,
$\mathrm{d}\theta_{i}(u,v,w):=\theta_{i}([u,v],w)+\theta_{i}([v,w],u)+\theta_{i}([w,u],v)=0$,

  3. $(iii)$

${\mathfrak{h}}$ is totally isotropic with respect to
$(\theta_{1},\ldots,\theta_{k})$, i.e., $\theta_{i}(u,v)=0$ for any
$u,v\in{\mathfrak{h}}$ and for any $i=1,\ldots,k$.

There is a correspondence between left invariant $k$-symplectic structures on
a Lie group and $k$-symplectic structures on its Lie algebra.

Note that a symplectic Lie algebra is a Lie algebra endowed with a
nondegenerate 2-cocycle and a 1-symplectic Lie algebra is a symplectic Lie
algebra $(\mathfrak{g},\theta)$ having a Lagrangian subalgebra
${\mathfrak{h}}$, i.e., $\dim{\mathfrak{h}}=\frac{1}{2}\dim\mathfrak{g}$ and
$\theta({\mathfrak{h}},{\mathfrak{h}})=0$.

######  Example 1.

  1. 1.

Let $\mathfrak{g}$ be an abelian Lie algebra of dimension $n(k+1)$,
$(e_{pi},e_{i})_{1\leq p\leq k,1\leq i\leq n}$ a basis of $\mathfrak{g}$ and
$(\omega^{pi},\omega^{i})_{1\leq p\leq k,1\leq i\leq n}$ its dual basis. For
any $\alpha\in\\{1,\ldots,k\\}$, put

$\theta_{\alpha}=\sum_{i=1}^{n}\omega^{\alpha
i}\wedge\omega^{i}\quad\mbox{and}\quad{\mathfrak{h}}=\ker\omega^{1}\cap\ldots\cap\ker\omega^{n}.$

Then $({\mathfrak{h}},\theta_{1},\ldots,\theta_{k})$ is a $k$-symplectic
structure of $\mathfrak{g}$. Thus any abelian Lie algebra of dimension
$n(k+1)$ has a $k$-symplectic structure.

  2. 2.

Let $\mathfrak{g}=\mathrm{span}\\{e_{1},e_{2},f_{1},f_{2},f_{3},f_{4}\\}$ be
the 6-dimensional Lie algebra where the non vanishing Lie brackets are given
by

$\begin{cases}[f_{1},f_{2}]=-f_{1},\ [f_{1},f_{4}]=-af_{1},\
[f_{2},f_{3}]=f_{3},\ [f_{3},f_{4}]=-af_{3},\\\ [f_{2},e_{1}]=-e_{1},\
[f_{2},e_{2}]=-c(af_{2}-f_{4}),\ [f_{4},e_{1}]=-ae_{1},\
[f_{4},e_{2}]=-b(af_{2}-f_{4}),\quad a,b,c\in\hbox{\bb R}\end{cases}$

The triple

${\mathfrak{h}}=\mathrm{span}\\{f_{1},f_{2},f_{3},f_{4}\\},\;\theta_{1}=f_{1}^{*}\wedge
e_{1}^{*}+f_{2}^{*}\wedge
e_{2}^{*}\quad\mbox{and}\quad\theta_{2}=f_{3}^{*}\wedge
e_{1}^{*}+f_{4}^{*}\wedge e_{2}^{*}$

defines a 2-symplectic structure on $\mathfrak{g}$.

Let $G$ be a Lie group and $(\mathfrak{g}=T_{e}G,[\;,\;])$ its Lie algebra. A
pseudo-Riemannian metric $g$ on $G$ is called bi-invariant if its invariant by
the left and the right multiplications of $G$. This is equivalent to
$\langle\;,\;\rangle=h(e)$ satisfying

$\langle[u,v],w\rangle+\langle[u,w],v\rangle=0,$ (1)

for any $u,v,w\in\mathfrak{g}$. This means that, for any $u\in\mathfrak{g}$,
${\mathrm{ad}}_{u}$ is skew-symmetric with respect to $\langle\;,\;\rangle$
where ${\mathrm{ad}}_{u}(v)=[u,v]$.

A Lie group endowed with a bi-invariant metric will be called quadratic and a
real Lie algebra endowed with a bilinear symmetric nondegenerate form
satisfying (1) will be called a quadratic Lie algebra. A metric
$\langle\;,\;\rangle$ satisfying (1) will be called quadratic or invariant.
There is a correspondence between connected quadratic Lie groups and quadratic
Lie algebras so we will work at the level of Lie algebras.

There are three important classes of quadratic Lie algebra: semi-simple Lie
algebras endowed with their Killing form, $T^{*}$-extensions of Lie algebras
and quadratic Lie algebras obtained by the process of double extension (see
[13] for more details).

  1. $(a)$

Recall that a semi-simple Lie algebra is a Lie algebra $\mathfrak{g}$ such
that its Killing form given by
$k(u,v)={\mathrm{tr}}({\mathrm{ad}}_{u}\circ{\mathrm{ad}}_{v})$ is
nondegenerate. Since $k$ is invariant, $(\mathfrak{g},k)$ is a quadratic Lie
algebra.

  2. $(b)$

The $T^{*}$-extension of a Lie algebra $\mathfrak{g}$ is the Lie algebra
$T^{*}\mathfrak{g}=\mathfrak{g}\oplus\mathfrak{g}^{*}$ where the Lie bracket
is given by

$[u+\alpha,v+\beta]=[u,v]+{\mathrm{ad}}_{u}^{*}\beta-{\mathrm{ad}}_{v}^{*}\alpha,\quad
u,v\in\mathfrak{g},\alpha,\beta\in\mathfrak{g}^{*},$ (2)

where ${\mathrm{ad}}_{u}^{*}\alpha(v)=-\alpha([u,v])$. The bilinear symmetric
form $\langle\;,\;\rangle$ on $T^{*}\mathfrak{g}$ given by

$\langle u+\alpha,v+\beta\rangle=\alpha(v)+\beta(u)$ (3)

is nondegenerate and invariant and hence
$(T^{*}\mathfrak{g},\langle\;,\;\rangle)$ is a quadratic Lie algebra.

  3. $(c)$

Let $(\mathfrak{g},[\;,\;]_{\mathfrak{g}},\langle\;,\;\rangle_{\mathfrak{g}})$
be a quadratic Lie algebra. The double extension of $\mathfrak{g}$ by the mean
of $A\in\mathrm{so}(\mathfrak{g},\langle\;,\;\rangle_{\mathfrak{g}})$ is the
Lie algebra $\delta_{A}(\mathfrak{g})=\hbox{\bb
R}e\oplus\mathfrak{g}\oplus\hbox{\bb R}\bar{e}$ where the non vanishing Lie
brackets and the metric are given, for any $u,v\in\mathfrak{g}$, by

$[\bar{e},u]=Au,\;[u,v]=\langle
Au,v\rangle_{\mathfrak{g}}e+[u,v]_{\mathfrak{g}},\;\langle
xe+u+\bar{x}\bar{e},xe+u+\bar{x}\bar{e}\rangle=2x\bar{x}+\langle
u,u\rangle_{\mathfrak{g}}.$

$(\delta_{A}(\mathfrak{g}),[\;,\;],\langle\;,\;\rangle)$ is a quadratic Lie
algebra.

Let us start by a well-known result.

######  Proposition 2.1.

Let $(\mathfrak{g},[\;,\;],\langle\;,\;\rangle)$ be a quadratic Lie algebra.
If $\mathfrak{g}$ has a symplectic form $\theta$ then $\mathfrak{g}$ is
nilpotent.

###### Proof.

It a consequence of the fact that the endomorphism $D$ given by
$\theta(u,v)=\langle Du,v\rangle$ is an invertible derivation and, by virtue
of Jacobson’s Theorem [11], a Lie algebra with an invertible derivation must
be nilpotent. ∎

However, non nilpotent quadratic Lie algebras can carry a $k$-symplectic
structure with $k\geq 2$ as the following example shows.

######  Example 2.

We consider $\mathrm{sl}(2,\hbox{\bb R})$ with its basis
$\left\\{e_{1}=\left(\begin{matrix}0&1\\\
0&0\end{matrix}\right),e_{2}=\left(\begin{matrix}0&0\\\
1&0\end{matrix}\right),e_{3}=\left(\begin{matrix}1&0\\\
0&-1\end{matrix}\right)\right\\},$ where

$[e_{3},e_{1}]=2e_{1},\;[e_{3},e_{2}]=-2e_{2}\quad\mbox{and}\quad[e_{1},e_{2}]=e_{3}.$

Then ${\mathfrak{h}}_{0}=\mathrm{span}\\{h,g\\}$, $\theta_{1}=e_{3}^{*}\wedge
e_{2}^{*}+be_{1}^{*}\wedge e_{2}^{*}$ and $\theta_{2}=e_{1}^{*}\wedge
e_{2}^{*}$ is a 2-symplectic structure on $\mathrm{sl}(2,\hbox{\bb R})$.

Let $(\mathfrak{g},{\mathfrak{h}},\theta_{1},\ldots,\theta_{k})$ be a
$k$-symplectic Lie algebra of dimension $n(k+1)$. The linear map
$\Theta:{\mathfrak{h}}\longrightarrow(\mathfrak{g}/{\mathfrak{h}})^{*}\times\ldots\times(\mathfrak{g}/{\mathfrak{h}})^{*}$,
$h\mapsto(\Theta_{1}(h),\ldots,\Theta_{k}(h))$ where, for any
$p\in\mathfrak{g}$,

$\Theta_{\alpha}(h)([p])=\theta_{\alpha}(h,p)$

is well-defined, injective and for dimensional reasons it is an isomorphism.
For any $\alpha\in\\{1,\ldots,k\\}$, the vector subspace
${\mathfrak{h}}^{\alpha}$ of ${\mathfrak{h}}$ given by

${\mathfrak{h}}^{\alpha}=\\{h\in{\mathfrak{h}},\Theta_{\beta}(h)=0,\beta=1,\ldots,k,\beta\not=\alpha\\}.$

has dimension $n$ and
$\displaystyle{\mathfrak{h}}=\oplus_{\alpha=1}^{k}{\mathfrak{h}}^{\alpha}$.

Suppose now that $\mathfrak{g}$ carries an invariant metric
$\langle\;,\;\rangle$. Then, for any $\alpha=1,\ldots,k$, there exists a skew-
symmetric endomorphism $D_{\alpha}:\mathfrak{g}\longrightarrow\mathfrak{g}$
such that, for any $u,v\in\mathfrak{g}$,

$\theta_{\alpha}(u,v)=\langle D_{\alpha}u,v\rangle.$

The fact that $\theta_{\alpha}$ is closed is equivalent to $D_{\alpha}$ is a
derivation of $(\mathfrak{g},[\;,\;])$ and the fact that
$(\theta_{1},\ldots,\theta_{k})$ is nondegenerate is equivalent to
$\bigcap_{\alpha}\ker D_{\alpha}=\\{0\\}$. Moreover,
$\theta_{\alpha}({\mathfrak{h}},{\mathfrak{h}})=0$ is equivalent to
$D_{\alpha}({\mathfrak{h}})\subset{\mathfrak{h}}^{\perp}$, where
${\mathfrak{h}}^{\perp}$ is the orthogonal of ${\mathfrak{h}}$ with respect to
$\langle\;,\;\rangle$.

In conclusion, a $k$-symplectic structure on a quadratic Lie algebra
$(\mathfrak{g},\langle\;,\;\rangle)$ of dimension $n(k+1)$ is given by a
subalgebra ${\mathfrak{h}}$ of dimension $nk$ and a family of skew-symmetric
derivations $(D_{1},\ldots,D_{k})$ such that $\bigcap_{\alpha}\ker
D_{\alpha}=\\{0\\}$ and
$D_{\alpha}({\mathfrak{h}})\subset{\mathfrak{h}}^{\perp}$ for any
$\alpha\in\\{1,\ldots,k\\}$.

Note that in a quadratic Lie algebra $(\mathfrak{g},\langle\;,\;\rangle)$,

$H^{2}(\mathfrak{g})\simeq\mathrm{Der}(\mathfrak{g})\cap\mathrm{so}(\mathfrak{g},\langle\;,\;\rangle)/\\{{\mathrm{ad}}_{u},u\in\mathfrak{g}\\},$
(4)

where $\mathrm{Der}(\mathfrak{g})$ is the Lie algebra of derivations of
$\mathfrak{g}$ and $\mathrm{so}(\mathfrak{g},\langle\;,\;\rangle)$ is the Lie
algebra of skew-symmetric endomorphisms.

######  Proposition 2.2.

Let $(\mathfrak{g},[\;,\;],\langle\;,\;\rangle)$ be a quadratic Lie algebra.
Then the following assertions hold.

  1. 1.

If $H^{2}(\mathfrak{g})=\\{0\\}$ and the center $Z(\mathfrak{g})\not=\\{0\\}$
then $G$ has no $k$-symplectic structure.

  2. 2.

If $\dim Z(\mathfrak{g})=1$ and $Z(\mathfrak{g})$ is nondegenerate then, for
any skew-symmetric derivation $D$, $D(Z(\mathfrak{g}))=0$. In particular,
$\mathfrak{g}$ has no $k$-symplectic structure.

###### Proof.

  1. 1.

If $H^{2}(\mathfrak{g})=\\{0\\}$ then any slew-symmetric derivation is inner
and hence trivial on the center.

  2. 2.

Put $Z(\mathfrak{g})=\hbox{\bb R}e_{0}$ and let $D$ be a skew-symmetric
derivation. Since the derivations preserve the center, then $D(e_{0})=\alpha
e_{0}$. If $\langle e_{0},e_{0}\rangle\not=0$ then $\alpha=0$.∎

######  Remark 1.

Note that a quadratic Lie algebra $(\mathfrak{g},\langle\;,\;\rangle)$
satisfies $[\mathfrak{g},\mathfrak{g}]^{\perp}=Z(\mathfrak{g})$ and hence a
solvable quadratic Lie algebra satisfies $Z(\mathfrak{g})\not=\\{0\\}$.

######  Proposition 2.3.

Let $(\mathfrak{g},[\;,\;],\langle\;,\;\rangle)$ be a quadratic nilpotent Lie
algebra of dimension $n$. If $\mathfrak{g}$ admits a $n-1$-symplectic
structure then $\mathfrak{g}$ is abelian.

###### Proof.

Suppose that $\mathfrak{g}$ carries a $n-1$-symplectic structure. Then there
exists a $n-1$-dimensional Lie subalgebra ${\mathfrak{h}}$, a family
$D_{1},\ldots,D_{n-1}$ of skew-symmetric derivations such that
$D_{i}({\mathfrak{h}})\subset{\mathfrak{h}}^{\perp}$ and $\bigcap_{i}\ker
D_{i}=\\{0\\}$. Since ${\mathfrak{h}}$ is a subalgebra and
$\langle\;,\;\rangle$ is quadratic then
$[{\mathfrak{h}},{\mathfrak{h}}^{\perp}]\subset{\mathfrak{h}}^{\perp}$. But
$\dim{\mathfrak{h}}^{\perp}=1$ and ${\mathrm{ad}}_{u}$ is nilpotent for any
$u$ and hence $[{\mathfrak{h}},{\mathfrak{h}}^{\perp}]=\\{0\\}$. Now the facts
that $D_{i}({\mathfrak{h}})\subset{\mathfrak{h}}^{\perp}$,
$[{\mathfrak{h}},{\mathfrak{h}}^{\perp}]=\\{0\\}$ and $D_{i}$ a derivation
implies that $D_{i}([{\mathfrak{h}},{\mathfrak{h}}])=\\{0\\}$ for
$i=1,\ldots,n-1$ and hence $[{\mathfrak{h}},{\mathfrak{h}}]=\\{0\\}$. Put
$\mathfrak{g}={\mathfrak{h}}\oplus\hbox{\bb R}e$. We have
$[\mathfrak{g},\mathfrak{g}]=[{\mathfrak{h}},{\mathfrak{h}}]+[{\mathfrak{h}},e]=[{\mathfrak{h}},e]$
and for any $x,y\in{\mathfrak{h}}$,

$\langle[e,x],y\rangle=\langle e,[x,y]\rangle=0.$

So $[\mathfrak{g},\mathfrak{g}]\subset{\mathfrak{h}}^{\perp}$ and hence
${\mathfrak{h}}\subset[\mathfrak{g},\mathfrak{g}]^{\perp}=Z(\mathfrak{g})$.
This implies that $\mathfrak{g}$ is abelian which completes the proof. ∎

Let us state our first main result.

######  Theorem 2.1.

Let $(\mathfrak{g},\langle\;,\;\rangle)$ be a quadratic Lie algebra and
$({\mathfrak{h}},\theta_{1},\ldots,\theta_{k})$ a $k$-symplectic structure on
$\mathfrak{g}$. If ${\mathfrak{h}}$ is nondegenerate with respect to
$\langle\;,\;\rangle$ then $([\theta_{1}],\ldots,[\theta_{k}])$ are linearly
independent in $H^{2}(\mathfrak{g})$. In particular, if $k>\dim
H^{2}(\mathfrak{g})$ then ${\mathfrak{h}}$ is degenerate.

###### Proof.

For $\alpha=1,\ldots,k$,

$\theta_{\alpha}(u,v)=\langle D_{\alpha}u,v\rangle$

where $D_{\alpha}$ is a skew-symmetric derivation of $\mathfrak{g}$. Put
${\mathfrak{h}}^{\alpha}=\bigcap_{\beta\not=\alpha}(\ker
D_{\alpha}\cap{\mathfrak{h}})$. Then

${\mathfrak{h}}={\mathfrak{h}}^{1}\oplus\ldots\oplus{\mathfrak{h}}^{k}$

and
$(D_{\alpha})_{|{\mathfrak{h}}^{\alpha}}:{\mathfrak{h}}^{\alpha}\longrightarrow{\mathfrak{h}}^{\perp}$
is an isomorphism. Suppose that ${\mathfrak{h}}$ is nondegenerate and
$([\theta_{1}],\ldots,[\theta_{k}])$ are linearly dependent in
$H^{2}(\mathfrak{g})$. Then there exists
$(a_{1},\ldots,a_{k})\not=(0,\ldots,0)$ and $x\in\mathfrak{g}$ such that

${\mathrm{ad}}_{x}=a_{1}D_{1}+\ldots+a_{k}D_{k}.$

Suppose that $a_{1}\not=0$ and put $x=z+t$ where $z\in{\mathfrak{h}}$ and
$t\in{\mathfrak{h}}^{\perp}$. Since ${\mathfrak{h}}$ is a subalgebra and the
metric is invariant then, for any $h\in{\mathfrak{h}}^{1}$,
$[h,{\mathfrak{h}}^{\perp}]\subset{\mathfrak{h}}^{\perp}$ and hence

$[x,h]=a_{1}D_{1}(h)=[z,h]+[t,h]=[t,h].$

Then $t\not=0$ and
$(D_{1})_{|{\mathfrak{h}}^{1}}=\frac{1}{a_{1}}({\mathrm{ad}}_{t})_{|{\mathfrak{h}}^{1}}$.
But $\langle[t,h],t\rangle=0$ so
$(D_{1})_{|{\mathfrak{h}}^{1}}({\mathfrak{h}}^{1})\subset
t^{\perp}\cap{\mathfrak{h}}^{\perp}$. But this is impossible since

$\dim(t^{\perp}\cap{\mathfrak{h}}^{\perp})=\dim\mathfrak{g}-1-\dim{\mathfrak{h}}=\dim{\mathfrak{h}}^{\perp}-1.$

This completes the proof. ∎

This theorem has two important corollaries.

######  Corollary 2.1.

Let $(\mathfrak{g},\langle\;,\;\rangle)$ be a quadratic Lie Lie algebra with
$H^{2}(\mathfrak{g})=\\{0\\}$ and
$({\mathfrak{h}},\theta_{1},\ldots,\theta_{k})$ a $k$-symplectic structure on
$\mathfrak{g}$. Then ${\mathfrak{h}}$ is degenerate with respect to
$\langle\;,\;\rangle$.

Recall that a the connected and simply connected Lie group associated to a
semi-simple Lie algebra is compact if and only if the Killing form is definite
negative.

######  Corollary 2.2.

Let $\mathfrak{g}$ be a semi-simple Lie algebra and
$({\mathfrak{h}},\theta_{1},\ldots,\theta_{k})$ a $k$-symplectic structure on
$\mathfrak{g}$. Then ${\mathfrak{h}}$ is degenerate with respect to the
Killing form of $\mathfrak{g}$. In particular, a compact semi-simple Lie group
has no left invariant $k$-symplectic structure.

The following theorem shows that, in addition to the class of semi-simple Lie
algebras, there is a large class of quadratic Lie algebras satisfying
$H^{2}(\mathfrak{g})=\\{0\\}$.

######  Theorem 2.2.

If $\mathfrak{g}$ is a simple Lie algebra then $H^{2}(T^{*}\mathfrak{g})=0$.

###### Proof.

Recall that the quadratic structure of
$T^{*}\mathfrak{g}=\mathfrak{g}\oplus\mathfrak{g}^{*}$ is given by (2) and
(3). For $\omega\in\wedge^{2}\mathfrak{g}^{*}$ and
$\pi\in\wedge^{2}\mathfrak{g}$ we denote
$\omega^{\flat}:\mathfrak{g}\longrightarrow\mathfrak{g}^{*}$ and
$\pi_{\\#}:\mathfrak{g}^{*}\longrightarrow\mathfrak{g}$ given by
$\prec\omega(x),y\succ=\omega(x,y)$ and
$\prec\beta,\pi_{\\#}(\alpha)\succ=\pi(\alpha,\beta)$. According to (4), we
must show that any skew-symmetric derivation of $T^{*}\mathfrak{g}$ is inner.
Note first that for any $x+\alpha\in T^{*}\mathfrak{g}$, we have

${\mathrm{ad}}_{x+\alpha}=\left(\begin{matrix}{\mathrm{ad}}_{x}&0\\\
-(d\alpha)^{\flat}&{\mathrm{ad}}_{x}^{*}\end{matrix}\right),$

where $d\alpha(u,v)=-\alpha([u,v])$ and, for any skew-symmetric derivation
$D$, there exists $\omega\in\wedge^{2}\mathfrak{g}^{*}$,
$\pi\in\wedge^{2}\mathfrak{g}$ and
$D_{1}:\mathfrak{g}\longrightarrow\mathfrak{g}$ such that

$D=\left(\begin{matrix}D_{1}&\pi_{\\#}\\\
\omega^{\flat}&-D_{1}^{*}\end{matrix}\right),$

where $D_{1}^{*}$ is the dual of $D_{1}$. One can see easily that $D$ is a
derivation if and only if $D_{1}$ is a derivation of $\mathfrak{g}$, $\omega$
is a 2-cocycle of $\mathfrak{g}$ and $\pi$ is ${\mathrm{ad}}^{*}$-invariant,
i.e., for any $x\in\mathfrak{g}$ and $\alpha,\beta\in\mathfrak{g}^{*}$,

$\pi({\mathrm{ad}}_{x}^{*}\alpha,\beta)+\pi(\alpha,{\mathrm{ad}}_{x}^{*}\beta)=0.$

Since $\mathfrak{g}$ is simple then $\omega=d\alpha$ and
$D_{1}={\mathrm{ad}}_{x}$. On the other hand, let
$A=\pi_{\\#}\circ\kappa^{\flat}$. Since both $\pi$ and $\kappa$ are
${\mathrm{ad}}$-invariant then $[A,{\mathrm{ad}}_{x}]=0$ for any
$x\in\mathfrak{g}$. The complexification $A^{\mathbb{C}}$ of $A$ is an
endomorphism of $\mathfrak{g}^{\mathbb{C}}$ which commutes with
${\mathrm{ad}}_{x}$ for any $x\in\mathfrak{g}^{\mathbb{C}}$ and any eigenspace
of $A^{\mathbb{C}}$ is an ideal which implies that
$A=\lambda\mathrm{Id}_{\mathfrak{g}}$ and since $A$ is skew-symmetric then
$\lambda=0$ which completes the proof. ∎

######  Theorem 2.3.

Let $\delta_{A}(\mathfrak{g})=\hbox{\bb R}e\oplus\mathfrak{g}\oplus\bar{e}$ be
the double extension of an abelian quadratic Lie algebra
$(\mathfrak{g},\langle\;,\;\rangle_{\mathfrak{g}})$ with $A$ non nilpotent and
let $({\mathfrak{h}},\langle\;,\;\rangle_{\mathfrak{h}})$ be an abelian
quadratic Lie algebra. Then, for any skew-symmetric derivation $D$ of
$\delta_{A}(\mathfrak{g})\oplus{\mathfrak{h}}$ (product of the quadratic Lie
algebras $\delta_{A}(\mathfrak{g})$ and ${\mathfrak{h}}$), $D(e)=0$. In
particular, $\delta_{A}(\mathfrak{g})\oplus{\mathfrak{h}}$ has no
$k$-symplectic structure.

###### Proof.

Let $D$ be a skew-symmetric derivation of
$\delta_{A}(\mathfrak{g})\oplus{\mathfrak{h}}$. We have
$Z(\delta_{A}(\mathfrak{g})\oplus{\mathfrak{h}})=\hbox{\bb R}e\oplus\ker
A\oplus{\mathfrak{h}}$. Then $D(e)=\alpha e+z$ where $z\in\ker
A\oplus{\mathfrak{h}}$. For any $u\in\mathfrak{g}$,

$Du=\langle Du,\bar{e}\rangle e+D_{1}u+\langle
Du,e\rangle\bar{e}+D_{2}u=\langle Du,\bar{e}\rangle e+D_{1}u+D_{2}u,$

where $D_{1}u\in\mathfrak{g}$ and $D_{2}u\in{\mathfrak{h}}$. Note that
$D_{1}:\mathfrak{g}\longrightarrow\mathfrak{g}$ is skew-symmetric with respect
to $\langle\;,\;\rangle_{\mathfrak{g}}$. We have

$\displaystyle D[\bar{e},u]$ $\displaystyle=DAu=\langle DAu,\bar{e}\rangle
e+D_{1}Au+D_{2}Au$ $\displaystyle=[D\bar{e},u]+[\bar{e},Du]$
$\displaystyle=\langle D\bar{e},e\rangle[\bar{e},u]+[D_{1}\bar{e},u]+AD_{1}u$
$\displaystyle=-\alpha Au+\langle
D_{1}\bar{e},u\rangle_{\mathfrak{g}}e+AD_{1}u.$

So $\alpha A=[A,D_{1}].$ On the other hand, for any $u,v\in\mathfrak{g}$,

$\displaystyle D[u,v]$ $\displaystyle=\langle Au,v\rangle_{\mathfrak{g}}D(e)$
$\displaystyle=[Du,v]+[u,Dv]$ $\displaystyle=\left(\langle
AD_{1}u,v\rangle_{\mathfrak{g}}+\langle
u,AD_{1}v\rangle_{\mathfrak{g}}\right)e$
$\displaystyle=\langle[A,D_{1}]u,v\rangle e$ $\displaystyle=\alpha\langle
Au,v\rangle e.$

So $D(e)=\alpha e$. If $A$ is not nilpotent, then there exists $n\geq 2$ such
that ${\mathrm{tr}}(A^{n})\not=0$. But from the relation $\alpha A=[A,D_{1}]$
we can deduce that $\alpha{\mathrm{tr}}(A^{n})=0$ and hence $D(e)=0$ which
completes the proof. ∎

We apply Theorem 2.3 to an important class of quadratic Lie algebras, namely,
the oscillator Lie algebras.

The oscillator group named so by Streater in [14], is a four-dimensional
connected, simply connected Lie group, whose Lie algebra (known as the
oscillator algebra) coincides with the one generated by the differential
operators, acting on functions of one variable, associated to the harmonic
oscillator problem. The oscillator group has been generalized to any even
dimension $2n\geq 4$, there are the only solvable Lie groups which carry an
invariant Lorentzian metric.

For $n\in\hbox{\bb N}^{*}$ and
$\lambda=(\lambda_{1},\ldots,\lambda_{n})\in\hbox{\bb R}^{n}$ with
$0<\lambda_{1}\leq\cdots\leq\lambda_{n}$, the $\lambda$-oscillator group,
denoted by $\mathrm{0sc}_{\lambda}$, is the Lie group which the underlying
manifold $\hbox{\bb R}^{2n+2}=\hbox{\bb R}\times\hbox{\bb
R}\times\mathbb{C}^{n}$ and product

$(t,s,z).(t^{\prime},s^{\prime},z^{\prime})=\left(t+t^{\prime},s+s^{\prime}+\frac{1}{2}\sum_{j=1}^{n}Im\bar{z}_{j}exp(it\lambda_{j})z_{j}^{\prime},\ldots,z_{j}+exp(it\lambda_{j})z^{\prime}_{j},\ldots\right).$

The Lie algebra of $\mathrm{0sc}_{\lambda}$, denoted by
$\mathrm{osc}(\lambda)$, admits a basis $\hbox{\bb
B}=\left\\{e_{-1},e_{0},e_{i},\check{e}_{i}\right\\}_{i=1,\ldots,n}$ where the
brackets are given by

$[e_{-1},e_{i}]=\lambda_{i}\check{e}_{i},\qquad[e_{-1},\check{e}_{i}]=-\lambda_{i}e_{i},\qquad[e_{i},\check{e}_{i}]=e_{0},$

the unspecified brackets are either zero or given by antisymmetry. We note
that the center of $\mathrm{osc}(\lambda)$ is
${Z}(\mathrm{osc}(\lambda))=\hbox{\bb R}e_{0}$ and
$[\mathrm{osc}(\lambda),\mathrm{osc}(\lambda)]=\mathrm{span}_{\hbox{\bb
R}}\left\\{e_{0},e_{i},\check{e}_{i}\right\\}_{i=1,\ldots,n}.$

For $x=x_{-1}e_{-1}+x_{0}e_{0}+\sum_{i=1}^{n}(x_{i}e_{i}+y_{i}\check{e}_{i})$,
the quadratic metric on $\mathrm{osc}(\lambda)$ is given by

$\langle
x,x\rangle=2x_{-1}x_{0}+\sum_{i=1}^{n}\frac{1}{\lambda_{i}}(x_{i}^{2}+y_{i}^{2}).$

Hence $\mathrm{osc}(\lambda)$ is the double extension of $\hbox{\bb R}^{2{n}}$
by mean of $A=\mathrm{Diag}(A_{\alpha_{1}},\ldots,A_{\alpha_{n}})$ where
$A_{\lambda_{i}}=\left(\begin{matrix}0&-\lambda_{i}\\\
\lambda_{i}&0\end{matrix}\right).$ So as a corollary of Theorem 2.3 we get the
following result.

######  Corollary 2.3.

Let $({\mathfrak{h}},\langle\;,\;\rangle_{\mathfrak{h}})$ be an abelian
quadratic Lie algebra and $\mathrm{osc}(\lambda)\times{\mathfrak{h}}$ endowed
with the quadratic metric product. If $D$ is a skew-symmetric derivation of
$\mathrm{osc}(\lambda)\times{\mathfrak{h}}$ then $D(e_{0})=0$. In particular,
there is no $k$-symplectic structure on
$\mathrm{osc}(\lambda)\times{\mathfrak{h}}$.

We end this section, by building a $n$-symplectic structure on
$\mathrm{sl}(n,\hbox{\bb R})$.

For any $i,j$, we denote by $E_{i,j}$ the $n$-matrix with 1 in the $i$-row and
the $j$-column and $0$ elsewhere and we denote by $(E_{i,j}^{*})$ the dual
basis of the basis $(E_{i,j})$ of $\mathrm{gl}(n,\hbox{\bb R})$. We define
$({\mathfrak{h}},\theta_{1},\ldots,\theta_{n})$ by

${\mathfrak{h}}=\left\\{A\in\mathrm{sl}(n,\mathbb{R}),\exists\lambda\in\hbox{\bb
R}\;\mbox{such that}\;A(e_{n})=\lambda
e_{n}\right\\}\quad\mbox{and}\quad\theta_{\alpha}=(dE_{\alpha,n}^{*})_{|\mathrm{sl}(n,\hbox{\bb
R})}.$

######  Theorem 2.4.

$(\mathrm{sl}(n,\hbox{\bb R}),{\mathfrak{h}},\theta_{1},\ldots,\theta_{n})$ is
a $n$-symplectic Lie algebra.

###### Proof.

We denote by $(e_{1},\ldots,e_{n})$ the canonical basis of $\hbox{\bb R}^{n}$
and $(e_{1}^{*},\ldots,e_{n}^{*})$ its dual basis. We have
$\dim\mathrm{sl}(n,\hbox{\bb R})=(n-1)(n+1)$, ${\mathfrak{h}}$ is obviously a
subalgebra of dimension $n(n-1)$ and the $\theta_{i}$ are obviously closed.
Moreover, for any $A,B\in{\mathfrak{h}}$, $[A,B](e_{n})=0$ and hence, for any
$\alpha=1,\ldots,n$,

$\theta_{i}(A,B)=-E_{\alpha,n}^{*}([A,B])=-\prec
e_{\alpha}^{*},[A,B](e_{n})\succ=0.$

Let $X\in\mathrm{sl}(n,\mathbb{R})$ such that $i_{X}\theta_{\alpha}=0$ for any
$\alpha=1,\ldots,n$. This equivalent to

$\prec e_{\alpha}^{*},[X,E_{i,j}](e_{n})\succ=0\quad
i,j=1,\ldots,n,\alpha=1,\ldots,n.$

We have

$\displaystyle\prec e_{\alpha}^{*},[X,E_{i,j}](e_{n})\succ$
$\displaystyle=\prec e_{\alpha}^{*},\delta_{jn}X(e_{i})-X_{jn}e_{i}\succ$
$\displaystyle=\delta_{jn}X_{\alpha i}-X_{jn}\delta_{\alpha i}.$

For $j=n$ and $\alpha\not=i$, we get $X_{\alpha i}=0$. Otherwise, we get

$X_{\alpha\alpha}-X_{nn}=0.$

So $X=X_{nn}I_{n}$. But ${\mathrm{tr}}(X)=0$ implies $X=0$. This completes the
proof.∎

##  3 k-symplectic structures on indecomposable quadratic Lie algebras of
dimension $\leq 6$

In this section, we enumerate all indecomposable quadratic Lie algebras of
dimension $\leq 6$ and for each of them we investigate if yes or not it admits
a $k$-symplectic structure and we give such structure when it exists.
Indecomposable quadratic Lie algebras of dimension $\leq 6$, up to an
automorphism, are given in [13] and [6] (see Tables 1-4).

### Dimension 3.

There are two quadratic non abelian Lie algebras in dimension 3,
$\mathfrak{su}(2)$ which has no $k$-symplectic structure by virtue of
Corollary 2.2 and $\mathrm{sl}(2,\hbox{\bb R})$ which has a 2-symplectic
structure given in Example 2 and Theorem 2.4.

### Dimension 4.

In dimension 4, there are two non abelian quadratic Lie algebras, the
oscillator Lie algebra $\mathrm{osc}(4,\lambda)$ and $\mathfrak{g}_{1,4}$.
Both are obtained by double extension from an abelian Lie algebra by mean of
an invertible endomorphism and hence both of them have no $k$-symplectic
structure by virtue of Theorem 2.3.

Note that , $\mathfrak{su}(2)\times\hbox{\bb R}$ and $\mathrm{sl}(2,\hbox{\bb
R})\times\hbox{\bb R}$ are the only decomposable non abelian four dimensional
quadratic Lie algebras and both of theme have no $k$-symplectic structure by
virtue of Proposition 2.2.

### Dimension 5.

In dimension 5, there is only one indecomposable non abelian quadratic Lie
algebra, namely, $\mathfrak{g}_{1,5}$. Since $\mathfrak{g}_{1,5}$ is
nilpotent, it has no 5-symplectic structure by virtue of Proposition 2.3.

Note that the 5-dimensional decomposable quadratic Lie algebras
$\mathrm{osc}(4,\lambda)\times\hbox{\bb R}$ and
$\mathfrak{g}_{1,4}\times\hbox{\bb R}$ have no $k$-symplectic structure by
virtue of Theorem 2.3.

### Dimension 6.

The list of indecomposable quadratic Lie algebras of dimension 6 is given in
Table 4. The Lie algebras $\mathrm{osc}(6,\lambda_{1},\lambda_{2})$,
$\mathfrak{l}_{2,\lambda}$ and $\mathfrak{n}_{k}(2,2)$ for $k=2,\ldots,6$ are
obtained by double extension from an abelian Lie algebra by mean of an
invertible endomorphism and hence all of them have no $k$-symplectic structure
by virtue of Theorem 2.3.

However, $\mathfrak{n}_{1}(2,2)$ is obtained by double extension by the mean
of a nilpotent endomorphism. It is 2-step nilpotent and doesn’t carry a
5-symplectic structure by virtue of Proposition 2.3. However, it carries a
2-symplectic structure. For instance if we take
${\mathfrak{h}}={\mathrm{span}}(e_{1},e_{2},e_{4},e_{3}+e_{6})$,
$({\mathfrak{h}},\theta_{1},\theta_{2})$ is a 2-symplectic structure on
$\mathfrak{n}_{1}(2,2)$ where

$\begin{cases}\theta_{1}=e_{1}^{*}\wedge e_{3}^{*}-e_{1}^{*}\wedge
e_{5}^{*}-e_{1}^{*}\wedge e_{6}^{*}+e_{3}^{*}\wedge e_{4}^{*}+e_{3}^{*}\wedge
e_{6}^{*}+e_{4}^{*}\wedge e_{6}^{*},\\\ \theta_{2}=e_{1}^{*}\wedge
e_{3}^{*}-e_{1}^{*}\wedge e_{5}^{*}-e_{1}^{*}\wedge e_{6}^{*}+2e_{2}^{*}\wedge
e_{5}^{*}-e_{3}^{*}\wedge e_{4}^{*}+e_{3}^{*}\wedge e_{6}^{*}-e_{4}^{*}\wedge
e_{6}^{*}.\end{cases}$

Moreover, $(Z(\mathfrak{n}_{1}(2,2)),\theta)$ is a 1-symplectic structure on
$\mathfrak{n}_{1}(2,2)$ where

$\theta=e_{1}^{*}\wedge e_{3}^{*}-e_{1}^{*}\wedge e_{6}^{*}+e_{2}^{*}\wedge
e_{5}^{*}+e_{2}^{*}\wedge e_{6}^{*}+e_{4}^{*}\wedge e_{5}^{*}.$

On the other hand, $\mathrm{so}(3,1)$ has no 1-symplectic structure by virtue
of Proposition 2.1. However,
$({\mathfrak{h}}=\mathrm{span}(e_{3},e_{4},e_{1}+e_{5},e_{2}+e_{6}),\theta_{1},\theta_{2})$
where

$\begin{cases}\theta_{1}=-e_{1}^{*}\wedge e_{2}^{*}-e_{1}^{*}\wedge
e_{4}^{*}+e_{2}^{*}\wedge e_{3}^{*}+e_{3}^{*}\wedge e_{6}^{*}-e_{4}^{*}\wedge
e_{5}^{*}+e_{5}^{*}\wedge e_{6}^{*},\\\ \theta_{2}=e_{1}^{*}\wedge
e_{3}^{*}-e_{1}^{*}\wedge e_{5}^{*}+e_{2}^{*}\wedge e_{4}^{*}-e_{2}^{*}\wedge
e_{6}^{*}+e_{3}^{*}\wedge e_{5}^{*}+e_{4}^{*}\wedge e_{6}^{*}\end{cases}$

is a 2-symplectic structure on $\mathrm{so}(3,1)$. However, $\mathrm{so}(3,1)$
does not possess any five-dimensional Lie subalgebras [9] and hence has no
5-symplectic structure.

Finally, we will show that $T^{*}\mathrm{sl}(2,\hbox{\bb R})$ and
$T^{*}\mathfrak{su}(2)$ have no $k$-symplectic structure. This needs some work
based on the following two lemmas. But first let’s recall the definitions of
$\mathfrak{su}(2)$ and $\mathrm{sl}(2,\hbox{\bb R})$ and set some notations.

  1. 1.

The Lie algebra $\mathfrak{su}(2)=\left\\{\left(\begin{matrix}iz&y+ix\\\
-y+xi&-zi\end{matrix}\right),x,y,z\in\hbox{\bb R}\right\\}.$ It has a basis
$\hbox{\bb B}_{0}=(e_{1},e_{2},e_{3})$

$e_{1}=\frac{1}{2}\left(\begin{matrix}0&i\\\
i&0\end{matrix}\right),\;e_{2}=\frac{1}{2}\left(\begin{matrix}0&1\\\
-1&0\end{matrix}\right)\quad\mbox{and}\quad
e_{3}=\frac{1}{2}\left(\begin{matrix}-i&0\\\ 0&i\end{matrix}\right)$

where

$[e_{1},e_{2}]=e_{3},\;[e_{2},e_{3}]=e_{1}\quad\mbox{and}\quad[e_{3},e_{1}]=e_{2}.$

The group of automorphisms of $\mathfrak{su}(2)$ is generated by the three
rotations

$\mathrm{Rot}_{xy}=\left(\begin{array}[]{ccc}\cos(a)&\sin(a)&0\\\
-\sin(a)&\cos(a)&0\\\
0&0&1\end{array}\right),\mathrm{Rot}_{xz}=\left(\begin{array}[]{ccc}\cos(a)&0&\sin(a)\\\
0&1&0\\\
-\sin(a)&0&\cos(a)\end{array}\right),\mathrm{Rot}_{yz}=\left(\begin{array}[]{ccc}1&0&0\\\
0&\cos(a)&\sin(a)\\\ 0&-\sin(a)&\cos(a)\end{array}\right).$

  2. 2.

The Lie algebra $\mathrm{sl}(2,\hbox{\bb R})$ has a basis
$\left\\{e_{1}=\left(\begin{matrix}0&1\\\
0&0\end{matrix}\right),e_{2}=\left(\begin{matrix}0&0\\\
1&0\end{matrix}\right),e_{3}=\left(\begin{matrix}1&0\\\
0&-1\end{matrix}\right)\right\\},$ where

$[e_{3},e_{1}]=2e_{1},\;[e_{3},e_{2}]=-2e_{2}\quad\mbox{and}\quad[e_{1},e_{2}]=e_{3}.$

In the basis $(X_{1},X_{2},X_{3})$ with $X_{1}=\frac{1}{2}(e_{1}+e_{2})$,
$X_{2}=\frac{1}{2}e_{3}$ and $X_{3}=\frac{1}{2}(e_{1}-e_{2})$ the Lie brackets
are given by

$[X_{1},X_{2}]=-X_{3},\;[X_{2},X_{3}]=X_{1}\quad\mbox{and}\quad[X_{3},X_{1}]=X_{2}.$

The group of automorphisms of $\mathrm{sl}(2,\hbox{\bb R})$ is generated by
the automorphisms with the matrix given in the basis $(X_{1},X_{2},X_{3})$

$\mathrm{Rot}_{xy}=\left(\begin{array}[]{ccc}\cos(a)&\sin(a)&0\\\
-\sin(a)&\cos(a)&0\\\
0&0&1\end{array}\right),\;\mathrm{Boost}_{xz}=\left(\begin{array}[]{ccc}\cosh(a)&0&\sinh(a)\\\
0&1&0\\\
\sinh(a)&0&\cosh(a)\end{array}\right),\mathrm{Boost}_{yz}=\left(\begin{array}[]{ccc}1&0&0\\\
0&\cosh(a)&\sinh(a)\\\ 0&\sinh(a)&\cosh(a)\end{array}\right).$

  3. 3.

If $\mathfrak{g}$ is a Lie algebra and $F$ is an automorphism of
$\mathfrak{g}$ we denote by $T^{*}F$ the automorphism of $T^{*}\mathfrak{g}$
given by $T^{*}F=(F,(F^{-1})^{*})$. This automorphism preserves the quadratic
metric of $T^{*}\mathfrak{g}$.

######  Lemma 3.1.

  1. 1.

Let $X$ be a non null vector in $\mathfrak{su}(2)$. Then there exists an
automorphism of $\mathfrak{su}(2)$ which maps $X$ to $\alpha e_{1}$.

  2. 2.

Let $X$ be a non null vector in $\mathrm{sl}(2,\hbox{\bb R})$ which is non
colinear to $e_{3}$. Then there exists an automorphism of
$\mathrm{sl}(2,\hbox{\bb R})$ which maps $X$ to $ae_{1}+be_{2}$ with
$(a,b)\not=(0,0)$.

###### Proof.

It is straightforward by using the automorphisms given above. ∎

######  Lemma 3.2.

There is no 5-dimensional subalgebra of $T^{*}\mathfrak{su}(2)$ and if
${\mathfrak{h}}$ is a 4-dimensional degenerate subalgebra of
$T^{*}\mathfrak{su}(2)$ then there exists an automorphism $F$ of
$\mathfrak{su}(2)$ such that $T^{*}F({\mathfrak{h}})=\hbox{\bb
R}e_{1}\oplus\mathfrak{su}(2)^{*}$.

###### Proof.

Let ${\mathfrak{h}}$ be a 5-dimensional Lie subalgebra of
$T^{*}\mathfrak{su}(2)$. Then $\dim{\mathfrak{h}}\cap\mathfrak{su}(2)\geq 2$.
If $\dim{\mathfrak{h}}\cap\mathfrak{su}(2)=2$ then
${\mathfrak{h}}\cap\mathfrak{su}(2)$ is a subalgebra of $\mathfrak{su}(2)$.
But $\mathfrak{su}(2)$ has no 2-dimensional Lie algebra (see [7, Proposition
5.2]). If $\dim{\mathfrak{h}}\cap\mathfrak{su}(2)=3$ then
$\mathfrak{su}(2)\subset{\mathfrak{h}}$ and
${\mathfrak{h}}^{\perp}\subset\mathfrak{su}(2)^{\perp}=\mathfrak{su}(2)$. But
$[{\mathfrak{h}},{\mathfrak{h}}^{\perp}]\subset{\mathfrak{h}}^{\perp}$ an
hence ${\mathfrak{h}}^{\perp}$ is an ideal in $\mathfrak{su}(2)$ which is
impossible.

Let ${\mathfrak{h}}$ be a degenerate 4-dimensional Lie subalgebra of
$T^{*}\mathfrak{su}(2)$. Then we have three possibilities
$\dim{\mathfrak{h}}\cap\mathfrak{su}(2)=1,2$ or 3.

$\bullet$ If $\dim{\mathfrak{h}}\cap\mathfrak{su}(2)=3$ then
${\mathfrak{h}}^{\perp}\subset\mathfrak{su}(2)$ and hence
${\mathfrak{h}}^{\perp}$ is an ideal of $\mathfrak{su}(2)$ which is
impossible.

$\bullet$ If $\dim{\mathfrak{h}}\cap\mathfrak{su}(2)=2$ then
${\mathfrak{h}}\cap\mathfrak{su}(2)$ is a 2-dimensional subalgebra of
$\mathfrak{su}(2)$. But $\mathfrak{su}(2)$ has no 2-dimensional Lie algebra
(see [7, Proposition 5.2])

$\bullet$ If $\dim{\mathfrak{h}}\cap\mathfrak{su}(2)=1$ then, according to
Lemma 3.1, we can transform ${\mathfrak{h}}$ by $T^{*}F$ where $F$ is an
automorphism of $\mathfrak{su}(2)$ and choose $e_{1}$ as a generator of
${\mathfrak{h}}\cap\mathfrak{su}(2)$. On the other hand,

$\dim({\mathfrak{h}}^{\perp}\cap\mathfrak{su}(2))=6-\dim({\mathfrak{h}}+\mathfrak{su}(2))=6-4-3+1=0.$

Thus ${\mathfrak{h}}^{\perp}\cap\mathfrak{su}(2)=\\{0\\}$. There exists
$\alpha_{1}\not=0,\alpha_{2}\not=0$ and $y_{1},y_{2}\in\mathfrak{su}(2)$ such
that

${\mathfrak{h}}^{\perp}=\mathrm{span}\left(y_{1}+\alpha_{1},y_{2}+\alpha_{2}\right).$

We must have $\\{\alpha_{1},\alpha_{2}\\}$ linearly independent otherwise, we
can find a non null element of ${\mathfrak{h}}^{\perp}$ in $\mathfrak{su}(2)$.
We have $\alpha_{1}(e_{1})=\alpha_{2}(e_{1})=0$. So
$\mathrm{span}(\alpha_{1},\alpha_{2})=\mathrm{span}(e_{2}^{*},e_{3}^{*})$ and
hence there exists $x_{1},x_{2}\in\mathfrak{su}(2)$ such that

${\mathfrak{h}}^{\perp}=\mathrm{span}\left(x_{1}+e_{2}^{*},x_{2}+e_{3}^{*}\right).$

Now the fact that ${\mathrm{ad}}_{e_{1}}$ leaves ${\mathfrak{h}}^{\perp}$
invariant implies that

$\begin{cases}[e_{1},x_{1}]+[e_{1},e_{2}^{*}]=[e_{1},x_{1}]+e_{3}^{*}=\mu_{1}(x_{1}+e_{2}^{*})+\nu_{1}(x_{2}+e_{3}^{*}),\\\
[e_{1},x_{2}]+[e_{1},e_{3}^{*}]=[e_{1},x_{2}]-e_{2}^{*}=\mu_{2}(x_{1}+e_{2}^{*})+\nu_{2}(x_{2}+e_{3}^{*}),\\\
\end{cases}$

So

$[e_{1},x_{1}]=x_{2}\quad\mbox{and}\quad[e_{1},x_{2}]=-x_{1}.$

This is equivalent to $x_{1}=ae_{2}-be_{3}$ and $x_{2}=be_{2}+ae_{3}$ and
hence

${\mathfrak{h}}^{\perp}=\\{ae_{2}-be_{3}+e_{2}^{*},be_{2}+ae_{3}+e_{3}^{*}\\}$

The restriction of the metric to ${\mathfrak{h}}^{\perp}$ is degenerate if and
only if $a=0$. So
${\mathfrak{h}}=\mathrm{span}(e_{1},e_{1}^{*},-be_{3}+e_{2}^{*},be_{2}+e_{3}^{*})$.
But $[e_{1}^{*},-be_{3}+e_{2}^{*}]=be_{2}^{*}\in{\mathfrak{h}}$ if and only if
$b=0$. This completes the proof. ∎

######  Proposition 3.1.

There is non $k$-symplectic structure on $T^{*}\mathfrak{su}(2)$.

###### Proof.

According to Proposition 2.1 and Lemma 3.2, $T^{*}\mathfrak{su}(2)$ has no
$k$-symplectic structure with $k=1$ or $k=5$.

Let $({\mathfrak{h}},D_{1},D_{2})$ be a 2-symplectic structure on
$T^{*}\mathfrak{su}(2)$ where ${\mathfrak{h}}$ is a 4-dimensional subalgebra
and $D_{1},D_{2}$ two skew-symmetric derivations satisfying
$D_{i}({\mathfrak{h}})\subset{\mathfrak{h}}^{\perp}$ and $\ker D_{1}\cap\ker
D_{2}=\\{0\\}$. By virtue of Corollary 2.1 and Theorem 2.2, ${\mathfrak{h}}$
must be degenerate, and according to Lemma 3.2, we can suppose that
${\mathfrak{h}}=\mathrm{span}(e_{1},e_{1}^{*},e_{2}^{*},e_{3}^{*})$ and
${\mathfrak{h}}^{\perp}=\mathrm{span}(e_{2}^{*},e_{3}^{*})$. A skew-symmetric
derivation $D$ of $T^{*}\mathfrak{su}(2)$ is inner (see Theorem 2.2), i.e.,
$D={\mathrm{ad}}_{X}$ where $X$ has coordinate $(x_{1},\ldots,x_{6})$ and we
have

$D(e_{1}^{*})=x_{3}e_{2}^{*}-x_{2}e_{3}^{*},\;D(e_{2}^{*})=-x_{3}e_{1}^{*}+x_{1}e_{3}^{*}\quad\mbox{and}\quad
D(e_{3}^{*})=x_{2}e_{1}^{*}+x_{1}e_{2}^{*}.$

So $D(e_{i}^{*})\subset{\mathfrak{h}}^{\perp}$ for $i=1,\ldots,3$ if and only
if $D(e_{1}^{*})=0$. Thus $\ker D_{1}\cap\ker D_{2}\not=\\{0\\}$ which
completes the proof. ∎

######  Lemma 3.3.

  1. 1.

Let ${\mathfrak{h}}$ be a 5-dimensional degenerate subalgebra of
$T^{*}\mathrm{sl}(\hbox{\bb R},2)$ then there exists an automorphism $F$ of
$\mathrm{sl}(\hbox{\bb R},2)$ such that
$T^{*}F({\mathfrak{h}})=\mathrm{span}(e_{1},e_{3})\oplus\mathrm{sl}(\hbox{\bb
R},2)^{*}$.

  2. 2.

If ${\mathfrak{h}}$ is a 4-dimensional degenerate subalgebra of
$T^{*}\mathrm{sl}(\hbox{\bb R},2)$ then there exists an automorphism $F$ of
$\mathrm{sl}(\hbox{\bb R},2)$ such that $T^{*}F({\mathfrak{h}})$ has one of
the following forms

$\hbox{\bb R}e_{1}\oplus\mathrm{sl}(\hbox{\bb R},2)^{*},\;\hbox{\bb
R}e_{3}\oplus\mathrm{sl}(\hbox{\bb
R},2)^{*},\;\mathrm{span}\\{e_{1},e_{3},e_{1}^{*},e_{2}^{*}\\}\quad\mbox{or}\quad\mathrm{span}\\{e_{1},e_{3},e_{2}^{*},e_{3}^{*}\\}.$

###### Proof.

  1. 1.

Let ${\mathfrak{h}}$ be a degenerate 5-dimensional Lie subalgebra of
$T^{*}\mathrm{sl}(\hbox{\bb R},2)$. Then
$\dim{\mathfrak{h}}\cap\mathrm{sl}(\hbox{\bb R},2)\geq 2$.

If $\dim{\mathfrak{h}}\cap\mathrm{sl}(\hbox{\bb R},2)=3$ then
$\mathrm{sl}(\hbox{\bb R},2)\subset{\mathfrak{h}}$ and
${\mathfrak{h}}^{\perp}\subset\mathrm{sl}(\hbox{\bb
R},2)^{\perp}=\mathrm{sl}(\hbox{\bb R},2)$. But
$[{\mathfrak{h}},{\mathfrak{h}}^{\perp}]\subset{\mathfrak{h}}^{\perp}$ an
hence ${\mathfrak{h}}^{\perp}$ is an ideal in $\mathrm{sl}(\hbox{\bb R},2)$
which is impossible.

If $\dim{\mathfrak{h}}\cap\mathrm{sl}(\hbox{\bb R},2)=2$ then
${\mathfrak{h}}\cap\mathrm{sl}(\hbox{\bb R},2)$ is a subalgebra of
$\mathrm{sl}(\hbox{\bb R},2)$. According to [7, Proposition 5.2]), up to an
automorphism, ${\mathfrak{h}}\cap\mathrm{sl}(\hbox{\bb
R},2)=\mathrm{span}(e_{1},e_{3})$. So
${\mathfrak{h}}^{\perp}+\mathrm{sl}(\hbox{\bb R},2)=\mathrm{sl}(\hbox{\bb
R},2)\oplus\hbox{\bb R}e_{2}^{*}$. On the other hand,
$\dim{\mathfrak{h}}^{\perp}\cap\mathrm{sl}(\hbox{\bb R},2)=0$ and hence
${\mathfrak{h}}^{\perp}=\hbox{\bb R}(x+\alpha)$ with $\alpha\not=0$. We have
$\alpha(e_{1})=\alpha(e_{3})=0$ so we can choose $\alpha=e_{2}^{*}$. Since
${\mathfrak{h}}^{\perp}\subset{\mathfrak{h}}$ (${\mathfrak{h}}$ being
degenerate), we must have also $\alpha(x)=0$. Thus $x=ae_{1}+be_{3}$. But
$[{\mathfrak{h}},{\mathfrak{h}}^{\perp}]\subset{\mathfrak{h}}^{\perp}$ so
$[e_{1},x]=mx\quad\mbox{and}\quad[e_{3},x]=nx$ and hence $x=me_{1}$. Thus

${\mathfrak{h}}=\mathrm{span}(e_{1},e_{3},-me_{2}+e_{1}^{*},e_{2}^{*},e_{3}^{*}).$

But $[e_{3}^{*},-me_{2}+e_{1}^{*}]=me_{1}^{*}$ and
$me_{1}^{*}\in{\mathfrak{h}}$ if and only if $m=0$. Which complete the first
part of the proof.

  2. 2.

Let ${\mathfrak{h}}$ be a 4-dimensional Lie degenerate subalgebra of
$T^{*}\mathrm{sl}(\hbox{\bb R},2)$. We have three possibilities
$\dim{\mathfrak{h}}\cap\mathrm{sl}(\hbox{\bb R},2)=1,2$ or 3. The same
argument as in the proof of Lemma 3.2 shows that the case
$\dim{\mathfrak{h}}\cap\mathrm{sl}(\hbox{\bb R},2)=3$ is impossible.

$\bullet$ Suppose that
$\mathfrak{g}_{1}={\mathfrak{h}}\cap\mathrm{sl}(\hbox{\bb R},2)$ is a two
dimensional subalgebra of $\mathrm{sl}(\hbox{\bb R},2)$. According to [7,
Proposition 5.1], we can suppose that
$\mathfrak{g}_{1}=\mathrm{span}(e_{1},e_{3})$. On the other hand,
$\dim{\mathfrak{h}}^{\perp}\cap\mathrm{sl}(\hbox{\bb
R},2)=6-\dim({\mathfrak{h}}+\dim\mathrm{sl}(\hbox{\bb R},2))=1$. Denote by
$\alpha$ a generator of ${\mathfrak{h}}^{\perp}\cap\mathrm{sl}(\hbox{\bb
R},2)$. Then
${\mathfrak{h}}\subset\alpha^{\perp}=\alpha^{0}\oplus\mathrm{sl}(\hbox{\bb
R},2)^{*}$ where $\alpha^{0}$ is the annihilator of $\alpha$. This implies
that $\mathfrak{g}_{2}={\mathfrak{h}}\cap\mathrm{sl}(\hbox{\bb R},2)^{*}$ is
2-dimensional. So finally,
${\mathfrak{h}}=\mathfrak{g}_{1}\oplus\mathfrak{g}_{2}$ and $\mathfrak{g}_{2}$
is invariant by ${\mathrm{ad}}_{e_{1}}^{*}$ and ${\mathrm{ad}}_{e_{3}}^{*}$.
In the dual basis, we have

${\mathrm{ad}}_{e_{1}}^{*}=\left[\begin{array}[]{ccc}0&0&0\\\ 0&0&1\\\
-2&0&0\end{array}\right]\quad\mbox{and}\quad{\mathrm{ad}}_{e_{3}}^{*}=\left[\begin{array}[]{ccc}2&0&0\\\
0&-2&\\\ 0&0&0\end{array}\right].$

We have ${\mathrm{ad}}_{e_{1}}^{*}$ leaves invariant
$\mathrm{span}(e_{2}^{*},e_{3}^{*})$ and $\mathfrak{g}_{2}$ so it leaves
invariant their intersection and since 0 is the only real eigenvalue of
${\mathrm{ad}}_{e_{1}}^{*}$, we get that $e_{2}^{*}\in\mathfrak{g}_{2}$. We
have also that ${\mathrm{ad}}_{e_{3}}^{*}$ leaves invariant
$\mathrm{span}(e_{1}^{*},e_{3}^{*})$ and $\mathfrak{g}_{2}$ so it leaves
invariant their intersection and hence either $e_{1}^{*}\in\mathfrak{g}_{2}$
or $e_{3}^{*}\in\mathfrak{g}_{2}$. Finally,

${\mathfrak{h}}=\mathrm{span}\\{e_{1},e_{3},e_{1}^{*},e_{2}^{*}\\}\quad\mbox{or}\quad{\mathfrak{h}}=\mathrm{span}\\{e_{1},e_{3},e_{2}^{*},e_{3}^{*}\\}.$

$\bullet$ $\dim{\mathfrak{h}}\cap\mathrm{sl}(\hbox{\bb R},2)=1$. According to
Lemma 3.1, we can suppose that a generator of
${\mathfrak{h}}\cap\mathrm{sl}(\hbox{\bb R},2)$ is either $ae_{1}+be_{2}$ or
$e_{3}$ with $(a,b)\not=0$.

Suppose that $a\not=0$ and hence we can choose $e_{1}+be_{2}$ as a generator
of ${\mathfrak{h}}\cap\mathrm{sl}(\hbox{\bb R},2)$. We must have
${\mathfrak{h}}^{\perp}\cap\mathfrak{su}(2)=\\{0\\}$ and hence there exists
$\alpha_{1}\not=0,\alpha_{2}\not=0$ and $x_{1},x_{2}\in\mathfrak{su}(2)$ such
that

${\mathfrak{h}}^{\perp}=\mathrm{span}\left(x_{1}+\alpha_{1},x_{2}+\alpha_{2}\right).$

We must have $\\{\alpha_{1},\alpha_{2}\\}$ linearly independent. We have
$\alpha_{1}(e_{1}+be_{2})=\alpha_{2}(e_{1}+be_{2})=0$ and hence we can choose
$\alpha_{1}=-be_{1}^{*}+e_{2}^{*}$ and $\alpha_{2}=e_{3}^{*}$. Now the fact
${\mathrm{ad}}_{e_{1}}$ leaves ${\mathfrak{h}}^{\perp}$ invariant implies that

$\begin{cases}[e_{1},x_{1}]+[e_{1},e_{2}^{*}]-b[e_{1},e_{1}^{*}]=[e_{1},x_{1}]-2be_{3}^{*}=\mu_{1}(x_{1}+e_{2}^{*}-be_{1}^{*})+\nu_{1}(x_{2}+e_{3}^{*}),\\\
[e_{1},x_{2}]+[e_{1},e_{3}^{*}]=[e_{1},x_{2}]-e_{2}^{*}=\mu_{2}(x_{1}+e_{2}^{*}-be_{1}^{*})+\nu_{2}(x_{2}+e_{3}^{*}),\\\
\end{cases}$

So $b=0$ and

$[e_{1},x_{1}]=0\quad\mbox{and}\quad[e_{1},x_{2}]=-x_{1}.$

This is equivalent to $x_{1}=\alpha e_{1}$ and
$x_{2}=\frac{\alpha}{2}e_{3}+\beta e_{1}$ and hence

${\mathfrak{h}}^{\perp}=\\{\alpha e_{1}+e_{2}^{*},\beta
e_{1}+\frac{\alpha}{2}e_{3}+e_{3}^{*}\\}$

Then

${\mathfrak{h}}=\mathrm{span}\left(e_{1},e_{2}^{*},ae_{2}-e_{1}^{*},\frac{a}{2}e_{3}-e_{3}^{*}\right).$

An one can see that ${\mathfrak{h}}$ is a subalgebra if and only if $a=0$.
Hence ${\mathfrak{h}}=\hbox{\bb R}e_{1}\oplus\mathrm{sl}(2,\hbox{\bb R})^{*}$
and ${\mathfrak{h}}^{\perp}=\mathrm{span}(e_{2}^{*},e_{3}^{*})$.

Suppose that $e_{3}$ is a generator of
${\mathfrak{h}}\cap\mathrm{sl}(\hbox{\bb R},2)$. Then
${\mathfrak{h}}^{\perp}\cap\mathrm{sl}(\hbox{\bb R},2)=\\{0\\}$ and there
exists $\alpha_{1}\not=0,\alpha_{2}\not=0$ and
$y_{1},y_{2}\in\mathfrak{su}(2)$ such that

${\mathfrak{h}}^{\perp}=\mathrm{span}\left(x_{1}+\alpha_{1},x_{2}+\alpha_{2}\right).$

We must have $\\{\alpha_{1},\alpha_{2}\\}$ linearly independent. We have
$\alpha_{1}(e_{3})=\alpha_{2}(e_{3})=0$ and hence we can choose
$\alpha_{1}=e_{1}^{*}$ and $\alpha_{2}=e_{2}^{*}$. Now the fact
${\mathrm{ad}}_{e_{3}}$ leaves ${\mathfrak{h}}^{\perp}$ invariant implies that

$\begin{cases}[e_{3},x_{1}]+[e_{3},e_{1}^{*}]=[e_{1},x_{1}]-2e_{1}^{*}=\mu_{1}(x_{1}+e_{1}^{*})+\nu_{1}(x_{2}+e_{2}^{*}),\\\
[e_{3},x_{2}]+[e_{3},e_{2}^{*}]=[e_{1},x_{2}]+2e_{2}^{*}=\mu_{2}(x_{1}+e_{1}^{*})+\nu_{2}(x_{2}+e_{2}^{*}),\\\
\end{cases}$

So

$[e_{3},x_{1}]=-2x_{1}\quad\mbox{and}\quad[e_{3},x_{2}]=2x_{2}.$

So $x_{1}=\alpha e_{2}$ and $x_{2}=\beta e_{1}$. So

${\mathfrak{h}}^{\perp}=\\{\alpha e_{2}+e_{1}^{*},\beta e_{1}+e_{2}^{*}\\}$

Hence ${\mathfrak{h}}=\hbox{\bb R}e_{3}\oplus\mathrm{sl}(2,\hbox{\bb R})^{*}$
and ${\mathfrak{h}}^{\perp}=\mathrm{span}(e_{2}^{*},e_{3}^{*})$.∎

######  Proposition 3.2.

There is non $k$-symplectic structure on $T^{*}\mathrm{sl}(\hbox{\bb R},2)$.

###### Proof.

By virtue of Proposition 2.1, $T^{*}\mathrm{sl}(\hbox{\bb R},2)$ has no
1-symplectic structure.

Let ${\mathfrak{h}}$ be a degenerate 5-dimensional Lie subalgebra of
$T^{*}\mathrm{sl}(\hbox{\bb R},2)$ and $D$ a skew-symmetric derivation of
$T^{*}\mathrm{sl}(\hbox{\bb R},2)$ such that
$D({\mathfrak{h}})\subset{\mathfrak{h}}^{\perp}$. According to Lemma 3.3,
${\mathfrak{h}}={\mathrm{span}}(e_{2},e_{3})\oplus\mathrm{sl}(n,\mathbb{R})^{*}$
and ${\mathfrak{h}}^{\perp}=\hbox{\bb R}e_{1}^{*}$. Moreover, according to
Theorem 2.2, $D={\mathrm{ad}}_{x+\alpha}$. So we must have
${\mathrm{ad}}_{x}^{*}(\mathrm{sl}(n,\mathbb{R})^{*})\subset\hbox{\bb
R}e_{1}^{*}$ this implies that ${\mathrm{ad}}_{x}$ in restriction to
$\mathrm{sl}(\hbox{\bb R},2)$ has rank $\leq 1$ which implies that
${\mathrm{ad}}_{x}=0$. So $D$ vanishes in restriction to
$\mathrm{sl}(n,\mathbb{R})^{*}$. This shows that $T^{*}\mathrm{sl}(\hbox{\bb
R},2)$ has no 5-symplectic structure.

Let ${\mathfrak{h}}$ be a nondegenerate 4-dimensional Lie subalgebra of
$T^{*}\mathrm{sl}(\hbox{\bb R},2)$. According to Lemma 3.3, we have four
cases:

  1. 1.

${\mathfrak{h}}=\hbox{\bb R}e_{1}\oplus\mathrm{sl}(\hbox{\bb R},2)^{*}$. Let
$D={\mathrm{ad}}_{X}$ be a skew-symmetric derivation of
$T^{*}\mathrm{sl}(2,\hbox{\bb R})$ with $X=[x_{1},\ldots,x_{6}]$ then
$D(e_{1}^{*}),D(e_{3}^{*})\in{\mathfrak{h}}^{\perp}$ if and only if
$D(e_{2}^{*})=0)$.

  2. 2.

${\mathfrak{h}}=\hbox{\bb R}e_{3}\oplus\mathrm{sl}(\hbox{\bb R},2)^{*}$. Let
$D={\mathrm{ad}}_{X}$ be a skew-symmetric derivation of
$T^{*}\mathrm{sl}(2,\hbox{\bb R})$ with $X=[x_{1},\ldots,x_{6}]$ then
$D(e_{1}^{*})\in{\mathfrak{h}}^{\perp}$ if and only if $D(e_{1}^{*})=0$

  3. 3.

${\mathfrak{h}}=\mathrm{span}\\{e_{1},e_{3},e_{1}^{*},e_{2}^{*}\\}$ or
$\mathrm{span}\\{e_{1},e_{3},e_{2}^{*},e_{3}^{*}\\}$. Let $D$ be a skew-
symmetric derivation of $T^{*}\mathrm{sl}(\hbox{\bb R},2)$. Then
$D={\mathrm{ad}}_{X}$ where $X=(x_{1},\ldots,x_{6})$ and

$D(e_{2})=\left[\begin{array}[]{cccccc}0&-2x_{3}&x_{1}&-x_{6}&0&2x_{5}\end{array}\right]\quad\mbox{and}\quad
D(e_{1}^{*})=\left[\begin{array}[]{cccccc}0&0&0&-2x_{3}&0&2x_{1}\end{array}\right].$

If ${\mathfrak{h}}=\mathrm{span}\\{e_{1},e_{3},e_{1}^{*},e_{2}^{*}\\}$ then
${\mathfrak{h}}^{\perp}=\mathrm{span}(e_{3},e_{2}^{*})$ and
$D(e_{2})\in{\mathfrak{h}}^{\perp}$ if and only if $D(e_{2})=0$. If
${\mathfrak{h}}=\mathrm{span}\\{e_{1},e_{3},e_{2}^{*},e_{3}^{*}\\}$ then
${\mathfrak{h}}^{\perp}=\mathrm{span}(e_{1},e_{2}^{*})$ and
$D(e_{1}^{*})\in{\mathfrak{h}}^{\perp}$ if and only if $D(e_{1}^{*})=0$.

This shows that ${\mathfrak{h}}$ cannot be the subalgebra of a 2-symplectic
structure and hence $T^{*}\mathrm{sl}(n,\mathbb{R})$ has no 2-symplectic
structure. ∎

Lie algebra | The non vanishing bracket | The metric
---|---|---
$\mathfrak{su}(2)$ | $[e_{1},e_{2}]=e_{3},[e_{2},e_{3}]=e_{1},[e_{3},e_{1}]=e_{2}.$ | $c\mathrm{id}$
$\mathrm{sl}(2,\hbox{\bb R})$ | $[e_{1},e_{2}]=e_{3},[e_{3},e_{1}]=2e_{1},[e_{3},e_{2}]=-2e_{2}.$ | $c\\{(1,2)=1,(3,3)=2\\}$

Table 1: Indecomposable quadratic Lie algebras of dimension 3

Lie algebra | The non vanishing bracket | The metric
---|---|---
$\mathrm{os}(4,\lambda)$ | $[e_{2},e_{3}]=e_{1},[e_{4},e_{2}]=\lambda e_{3},[e_{4},e_{3}]=-\lambda e_{2}$ | $\\{(2,2)=(3,3)=\frac{1}{\lambda},(1,4)=1\\}$
$\mathfrak{g}_{1,4}$ | $[e_{4},e_{2}]=e_{2},\;[e_{4},e_{3}]=-e_{3},\;[e_{2},e_{3}]=e_{1}$ | $\\{(1,4)=(2,3)=1\\}$

Table 2: Indecomposable non abelian quadratic Lie algebras of dimension 4

Lie algebra | The non vanishing bracket | The metric
---|---|---
$\mathfrak{g}_{1,5}$ | $[e_{2},e_{3}]=e_{1},[e_{3},e_{4}]=-e_{1},[e_{5},e_{2}]=e_{3},$ | $\\{(1,5)=-(2,2)=(3,3)=(4,4)=1\\}$
| $[e_{5},e_{3}]=e_{2}-e_{4},[e_{5},e_{4}]=e_{3}$ |

Table 3: Indecomposable non abelian quadratic Lie algebras of dimension 5

Lie algebra | The non vanishing bracket | The metric
---|---|---
$\mathrm{osc}(6,(\lambda_{1},\lambda_{2}))$ | $[e_{2},e_{4}]=[e_{3},e_{5}]=e_{1},[e_{6},e_{2}]=\lambda_{1}e_{4}$ | $\\{(2,2)=(3,3)=(4,4)=(5,5)=\frac{1}{\lambda}$
| $[e_{6},e_{4}]=-\lambda_{1}e_{2},[e_{6},e_{3}]=\lambda_{2}e_{5},[e_{6},e_{5}]=-\lambda_{2}e_{3}$ | $(1,6)=1\\}$
$\mathfrak{l}_{2,\lambda}$ | $[e_{2},e_{3}]=e_{1},[e_{4},e_{5}]=\lambda e_{1},[e_{6},e_{2}]=e_{3},$ | $\\{-(2,2)=(3,3)=(4,4)=(5,5)=1$
| $[e_{6},e_{3}]=e_{2},[e_{6},e_{4}]=\lambda e_{5},[e_{6},e_{5}]=-\lambda e_{4}$ | $(1,6)=1\\}$
$\mathrm{so}(3,1)$ | $[e_{1},e_{2}]=e_{3},[e_{1},e_{3}]=-e_{2},[e_{1},e_{4}]=e_{5},[e_{1},e_{5}]=-e_{4},$ | $\\{(1,1)=(2,2)=(3,3)=1$
| $[e_{2},e_{3}]=e_{1},[e_{2},e_{4}]=e_{6},[e_{2},e_{6}]=-e_{4},[e_{3},e_{5}]=e_{6},$ | $(4,4)=(5,5)=(6,6)=-1\\}$
| $[e_{3},e_{6}]=-e_{5},[e_{4},e_{5}]=-e_{1},[e_{4},e_{6}]=-e_{2},[e_{5},e_{6}]=-e_{3}.$ |
$T^{*}\mathrm{sl}(2,\hbox{\bb R})$ | $[e_{1},e_{2}]=e_{3},[e_{3},e_{1}]=2e_{1},[e_{3},e_{2}]=-2e_{2}$ | $\\{(1,4)=(2,5)=(3,6)=1\\}$
| $[e_{1},e_{1}^{*}]=2e_{3}^{*},[e_{1},e_{3}^{*}]=-e_{2}^{*},[e_{2},e_{2}^{*}]=-2e_{3}^{*}$ |
| $[e_{2},e_{3}^{*}]=e_{1}^{*},[e_{3},e_{1}^{*}]=-2e_{1}^{*},[e_{3},e_{2}^{*}]=2e_{2}^{*}$ |
$T^{*}\mathfrak{su}(2)$ | $[e_{1},e_{2}]=e_{3},[e_{2},e_{3}]=e_{1},[e_{3},e_{1}]=e_{2}$ | $\\{(1,4)=(2,5)=(3,6)=1\\}$
| $[e_{1},e_{2}^{*}]=e_{3}^{*},[e_{1},e_{3}^{*}]=-e_{2}^{*},[e_{2},e_{1}^{*}]=-e_{3}^{*},$ |
| $[e_{2},e_{3}^{*}]=e_{1}^{*},[e_{3},e_{1}^{*}]=e_{2}^{*},[e_{3},e_{2}^{*}]=-e_{1}^{*}.$ |
$\mathfrak{n}_{1}(2,2)$ | $[e_{6},e_{3}]=e_{2},[e_{6},e_{5}]=e_{4},[e_{3},e_{5}]=e_{1}.$ | $\\{(1,6)=(2,5)=-(3,4)=1\\}$
$\mathfrak{n}_{2}(2,2)$ | $[e_{6},e_{2}]=e_{2}+te_{3},[e_{6},e_{3}]=-te_{2}+e_{3},[e_{6},e_{4}]=-e_{4}+te_{5}$ | $\\{(1,6)=(2,5)=-(3,4)=1\\}$
| $[e_{6},e_{5}]=-te_{4}-e_{3},[e_{2},e_{4}]=-te_{1},[e_{2},e_{5}]=e_{1},$ |
| $[e_{3},e_{4}]=-e_{1},[e_{3},e_{5}]=-te_{1}.t>0$ |
$\mathfrak{n}_{3}(2,2)$ | $[e_{6},e_{2}]=e_{3},[e_{6},e_{3}]=-e_{2},[e_{6},e_{4}]=\epsilon e_{2}+e_{5},\epsilon^{2}=1,$ | $\\{(1,6)=(2,5)=-(3,4)=1\\}$
| $[e_{6},e_{5}]=\epsilon e_{3}-e_{4},[e_{2},e_{4}]=-e_{1},[e_{3},e_{5}]=-e_{1},[e_{4},e_{5}]=\epsilon e_{1}.$ |
$\mathfrak{n}_{4}(2,2)$ | $[e_{6},e_{2}]=e_{3},[e_{6},e_{3}]=-e_{2},[e_{6},e_{4}]=te_{5},$ | $\\{-(2,2)=-(3,3)=(4,4)=(5,5)=1$
| $[e_{6},e_{5}]=-te_{4},[e_{2},e_{3}]=-e_{1},[e_{4},e_{5}]=te_{1}.\quad(t>0)$ | $(1,6)=1\\}$
$\mathfrak{n}_{5}(2,2)$ | $[e_{6},e_{2}]=e_{2},[e_{6},e_{3}]=e_{2}+e_{3},[e_{6},e_{4}]=-e_{4},[e_{6},e_{5}]=e_{4}-e_{5},$ | $\\{(1,6)=(2,5)=-(3,4)=1\\}$
| $[e_{2},e_{5}]=e_{1},[e_{3},e_{4}]=-e_{1},[e_{3},e_{5}]=e_{1}.$ |
$\mathfrak{n}_{6}(2,2)$ | $[e_{6},e_{2}]=e_{2},[e_{6},e_{3}]=-e_{3},[e_{6},e_{4}]=te_{4},$ | $\\{(1,6)=(2,3)=(4,5)=1\\}$
| $[e_{6},e_{5}]=-te_{5},[e_{2},e_{3}]=e_{1},[e_{4},e_{5}]=te_{1},t\geq 1$ |

Table 4: Indecomposable non abelian quadratic Lie algebras of dimension 6

## References

  * [1] H. Abchir, Ilham Ait Brik & Mohamed Boucetta, On $k$-para-Kähler Lie algebras, a subclass of $k$-symplectic Lie algebras, Communication in Algebra, https://doi.org/10.1080/00927872.2021.1922697.
  * [2] A. Awane, A. Chkiriba, M. Goze, $k$-symplectic affine Lie algebras, African Journal of Mathematical Physics Vol. 2 N. 1 (2005) 77-85.
  * [3] A. Awane, $G$-espaces $k$-symplectiques homogènes, J. Geom. Phys. 13, 2, pp. 139-157 (1994).
  * [4] A. Awane & M. Goze, Pfaffian Systems, k-Symplectic Systems, Kluwer Academic Publishers 2000.
  * [5] A. Awane, $k$-symplectic structures, Journal of Mathematical Physics 33, 4046 (1992).
  * [6] H. Baum, I. Kath, Doubly extended Lie groups curvature, holonomy and parallel spinors, Differ. Geom. Appl. 19 (3) (2003), 253-280 .
  * [7] M. Boucetta, A. Meliani, A. Zeghib, Kundt three-dimensional left invariant spacetimes, J. Math. Phys. 63, 112501 (2022); https://doi.org/10.1063/5.0091202
  * [8] J. Figueroa-Ofarrill, S. Stancu, On the structure of symmetric self-dual Lie algebras,J.Math. Phys. 37 (8) (1996), 4121-4134.
  * [9] Ryad Ghanam, Gerard Thompson, Narayana Bandara, Lie subalgebras of $\mathrm{so}(3,1)$ up to conjugacy, Arab Journal of Mathematical Sciences Vol. 28 No. 2, 2022 pp. 253-261.
  * [10] Günther, C., The polysymplectic Hamiltonian formalism in field theory and calculus of variations. I. The local case, J. Differential Geom. 25, 1, pp. 23-53 (1987).
  * [11] N. Jacobson, A note on automorphisms and derivations of Lie algebras, Proc. Amer. Math. Soc. vol. 6 (1955) pp. 281-283.
  * [12] M. de Leòn, M. Salgado, S. Vilarino, Methods of Differential Geometry in Classical Field Theories: $k$-symplectic and $k$-cosymplectic approaches, World scientific publishing (2016).
  * [13] G. P. Ovando, Lie algebras with ad-invariant metrics: A survey-Guide, Rendiconti Seminario Matematico Univ. Pol. Torino Workshop for Sergio Console Vol. 74, 1 (2016), 243 - 268.
  * [14] R. F. Streater, The representations of the oscillator group, Comm. Math. Phys. 4, 217 (1967).