#  Complete Description of Invariant, Associative Pseudo-Euclidean Metrics on
Left Leibniz Algebras via Quadratic Lie Algebras

Fatima-Ezzahrae Abid  Université Cadi-Ayyad
Faculté des sciences et techniques
BP 549 Marrakech Maroc
e-mail<EMAIL_ADDRESS>Mohamed Boucetta  Université Cadi-Ayyad
Faculté des sciences et techniques
BP 549 Marrakech Maroc
e-mail<EMAIL_ADDRESS>
###### Abstract

A pseudo-Euclidean non-associative algebra $(\mathfrak{g},\bullet)$ is a real
algebra of finite dimension that has a metric, i.e., a bilinear, symmetric,
and non-degenerate form $\langle\;,\;\rangle$. The metric is considered
${\mathrm{L}}$-invariant (resp. ${\mathrm{R}}$-invariant) if all left
multiplications (resp. right multiplications) are skew-symmetric. The metric
is called associative if $\langle u\bullet v,w\rangle=\langle u,v\bullet
w\rangle$ for all $u,v,w\in\mathfrak{g}$. These three notions coincide when
$\mathfrak{g}$ is a Lie algebra and in this case $\mathfrak{g}$ endowed with
the metric is known as a quadratic Lie algebra.

This paper provides a complete description of ${\mathrm{L}}$-invariant,
${\mathrm{R}}$-invariant, or associative pseudo-Euclidean metrics on left
Leibniz algebras. It shows that a left Leibniz algebra with an associative
metric is also right Leibniz and can be obtained easily from its underlying
Lie algebra, which is a quadratic Lie algebra. Additionally, it shows that at
the core of a left Leibniz algebra endowed with a ${\mathrm{L}}$-invariant or
${\mathrm{R}}$-invariant metric, there are two Lie algebras with one quadratic
and the left Leibniz algebra can be built from these Lie algebras. We derive
many important results from these complete description. Finally, the paper
provides a list of left Leibniz algebras with an associative metric up to
dimension 6, as well as a list of left Leibniz algebras with an
${\mathrm{L}}$-invariant metric, up to dimension 4, and
${\mathrm{R}}$-invariant metric up to dimension 5.

Keywords: Leibniz algebras, invariant metrics, associative metrics, double
extension, $T^{*}$-extension.

##  1 Introduction

Lie groups having a bi-invariant pseudo-Riemannian metric is a very large
class which 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 [20]. The Lie algebra of a Lie group having
a bi-invariant pseudo-Riemannian is called quadratic. It is a real Lie algebra
$(\mathfrak{g},[\;,\;])$ endowed with a bilinear non-degenerate form
$\langle\;,\;\rangle$ such that, for any $u\in\mathfrak{g}$,
${\mathrm{ad}}_{u}:\mathfrak{g}\longrightarrow\mathfrak{g}$, $v\mapsto[u,v]$
is skew-symmetric with respect to $\langle\;,\;\rangle$. This class of pseudo-
Euclidean Lie algebras has been the focus of intensive study by numerous
authors (refer to [28] and the bibliography therein).

There are three ways of generalizing the notion of bi-invariant pseudo-
Euclidean metric on Lie algebras. Let $(\mathfrak{g},\bullet)$ be a
nonassociative algebra and $\langle\;,\;\rangle$ a pseudo-Euclidean metric on
$\mathfrak{g}$. For any $u\in\mathfrak{g}$, we denote by
${\mathrm{L}}_{u},{\mathrm{R}}_{u}:\mathfrak{g}\longrightarrow\mathfrak{g}$,
respectively, the left and the right multiplication by $u$, i.e.,
${\mathrm{L}}_{u}v={\mathrm{R}}_{v}u=u\bullet v$. The metric is called
${\mathrm{L}}$-invariant (resp. ${\mathrm{R}}$-invariant) if, for any
$u\in\mathfrak{g}$, ${\mathrm{L}}_{u}$ (resp. ${\mathrm{R}}_{u}$) is skew-
symmetric. It is called associative if, for any $u\in\mathfrak{g}$,
${\mathrm{R}}_{u}^{*}={\mathrm{L}}_{u}$, i.e., $\langle u\bullet
v,w\rangle=\langle u,v\bullet w\rangle$. In the first case (resp. the second
case), we call $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$
${\mathrm{L}}$-quadratic (resp. ${\mathrm{R}}$-quadratic) and, in the last
case, we call $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ metrised. All these
notions coincide when the product $\bullet$ is anti-commutative. In the
previous decade, various nonassociative algebras endowed with an associative
pseudo-Euclidean metric have been studied, for example, general nonassociative
algebras ([17]), quadratic Lie algebras (see [28] for a large bibliography),
quadratic super-Lie algebras ([2, 15]) or quadratic Malcev super-algebras
([3]), Leibniz algebras ([25, 12]). However, ${\mathrm{L}}$-quadratic or
${\mathrm{R}}$-quadratic nonassociative algebras have attracted little
attention. There are some results on ${\mathrm{L}}$-quadratic or
${\mathrm{R}}$-quadratic Leibniz algebras in [14, 25].

Leibniz algebras are a non-commutative generalization of Lie algebras and were
first introduced and investigated in the papers of Bloh [18, 16] under the
name of D-algebras. Then they were rediscovered by Loday [26] who called them
Leibniz algebras. A left Leibniz algebra (resp. right Leibniz algebra) is an
algebra $(\mathfrak{g},\bullet)$ over a field $\mathbb{K}$ such that, for any
$u\in\mathfrak{g}$, ${\mathrm{L}}_{u}$ (resp. ${\mathrm{R}}_{u}$) is a
derivation of $(\mathfrak{g},\bullet)$. An algebra which is both left and
right Leibniz is called symmetric Leibniz algebra. A Lie algebra is obviously
a symmetric Leibniz algebra. Many results of the theory of Lie algebras can be
extended to left Leibniz algebras (see [6, 7, 8]). Moreover, real left Leibniz
algebras are the infinitesimal version of Lie racks. In 2004, Kinyon [22]
proved that if $(X,e)$ is a pointed Lie rack, $T_{e}X$ carries a structure of
left Leibniz algebra.

In this paper, we undertake a complete study of ${\mathrm{L}}$-quadratic,
${\mathrm{R}}$-quadratic or metrised left Leibniz algebras. Our results can be
translated to right Leibniz algebras since there is a correspondence between
left and right Leibniz algebras.

Let us explain the two main ideas at the heart of this paper. First, it is
known that a metrised left Leibniz algebra is also a right Leibniz algebra and
hence it is Lie admissible. We show that the underlying Lie algebra is a
quadratic Lie algebra which can be used to build the metrised left Leibniz
algebra straightforwardly.

Secondly, let $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ be a
${\mathrm{L}}$-quadratic or ${\mathrm{R}}$-quadratic left Leibniz algebra. The
vector space $\mathrm{Leib}(\mathfrak{g})={\mathrm{span}}\\{u\bullet
v+v\bullet u\\}$ is an ideal known as the Leibniz ideal. Denote by
$\mathrm{Leib}(\mathfrak{g})^{\perp}$ its orthogonal. When
$\mathrm{Leib}(\mathfrak{g})\cap\mathrm{Leib}(\mathfrak{g})^{\perp}=\\{0\\}$,
$\mathfrak{g}$ is a Lie algebra when the metric is ${\mathrm{R}}$-invariant
and, when the metric is ${\mathrm{L}}$-invariant,
$\mathfrak{g}=\mathrm{Leib}(\mathfrak{g})\oplus\mathrm{Leib}(\mathfrak{g})^{\perp}$
and $\mathrm{Leib}(\mathfrak{g})^{\perp}$ is a quadratic Lie algebra.

When
$I=\mathrm{Leib}(\mathfrak{g})\cap\mathrm{Leib}(\mathfrak{g})^{\perp}\not=\\{0\\}$,
we show that ${\mathfrak{h}}=\mathfrak{g}/I^{\perp}$ has a structure of a Lie
algebra and $A=I^{\perp}/I\cap I^{\perp}$ has a structure of a
${\mathrm{L}}$-quadratic left Leibniz algebra with nondegenerate Leibniz ideal
when the metric is ${\mathrm{L}}$-invariant, and a structure of a quadratic
Lie algebra when the metric is ${\mathrm{R}}$-invariant . We call $A$ and
${\mathfrak{h}}$ endowed with their structures the core of
$(\mathfrak{g},\bullet,\langle\;,\;\rangle)$. Our main result is that
$(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ can be reconstructed from its
core in a way which combine the double extension process and the
$T^{*}$-extension. The double extension is a well-known process which appeared
first in [27] in the study of quadratic Lie algebras while the
$T^{*}$-extension was introduced in [17] in the study of metrised
nonassociative algebras.

Let us enumerate our main results and give the organization of the paper.

  1. 1.

In Section 2, we study the properties of ${\mathrm{L}}$-invariant,
${\mathrm{R}}$-invariant or associative bilinear forms (not necessary
nondegenerate) on left Leibniz algebras. The main result of this section is
the introduction of the accurate notion of Killing form on a left Leibniz
algebras. It permits the generalization of Cartan’s criterions valid for Lie
algebras to left Leibniz algebras (see Propositions 2.2 and 2.3).

  2. 2.

In Section 3, we undertake the study of metrised left Leibniz algebras. In
[12], it was pointed out that if a left Leibniz algebra has an associative
metric then it is also right Leibniz. Based on this and on the
characterization of symmetric Leibniz algebras given in [11], we show that
metrised left Leibniz algebras can be obtained from quadratic Lie algebras in
an easy way (see Theorems 3.1, 3.2 and 3.3). Since quadratic Lie algebras are
known up to dimension 6, we get almost without computation all metrised left
Leibniz algebras up to dimension 6 (see Tables 1)

  3. 3.

In Sections 4, 5 and 6, we give a complete description of
${\mathrm{L}}$-quadratic and ${\mathrm{R}}$-quadratic left Leibniz algebras.
The main results of Sections 5 and 6 are Theorems 5.1 and 6.1 which show that
both ${\mathrm{L}}$-quadratic and ${\mathrm{R}}$-quadratic left Leibniz
algebras can be obtained by a combination of the double extension process and
the $T^{*}$-extension from its core $A$ and ${\mathfrak{h}}$. These two
theorems have many interesting corollaries. When $A$ is a Lie algebra with a
trivial center and $H^{2}(A)=0$ then the original algebra can be described
easily (See Theorems 5.2 and 6.2). We derive a complete description of
${\mathrm{L}}$-quadratic simple left Leibniz algebras. We show that a
${\mathrm{R}}$-quadratic left Leibniz algebra which is semi-simple or for
which the metric is definite positive or Lorentzian must be a Lie algebra. We
derive also a complete description of ${\mathrm{L}}$-quadratic symmetric
Leibniz algebra (see Theorem 6.3) and we deduce that any Lorentzian
${\mathrm{L}}$-quadratic symmetric Leibniz algebra is actually a Lie algebra.

  4. 4.

Section 7 is an appendix where we give the details of the computations needed
in the proofs of Theorems 5.1 and 6.1.

  5. 5.

Finally, we provide a list of left Leibniz algebras with an associative metric
up to dimension 6, as well as a list of left Leibniz algebras with an
${\mathrm{L}}$-invariant metric, up to dimension 4, and
${\mathrm{R}}$-invariant metric up to dimension 5 (see Tables 1-3).

#### Notations

For a non associative algebra $(\mathfrak{g},\bullet)$, we denote by
${\mathrm{L}}_{u}$ and ${\mathrm{R}}_{u}$, respectively, the left and the
right multiplication by $u$, i.e., ${\mathrm{L}}_{u}(v)=u\bullet v$ and
${\mathrm{R}}_{u}(v)=v\bullet u$. If $F,G$ are two vector subspaces of
$\mathfrak{g}$, we denote by $F\bullet G$ the subspace generated by $u\bullet
v$ for any $u\in G$ and $v\in G$. A vector subspace $I$ is a left ideal (resp.
right ideal) if $\mathfrak{g}\bullet I\subset I$ (resp.
$I\bullet\mathfrak{g}\subset I$) and $I$ is an ideal if it is both a left
ideal and a right ideal. Finally, we denote

$Z^{\ell}(\mathfrak{g})=\\{u\in\mathfrak{g},{\mathrm{L}}_{u}=0\\},\;Z^{r}(\mathfrak{g})=\\{u\in\mathfrak{g},{\mathrm{R}}_{u}=0\\}\quad\mbox{and}\quad
Z(\mathfrak{g})=Z^{\ell}(\mathfrak{g})\cap Z^{r}(\mathfrak{g}).$

A metric on $\mathfrak{g}$ is a nondegenerate bilinear symmetric form
$\langle\;,\;\rangle$. For any endomorphism
$F:\mathfrak{g}\longrightarrow\mathfrak{g}$ and a vector subspace $I$, we
denote by $F^{*}$ the adjoint with respect to $\langle\;,\;\rangle$ and by
$I^{\perp}$ the orthogonal with respect to $\langle\;,\;\rangle$. The vector
subspace $I$ is called non-degenerate if $I\cap I^{\perp}=\\{0\\}$ and it is
called totally isotropic if $I\subset I^{\perp}$. We call a metric Euclidean
if it is definite positive and Lorentzian if it has the signature $(n-1,1)$.

##  2 Invariant symmetric bilinear forms and the Killing form of left Leibniz
algebras

In the literature, there are three notions of invariant symmetric bilinear
forms on a nonassociative algebra. We recall these notions and give some of
their properties. The Killing form of a Lie algebra is an invariant bilinear
form which plays an important role and can be used to give a criterion for a
Lie algebra to be semi-simple or solvable. Left Leibniz algebras generalize
Lie algebras but, as we will show in this section, lack a good notion of
Killing form. We introduce a bilinear symmetric form on any Leibniz algebra
$\mathfrak{g}$ and we show that it is the accurate notion of the Killing form
of $\mathfrak{g}$ since, in particular, it can give an analogue of Cartan’s
criterions for semi-simple and solvable Leibniz algebras.

Let $(A,\bullet)$ be a nonassociative algebra and $S$ a symmetric bilinear
form on $A$. $S$ is called

  1. 1.

${\mathrm{L}}$-invariant if for any $u,v,w\in A$, $S(u\bullet
v,w)+S(v,u\bullet w)=0$,

  2. 2.

${\mathrm{R}}$-invariant if for any $u,v,w\in A$, $S(u\bullet
v,w)+S(u,w\bullet v)=0$,

  3. 3.

associative if for any $u,v,w\in A$, $S(u\bullet v,w)=S(u,v\bullet w)$.

It is clear that these three notions coincide if $\bullet$ is anti-
commutative. However, when $\bullet$ is commutative and $S$ is
${\mathrm{L}}$-invariant or ${\mathrm{R}}$-invariant then $S(u\bullet v,w)=0$
for any $u,v\in A$.

######  Proposition 2.1.

Let $(A,\bullet)$ be a nonassociative algebra and $S$ a symmetric bilinear
form on $A$. If $S$ satisfies two of the properties above then it satisfies
the third one and, for any $u,v\in A$, $u\bullet v+v\bullet u\in\ker S$. In
this case, we call $S$ bi-invariant.

###### Proof.

Suppose, for instance, that $S$ is ${\mathrm{L}}$-invariant and
${\mathrm{R}}$-invariant. Then, for any $u,v,w\in A$,

$S(u\bullet v,w)=-S(v,u\bullet w)=S(v\bullet w,u)$

and hence $S$ is associative. Moreover,

$S(u\bullet v,w)=-S(v,u\bullet w)=S(v\bullet w,u)=-S(w,v\bullet u)$

which shows that $u\bullet v+v\bullet u\in\ker S$. The other cases can be
treated in a similar way. ∎

Recall that $(\mathfrak{g},\bullet)$ is a left Leibniz algebra if for any
$u,v,w\in\mathfrak{g}$,

$u\bullet(v\bullet w)=(u\bullet v)\bullet w+v\bullet(u\bullet
w)\quad\mbox{or}\quad\mathrm{L}_{u\bullet v}=[\mathrm{L}_{u},\mathrm{L}_{v}].$
(1)

$(\mathfrak{g},\bullet)$ is a right Leibniz algebra if for any
$u,v,w\in\mathfrak{g}$,

$(v\bullet w)\bullet u=(v\bullet u)\bullet w+v\bullet(w\bullet
u)\quad\mbox{or}\quad\mathrm{R}_{w\bullet
u}=-[\mathrm{R}_{w},\mathrm{R}_{u}].$ (2)

It is obvious that if $(\mathfrak{g},\bullet)$ is a left Leibniz algebra then
$(\mathfrak{g},\circ)$ is a right Leibniz algebra when $u\circ v=u\bullet v$.

$(\mathfrak{g},\bullet)$ is called a symmetric Leibniz algebra if it is both a
left and right Leibniz algebra.

Let $(\mathfrak{g},\bullet)$ be a left Leibniz algebra. We denote
$\mathrm{Leib}(\mathfrak{g})=\mathrm{span}\left\\{u\bullet v+v\bullet
u,u,v\in\mathfrak{g}\right\\}$. It is well-known [9] that
$\mathrm{Leib}(\mathfrak{g})$ is an ideal of $\mathfrak{g}$, for any
$u\in\mathrm{Leib}(\mathfrak{g})$, $\mathrm{L}_{u}=0$,
$\mathrm{Lie}(\mathfrak{g})=\mathfrak{g}/\mathrm{Leib}(\mathfrak{g})$ is a Lie
algebra and $\mathfrak{g}$ is a Lie algebra if and only if
$\mathrm{Leib}(\mathfrak{g})=\\{0\\}$. We call $\mathrm{Leib}(\mathfrak{g})$
the Leibniz ideal of $\mathfrak{g}$.

We consider $B$ and $K$ the two bilinear symmetric forms on $\mathfrak{g}$
given by

$B(u,v)={\mathrm{tr}}(\mathrm{L}_{u}\circ\mathrm{L}_{v})\quad\mbox{and}\quad
K(u,v)=-\frac{1}{2}\left({\mathrm{tr}}(\mathrm{L}_{u}\circ\mathrm{R}_{v})+{\mathrm{tr}}(\mathrm{L}_{v}\circ\mathrm{R}_{u})\right).$

The following proposition is our first main result.

######  Proposition 2.2.

Let $(\mathfrak{g},\bullet)$ be a left Leibniz algebra. Then the 2-forms $B$
and $K$ are bi-invariant and hence $\ker B$ and $\ker K$ contain
$\mathrm{Leib}(\mathfrak{g})$. Moreover, for any $u,v\in\mathfrak{g}$,

$K(u,v)=\widehat{K}(\pi(u),\pi(v))$ (3)

where
$\pi:\mathfrak{g}\longrightarrow\mathfrak{g}/\mathrm{Leib}(\mathfrak{g})$ and
$\widehat{K}$ is the Killing form of the Lie algebra
$\mathfrak{g}/\mathrm{Leib}(\mathfrak{g})$. Furthermore, $\ker K$ is a
solvable ideal of $\mathfrak{g}$ and hence its contained in the radical of
$\mathfrak{g}$ (the largest solvable ideal in $\mathfrak{g}$).

###### Proof.

The fact that $B$ is bi-invariant is an immediate consequence of (1). To show
that $K$ is bi-invariant, it suffices to show (3). Consider a complement $V$
of $\mathrm{Leib}(\mathfrak{g})$, i.e.,
$\mathfrak{g}=\mathrm{Leib}(\mathfrak{g})\oplus V$ and put, for any
$u,v\in\mathfrak{g}$,

$u\bullet v=[u,v]_{V}+\omega(u,v)$

where $[u,v]_{V}\in V$ and $\omega(u,v)\in\mathrm{Leib}(\mathfrak{g})$. Since,
for any $u,v\in\mathfrak{g}$, $u\bullet v+v\bullet
u\in\mathrm{Leib}(\mathfrak{g})$ then $[\;,\;]_{V}$ is skew-symmetric and, by
using the Leibniz identity, one can check easily that $(V,[\;,\;]_{V})$ is a
Lie algebra. For any $u\in V$, denote by
${\mathrm{ad}}_{u}^{V}:V\longrightarrow V$ the adjoint. For any
$u\in\mathrm{Leib}(\mathfrak{g})$ and $v\in V$,

$\mathrm{L}_{u}=0,\;\mathrm{R}_{u}=\left(\begin{array}[]{cc}0&\omega(.,u)\\\
0&0\end{array}\right),\;\mathrm{L}_{v}=\left(\begin{array}[]{cc}D_{v}&\omega(v,.)\\\
0&{\mathrm{ad}}_{v}^{V}\end{array}\right)\quad\mbox{and}\quad\mathrm{R}_{v}=\left(\begin{array}[]{cc}0&\omega(.,v)\\\
0&-{\mathrm{ad}}_{v}^{V}\end{array}\right).$

So

$K(u,v)=\begin{cases}0\quad\mbox{if}\quad
u\in\mathrm{Leib}(\mathfrak{g})\;\mbox{or}\;v\in\mathrm{Leib}(\mathfrak{g}),\\\
{\mathrm{tr}}_{V}({\mathrm{ad}}_{u}^{V}\circ{\mathrm{ad}}_{v}^{V})=K_{V}(u,v)\quad\mbox{if}\quad
u,v\in V,\end{cases}$

where $K_{V}$ is the Killing form of $(V,[\;,\;]_{V})$. From this formula, we
can deduce that $\pi:V\longrightarrow\mathfrak{g}/\mathrm{Leib}(\mathfrak{g})$
is an isomorphism of Lie algebras which sends $K_{V}$ to the quotient of $K$.

Since $K$ is bi-invariant then $\ker K$ is obviously an ideal. We have
$\mathrm{Leib}(\mathfrak{g})$ is solvable and, according to (3), $\ker
K/\mathrm{Leib}(\mathfrak{g})=\ker\widehat{K}$. It is a well-known fact [24]
that the kernel of the Killing form of a Lie algebra is solvable, hence $\ker
K$ is solvable. ∎

The 2-form $B$ was presented by many authors as the Killing form of the
Leibniz algebra (see [9, 5]). However, it presents a serious inconvenient,
namely, the quotient $\widetilde{B}$ of $B$ on
$\mathfrak{g}/\mathrm{Leib}(\mathfrak{g})$ is not, in general, the Killing
form of $\mathfrak{g}/\mathrm{Leib}(\mathfrak{g})$ as the following example
shows.

######  Example 1.

Consider the 2-dimensional Leibniz algebra $\mathfrak{g}=\hbox{\bb R}^{2}$
endowed with the bracket

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

Then $B(e_{1},e_{1})=B(e_{1},e_{2})=0$ and $B(e_{2},e_{2})=1$. The quotient
$\mathfrak{g}/\mathrm{Leib}(\mathfrak{g})=\hbox{\bb R}e_{2}$ is abelian but
the quotient of $\widetilde{B}$ of $B$ satisfies
$\widetilde{B}(e_{2},e_{2})=1$. However, $K=0$.

######  Remark 1.

From the proof of the last proposition, one can see that, for any $u,v\in V$,

$B(u,v)={\mathrm{tr}}_{\mathrm{Leib}(\mathfrak{g})}(D_{u}\circ
D_{v})+{\mathrm{tr}}_{V}({\mathrm{ad}}_{u}^{V}\circ{\mathrm{ad}}_{v}^{V})$

which explains the difference between $B$ and $K$.

It is known [9] that a left Leibniz algebra is semi-simple (resp. solvable) if
and only if $\mathfrak{g}/\mathrm{Leib}(\mathfrak{g})$ is semi-simple (resp.
solvable). Thus we can deduce the following result which generalize Cartan’s
criterions and shows that $K$ is the good notion of Killing form.

######  Proposition 2.3.

  1. 1.

A left Leibniz algebra $\mathfrak{g}$ is semi-simple if and only if $\ker
K=\mathrm{Leib}(\mathfrak{g})$.

  2. 2.

A left Leibniz algebra $\mathfrak{g}$ is solvable if and only if
$\mathfrak{g}\bullet\mathfrak{g}\subset\ker K$.

##  3 Metrised left Leibniz algebras

Associative metrics on left Leibniz algebra were studied in [12, 14]. It was
shown that if a left Leibniz algebra has an associative metric then it is also
right Leibniz. In this section, we complete the results obtained in [12, 14]
by giving a simple and complete description of symmetric Leibniz algebras
having an associative metric. We call such pseudo-Euclidean symmetric Leibniz
algebras metrised.

Let $(\mathfrak{g},\bullet)$ be a nonassociative algebra. Denote,
respectively, by $[\;,\;]$ and $\circ$ the skew-symmetric and symmetric parts
of $\bullet$. That is

$[u,v]=\frac{1}{2}(u\bullet v-v\bullet u),\;u\circ v=\frac{1}{2}(u\bullet
v+v\bullet u),\quad\mbox{and}\quad u\bullet v=[u,v]+u\circ v.\;\forall
u,v\in\mathfrak{g}.$ (4)

Put, ${\mathrm{ad}}_{u}:\mathfrak{g}\longrightarrow\mathfrak{g}$,
$v\mapsto[u,v]$ and
${\mathrm{sd}}:\mathfrak{g}\longrightarrow\mathfrak{g}$,$v\mapsto u\circ v$
for any $u\in\mathfrak{g}$. Hence,

${\mathrm{L}}_{u}={\mathrm{ad}}_{u}+{\mathrm{sd}}_{u}\quad\mbox{and}\quad{\mathrm{R}}_{u}=-{\mathrm{ad}}_{u}+{\mathrm{sd}}_{u}.$

The following result, even if it is easy to establish, will have an important
consequence and to our knowledge doesn’t appear in the literature.

######  Proposition 3.1.

Let $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ be an algebra endowed with a
metric $\langle\;,\;\rangle$. Then $\langle\;,\;\rangle$ is associative if and
only if $\langle\;,\;\rangle$ is bi-invariant on $(\mathfrak{g},[\;,\;])$ and
associative on $(\mathfrak{g},\circ)$.

###### Proof.

The metric is associative if and only if
${\mathrm{L}}_{u}={\mathrm{R}}_{u}^{*}$ for any $u\in g$. That is, for any
$u\in\mathfrak{g}$,

${\mathrm{ad}}_{u}+\mathrm{sd}_{u}=-{\mathrm{ad}}_{u}^{*}+\mathrm{sd}_{u}^{*}.$

Thus,
${\mathrm{ad}}_{u}+{\mathrm{ad}}_{u}^{*}={\mathrm{sd}}_{u}^{*}-{\mathrm{sd}}_{u}$.
Since ${\mathrm{ad}}_{u}+{\mathrm{ad}}_{u}^{*}$ is symmetric and
${\mathrm{sd}}_{u}^{*}-{\mathrm{sd}}_{u}$ skew-symmetric then
${\mathrm{ad}}_{u}+{\mathrm{ad}}_{u}^{*}={\mathrm{sd}}_{u}^{*}-{\mathrm{sd}}_{u}=0$,
for any $u\in\mathfrak{g}$ which completes the proof. ∎

Let $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ be a left Leibniz algebra
endowed with an associative metric. Then, for any $u\in\mathfrak{g}$,
${\mathrm{L}}_{u}^{*}={\mathrm{R}}_{u}$ and by using (1) and $(2)$, we deduce
that $(\mathfrak{g},\bullet)$ is also right Leibniz algebra. This remark
appeared in [12, Theorem 2.4]. We conclude that if a left Leibniz algebra has
an associative metric then it is a symmetric Leibniz algebra.

On the other hand, according to [11, Proposition 2.11], any symmetric Leibniz
algebra $(\mathfrak{g},\bullet)$ is given by

$u\bullet b=[a,b]+\omega(u,v),$

where $[\;,\;]$ is a Lie bracket on $\mathfrak{g}$ and $\omega$ is a bilinear
symmetric map $\omega\,:\mathfrak{g}\times\mathfrak{g}\longrightarrow
Z(\mathfrak{g})$ ($Z(\mathfrak{g})$ is the center of
$(\mathfrak{g},[\;,\;])$), $(x,y)\mapsto\omega(x,y)=x\circ y$ such that, for
any $x,\,y,\,z\in\mathfrak{g}$,

$\omega([x,y],z)=\omega(\omega(x,y),z)=0.$ (5)

We call $(\mathfrak{g},[\;,\;]+\omega)$ the symmetric Leibniz algebra obtained
from the Lie algebra $(\mathfrak{g},[\;,\;])$ by means of $\omega$.

By using this characterization of symmetric Leibniz algebras an Proposition
3.1, we get a new and complete description of symmetric Leibniz algebras
endowed with an associative metric. But before, we can give a complete
description of metrised symmetric Leibniz algebras whose underlying Lie
algebra is abelian.

######  Theorem 3.1.

Let $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ be a metrised commutative
symmetric Leibniz algebra. Then there exists a pseudo-Euclidean vector space
$(A,\langle\;,\;\rangle_{A})$, a vector space ${\mathfrak{h}}$ and a symmetric
3-linear form
$T:{\mathfrak{h}}\times{\mathfrak{h}}\times{\mathfrak{h}}\longrightarrow\hbox{\bb
R}$ such that $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ is isomorphic to
$({\mathfrak{h}}\oplus
A\oplus{\mathfrak{h}}^{*},\circ,\langle\;,\;\rangle_{0})$ where
${\mathfrak{h}}\circ{\mathfrak{h}}\subset{\mathfrak{h}}^{*}$ and for any
$X,Y,Z\in{\mathfrak{h}}$, $a\in A$, $\alpha\in{\mathfrak{h}}^{*}$,

$\langle X+a+\alpha,X+a+\alpha\rangle_{0}=2\prec\alpha,X\succ+\langle
a,a\rangle_{A},\;{\mathrm{L}}_{a}={\mathrm{L}}_{\alpha}=0\quad\mbox{and}\quad\prec
X\circ Y,Z\succ=T(X,Y,Z),$

###### Proof.

According to what above, a commutative symmetric Leibniz algebra is a
commutative algebra $(\mathfrak{g},\bullet)$ such that for any
$u,v\in\mathfrak{g}$, ${\mathrm{L}}_{u}\circ{\mathrm{L}}_{v}=0$. Suppose that
$\langle\;,\;\rangle$ is an associative metric on $\mathfrak{g}$. Then, for
any $u\in\mathfrak{g}$, ${\mathrm{L}}_{u}$ is symmetric with respect to
$\langle\;,\;\rangle$ and, if
$I=\sum_{u\in\mathfrak{g}}\mathrm{Im}{\mathrm{L}}_{u}$,
$I^{\perp}=\bigcap_{u\in\mathfrak{g}}\ker{\mathrm{L}}_{u}$ and $I\subset
I^{\perp}$. Write $I^{\perp}=I\oplus A$ for a complement $A$. The restriction
$\langle\;,\;\rangle_{A}$ of the metric to $A$ is nondegenerate and there
exists a totally isotropic space ${\mathfrak{h}}$ such that
$A^{\perp}=I\oplus{\mathfrak{h}}$ and $\mathfrak{g}=I\oplus
A\oplus{\mathfrak{h}}$. The metric identify $I$ to the dual
${\mathfrak{h}}^{*}$ and we can identify $(\mathfrak{g},\langle\;,\;\rangle)$
to $({\mathfrak{h}}^{*}\oplus A\oplus{\mathfrak{h}},\langle\;,\;\rangle_{0})$.

With this identification in mind, the bracket $\bullet$ satisfies
${\mathfrak{h}}^{*}\oplus A\subset Z({\mathfrak{h}}^{*}\oplus
A\oplus{\mathfrak{h}})$ and
${\mathfrak{h}}\bullet{\mathfrak{h}}\subset{\mathfrak{h}}^{*}$ and the
3-linear form given by

$T(X,Y,Z):=\prec X\bullet Y,Z\succ=\prec Y,X\bullet Z\rangle,$

for any $X,Y,Z\in{\mathfrak{h}}$ is symmetric and defines $\bullet$.∎

We state now an important result.

######  Theorem 3.2.

Let $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ be a pseudo-Euclidean
symmetric Leibniz algebra obtained from $(\mathfrak{g},[\;,\;])$ by means of
$\omega$. We denote by $Z(\mathfrak{g})$ the center of
$(\mathfrak{g},[\;,\;])$. Then $\langle\;,\;\rangle$ is associative if and
only if $(\mathfrak{g},[\;,\;],\langle\;,\;\rangle)$ is a quadratic Lie
algebra and there exists a 3-linear symmetric form
$T:\mathfrak{g}\times\mathfrak{g}\times\mathfrak{g}\longrightarrow\hbox{\bb
R}$ and a totally isotropic vector subspace $I\subset Z(\mathfrak{g})$ such
that $T(I^{\perp},.,.)=0$ and

$\langle\omega(x,y),z\rangle=T(x,y,z),\quad x,y,z\in\mathfrak{g}.$ (6)

###### Proof.

According to Proposition 3.1 and the description of symmetric Leibniz algebras
given above, if $\langle\;,\;\rangle$ is associative then
$(\mathfrak{g},[\;,\;],\langle\;,\;\rangle)$ is a quadratic Lie algebra and
$\omega$ satisfies $I=\mathrm{span}\\{\omega(x,y),x,y\in\mathfrak{g}\\}\subset
Z(\mathfrak{g})$,

$\omega([x,y],z)=\omega(\omega(x,y),z)=0\quad\mbox{and}\quad\langle\omega(x,y),z\rangle=\langle\omega(x,z),y\rangle,\quad
x,y,z\in\mathfrak{g}.$

For any $x,y,z,t\in\mathfrak{g}$,

$0=\langle\omega(\omega(x,y),z),t\rangle=\langle\omega(z,t),\omega(x,y)\rangle=0$

and hence $I$ is totally isotropic. Put
$T(x,y,z)=\langle\omega(x,y),z\rangle$. Then $T$ is symmetric and for any
$z\in I^{\perp}$ and $x,y\in\mathfrak{g}$,

$T(x,y,z)=\langle\omega(x,y),z\rangle=0.$

Conversely, let $(\mathfrak{g},[\;,\;],\langle\;,\;\rangle)$ be a quadratic
Lie algebra, $I\subset Z(\mathfrak{g})$ a totally isotropic vector subspace
and
$T:\mathfrak{g}\times\mathfrak{g}\times\mathfrak{g}\longrightarrow\hbox{\bb
R}$ a 3-linear symmetric form such that $T(I^{\perp},.,.)=0$. Define the
symmetric bilinear form
$\omega:\mathfrak{g}\times\mathfrak{g}\longrightarrow\mathfrak{g}$ by

$T(x,y,z)=\langle\omega(x,y),z\rangle=0,\;x,y,z\in\mathfrak{g}.$

It is obvious that $\omega$ takes its values in $I$ and, for any
$x,y,z,t\in\mathfrak{g}$,

$\langle\omega(\omega(x,y),z),t\rangle=T(\omega(x,y),z,t)=0\quad\mbox{and}\quad\langle\omega([x,y],z),t\rangle=\langle\omega(z,t),[x,y]\rangle=-\langle[x,\omega(z,t)],y\rangle=0.$

So $(\mathfrak{g},[\;,\;]+\omega,\langle\;,\;\rangle)$ is a metrised symmetric
Leibniz algebra. ∎

######  Remark 2.

Let $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ be a metrised symmetric
Leibniz algebra obtained from $(\mathfrak{g},[\;,\;])$ by means of $\omega$.
If $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$ then
$Z(\mathfrak{g})=[\mathfrak{g},\mathfrak{g}]^{\perp}=\\{0\\}$. Thus $\omega=0$
and hence $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ is a quadratic Lie
algebra.

In Theorem 3.2, the determination of $I$ and $T$ is an easy task so the
determination of metrised non Lie symmetric Leibniz algebras reduces to the
determination of quadratic Lie algebras with non trivial center. If the center
is nondegenerate, we will see that they can be described easily. Moreover,
quadratic Lie algebras with non trivial degenerate center are determined by
the double extension process (see [28, 19]). Let us recall the process of
double extension (see [28] for more details).

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

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

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

######  Theorem 3.3.

Let $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ be a metrised symmetric
Leibniz algebra obtained from $(\mathfrak{g},[\;,\;])$ by means of $\omega$
such that $[\mathfrak{g},\mathfrak{g}]\not=\mathfrak{g}$. Then the following
assertions hold.

  1. 1.

If $Z(\mathfrak{g})$ is non-degenerate then
$\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]\oplus Z(\mathfrak{g})$ and there
exists a totally isotropic subspace $I\in Z(\mathfrak{g})$ and a 3-linear form
$T:Z(\mathfrak{g})\times Z(\mathfrak{g})\times
Z(\mathfrak{g})\longrightarrow\hbox{\bb R}$ such that $T(I^{\perp}\cap
Z(\mathfrak{g}),.,.)=0$ and

$x\bullet y=\begin{cases}[x,y]\quad\mbox{if}\quad
x\in[\mathfrak{g},\mathfrak{g}]\;\mbox{or}\;y\in[\mathfrak{g},\mathfrak{g}],\\\
\omega(x,y)\quad\mbox{if}\quad x,y\in Z(\mathfrak{g}).\end{cases}$

where the restriction of $\omega$ to $Z(\mathfrak{g})$ is given by (6).

  2. 2.

If $Z(\mathfrak{g})$ is degenerate then
$(\mathfrak{g},[\;,\;],\langle\;,\;\rangle)=\delta_{A}({\mathfrak{h}})$ and
when $A$ is invertible or
$({\mathfrak{h}},[\;,\;]_{\mathfrak{h}},\langle\;,\;\rangle_{\mathfrak{h}})$
is Euclidean then $\omega$ satisfies $\omega({\mathfrak{h}}\oplus\hbox{\bb
R}e,.)=0$ and $\omega(\bar{e},\bar{e})=\mu e$ with $\mu\in\hbox{\bb R}$.

###### Proof.

The first assertion is an immediate consequence of Theorem 3.2 and the fact
that $[\mathfrak{g},\mathfrak{g}]^{\perp}=Z(\mathfrak{g})$.

If If $Z(\mathfrak{g})$ is degenerate then
$(\mathfrak{g},[\;,\;],\langle\;,\;\rangle)$ is obtained by the double
extension process and hence
$(\mathfrak{g},[\;,\;],\langle\;,\;\rangle)=\delta_{A}({\mathfrak{h}})$. In
this case, $Z(\mathfrak{g})=\hbox{\bb R}e\oplus Z({\mathfrak{h}})\cap\ker A$.
If $A$ is invertible then $Z(\mathfrak{g})=\hbox{\bb R}e$ and we get the
result as consequence of Theorem 3.2. If the metric on ${\mathfrak{h}}$ is
definite positive then $\hbox{\bb R}e$ is the only non trivial totally
isotropic subspace of $Z(\mathfrak{g})$ and the result follows. ∎

######  Corollary 3.1.

Let $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ be a metrised symmetric
Leibniz algebra such the restriction of $\langle\;,\;\rangle$ to
$Z(\mathfrak{g})$ is definite positive. Then
$(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ is a quadratic Lie algebra.

The lists of indecomposable quadratic Lie algebras of dimension $\leq 6$, up
to an automorphism, are given in [28] and [13]. From these lists, we can find
all quadratic Lie algebras of dimension $\leq 6$, up to an automorphism. By
using Theorems 3.1-3.3 and Corollary 3.1, we get almost without any
computation the following result.

######  Theorem 3.4.

  1. 1.

In dimension 2 there is only one metrised non Lie symmetric Leibniz algebra up
to an isomorphism, namely $\hbox{\bb R}^{2}$ endowed with the bracket and the
metric

$e_{1}\bullet e_{1}=\pm e_{2}\quad\mbox{and}\quad
m=\left(\begin{matrix}0&\alpha\\\ \alpha&0\end{matrix}\right),\quad\alpha>0.$

  2. 2.

In dimension 3 there is only one metrised non Lie symmetric Leibniz algebra up
to an isomorphism, namely $\hbox{\bb R}^{3}$ endowed with the bracket and the
metrics

$e_{1}\bullet e_{1}=\pm e_{3}\quad\mbox{and}\quad
m=\left(\begin{matrix}0&\alpha&0\\\ \alpha&0&0\\\ 0&0&\pm
1\end{matrix}\right),\quad\alpha>0.$

  3. 3.

In dimension 4,5 and 6, up to an isomorphism, all symmetric Leibniz algebra
carrying an associative metric and obtained from an indecomposable quadratic
Lie algebra are given in 1.

##  4 ${\mathrm{L}}$-quadratic and ${\mathrm{R}}$-quadratic left Leibniz
algebras

In this section, we give some immediate properties of ${\mathrm{L}}$-quadratic
and ${\mathrm{R}}$-quadratic left Leibniz algebras, we describe
${\mathrm{L}}$-quadratic left Leibniz algebras with nondegenerate Leibniz
ideal and we define the core of ${\mathrm{L}}$-quadratic and
${\mathrm{R}}$-quadratic left Leibniz algebras.

###  4.1 General properties of ${\mathrm{L}}$-quadratic and
${\mathrm{R}}$-quadratic left Leibniz algebras

A ${\mathrm{L}}$-quadratic (resp. ${\mathrm{R}}$-quadratic) left Leibniz
algebra is a pseudo-Euclidean left Leibniz algebra
$(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ such that the metric is
${\mathrm{L}}$-invariant (resp. ${\mathrm{R}}$-invariant). This means that for
any $u\in\mathfrak{g}$, ${\mathrm{L}}_{u}^{*}=-{\mathrm{L}}_{u}$ (resp.
${\mathrm{R}}_{u}^{*}=-{\mathrm{R}}_{u}$).

According to Proposition 2.1, if a pseudo-Euclidean left Leibniz algebra is
both ${\mathrm{L}}$-quadratic and ${\mathrm{R}}$-quadratic then it is a Lie
algebra.

The following example shows that the classes of ${\mathrm{L}}$-quadratic or
${\mathrm{R}}$-quadratic left Leibniz algebras are large.

######  Example 2.

  1. 1.

Let $(\mathfrak{g},[\;,\;])$ be a Lie algebra. We denote by
$\mathrm{ad}^{*}:\mathfrak{g}\longrightarrow\mathrm{End}(\mathfrak{g}^{*})$
its co-adjoint representation, i.e.,
$\prec\mathrm{ad}^{*}_{u}\alpha,v\succ=-\prec\alpha,[u,v]\succ$. We endow
$T^{*}\mathfrak{g}=\mathfrak{g}\oplus\mathfrak{g}^{*}$ by the bracket

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

and its canonical neutral metric given by

$\langle u+\alpha,v+\beta\rangle=\prec\alpha,v\succ+\prec\beta,u\succ.$

Then $(T^{*}\mathfrak{g},\bullet,\langle\;,\;\rangle)$ is a
${\mathrm{L}}$-quadratic left Leibniz algebra.

  2. 2.

Let, $(\mathfrak{g},[\;,\;])$ be a 2-step nilpotent Lie algebra. . We endow
$T^{*}\mathfrak{g}=\mathfrak{g}\oplus\mathfrak{g}^{*}$ by the bracket

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

and its canonical neutral metric $\langle\;,\;\rangle$. Then
$(T^{*}\mathfrak{g},\bullet,\langle\;,\;\rangle)$ is a
${\mathrm{R}}$-quadratic left Leibniz algebra.

The following proposition can be easily established.

######  Proposition 4.1.

Let $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ be a pseudo-Euclidean left
Leibniz algebra and $I$ an ideal of $\mathfrak{g}$. Then the following
assertions hold.

  1. 1.

If $\langle\;,\;\rangle$ is ${\mathrm{L}}$-invariant
$($resp.${\mathrm{R}}$-invariant) then
$Z^{r}(\mathfrak{g})=(\mathfrak{g}\bullet\mathfrak{g})^{\perp}$ (resp.
$Z^{l}(\mathfrak{g})=(\mathfrak{g}\bullet\mathfrak{g})^{\perp})$.

  2. 2.

If $\langle\;,\;\rangle$ is ${\mathrm{L}}$-invariant, then $I^{\perp}$ is a
left ideal of $\mathfrak{g}$ and $I\bullet I^{\perp}=0$. Moreover, $I^{\perp}$
is an ideal of $\mathfrak{g}$ if and only if $I^{\bot}\bullet I=0$.

  3. 3.

If $\langle\;,\;\rangle$ is ${\mathrm{R}}$-invariant, then $I^{\perp}$ is a
right ideal of $\mathfrak{g}$ and $I^{\perp}\bullet I=0$. Moreover,
$I^{\perp}$ is an ideal of $\mathfrak{g}$ if and only if $I\bullet
I^{\perp}=0$.

The following proposition will be useful later.

######  Proposition 4.2.

Let $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ be a pseudo-Euclidean left
Leibniz algebra. Then the following assertions hold.

  1. 1.

If $\langle\;,\;\rangle$ is ${\mathrm{L}}$-invariant then
$\mathrm{Leib}(\mathfrak{g})^{\perp}=\\{u\in\mathfrak{g},\mathrm{R}_{u}+\mathrm{R}_{u}^{*}=0\\}.$
Moreover, $\mathrm{Leib}(\mathfrak{g})^{\perp}$ is a left ideal,
$I=\mathrm{Leib}(\mathfrak{g})+\mathrm{Leib}(\mathfrak{g})^{\perp}$ is an
ideal and $I\bullet I^{\perp}=0$

  2. 2.

If $\langle\;,\;\rangle$ is ${\mathrm{R}}$-invariant then
$\mathrm{Leib}(\mathfrak{g})^{\perp}=\\{u\in\mathfrak{g},\mathrm{L}_{u}+\mathrm{L}_{u}^{*}=0\\}.$
Moreover,
$\mathrm{Leib}(\mathfrak{g})\subset\mathfrak{g}\bullet\mathfrak{g}\subset\mathrm{Leib}(\mathfrak{g})^{\perp}$.
Thus, $\mathrm{Leib}(\mathfrak{g})$ is a totally isotropic ideal of
$\mathfrak{g}$, $\mathrm{Leib}(\mathfrak{g})^{\perp}$ is an ideal and
$\mathrm{Leib}(\mathfrak{g})^{\perp}\bullet\mathrm{Leib}(\mathfrak{g})=0$.

###### Proof.

  1. 1.

Let $u\in\mathrm{Leib}(\mathfrak{g})^{\perp}$. We have, for any
$v,w\in\mathfrak{g}$,

$0=\langle u,v\bullet w+w\bullet v\rangle\\\ =-\langle v\bullet
u,w\rangle-\langle w\bullet u,v\rangle\\\
=-\langle{\mathrm{R}}_{u}v,w\rangle-\langle{\mathrm{R}}_{u}w,v\rangle$

and hence ${\mathrm{R}}_{u}^{*}=-{\mathrm{R}}_{u}$. According to Proposition
4.1, $\mathrm{Leib}(\mathfrak{g})^{\perp}$ is a left ideal. Put
$I=\mathrm{Leib}(\mathfrak{g})+\mathrm{Leib}(\mathfrak{g})^{\perp}$. Then $I$
is a left ideal,
$I^{\perp}=\mathrm{Leib}(\mathfrak{g})\cap\mathrm{Leib}(\mathfrak{g})^{\perp}$
and, for any $u\in\mathfrak{g}$, $v\in I$ and $w\in I^{\perp}$,

$\langle v\bullet u,w\rangle=-\langle u,v\bullet w\rangle=\langle u\bullet
w,v\rangle=0$

since $I^{\perp}$ is a left ideal and
${\mathrm{R}}_{w}^{*}=-{\mathrm{R}}_{w}$. Hence $I$ is an ideal and $I\bullet
I^{\perp}=0$, according to Proposition 4.1.

  2. 2.

The first part of the second assertion can be obtained similarly. Since
$\mathrm{Leib}(\mathfrak{g})$ is an ideal,
$\mathrm{Leib}(\mathfrak{g})^{\perp}\bullet\mathrm{Leib}(\mathfrak{g})=0$ by
virtue of Proposition 4.1.

On the other hand, recall that for any $u\in\mathrm{Leib}(\mathfrak{g})$,
${\mathrm{L}}_{u}=0$. Then for $v,w\in\mathfrak{g}$ and
$u\in\mathrm{Leib}(\mathfrak{g})$,

$\langle u,v\bullet w\rangle=-\langle u\bullet w,v\rangle=0$

so
$\mathrm{Leib}(\mathfrak{g})\subset(\mathfrak{g}\bullet\mathfrak{g})^{\perp}$
which completes the proof since obviously
$\mathrm{Leib}(\mathfrak{g})\subset\mathfrak{g}\bullet\mathfrak{g}$.∎

The following result is an immediate consequence of Proposition 4.2.

######  Proposition 4.3.

A ${\mathrm{R}}$-quadratic Euclidean left Leibniz algebra is necessarily a Lie
algebra.

###  4.2 ${\mathrm{L}}$-quadratic left Leibniz algebras with nondegenerate
Leibniz ideal

${\mathrm{L}}$-quadratic left Leibniz algebras with nondegenerate Leibniz
ideal are quite easy to describe. We call them nondegenerate
${\mathrm{L}}$-quadratic left Leibniz algebras.

Let $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ be a ${\mathrm{L}}$-quadratic
left Leibniz algebra such that $\mathrm{Leib}(\mathfrak{g})$ is nondegenerate.
According to Proposition 4.2,
$\mathfrak{g}=\mathrm{Leib}(\mathfrak{g})\oplus\mathrm{Leib}(\mathfrak{g})^{\perp}$,
where $(\mathrm{Leib}(\mathfrak{g})^{\perp},\bullet,\langle\;,\;\rangle)$ is a
quadratic Lie algebra and there is a representation
$\rho:\mathrm{Leib}(\mathfrak{g})^{\perp}\longrightarrow\mathrm{so}(\mathrm{Leib}(\mathfrak{g}),\langle\;,\;\rangle)$
given by $\rho(u)(v)=u\bullet v$.

Conversely, let $({\mathfrak{h}},\langle\;,\;\rangle_{\mathfrak{h}})$ be a
quadratic Lie algebra, $(N,\langle\;,\;\rangle_{N})$ a pseudo-Euclidean vector
space and
$\rho:{\mathfrak{h}}\longrightarrow\mathrm{so}(N,\langle\;,\;\rangle_{N})$ a
representation of ${\mathfrak{h}}$. We define on
$\mathfrak{g}={\mathfrak{h}}\oplus N$ the product

$u\bullet v=\begin{cases}0,\;\mbox{if}\;u\in N,\\\
[u,v]_{\mathfrak{h}}\;\mbox{if}\;u,v\in{\mathfrak{h}},\\\
\rho_{u}(v)\;\mbox{if}\;u\in{\mathfrak{h}},v\in N,\end{cases}$

and the metric
$\langle\;,\;\rangle=\langle\;,\;\rangle_{\mathfrak{h}}\oplus\langle\;,\;\rangle_{N}$.
Then $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ is a
${\mathrm{L}}$-quadratic left Leibniz algebra.

###  4.3 The core of ${\mathrm{L}}$-quadratic and ${\mathrm{R}}$-quadratic
left Leibniz algebras

In this subsection, we construct two associated algebras for any
${\mathrm{L}}$-quadratic (resp. ${\mathrm{R}}$-quadratic) left Leibniz
algebra: a Lie algebra and a nondegenerate ${\mathrm{L}}$-quadratic left
Leibniz algebra (resp. a quadratic Lie algebra). We call these algebras the
core of the ${\mathrm{L}}$-quadratic (resp. ${\mathrm{R}}$-quadratic) left
Leibniz algebra and they will be used in Sections 5 and 6 to reconstruct the
original algebra.

######  Proposition 4.4.

Let $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ be a ${\mathrm{L}}$-quadratic
(resp. ${\mathrm{R}}$-quadratic) left Leibniz algebra and denote by
$I=\mathrm{Leib}(\mathfrak{g})\cap\mathrm{Leib}(\mathfrak{g})^{\perp}$ (resp.
$I=\mathrm{Leib}(\mathfrak{g})$), $A=I^{\perp}/I$ and
${\mathfrak{h}}=\mathfrak{g}/I^{\perp}$. Let $\pi_{A}:I^{\perp}\longrightarrow
A$ and $\pi_{\mathfrak{h}}:\mathfrak{g}\longrightarrow{\mathfrak{h}}$ the
canonical projections. Then the following assertions hold.

  1. 1.

If the metric is ${\mathrm{L}}$-quadratic then $A$ has a structure of
${\mathrm{L}}$-quadratic left Leibniz algebra given by

$\pi_{A}(u)\bullet_{A}\pi_{A}(v)=\pi_{A}(u\bullet
v)\quad\mbox{and}\quad\langle\pi_{A}(u),\pi_{A}(v)\rangle_{A}=\langle
u,v\rangle$

and $\mathrm{Leib}(A)=\pi_{A}(\mathrm{Leib}(\mathfrak{g}))$ is nondegenerate
and

$A=\mathrm{Leib}(A)\stackrel{{\scriptstyle\perp}}{{\oplus}}\pi_{A}(\mathrm{Leib}(\mathfrak{g})^{\perp})$

where $\pi_{A}(\mathrm{Leib}(\mathfrak{g})^{\perp})$ is a quadratic Lie
algebra.

  2. 2.

If the metric is ${\mathrm{R}}$-quadratic then $A$ has a structure of a
quadratic Lie algebra given by

$[\pi_{A}(u),\pi_{A}(v)]_{A}=\pi_{A}(u\bullet
v)\quad\mbox{and}\quad\langle\pi_{A}(u),\pi_{A}(v)\rangle_{A}=\langle
u,v\rangle.$

  3. 3.

If the metric is ${\mathrm{L}}$-invariant then $I^{\perp}$ is an ideal of
$\mathfrak{g}$ and ${\mathfrak{h}}=\mathfrak{g}/I^{\perp}$ has a structure
$[\;,\;]_{\mathfrak{h}}$ of a Lie algebra.

  4. 4.

If the metric is ${\mathrm{R}}$-invariant then the bracket
$[\;,\;]_{\mathfrak{h}}$ on ${\mathfrak{h}}$ given by the formula

$[\pi_{\mathfrak{h}}(u),\pi_{\mathfrak{h}}(v)]_{\mathfrak{h}}=-\pi_{\mathfrak{h}}({\mathrm{L}}_{v}^{*}u)$
(7)

is well-defined and it is a Lie bracket and for any $X,Y,Z\in{\mathfrak{h}}$,

$[[X,[Y,Z]_{\mathfrak{h}}]]_{\mathfrak{h}}=0.$

###### Proof.

  1. 1.

According to Proposition 4.2, $\mathrm{Leib}(\mathfrak{g})$ is an ideal and
$\mathrm{Leib}(\mathfrak{g})^{\perp}$ is a left ideal and hence $I^{\perp}$ is
a subalgebra of $\mathfrak{g}$ and $I$ is an ideal in $I^{\perp}$. This
implies that $A$ has a structure of left Leibniz algebra and the other claims
follow immediately.

  2. 2.

According to Proposition 4.2, both $\mathrm{Leib}(\mathfrak{g})$ and
$\mathrm{Leib}(\mathfrak{g})^{\perp}$ are ideal and
$\mathrm{Leib}(\mathfrak{g})\subset\mathrm{Leib}(\mathfrak{g})^{\perp}$ this
shows that $A$ has a structure of left Leibniz algebra. Moreover, for any
$u\in\mathrm{Leib}(\mathfrak{g})^{\perp}$, $u\bullet
u\in\mathrm{Leib}(\mathfrak{g})$ and hence $\pi_{A}(u\bullet u)=0$ which shows
that $A$ is actually a Lie algebra.

  3. 3.

We have seen in Proposition 4.2 that $I^{\perp}$ is an ideal and hence
$\mathfrak{g}/I^{\perp}$ has a structure of left Leibniz algebra. Moreover,
for any $u\in\mathfrak{g}$, $u\bullet u\in\mathrm{Leib}(\mathfrak{g})\subset
I^{\perp}$ and hence $\pi_{\mathfrak{h}}(u\bullet u)=0$ so the quotient is
actually a Lie algebra.

  4. 4.

We have seen in Proposition 4.2 that $\mathrm{Leib}(\mathfrak{g})^{\perp}$ is
an ideal which contains $\mathfrak{g}\bullet\mathfrak{g}$ and hence the
quotient $\mathfrak{g}/\mathrm{Leib}(\mathfrak{g})^{\perp}$ is trivial.
However, we will show that the bracket $[\;,\;]_{\mathfrak{h}}$ is a Lie
bracket which is not, in general, the quotient bracket.

Note first that for any $u,v,w\in\mathfrak{g}$,

$\langle{\mathrm{L}}_{u}^{*}v,w\rangle=\langle v,u\bullet w\rangle=-\langle
v\bullet w,u\rangle=-\langle{\mathrm{L}}_{v}^{*}u,w\rangle$

and hence ${\mathrm{L}}_{u}^{*}v=-{\mathrm{L}}_{v}^{*}u$. Moreover, for any
$u\in I^{\perp}$, $v\in\mathfrak{g}$ and $w\in I$,

$\langle{\mathrm{L}}_{u}^{*}v,w\rangle=\langle v,u\bullet w\rangle=-\langle
v\bullet w,u\rangle=0$

since $I=\mathrm{Leib}(\mathfrak{g})$ is an ideal. So we have shown that
$[\;,\;]_{\mathfrak{h}}$ is well-defined and skew-symmetric.

Let $X=\pi_{\mathfrak{h}}(u),Y=\pi_{\mathfrak{h}}(v),Z=\pi_{\mathfrak{h}}(w)$.
We have

$[X,[Y,Z]_{\mathfrak{h}}]_{\mathfrak{h}}=\pi_{\mathfrak{h}}({\mathrm{L}}_{u}^{*}\circ{\mathrm{L}}_{v}^{*}w).$

By virtue of (1),

${\mathrm{L}}_{u}^{*}\circ{\mathrm{L}}_{v}^{*}w={\mathrm{L}}_{v}^{*}\circ{\mathrm{L}}_{u}^{*}w-{\mathrm{L}}_{v\bullet
u}^{*}w.$

But for any $t\in I=\mathrm{Leib}(\mathfrak{g})$, we have

$\langle{\mathrm{L}}_{v\bullet u}^{*}w,t\rangle=\langle w,(v\bullet u)\bullet
t\rangle=-\langle w\bullet t,v\bullet u\rangle=0$

since we have seen in Proposition 4.2 that $I$ is an ideal and
$I\subset(\mathfrak{g}\bullet\mathfrak{g})^{\perp}$. So we get that for any
$X,Y,Z\in{\mathfrak{h}}$,

$[X,[Y,Z]_{\mathfrak{h}}]_{\mathfrak{h}}=[Y,[X,Z]_{\mathfrak{h}}]_{\mathfrak{h}}.$

This relation combined to the skew-symmetry of $[\;,\;]_{\mathfrak{h}}$ will
show that $[X,[Y,Z]_{\mathfrak{h}}]_{\mathfrak{h}}=0$. Indeed,

$\displaystyle[X,[Y,Z]_{\mathfrak{h}}]_{\mathfrak{h}}$
$\displaystyle=[Y,[X,Z]_{\mathfrak{h}}]_{\mathfrak{h}}$
$\displaystyle=-[Y,[Z,X]_{\mathfrak{h}}]_{\mathfrak{h}}$
$\displaystyle=-[Z,[Y,X]_{\mathfrak{h}}]_{\mathfrak{h}}$
$\displaystyle=[Z,[X,Y]_{\mathfrak{h}}]_{\mathfrak{h}}$
$\displaystyle=[X,[Z,Y]_{\mathfrak{h}}]_{\mathfrak{h}}$
$\displaystyle=-[X,[Y,Z]_{\mathfrak{h}}]_{\mathfrak{h}}$

which completes the proof.∎

######  Definition 4.1.

Let $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ be a ${\mathrm{L}}$-quadratic
(resp. ${\mathrm{R}}$-quadratic) left Leibniz algebras. We call the
${\mathrm{L}}$-quadratic left Leibniz algebra
$(A,\bullet_{A},\langle\;,\;\rangle_{A})$ and the Lie algebra
$({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$ (resp. the quadratic Lie algebra
$(A,[\;,\;]_{A},\langle\;,\;\rangle_{A})$ and the Lie algebra
$({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$) defined in Proposition 4.4 the core
of $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$.

Actually, we can go further and give more information about the core of a
${\mathrm{L}}$-quadratic or ${\mathrm{R}}$-quadratic left Leibniz algebra.

######  Proposition 4.5.

Let $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ be a pseudo-Euclidean left
Leibniz algebra. Denote by
$\mathrm{Lie}(\mathfrak{g})=\mathfrak{g}/\mathrm{Leib}(\mathfrak{g})$ the
quotient Lie algebra and
$\pi:\mathfrak{g}\longrightarrow\mathrm{Lie}(\mathfrak{g})$ the canonical
projection.

  1. 1.

If $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ is ${\mathrm{L}}$-quadratic
then the map $\kappa:A\longrightarrow\mathrm{Lie}(\mathfrak{g})$ is a well-
defined morphism of left algebras,
$\ker\kappa=\pi_{A}(\mathrm{Leib}(\mathfrak{g}))$ and
$\pi_{A}(\mathrm{Leib}(\mathfrak{g})^{\perp})$ with an ideal of
$\mathrm{Lie}(\mathfrak{g})$ and the Lie algebra quotient
$\mathrm{Lie}(\mathfrak{g})/\pi_{A}(\mathrm{Leib}(\mathfrak{g})^{\perp})$ is
isomorphic to $({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$

  2. 2.

If $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ is ${\mathrm{R}}$-quadratic
then the map $\kappa:A\longrightarrow\mathrm{Lie}(\mathfrak{g})$ is a well-
defined into morphism of Lie algebras and $\kappa(A)$ is an ideal of
$\mathrm{Lie}(\mathfrak{g})$ which contains
$[\mathrm{Lie}(\mathfrak{g}),\mathrm{Lie}(\mathfrak{g})]$.

###### Proof.

  1. 1.

The first part of statement is obvious and the second part is a consequence of
the fact that the map
$m:\mathrm{Lie}(\mathfrak{g})\longrightarrow{\mathfrak{h}}$,
$\pi(u)\mapsto\pi_{\mathfrak{h}}(u)$ is an onto morphism of Lie algebras and
$\ker m=\pi_{A}(\mathrm{Leib}(\mathfrak{g})^{\perp})$.

  2. 2.

It is clear if one has in mind the fact
$\mathrm{Leib}(\mathfrak{g})\subset\mathfrak{g}\bullet\mathfrak{g}\subset\mathrm{Leib}(\mathfrak{g})^{\perp}$
established in Proposition 4.2.∎

Recall that a left Leibniz algebra $\mathfrak{g}$ is called simple if, by
definition, the only ideals of $\mathfrak{g}$ are $\\{0\\}$,
$\mathrm{Leib}(\mathfrak{g})$ and $\mathfrak{g}$ and
$\mathfrak{g}\bullet\mathfrak{g}\not=\mathrm{Leib}(\mathfrak{g})$. It is
called semi-simple if its radical is equal to $\mathrm{Leib}(\mathfrak{g})$
which is equivalent to $\mathrm{Leib}(\mathfrak{g})$ is semi-simple. One can
see [5] for details.

From what above we can derive the following important results.

######  Proposition 4.6.

Let $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ be a pseudo-Euclidean left
Leibniz algebra. Then the following assertions hold.

  1. 1.

If $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ is ${\mathrm{L}}$-quadratic
and simple then either
$\mathrm{Leib}(\mathfrak{g})=\mathrm{Leib}(\mathfrak{g})^{\perp}$, $A=\\{0\\}$
and ${\mathfrak{h}}=\mathrm{Lie}(\mathfrak{g})$ or
$\mathrm{Leib}(\mathfrak{g})$ is nondegenerate,
$\mathrm{Leib}(\mathfrak{g})^{\perp}$ is a simple quadratic Lie algebra and
$\mathfrak{g}=\mathrm{Leib}(\mathfrak{g})\oplus\mathrm{Leib}(\mathfrak{g})^{\perp}$.

  2. 2.

If $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ is ${\mathrm{R}}$-quadratic
and semi-simple then $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ is a
quadratic Lie algebra.

###### Proof.

  1. 1.

According to Proposition 4.2,
$\mathrm{Leib}(\mathfrak{g})+\mathrm{Leib}(\mathfrak{g})^{\perp}$ is an ideal
and hence
$\mathrm{Leib}(\mathfrak{g})+\mathrm{Leib}(\mathfrak{g})^{\perp}=\mathrm{Leib}(\mathfrak{g})$
or
$\mathrm{Leib}(\mathfrak{g})+\mathrm{Leib}(\mathfrak{g})^{\perp}=\mathfrak{g}$.
In the first case
$\mathrm{Leib}(\mathfrak{g})^{\perp}\subset\mathrm{Leib}(\mathfrak{g})$ and
$\mathrm{Leib}(\mathfrak{g})^{\perp}$ becomes an ideal and
$\mathrm{Leib}(\mathfrak{g})^{\perp}=\mathrm{Leib}(\mathfrak{g})$. Thus
$A=\\{0\\}$ and ${\mathfrak{h}}=\mathrm{Lie}(\mathfrak{g})$. In the second
case $\mathrm{Leib}(\mathfrak{g})$ is nondegenerate,
$\mathfrak{g}=\mathrm{Leib}(\mathfrak{g})\oplus\mathrm{Leib}(\mathfrak{g})^{\perp}$
and $\mathrm{Leib}(\mathfrak{g})^{\perp}$ is a simple quadratic Lie algebra.

  2. 2.

If $\mathfrak{g}$ is semi-simple then
$\mathfrak{g}\bullet\mathfrak{g}=\mathfrak{g}$ but, according to Proposition
4.2,
$\mathfrak{g}\bullet\mathfrak{g}\subset\mathrm{Leib}(\mathfrak{g})^{\perp}$
and hence $\mathrm{Leib}(\mathfrak{g})=\\{0\\}$ which completes the proof.∎

##  5  Description of ${\mathrm{L}}$-quadratic left Leibniz algebras with
degenerate Leibniz ideal

In this section, we utilize the core of a ${\mathrm{L}}$-quadratic left
Leibniz algebra, as defined in Definition 4.1, to reconstruct the original
algebra using a process that resembles a combination of double extension and
$T^{*}$-extension. This reconstruction method enables us to derive several
significant results. The double extension is a well-known process which
appeared first in [27] in the study of quadratic Lie algebras while the
$T^{*}$-extension was introduced in [17] in the study of metrised
nonassociative algebras.

For an algebra $(A,\bullet)$, $\mathrm{Der}(A)$ is the Lie algebra of
derivations of $A$, and if $A$ carries a pseudo-Euclidean metric
$\langle\;,\;\rangle_{A}$, $\mathrm{so}(A)$ is the Lie algebra of skew-
symmetric endomorphisms of $A$ and, for any vector space ${\mathfrak{h}}$, we
call natural the metric $\langle\;,\;\rangle_{n}$ on ${\mathfrak{h}}\oplus
A\oplus{\mathfrak{h}}^{*}$ given by

$\langle X+a+\alpha,X+a+\alpha\rangle_{n}=2\prec\alpha,X\rangle+\langle
a,a\rangle_{A},\quad X\in{\mathfrak{h}},a\in A,\alpha\in{\mathfrak{h}}^{*}.$
(8)

For a Lie algebra $(\mathfrak{g},[\;,\;])$,
${\mathrm{ad}}:\mathfrak{g}\longrightarrow\mathrm{End}(\mathfrak{g})$ is the
adjoint representation,
${\mathrm{ad}}^{*}:\mathfrak{g}\longrightarrow\mathrm{End}(\mathfrak{g}^{*})$
the coadjoint representation give by
$\prec{\mathrm{ad}}_{X}^{*}\alpha,Y\succ=-\prec\alpha,[X,Y]\succ.$ The
coadjoint representation extends to $\wedge^{2}\mathfrak{g}^{*}$ by the
formula

${\mathrm{ad}}_{X}^{*}\omega(Y,Z)=-\omega([X,Y],Z)-\omega(Y,[X,Z]),\quad\omega\in\wedge^{2}\mathfrak{g}^{*},X,Y,Z\in\mathfrak{g}.$

For $\Omega:\mathfrak{g}\longrightarrow\wedge^{2}\mathfrak{g}^{*}$, the
differential of $\Omega$ is the skew-symmetric bilinear map
$\Delta\Omega:\mathfrak{g}\times\mathfrak{g}\longrightarrow\wedge^{2}\mathfrak{g}^{*}$
given by

$\Delta(\Omega)(X,Y)={\mathrm{ad}}_{X}^{*}\Omega(Y)-{\mathrm{ad}}_{Y}^{*}\Omega(X)-\Omega([X,Y])$
(9)

Finally, for any $\omega\in\wedge^{2}\mathfrak{g}^{*}$, we denote by
$\omega^{\flat}:\mathfrak{g}\longrightarrow\mathfrak{g}^{*}$ the linear map
given by $\prec\omega^{\flat}(X),Y\succ=\omega(X,Y)$.

Let us state one of our main result.

######  Theorem 5.1.

Let $(\mathfrak{g},\bullet_{\mathfrak{g}},\langle\;,\;\rangle_{\mathfrak{g}})$
be a ${\mathrm{L}}$-quadratic left Leibniz algebra such that
$\mathrm{Leib}(\mathfrak{g})$ is degenerate and let
$({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$ and
$(A,\bullet_{A},\langle\;,\;\rangle_{A})$ be its core. Then
$(\mathfrak{g},\bullet_{\mathfrak{g}},\langle\;,\;\rangle_{\mathfrak{g}})$ is
isomorphic to $({\mathfrak{h}}\oplus
A\oplus{\mathfrak{h}}^{*},\bullet,\langle\;,\;\rangle_{n})$ where, for any
$X,Y\in{\mathfrak{h}}$, $a,b\in A$, $\alpha\in{\mathfrak{h}}^{*}$, the non
vanishing Leibniz products are given by

$\begin{cases}X\bullet
Y=[X,Y]_{\mathfrak{h}}+\theta(X,Y)+\Omega(X)^{\flat}(Y),\;a\bullet
b=a\bullet_{A}b+\mu(a,b),,\\\ X\bullet
a=F(X)a+\nu(X,a),\;X\bullet\alpha=(\mathrm{ad}^{{\mathfrak{h}}})_{X}^{*}\alpha,\;a\bullet
X=G(X)a+\rho(a,X),\end{cases}$ (10)

and $F:{\mathfrak{h}}\longrightarrow\mathrm{so}(A)\cap\mathrm{Der}(A)$,
$G:{\mathfrak{h}}\longrightarrow\mathrm{End}(A)$,
$\Omega:{\mathfrak{h}}\longrightarrow\wedge^{2}{\mathfrak{h}}^{*}$ are linear
maps, $\theta,\omega:{\mathfrak{h}}\times{\mathfrak{h}}\longrightarrow A$ are
bilinear maps such that $\omega$ is skew-symmetric and $\mu,\rho,\nu$ are
defined by the following relations:

$\prec\rho(a,X),Y\succ=-\langle\omega(X,Y),a\rangle_{A},\;\prec\mu(a,b),X\succ=-\langle
G(X)a,b\rangle_{A},\;\prec\nu(X,a),Y\succ=-\langle\theta(X,Y),a\rangle_{A}.$
(11)

Moreover, $(F,G,\omega,\theta,\Omega)$ satisfy the following equations, for
any $X,Y,Z,T\in{\mathfrak{h}}$ and $a\in A$,

$\displaystyle\begin{cases}[{\mathrm{L}}_{a}^{A},F(X)]={\mathrm{L}}_{G(X)a}^{A},\;[{\mathrm{L}}_{a}^{A},G(X)]={\mathrm{R}}^{A}_{G(X)a}\\\
{\mathrm{L}}_{\omega(X,Y)}^{A}=G^{*}(X)G(Y)-G^{*}(Y)G(X),\\\
{\mathrm{R}}_{\theta(X,Y)}=[F(X),G(Y)]-G([X,Y]_{\mathfrak{h}})=F(X)G(Y)+G(Y)G(X)-G([X,Y]_{{\mathfrak{h}}}),\\\
{\mathrm{L}}^{A}_{\theta(X,Y)}=[F(X),F(Y)]-F([X,Y]_{\mathfrak{h}}),\\\
\theta(X,[Y,Z]_{\mathfrak{h}})-\theta([X,Y]_{\mathfrak{h}},Z)-\theta(Y,[X,Z]_{\mathfrak{h}})+F(X)\theta(Y,Z)-F(Y)\theta(X,Z)-G(Z)\theta(X,Y)=0,\\\
G(Z)^{*}\theta(X,Y)-G(Y)^{*}\theta(X,Z)-F(X)\omega(Y,Z)+\omega(Y,[X,Z]_{\mathfrak{h}})+\omega([X,Y]_{\mathfrak{h}},Z)=0,\\\
(G(X)^{*}+F(X))\omega(Y,Z)=0,\\\
\Delta(\Omega)(X,Y)(Z,T)=\langle\theta(X,T),\theta(Y,Z)\rangle_{A}-\langle\theta(Y,T),\theta(X,Z)\rangle_{A}-\langle\omega(Z,T),\theta(X,Y)\rangle_{A},\end{cases}$
(12)

where $\Delta$ is given by (9).

Conversely, if $(A,\bullet_{A},\langle\;,\;\rangle_{A})$ is a
${\mathrm{L}}$-quadratic left Leibniz algebra,
$({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$ is a Lie algebra and
$(F,G,\omega,\theta,\Omega)$ a list as above satisfying (11) and (12) then
$({\mathfrak{h}}\oplus
A\oplus{\mathfrak{h}}^{*},\bullet,\langle\;,\;\rangle_{n})$ where $\bullet$ is
given by (10) is a ${\mathrm{L}}$-quadratic left Leibniz algebra. We call it
the ${\mathrm{L}}$-quadratic left algebra obtained form
$(A,\bullet_{A},\langle\;,\;\rangle_{A})$ and
$({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$ by means of
$(F,G,\omega,\theta,\Omega)$.

###### Proof.

We prove the theorem in two steps.

First step. In this step we build all the data given in the theorem.

Let $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ be a degenerate
${\mathrm{L}}$-invariant Leibniz algebra. Let
$({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$ and
$(A,\bullet_{A},\langle\;,\;\rangle_{A})$ be its core defined in Definition
4.1. We denote by
$I=\mathrm{Leib}(\mathfrak{g})\cap\mathrm{Leib}(\mathfrak{g})^{\perp}$. Choose
a vector subspace $B$ such that $I^{\perp}=B\oplus I$. $B^{\perp}$ contains
$I$ and $\dim B^{\perp}=2\dim I$. So there exists a totally isotropic vector
subspace ${\mathfrak{h}}_{0}$ such that $B^{\perp}=I\oplus{\mathfrak{h}}_{0}$.
Now, the metric on $B^{\perp}$ permits the identification of $I$ to
${\mathfrak{h}}_{0}^{*}$ and hence $(B^{\perp},\langle\;,\;\rangle)$ is
isometric ${\mathfrak{h}}_{0}\oplus{\mathfrak{h}}_{0}^{*}$ endowed with its
canonical neutral metric. We obtain that
$\mathfrak{g}={\mathfrak{h}}_{0}\oplus B\oplus{\mathfrak{h}}_{0}^{*}$ and the
metric is $\langle\;,\;\rangle_{n}$.

Now, $I$ is a totally isotropic ideal such that for any $u\in I$,
$\mathrm{L}_{u}=0$,
$I^{\perp}=\mathrm{Leib}(\mathfrak{g})+\mathrm{Leib}(\mathfrak{g})^{\perp}$
and, by virtue of Proposition 4.2, $I^{\perp}$ is an ideal and
$I^{\perp}\bullet I=0$. So the Leibniz product can be written, for any
$X,Y\in{\mathfrak{h}}$, $a,b\in B$, $\alpha\in{\mathfrak{h}}^{*}$,

$\begin{cases}X\bullet
Y=[X,Y]_{{\mathfrak{h}}_{0}}+\theta(X,Y)+\theta_{2}(X,Y),\;a\bullet
b=a\bullet_{B}b+\mu(a,b),\;\alpha\bullet X=\alpha\bullet a=0,\\\ X\bullet
a=F(X)a+\nu(X,a),\;X\bullet\alpha=H(X)\alpha,\;a\bullet
X=G(X)a+\rho(a,X),\;a\bullet\alpha=0,\end{cases}$

where $\theta(X,Y)\in B$, $\theta_{2}(X,Y)\in{\mathfrak{h}}_{0}^{*}$,
$a\bullet_{B}b\in B$, $\mu(X,a),\nu(X,a),\rho(a,X)\in{\mathfrak{h}}_{0}^{*}$,
$F(X),G(X)\in\mathrm{End}(A)$ and
$H(X)\in\mathrm{End}({\mathfrak{h}}_{0}^{*})$.

Let
$\pi_{\mathfrak{h}}:\mathfrak{g}\longrightarrow{\mathfrak{h}}=\mathfrak{g}/I^{\perp}$
and $\pi_{A}:I^{\perp}\longrightarrow A=I^{\perp}/I$. It is clear that
$\pi_{\mathfrak{h}}$ identifies
$({\mathfrak{h}}_{0},[\;,\;]_{{\mathfrak{h}}_{0}})$ with
$({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$ and $\pi_{A}$ identifies
$(B,\bullet_{B},\langle\;,\;\rangle_{B})$ with
$(A,\bullet_{A},\langle\;,\;\rangle_{A})$. This shows that
$[\;,\;]_{{\mathfrak{h}}_{0}}$ is a Lie bracket and $\bullet_{B}$ is a left
Leibniz bracket. We identify ${\mathfrak{h}}_{0}$ to ${\mathfrak{h}}$ and $B$
to $A$. Furthermore, the relation $\langle
X\bullet\alpha,Y\rangle=-\langle\alpha,X\bullet Y\rangle$ gives that
$H(X)=({\mathrm{ad}}_{X}^{\mathfrak{h}})^{*}$. For any
$X,Y,Z\in{\mathfrak{h}}$

$\langle X\bullet Y,Z\rangle_{n}=\prec\theta_{2}(X,Y),Z\succ=-\langle X\bullet
Z,Y\rangle_{n}=-\prec\theta_{2}(X,Z),Y\succ.$

Define $\Omega:{\mathfrak{h}}\longrightarrow\wedge^{2}{\mathfrak{h}}^{*}$ by
the relation $\Omega(X)(Y,Z)=\prec\theta_{2}(X,Y),Z\rangle$ and hence
$\theta_{2}(X,Y)=\Omega(X)^{\flat}(Y)$. We introduce
$\omega\in\wedge^{2}{\mathfrak{h}}^{*}$ by the relation
$\prec\rho(a,X),Y\succ=-\langle\omega(X,Y),a\rangle_{A}$ and the other
relation in (11) are a direct consequence of the fact that the metric is
${\mathrm{L}}$-quadratic.

Second step. In this step we show that the product given by (10) is left
Leibniz and the metric is ${\mathrm{L}}$-quadratic if and only if (12) holds
and $F$ takes its values in $\mathrm{Der}(A)\cap\mathrm{so}(A)$. This is a
long and tedious computation we defer to the appendix.∎

######  Corollary 5.1.

Let $(\mathfrak{g},\bullet_{\mathfrak{g}},\langle\;,\;\rangle_{\mathfrak{g}})$
be a ${\mathrm{L}}$-quadratic left Leibniz algebra,
$({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$ and
$(A,\bullet_{A},\langle\;,\;\rangle_{A})$ its core. If
$\mathrm{Leib}(\mathfrak{g})^{\perp}=\mathrm{Leib}(\mathfrak{g})$ then
$A=\\{0\\}$, ${\mathfrak{h}}=\mathrm{Lie}(\mathfrak{g})$ and
$(\mathfrak{g},\bullet_{\mathfrak{g}},\langle\;,\;\rangle_{\mathfrak{g}})$ is
isomorphic to
$({\mathfrak{h}}\oplus{\mathfrak{h}}^{*},\bullet,\langle\;,\;\rangle_{n})$
where, for any $X,Y\in{\mathfrak{h}}$, $\alpha,\beta\in{\mathfrak{h}}^{*}$

$(X+\alpha)\bullet(Y+\beta)=[X,Y]_{\mathfrak{h}}+({\mathrm{ad}}^{{\mathfrak{h}}})_{X}^{*}\beta+\Omega(X)^{\flat}(Y),\;$
(13)

and $\Omega:{\mathfrak{h}}\longrightarrow\wedge^{2}{\mathfrak{h}}^{*}$ is a
1-cocycle, i.e., $\Delta(\Omega)=0$. Moreover, two such structures of
${\mathrm{L}}$-quadratic left Leibniz algebras on
${\mathfrak{h}}\oplus{\mathfrak{h}}^{*}$ associated to two 1-cocycles
$\Omega_{1}$ and $\Omega_{2}$ are isomorphic if $\Omega_{1}$ and $\Omega_{2}$
define the same cohomolgy class in
$H^{2}({\mathfrak{h}},\wedge^{2}{\mathfrak{h}}^{*})$.

###### Proof.

The first part is a consequence of Theorem 5.1. Let $({\mathfrak{h}},[\;,\;])$
be a Lie algebra and $\Omega_{1}$ and $\Omega_{2}$ two 1-cocycles on
${\mathfrak{h}}$ with values in $\wedge^{2}{\mathfrak{h}}^{*}$. Denote by
$\bullet_{i}$ the left Leibniz product on
${\mathfrak{h}}\oplus{\mathfrak{h}}^{*}$ associated to $\Omega_{i}$ and given
by (13). If $\Omega_{1}$ and $\Omega_{2}$ define the same cohomolgy class in
$H^{2}({\mathfrak{h}},\wedge^{2}{\mathfrak{h}}^{*})$ then there exists
$\omega\in\wedge^{2}{\mathfrak{h}}^{*}$ such that for any
$X\in{\mathfrak{h}}$,
$\Omega_{2}(X)=\Omega_{1}(X)+{\mathrm{ad}}_{X}^{*}\omega$. On can check easily
that the isomorphism
$\phi:{\mathfrak{h}}\oplus{\mathfrak{h}}^{*}\longrightarrow{\mathfrak{h}}\oplus{\mathfrak{h}}^{*}$,
$X+\alpha\mapsto X+\alpha-\omega^{\flat}(X)$ is an isomorphism of
${\mathrm{L}}$-quadratic left Leibniz algebras between
$({\mathfrak{h}}\oplus{\mathfrak{h}}^{*},\bullet_{1},\langle\;,\;\rangle_{n})$
and
$({\mathfrak{h}}\oplus{\mathfrak{h}}^{*},\bullet_{2},\langle\;,\;\rangle_{n})$.
∎

As a consequence we can give a complete description of
${\mathrm{L}}$-quadratic simple left Leibniz algebras.

######  Corollary 5.2.

$(\mathfrak{g},\bullet_{\mathfrak{g}},\langle\;,\;\rangle_{\mathfrak{g}})$ be
a ${\mathrm{L}}$-quadratic simple left Leibniz algebra. Then
$(\mathfrak{g},\bullet_{\mathfrak{g}},\langle\;,\;\rangle_{\mathfrak{g}})$ is
isomorphic to
$({\mathfrak{h}}\oplus{\mathfrak{h}}^{*},\bullet,\langle\;,\;\rangle_{n})$
where ${\mathfrak{h}}=\mathrm{Lie}(\mathfrak{g})$ and, for any
$X,Y\in{\mathfrak{h}}$, $\alpha,\beta\in{\mathfrak{h}}^{*}$

$(X+\alpha)\bullet(Y+\beta)=[X,Y]_{\mathfrak{h}}+({\mathrm{ad}}_{X}^{{\mathfrak{h}}})^{*}\beta.$

###### Proof.

According to Proposition 4.6, $A=0$ and
${\mathfrak{h}}=\mathrm{Leib}(\mathfrak{g})$ which is a simple Lie algebra. We
can apply Corollary 5.1 and in addition, since ${\mathfrak{h}}$ is simple,
$\Omega$ is a co-boundary which completes the proof. ∎

Let us recall a property of quadratic Lie algebras we will use in the
following Theorem.

Let $(\mathfrak{g},[\;,\;],\langle\;,\;\rangle)$ be a quadratic Lie algebra
and $\omega\in\wedge^{2}\mathfrak{g}^{*}$. Then there exists a skew-symmetric
endomorphism $D:\mathfrak{g}\longrightarrow\mathfrak{g}$ such that, for any
$x,y\in\mathfrak{g}$,

$\omega(x,y)=\langle Dx,y\rangle.$

It is a well-known fact which is easy to establish that $\omega$ is a
2-cocycle of $\mathfrak{g}$ if and only if $D$ is a derivation. Hence
$H^{2}(\mathfrak{g})=(\mathrm{so}(\mathfrak{g})\cap\mathrm{Der}(\mathfrak{g}))/\mathrm{Inner}(\mathfrak{g})$
where $\mathrm{Inner}(\mathfrak{g})=\\{{\mathrm{ad}}_{u},u\in\mathfrak{g}\\}$.

######  Theorem 5.2.

Let $({\mathfrak{h}}\oplus
A\oplus{\mathfrak{h}}^{*},\bullet,\langle\;,\;\rangle_{n})$ be the
${\mathrm{L}}$-quadratic left algebra obtained form
$(A,\bullet_{A},\langle\;,\;\rangle_{A})$ and
$({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$ by means of
$(F,G,\omega,\theta,\Omega)$.

If $A$ is a Lie algebra, $Z(A)=0$ and $H^{2}(A)=0$ then there exists a
1-cocycle $\Omega_{0}$ of ${\mathfrak{h}}$ with values in
$\wedge^{2}{\mathfrak{h}}^{*}$ and an endomorphism
$U:{\mathfrak{h}}\longrightarrow A$ such that $({\mathfrak{h}}\oplus
A\oplus{\mathfrak{h}}^{*},\bullet,\langle\;,\;\rangle_{n})$ is isomorphic to
$({\mathfrak{h}}\oplus A\oplus{\mathfrak{h}}^{*},\circ,\langle\;,\;\rangle)$
where, for any $X,Y\in{\mathfrak{h}}$, $a,b\in A$,
$\alpha,\beta\in{\mathfrak{h}}^{*}$, the non vanishing Leibniz products and
the metric are given by

$\displaystyle X\circ Y=[X,Y]_{\mathfrak{h}}+\Omega_{0}(X)^{\flat}(Y),\;a\circ
b=[a,b]_{A}+U^{*}([a,b]_{A}),\;X\circ\alpha=({\mathrm{ad}}_{X}^{\mathfrak{h}})^{*}\alpha,$
$\displaystyle\langle X+a+\alpha,Y+b+\beta\rangle=\langle
X+a+\alpha,Y+b+\beta\rangle_{n}+\langle U(X),U(Y)\rangle_{A}-\langle
a,U(Y)\rangle_{A}-\langle b,U(X)\rangle_{A},$ $\displaystyle\prec
U^{*}(a),X\succ=\langle a,U(X)\rangle_{A}.$

###### Proof.

The structure of ${\mathrm{L}}$-quadratic left Leibniz algebra of
${\mathfrak{h}}\oplus A\oplus{\mathfrak{h}}^{*}$ is given in Theorem 5.1 with
the conditions satisfied by $(F,G,\omega,\theta,\Omega)$. Suppose that $A$ is
a Lie algebra, $Z(A)=0$ and $H^{2}(A)=0$. Then $[A,A]_{A}=Z(A)^{\perp}=A$ and
any skew-symmetric derivation is inner. From the first equation in (12), the
fact that $F$ is a derivation and $Z(A)=0$ we deduce that $G=-F$. So exists a
unique endomorphism $U:{\mathfrak{h}}\longrightarrow A$ such that
$F(X)=-G(X)={\mathrm{ad}}_{U(X)}$ and from the third equation in (12), we must
have

$\theta(X,Y)=[U(X),U(Y)]_{A}-U([X,Y]_{\mathfrak{h}}).$

We consider now the isomorphism $\phi:{\mathfrak{h}}\oplus
A\oplus{\mathfrak{h}}^{*}\longrightarrow{\mathfrak{h}}\oplus
A\oplus{\mathfrak{h}}^{*}$ given by $\phi(X+a+\alpha)=X+a-U(X)+\alpha$. Under
this isomorphism, $({\mathfrak{h}}\oplus
A\oplus{\mathfrak{h}}^{*},\bullet,\langle\;,\;\rangle_{n})$ is isomorphic to
$({\mathfrak{h}}\oplus A\oplus{\mathfrak{h}}^{*},\circ,\langle\;,\;\rangle)$
where, for any $u,v\in{\mathfrak{h}}\oplus A\oplus{\mathfrak{h}}^{*}$,

$u\circ v=\phi^{-1}(\phi(u)\bullet\phi(v))\quad\mbox{and}\quad\langle
u,v\rangle=\langle\phi(u),\phi(v)\rangle_{n}.$

So

$\langle X+a+\alpha,Y+b+\beta\rangle=\langle
X+a+\alpha,X+b+\alpha\rangle_{n}+\langle U(X),U(Y)\rangle_{A}-\langle
a,U(Y)\rangle_{A}-\langle b,U(X)\rangle_{A}.$

For any $X,Y\in{\mathfrak{h}}$,

$\displaystyle\phi(X\circ Y)$ $\displaystyle=(X-U(X))\bullet(Y-U(Y))$
$\displaystyle=[X,Y]_{\mathfrak{h}}+\theta(X,Y)+\Omega(X)^{\flat}(Y)-[U(X),U(Y)]_{A}-\nu(X,U(Y))+[U(Y),U(X)]_{A}$
$\displaystyle-\rho(U(X),Y)+[U(X),U(Y)]_{A}+\mu(U(X),U(Y))$
$\displaystyle=[X,Y]_{\mathfrak{h}}-U([X,Y]_{\mathfrak{h}})+\Omega(X)^{\flat}(Y)-\rho(U(X),Y)+\mu(U(X),U(Y)).$

So there exist $\theta_{2}$ such that

$X\circ Y=[X,Y]_{\mathfrak{h}}+\theta_{2}(X,Y).$

From the relation

$\langle X\circ Y,Z\rangle=\prec\theta_{2}(X,Y),Z\succ=-\langle X\circ
Z,Y\rangle=-\prec\theta_{2}(X,Z),Y\succ$

we can define
$\Omega_{0}:{\mathfrak{h}}\longrightarrow\wedge^{2}{\mathfrak{h}}^{*}$ by the
relation $\Omega_{0}(X)(Y,Z)=\prec\theta_{2}(X,Y),Z\succ$ and write

$X\circ Y=[X,Y]_{\mathfrak{h}}+\Omega_{0}(X)^{\flat}(Y).$

In the same way, we can show that there exists, $\nu^{0},\rho^{0}$ such that
for any $X\in{\mathfrak{h}}$, $a,b\in A$, $\alpha\in{\mathfrak{h}}^{*}$,

$a\circ b=[a,b]_{a}+\mu(a,b),\;X\circ
a=\nu^{0}(X,a),\;X\circ\alpha=({\mathrm{ad}}_{X}^{\mathfrak{h}})^{*}\alpha\quad\mbox{and}\quad
a\circ X=\rho^{0}(a,X).$

From the relations

$\langle a\circ b,X\rangle=-\langle b,a\circ
X\rangle\quad\mbox{and}\quad\langle X\circ a,Y\rangle=-\langle a,X\circ
Y\rangle$

we get

$\nu^{0}=0\quad\mbox{and}\quad\prec\mu(a,b),X\succ=\langle[a,b]_{A},U(X)\rangle_{A}.$

The bracket $\circ$ has the same form as (10) with $F=G=0$, $\theta=0$ and
$\nu=0$ and (12) are the conditions for this bracket to be left Leibniz. So we
get from the equation (2) in (12) that $\omega_{0}=0$ and the last equation in
(12) gives that $\Omega_{0}$ is a 1-cocycle. This completes the proof. ∎

######  Remark 3.

The class of quadratic Lie algebras $\mathfrak{g}$ satisfying
$Z(\mathfrak{g})=\\{0\\}$ and $H^{2}(\mathfrak{g})=\\{0\\}$ is large. It
contains the class of semi-simple Lie algebras. It contains also the
cotangents of simple Lie algebras. Indeed, if
$(\mathfrak{g},[\;,\;]_{\mathfrak{g}})$ is a Lie algebra its cotangent
$T^{*}\mathfrak{g}$ is the Lie algebra $\mathfrak{g}\oplus\mathfrak{g}^{*}$
with the bracket

$[X+\alpha,Y+\beta]=[X,Y]_{\mathfrak{g}}+{\mathrm{ad}}_{X}^{*}\alpha-{\mathrm{ad}}_{Y}^{*}\beta,\quad
X,Y\in\mathfrak{g},\alpha,\beta\in\mathfrak{g}^{*}.$

The natural metric on $T^{*}\mathfrak{g}$ is quadratic. Moreover, if
$\mathfrak{g}$ is simple then $Z(T^{*}\mathfrak{g})=\\{0\\}$ and
$H^{2}(T^{*}\mathfrak{g})=\\{0\\}$ (see [1, Theorem 2.2]).

######  Corollary 5.3.

Let $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ be a ${\mathrm{L}}$-quadratic
left Leibniz algebra which is not a Lie algebra. Then
$\dim\mathrm{Leib}(\mathfrak{g})\geq 2$.

###### Proof.

Suppose that $\dim\mathrm{Leib}(\mathfrak{g})=1$. If
$\mathrm{Leib}(\mathfrak{g})$ is nondegenerate then
$\mathfrak{g}=\mathrm{Leib}(\mathfrak{g})\oplus\mathrm{Leib}(\mathfrak{g})^{\perp}$
and the action of $\mathrm{Leib}(\mathfrak{g})^{\perp}$ on
$\mathrm{Leib}(\mathfrak{g})$ is made by skew-symmetric endomorphisms so it is
trivial and hence $\mathfrak{g}$ is a Lie algebra.

Suppose tha $\mathrm{Leib}(\mathfrak{g})$ is degenerate then
$\mathrm{Leib}(\mathfrak{g})\subset\mathrm{Leib}(\mathfrak{g})^{\perp}$ and
hence $\dim{\mathfrak{h}}=1$. So $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$
is isomorphic to the ${\mathrm{L}}$-quadratic left Leibniz algebra
$({\mathfrak{h}}\oplus
A\oplus{\mathfrak{h}}^{*},\bullet,\langle\;,\;\rangle_{n})$ described in
Theorem 5.1 with the data $(F,G,\omega,\theta,\Omega)$. Since
$\dim{\mathfrak{h}}=1$, we have $\Omega=0$ and $\omega=0$. Moreover,
${\mathfrak{h}}^{*}=\mathrm{Leib}(\mathfrak{g})$ and hence for any
$X,Y\in{\mathfrak{h}}$ and $a\in A$,

$X\bullet Y+Y\bullet
X=\theta(X,Y)+\theta(Y,X)\in{\mathfrak{h}}^{*}\quad\mbox{and}\quad X\bullet
a+a\bullet X=F(X)a+G(X)a+\nu(X,a)\in{\mathfrak{h}}^{*}$

this implies that $F=-G$, $\theta=0$ and $\nu=0$. The product $\bullet$
becomes skew-symmetric and hence $\mathfrak{g}$ is a Lie algebra which is a
contradiction. ∎

We end this section by giving a complete description of
${\mathrm{L}}$-quadratic left Leibniz algebras such that
$\dim(\mathrm{Leib}(\mathfrak{g})\cap\mathrm{Leib}(\mathfrak{g})^{\perp})=1$.
This class contains the class of Lorentzian ${\mathrm{L}}$-quadratic left
Leibniz algebras with degenerate Leibniz ideal. However, let
$(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ be a Lorentzian
${\mathrm{L}}$-quadratic Leibniz algebra such that
$\mathrm{Leib}(\mathfrak{g})$ is nondegenerate. Then
$\mathfrak{g}=\mathrm{Leib}(\mathfrak{g})\oplus\mathrm{Leib}(\mathfrak{g})^{\perp}$
where $\mathrm{Leib}(\mathfrak{g})^{\perp}$ is either a quadratic Euclidean
Lie algebra or a quadratic Lorentzian Lie algebra (see Section 4).

######  Theorem 5.3.

Let $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ be a ${\mathrm{L}}$-quadratic
left Leibniz algebra such that
$\dim(\mathrm{Leib}(\mathfrak{g})\cap\mathrm{Leib}(\mathfrak{g})^{\perp})=1$.
Then $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ is isomorphic to $(\hbox{\bb
R}e\oplus A\oplus\hbox{\bb R}\bar{e},\bullet,\langle\;,\;\rangle_{n})$ where
$(A,\bullet_{A},\langle\;,\;\rangle_{A})$ is a ${\mathrm{L}}$-quadratic left
Leibniz algebra and the non vanishing Leibniz products are given by

$\bar{e}\bullet\bar{e}=\delta,\;a\bullet b=a\bullet_{A}b-\langle
Ga,b\rangle_{A}e,\;\bar{e}\bullet a=Fa-\langle\delta,a\rangle
e,\;a\bullet\bar{e}=Ga,$

where $a,b\in A$, $\delta\in A$, $F\in\mathrm{Der}(A)\cap\mathrm{so}(A)$,
$G\in\mathrm{End}(A)$ satisfying

$G(\delta)=0,\;{\mathrm{L}}_{\delta}^{A}=0,\;\mathrm{Im}(F+G)\in
Z^{l}(A),\;[{\mathrm{L}}_{a}^{A},G]={\mathrm{R}}_{Ga}^{A}\quad\mbox{and}\quad[G,F]=G^{2}+GF={\mathrm{R}}_{\delta}.$

###### Proof.

According to Theorem 5.1, $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ is
isomorphic to $\mathfrak{g}={\mathfrak{h}}\oplus A\oplus{\mathfrak{h}}^{*}$
where $(A,\bullet_{A},\langle\;,\;\rangle_{A})$ is a ${\mathrm{L}}$-quadratic
Euclidean left Leibniz algebra, ${\mathfrak{h}}=\hbox{\bb R}\bar{e}$,
${\mathfrak{h}}^{*}=\hbox{\bb R}e$ and the Leibniz product is given by (10).
Since $\dim{\mathfrak{h}}=1$, we have $\Omega=0$ and $\omega=0$. Put

$F=F(\bar{e}),\;G=G(e),\;\theta(\bar{e},\bar{e})=\delta\in
A,\;\mu(a,b)=-\langle Ga,b\rangle_{A}e.$

Then $\nu(\bar{e},a)=-\langle\delta,a\rangle e$ and we get the desired
expression for the Leibniz product. One can check easily that (12) reduces to
the equations given in the statement of the theorem. ∎

##  6 Description of ${\mathrm{R}}$-quadratic left Leibniz algebras

In this section, we adopt a similar approach as in the previous section to
give a complete description ${\mathrm{R}}$-quadratic left Leibniz algebras and
to derive some important results. We use the same notations as the previous
section.

The following theorem has many similarities with Theorem 5.1.

######  Theorem 6.1.

Let $(\mathfrak{g},\bullet_{\mathfrak{g}},\langle\;,\;\rangle_{\mathfrak{g}})$
be a ${\mathrm{R}}$-quadratic left Leibniz algebra,
$({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$ and
$(A,[\;,\;]_{A},\langle\;,\;\rangle_{A})$ its core. Then
$(\mathfrak{g},\bullet_{\mathfrak{g}},\langle\;,\;\rangle_{\mathfrak{g}})$ is
isomorphic to $({\mathfrak{h}}\oplus
A\oplus{\mathfrak{h}}^{*},[\;,\;],\langle\;,\;\rangle_{n})$ where for
$X,Y\in{\mathfrak{h}}$, $a,b\in A$, $\alpha\in{\mathfrak{h}}^{*}$, the non
vanishing products are given by

$\begin{cases}X\bullet Y=\theta(X,Y)+\Omega(Y)^{\flat}(X),\;a\bullet
b=[a,b]_{A}+\mu(a,b),\\\ X\bullet
a=F(X)a+\nu(X,a),\;X\bullet\alpha=({\mathrm{ad}}^{\mathfrak{h}})_{X}^{*}\alpha,\;a\bullet
X=-F(X)a+\rho(a,X),\;\end{cases}$ (14)

and $F:{\mathfrak{h}}\longrightarrow\mathrm{so}(A)\cap\mathrm{Der}(A)$,
$\Omega:{\mathfrak{h}}\longrightarrow\wedge^{2}{\mathfrak{h}}^{*}$ are linear
maps, $\theta,\omega:{\mathfrak{h}}\times{\mathfrak{h}}\longrightarrow A$ are
bilinear skew-symmetric, and $\mu,\rho,\nu$ are defined by the relations

$\prec\nu(X,a),Y\succ=\langle\omega(X,Y),a\rangle_{A},\;\prec\mu(a,b),X\succ=\langle
F(X)a,b\rangle_{A},\;\quad\mbox{and}\quad\prec\rho(a,X),Y\succ=\langle\theta(X,Y),a\rangle_{A}.$
(15)

Moreover, $(F,\omega,\theta,\Omega)$ satisfy the following equations, for any
$X,Y,Z,T\in{\mathfrak{h}}$,

$\begin{cases}{\mathrm{ad}}_{\omega(X,Y)}^{A}=F([X,Y]_{\mathfrak{h}})-[F(X),F(Y)],\\\
{\mathrm{ad}}^{A}_{\theta(X,Y)}=[F(X),F(Y)],\\\
F(X)\theta(Y,Z)+F(Y)\theta(Z,X)+F(Z)\theta(X,Y)=0,\\\
F(X)\omega(Y,Z)+F(Y)\omega(Z,X)-F(Z)\theta(X,Y)+\omega(X,[Y,Z]_{\mathfrak{h}})-\omega(Y,[X,Z]_{\mathfrak{h}})=0,\\\
F(X)\theta(Y,Z)-F(Y)\omega(Z,X)+F(Z)\theta(X,Y)-\theta(Y,[X,Z]_{\mathfrak{h}})=0,\\\
({\mathrm{ad}}^{\mathfrak{h}}_{T})^{*}\Omega(Z)(X,Y)=\langle\omega(Y,T),\theta(X,Z)\rangle_{A}+\langle\omega(X,T),\theta(Z,Y)\rangle_{A}+\langle\theta(Z,T),\theta(X,Y)\rangle_{A}.\end{cases}$
(16)

Conversely, if $(A,[\;,\;]_{A},\langle\;,\;\rangle_{A})$ is a quadratic Lie
algebra, $({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$ is a Lie algebra and
$(F,\omega,\theta,\Omega)$ a list as above satisfying (15) and (16) then
$({\mathfrak{h}}\oplus
A\oplus{\mathfrak{h}}^{*},[\;,\;],\langle\;,\;\rangle_{n})$ is a
${\mathrm{R}}$-quadratic left Leibniz algebra. We call it the
${\mathrm{R}}$-quadratic left algebra obtained form
$(A,[\;,\;]_{A},\langle\;,\;\rangle_{A})$ and
$({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$ by means of
$(F,\omega,\theta,\Omega)$.

###### Proof.

We follow the same steps as in the proof of Theorem 5.1.

Let $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ be a degenerate
${\mathrm{R}}$-invariant Leibniz algebra. Let
$({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$ and
$(A,[\;,\;]_{A},\langle\;,\;\rangle_{A})$ be its core defined in Definition
4.1.

We denote by $I=\mathrm{Leib}(\mathfrak{g})$. Choose a vector subspace $B$
such that $I^{\perp}=B\oplus I$ and as in the proof of Theorem 5.1 we have
$\mathfrak{g}={\mathfrak{h}}_{0}\oplus B\oplus{\mathfrak{h}}_{0}^{*}$ and the
metric is $\langle\;,\;\rangle_{n}$.

Now, $I$ is a totally isotropic ideal such that for any $u\in I$,
$\mathrm{L}_{u}=0$ and, by virtue of Proposition 4.2, $I^{\perp}$ is an ideal,
$I^{\perp}\bullet I=0$ and $\mathfrak{g}\bullet\mathfrak{g}\subset I^{\perp}$.
So the Leibniz product can be written, for any $X,Y\in{\mathfrak{h}}$, $a,b\in
B$, $\alpha\in{\mathfrak{h}}^{*}$,

$\begin{cases}X\bullet Y=\theta(X,Y)+\theta_{2}(X,Y),\;a\bullet
b=a\bullet_{B}b+\mu(a,b),\;\alpha\bullet X=\alpha\bullet a=0,\\\ X\bullet
a=F(X)a+\nu(X,a),\;X\bullet\alpha=H(X)\alpha,\;a\bullet
X=G(X)a+\rho(a,X),\;a\bullet\alpha=0,\end{cases}$

where $\theta(X,Y)\in B$, $\theta_{2}(X,Y)\in{\mathfrak{h}}_{0}^{*}$,
$a\bullet_{B}b\in B$, $\mu(X,a),\nu(X,a),\rho(a,X)\in{\mathfrak{h}}_{0}^{*}$,
$F(X),G(X)\in\mathrm{End}(A)$, $H(X)\in\mathrm{End}({\mathfrak{h}}_{0}^{*})$.
We have $X\bullet a+a\bullet X\in I$ and hence $G(X)=-F(X)$. We have also
$X\bullet Y+Y\bullet X\in I$ and hence $\theta$ is skew-symmetric.

Let
$\pi_{\mathfrak{h}}:\mathfrak{g}\longrightarrow{\mathfrak{h}}=\mathfrak{g}/I^{\perp}$
and $\pi_{A}:I^{\perp}\longrightarrow A=I^{\perp}/I$. It is clear that
$\pi_{\mathfrak{h}}$ identifies ${\mathfrak{h}}_{0}$ to ${\mathfrak{h}}$ and
$\pi_{A}$ identifies $(B,\bullet_{B},\langle\;,\;\rangle_{B})$ to
$(A,[\;,\;]_{A},\langle\;,\;\rangle_{A})$. This shows that $\bullet_{B}$ is a
Lie bracket. We identify ${\mathfrak{h}}_{0}$ to ${\mathfrak{h}}$ and $B$ to
$A$. Furthermore, the relation

$\prec
X\bullet\alpha,Y\succ=\prec\alpha,{\mathrm{L}}_{X}^{*}Y\succ\stackrel{{\scriptstyle\eqref{sub}}}{{=}}-\prec\alpha,[X,Y]_{\mathfrak{h}}\succ$

gives that $H(X)=({\mathrm{ad}}_{X}^{\mathfrak{h}})^{*}$. For any
$X,Y,Z\in{\mathfrak{h}}$

$\langle X\bullet Y,Z\rangle_{n}=\prec\theta_{2}(X,Y),Z\succ=-\langle Z\bullet
Y,X\rangle_{n}=-\prec\theta_{2}(Z,Y),X\succ.$

Define $\Omega:{\mathfrak{h}}\longrightarrow\wedge^{2}{\mathfrak{h}}^{*}$ by
the relation $\Omega(Y)(X,Z)=\prec\theta_{2}(X,Y),Z\rangle$ and hence
$\theta_{2}(X,Y)=\Omega(Y)^{\flat}(X)$. We introduce
$\omega\in\wedge^{2}{\mathfrak{h}}^{*}$ by the relation
$\prec\nu(X,a),Y\succ=\langle\omega(X,Y),a\rangle_{A}$ and the other relations
in (15) are a direct consequence of the fact that the metric is
${\mathrm{R}}$-quadratic.

The second step of the proof is similar to the second step of the proof of
Theorem 5.1 and we defer it also to the appendix. ∎

######  Corollary 6.1.

Let $(\mathfrak{g},\bullet_{\mathfrak{g}},\langle\;,\;\rangle_{\mathfrak{g}})$
be a ${\mathrm{R}}$-quadratic left Leibniz algebra such,
$({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$ and
$(A,[\;,\;]_{A},\langle\;,\;\rangle_{A})$ its core. If
$\mathrm{Leib}(\mathfrak{g})^{\perp}=\mathrm{Leib}(\mathfrak{g})$ then
$A=\\{0\\}$ and
$(\mathfrak{g},\bullet_{\mathfrak{g}},\langle\;,\;\rangle_{\mathfrak{g}})$ is
isomorphic to
$({\mathfrak{h}}\oplus{\mathfrak{h}}^{*},\bullet,\langle\;,\;\rangle_{n})$
where, for any $X,Y\in{\mathfrak{h}}$, $\alpha,\beta\in{\mathfrak{h}}^{*}$

$(X+\alpha)\bullet(Y+\beta)=({\mathrm{ad}}_{X}^{\mathfrak{h}})^{*}\beta+\Omega(Y)^{\flat}(X),$

and $\Omega:{\mathfrak{h}}\longrightarrow\wedge^{2}{\mathfrak{h}}^{*}$
satisfies
$\Omega(X)^{\flat}([{\mathfrak{h}},{\mathfrak{h}}]_{\mathfrak{h}})=0$ for any
$X\in{\mathfrak{h}}$.

######  Theorem 6.2.

Let $({\mathfrak{h}}\oplus
A\oplus{\mathfrak{h}}^{*},[\;,\;],\langle\;,\;\rangle_{n})$ be the
${\mathrm{R}}$-quadratic left algebra obtained form
$(A,[\;,\;]_{A},\langle\;,\;\rangle_{A})$ and
$({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$ by means of
$(F,\omega,\theta,\Omega)$.

If $Z(A)=0$ and $H^{2}(A)=0$ then there exists an endomorphism
$U:{\mathfrak{h}}\longrightarrow A$,
$\Omega_{0}:{\mathfrak{h}}\longrightarrow\wedge^{2}{\mathfrak{h}}^{*}$ and
$\omega_{0}\in\wedge^{2}{\mathfrak{h}}^{*}$ such that $({\mathfrak{h}}\oplus
A\oplus{\mathfrak{h}}^{*},[\;,\;],\langle\;,\;\rangle_{n})$ is isomorphic to
to $({\mathfrak{h}}\oplus
A\oplus{\mathfrak{h}}^{*},\bullet,\langle\;,\;\rangle)$ where, for any
$X,Y\in{\mathfrak{h}}$, $a,b\in A$, $\alpha,\beta\in{\mathfrak{h}}^{*}$, the
non vanishing Leibniz products

$\displaystyle X\bullet Y=\Omega(Y)^{\flat}(X),\;a\bullet
b=[a,b]_{A}+U^{*}([a,b]_{A}),X\bullet\alpha=({\mathrm{ad}}_{X}^{\mathfrak{h}})^{*}\alpha,$
$\displaystyle\langle X+a+\alpha,Y+b+\beta\rangle_{0}=\langle
X+a+\alpha,Y+b+\beta\rangle_{n}+\langle U(X),U(Y)\rangle_{A}-\langle
a,U(Y)\rangle_{A}-\langle b,U(X)\rangle_{A},$

where
$\Omega_{0}(X)^{\flat}([{\mathfrak{h}},{\mathfrak{h}}]_{\mathfrak{h}})=0$ for
any $X\in{\mathfrak{h}}$.

###### Proof.

Using the same arguments as those used in th the proof of Theorem 5.2, we can
show that $({\mathfrak{h}}\oplus
A\oplus{\mathfrak{h}}^{*},[\;,\;],\langle\;,\;\rangle_{n})$ is isomorphic to
$({\mathfrak{h}}\oplus A\oplus{\mathfrak{h}}^{*},[\;,\;],\langle\;,\;\rangle)$
where, for any $X,Y\in{\mathfrak{h}}$, $a,b\in A$,
$\alpha\in{\mathfrak{h}}^{*}$,

$X\bullet Y_{0}=\theta_{2}(X,Y),\;a\bullet b=[a,b]_{a}+\mu(a,b),\;X\bullet
a=\nu(X,a),\;X\bullet\alpha=({\mathrm{ad}}_{X}^{\mathfrak{h}})^{*}\alpha\quad\mbox{and}\quad
a\bullet X=\rho(a,X).$

The metric $\langle\;,\;\rangle$ is given by

$\langle X+a+\alpha,X+a+\alpha\rangle=2\prec\alpha,X\succ+\langle
a,a\rangle+\langle U(X),U(X)\rangle_{A}-2\langle a,U(X)\rangle_{A}.$

From the relations

$\langle a\bullet b,X\rangle=-\langle a,X\bullet
b\rangle\quad\mbox{and}\quad\langle a\bullet X,Y\rangle=-\langle a,Y\bullet
X\rangle$

we get

$\rho=0\quad\mbox{and}\quad\prec\mu(a,b),X\succ=\langle[a,b]_{A},U(X)\rangle_{A}.$

Now from the first equation in (16), we get $\nu=0$ and the last equation
gives the condition on $\Omega_{0}$. ∎

######  Corollary 6.2.

Let $(\mathfrak{g},[\;,\;],\langle\;,\;\rangle)$ be a ${\mathrm{R}}$-quadratic
left Leibniz which is not a Lie algebra. Then
$\dim\mathrm{Leib}(\mathfrak{g})\geq 2$.

In particular, a ${\mathrm{R}}$-quadratic left Leibniz algebra with the metric
Lorentzian is a Lie algebra.

###### Proof.

Suppose that $\dim\mathrm{Leib}(\mathfrak{g})=1$. Then $\dim{\mathfrak{h}}=1$
and by using Theorem 6.1, we get that $\mathfrak{g}$ is isomorphic to
${\mathfrak{h}}\oplus A\oplus{\mathfrak{h}}^{*}$ with $\theta=\omega=\Omega=0$
and the Leibniz product (14) becomes a Lie bracket which is a contradiction.

If $\mathfrak{g}$ is a ${\mathrm{R}}$-quadratic left Leibniz algebra with the
metric Lorentzian then, since the metric is Lorentzian and
$\mathrm{Leib}(\mathfrak{g})\subset\mathrm{Leib}(\mathfrak{g})^{\perp}$ then
$\dim\mathrm{Leib}(\mathfrak{g})\leq 1$ and hence
$\dim\mathrm{Leib}(\mathfrak{g})=0$. ∎

By using Theorem 5.1 and Theorem 6.1, we give now a complete description of
${\mathrm{L}}$-quadratic symmetric Leibniz algebras.

######  Theorem 6.3.

Let $(\mathfrak{g},\bullet_{\mathfrak{g}},\langle\;,\;\rangle_{\mathfrak{g}})$
be a ${\mathrm{L}}$-quadratic symmetric Leibniz algebra with
$({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$ and
$(A,\bullet_{A},\langle\;,\;\rangle_{A})$ its core. Then
$\mathrm{Leib}(\mathfrak{g})\subset\mathrm{Leib}(\mathfrak{g})^{\perp}$,
$(A,\bullet_{A},\langle\;,\;\rangle_{A})$ is a quadratic Lie algebra,
$({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$ is abelian and
$(\mathfrak{g},\bullet_{\mathfrak{g}},\langle\;,\;\rangle_{\mathfrak{g}})$ is
isomorphic to $({\mathfrak{h}}\oplus
A\oplus{\mathfrak{h}}^{*},\bullet,\langle\;,\;\rangle_{n})$ where for any
$X,Y\in{\mathfrak{h}}$, $a,b\in A$, $\alpha\in{\mathfrak{h}}^{*}$, the non
vanishing Leibniz products are given by

$\begin{cases}X\bullet Y=\theta(X,Y)+\Omega(X)^{\flat}(Y),\;a\bullet
b=[a,b]_{A}+\mu(a,b),\\\ X\bullet a=F(X)a+\nu(X,a),\;\;a\bullet
X=-F(X)a+\rho(a,X),\end{cases}$ (17)

and $F:{\mathfrak{h}}\longrightarrow\mathrm{so}(A)\cap\mathrm{Der}(A)$,
$\Omega:{\mathfrak{h}}\longrightarrow\wedge^{2}{\mathfrak{h}}^{*}$ are linear,
$\theta,\omega:{\mathfrak{h}}\times{\mathfrak{h}}\longrightarrow A$ are
bilinear skew-symmetric and $\mu,\rho,\nu$ are defined by the following
relations:

$\prec\rho(a,X),Y\succ=-\langle\omega(X,Y),a\rangle_{A},\;\prec\mu(a,b),X\succ=\langle
F(X)a,b\rangle_{A}\quad\mbox{and}\quad\prec\nu(X,a),Y\succ=-\langle\theta(X,Y),a\rangle_{A}.$

Moreover, $(F,\omega,\theta,\Omega)$ satisfy the following equations:

$\begin{cases}{\mathrm{ad}}^{A}_{\theta(X,Y)}=-{\mathrm{ad}}_{\omega(X,Y)}^{A}=[F(X),F(Y)],\\\
F(X)\theta(Y,Z)+F(Y)\theta(Z,X)+F(Z)\theta(X,Y)=0,\\\
F(X)\left[\theta(Y,Z)+\omega(Y,Z)\right]=0,\\\
\langle\theta(Y,T),\theta(Z,X)\rangle_{A}+\langle\theta(X,T),\theta(Y,Z)\rangle_{A}+\langle\theta(Z,T),\theta(X,Y)\rangle_{A}=0,\\\
\langle\theta(X,T)+\omega(X,T),\theta(Y,Z)\rangle_{A}=0,\end{cases}$ (18)

for any $X,Y,Z,T\in{\mathfrak{h}}$.

###### Proof.

According to Theorem 5.1, $(\mathfrak{g},\bullet,\langle\;,\;\rangle)$ is
isomorphic to $({\mathfrak{h}}\oplus A\oplus{\mathfrak{h}}^{*})$ with the non
vanishing Leibniz products are given, for any $X,Y\in{\mathfrak{h}}$, $a,b\in
A$, $\alpha\in{\mathfrak{h}}^{*}$, by

$\begin{cases}X\bullet
Y=[X,Y]_{\mathfrak{h}}+\theta(X,Y)+\Omega(X)^{\flat}(Y),\;a\bullet
b=a\bullet_{A}b+\mu(a,b),\\\ X\bullet
a=F(X)a+\nu(X,a),\;X\bullet\alpha=(\mathrm{ad}^{{\mathfrak{h}}})_{X}^{*}\alpha,\;a\bullet
X=G(X)a+\rho(a,X),\end{cases}$

and $(F,G,\omega,\theta,\Omega)$ satisfy (11) and (12).

$\mathfrak{g}$ is also a right Leibniz algebra and hence the product $\circ$
given by $X\circ Y=Y\bullet X$ is a left Leibniz product and
$(\mathfrak{g},\circ,\langle\;,\;\rangle)$ is ${\mathrm{R}}$-quadratic. So
$\mathrm{Leib}(\mathfrak{g})\subset\mathrm{Leib}(\mathfrak{g})^{\perp}$ and
hence the core of $(\mathfrak{g},\circ,\langle\;,\;\rangle)$ is
$(A,-\bullet_{A},\langle\;,\;\rangle_{A})$ and
$({\mathfrak{h}},[\;,\;]_{\circ})$. Thus implies that $\bullet_{A}$ is Lie
bracket, let denote it by $[\;,\;]_{A}$. Then, according to Theorem 6.1,
$(\mathfrak{g},\circ,\langle\;,\;\rangle)$ is isomorphic to
$({\mathfrak{h}}\oplus A\oplus{\mathfrak{h}}^{*})$ with the non vanishing
products

$\begin{cases}X\circ
Y=\theta^{\circ}(X,Y)+\Omega^{\circ}(Y)^{\flat}(X),\;a\circ
b=-[a,b]_{A}+\mu^{\circ}(a,b),\\\ X\circ
a=F^{\circ}(X)a+\nu^{\circ}(X,a),\;X\circ\alpha=(\mathrm{ad}^{\circ})_{X}^{*}\alpha,\;a\circ
X=-F^{\circ}(X)a+\rho^{\circ}(a,X),\end{cases}$

and $(F^{\circ},\omega^{\circ},\theta^{\circ},\Omega^{\circ})$ satisfy (16)
and (15). Moreover, $\omega^{\circ}$ and $\theta^{\circ}$ are skew-symmetric.

We deduce that $[\;,\;]_{\mathfrak{h}}=[\;,\;]_{\circ}=0$ and

$\theta^{\circ}=-\theta,\;\Omega^{\circ}=\Omega,F^{\circ}=-F=G\quad\mbox{and}\quad\omega^{0}=-\omega.$

If we replace in (5.1) and (6.1) we get the two systems of equations

$\begin{cases}{\mathrm{ad}}_{\omega(X,Y)}^{A}=-[F(X),F(Y)],\\\
{\mathrm{ad}}^{A}_{\theta(X,Y)}=[F(X),F(Y)],\\\
F(X)\theta(Y,Z)+F(Y)\theta(Z,X)+F(Z)\theta(X,Y)=0,\\\
F(Z)\theta(X,Y)+F(Y)\theta(Z,X)-F(X)\omega(Y,Z)=0,\\\
\langle\theta(X,T),\theta(Y,Z)\rangle_{A}-\langle\theta(Y,\star),\theta(X,Z)\rangle_{A}-\langle\omega(Z,T),\theta(X,Y)\rangle_{A}=0.\end{cases}$

$\begin{cases}{\mathrm{ad}}_{\omega(X,Y)}^{A}=-[F(X),F(Y)],\\\
{\mathrm{ad}}^{A}_{\theta(X,Y)}=[F(X),F(Y)],\\\
F(X)\theta(Y,Z)+F(Y)\theta(Z,X)+F(Z)\theta(X,Y)=0,\\\
F(X)\omega(Y,Z)+F(Y)\omega(Z,X)-F(Z)\theta(X,Y)=0,\\\
F(X)\theta(Y,Z)-F(Y)\omega(Z,X)+F(Z)\theta(X,Y)=0,\\\
\langle\omega(Y,T),\theta(X,Z)\rangle_{A}-\langle\omega(X,T),\theta(Y,Z)\rangle_{A}+\langle\theta_{1}(Z,T),\theta(X,Y)\rangle_{A}=0.\end{cases}$

These two systems are equivalent to

$\begin{cases}{\mathrm{ad}}_{\omega(X,Y)}^{A}=-[F(X),F(Y)],\\\
{\mathrm{ad}}^{A}_{\theta(X,Y)}=[F(X),F(Y)],\\\
F(X)\theta(Y,Z)+F(Y)\theta(Z,X)+F(Z)\theta(X,Y)=0,\\\
F(X)\omega(Y,Z)+F(Y)\omega(Z,X)-F(Z)\theta(X,Y)=0,\\\
F(X)\theta(Y,Z)-F(Y)\omega(Z,X)+F(Z)\theta(X,Y)=0,\\\
\langle\omega(Y,T),\theta(X,Z)\rangle_{A}-\langle\omega(X,T),\theta(Y,Z)\rangle_{A}+\langle\theta(Z,T),\theta(X,Y)\rangle_{A}=0,\\\
\langle\omega(Y,T),\theta(X,Z)\rangle_{A}-\langle\omega(X,T),\theta(Y,Z)\rangle_{A}+\langle\theta(Z,T),\theta(X,Y)\rangle_{A}=0.\end{cases}$

The equations (3), (4) and (5) in the above system are equivalent to

$F(X)\theta(Y,Z)+F(Y)\theta(Z,X)+F(Z)\theta(X,Y)=0\quad\mbox{and}\quad
F(X)\left[\theta(Y,Z)+\omega(Y,Z)\right]=0,$

which completes the proof. ∎

######  Corollary 6.3.

Any Lorentzian ${\mathrm{L}}$-quadratic symmetric Leibniz algebra is a
quadratic Lie algebra.

######  Remark 4.

In Theorem 6.3, if we take $\theta=\omega=0$, we can build for free many
examples of ${\mathrm{L}}$-quadratic symmetric Leibniz algebras. For instance
let ${\mathfrak{h}}$ be a vector space with a basis $(e_{1},\ldots,e_{r})$,
$(A,\langle\;,\;\rangle_{A})$ a 2-dimensional abelian Euclidean Lie algebra
with a basis $(f_{1},f_{2})$. For $i=1,\ldots,r$, take
$F(e_{i})=\left(\begin{matrix}0&\lambda_{i}\\\
-\lambda_{i}&0\end{matrix}\right)$. We take $\omega=\theta=0$ and
$\Omega(e_{i})=\left(a_{kl}^{i}\right)_{kl}$ a skew-symmetric matrix. So on
${\mathfrak{h}}\oplus A\oplus{\mathfrak{h}}^{*}$ we get a structure of
symmetric Leibniz algebra $\bullet$ and a ${\mathrm{L}}$-quadratic metric
$\langle\;,\;\rangle_{n}$ by

$e_{i}\bullet e_{j}=\sum_{k=1}^{r}a_{jk}^{i}e_{k}^{*},\;e_{i}\bullet
f_{1}=-\lambda_{i}f_{2},\;e_{i}\bullet f_{2}=\lambda_{i}f_{1},f_{1}\bullet
f_{2}=-f_{2}\bullet f_{1}=-\sum_{i=1}^{r}\lambda_{i}e_{i}^{*}.$

##  7 Appendix

In this appendix we give the conditions for which the pseudo-Euclidean left
Leibniz algebras $({\mathfrak{h}}\oplus
A\oplus{\mathfrak{h}}^{*},\bullet,\langle\;,\;\rangle_{n})$ appearing,
respectively, in Theorem 5.1 and Theorem 6.1 is a ${\mathrm{L}}$-quadratic
(resp. ${\mathrm{R}}$-quadratic) left Leibniz algebra in order to complete the
proof of these theorem.

The Leibniz products (10) and (14) can be written

$\begin{cases}X\bullet
Y=\epsilon[X,Y]_{\mathfrak{h}}+\theta(X,Y)+\theta_{2}(X,Y),\;a\bullet
b=a\bullet_{A}b+\mu(a,b),\\\ X\bullet
a=F(X)a+\nu(X,a),\;X\bullet\alpha=(\mathrm{ad}^{{\mathfrak{h}}})_{X}^{*}\alpha,\;a\bullet
X=G(X)a+\rho(a,X),\quad\epsilon\in\\{0,1\\},\end{cases}$

where

  1. 1.

$\epsilon=1$, $F(X)$ skew-symmetric, $\theta_{2}(X,Y)=\Omega(X)^{\flat}(Y)$,
$\omega$ skew-symmetric, and

$\prec\rho(a,X),Y\succ=-\langle\omega(X,Y),a\rangle_{A},\;\prec\mu(a,b),X\succ=-\langle
G(X)a,b\rangle_{A},\;\prec\nu(X,a),Y\succ=-\langle\theta(X,Y),a\rangle_{A}$

when the metric is ${\mathrm{L}}$-quadratic,

  2. 2.

$\epsilon=0$, $G=-F$, $F(X)$ skew-symmetric, $\theta$ and $\omega$ are skew-
symmetric, $\theta_{2}(X,Y)=\Omega(Y)^{\flat}(X)$ and

$\prec\nu(X,a),Y\succ=\langle\omega(X,Y),a\rangle_{A},\;\prec\mu(a,b),X\succ=\langle
F(X)a,b\rangle_{A},\;\prec\rho(a,X),Y\succ=\langle\theta(X,Y),a\rangle_{A}$

when the metric is ${\mathrm{R}}$-quadratic,

This product defines a left Leibniz structure if and only if, for any
$u,v,w\in\mathfrak{g}$,

$Q(u,v,w):=u\bullet(v\bullet w)-(u\bullet v)\bullet w-v\bullet(u\bullet w)=0.$

Since, for any $\alpha\in{\mathfrak{h}}^{*}$, ${\mathrm{L}}_{\alpha}=0$ and
${\mathfrak{h}}^{*}$ is a left ideal, we have $Q(\alpha,.,.)=Q(.,\alpha,.)=0$.
So the vanishing of $Q$ is equivalent to This is equivalent to

$\begin{cases}Q(X,Y,Z)=Q(X,Y,\alpha)=Q(X,Y,a)=Q(X,a,Y)=Q(X,a,\alpha)=Q(X,a,b)=0,\\\
Q(a,b,c)=Q(a,b,X)=Q(a,X,b)=Q(a,b,\alpha)=Q(a,X,Y)=Q(a,X,\alpha)=0,\end{cases}$

for any $X,Y,Z\in{\mathfrak{h}}$, $a,b,c\in A$, and
$\alpha,\beta\in{\mathfrak{h}}^{*}$.

On the other hand, when the metric is ${\mathrm{L}}$-quadratic, we have, for
any $r,u,v,w\in\mathfrak{g}$,

$\langle Q(u,v,w),r\rangle_{n}=-\langle Q(u,v,r),w\rangle_{n}.$

We will use this relation to ovoid redundant equations.

  1. 1.

We have

$\displaystyle X\bullet(Y\bullet Z)$
$\displaystyle=X\bullet(\epsilon[Y,Z]_{\mathfrak{h}}+\theta(Y,Z)+\theta_{2}(Y,Z))$
$\displaystyle=\epsilon[X,[Y,Z]_{\mathfrak{h}}]_{\mathfrak{h}}+\epsilon\theta(X,[Y,Z]_{\mathfrak{h}})+\epsilon\theta_{2}(X,[Y,Z]_{\mathfrak{h}})+F(X)\theta(Y,Z)+\nu(X,\theta(Y,Z))+({\mathrm{ad}}_{X}^{\mathfrak{h}})^{*}\theta_{2}(Y,Z)$
$\displaystyle(X\bullet Y)\bullet Z$
$\displaystyle=(\epsilon[X,Y]_{\mathfrak{h}}+\theta(X,Y)+\theta_{2}(X,Y))\bullet
Z$
$\displaystyle=\epsilon[[X,Y]_{\mathfrak{h}},Z]_{\mathfrak{h}}+\epsilon\theta([X,Y]_{\mathfrak{h}},Z)+\epsilon\theta_{2}([X,Y]_{\mathfrak{h}},Z)+G(Z)\theta(X,Y)+\rho(\theta(X,Y),Z)$
$\displaystyle Y\bullet(X\bullet Z)$
$\displaystyle=\epsilon[Y,[X,Z]_{\mathfrak{h}}]_{\mathfrak{h}}+\epsilon\theta(Y,[X,Z]_{\mathfrak{h}})+\epsilon\theta_{2}(Y,[X,Z]_{\mathfrak{h}})+F(Y)\theta(X,Z)+\nu(Y,\theta(X,Z))+({\mathrm{ad}}_{Y}^{\mathfrak{h}})^{*}\theta_{2}(X,Z).$

$\bullet$ If the metric is ${\mathrm{L}}$-quadratic, $Q(X,Y,Z)=0$ is
equivalent to $({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$ is a Lie algebra and

$\begin{cases}\theta(X,[Y,Z]_{\mathfrak{h}})+F(X)\theta(Y,Z)-\theta([X,Y]_{\mathfrak{h}},Z)-G(Z)\theta(X,Y)-\theta(Y,[X,Z]_{\mathfrak{h}})-F(Y)\theta(X,Z)=0,\\\
K:=\theta_{2}(X,[Y,Z]_{\mathfrak{h}})+\nu(X,\theta(Y,Z))+({\mathrm{ad}}_{X}^{\mathfrak{h}})^{*}\theta_{2}(Y,Z)-\theta_{2}([X,Y]_{\mathfrak{h}},Z)-\rho(\theta(X,Y),Z)-\theta_{2}(Y,[X,Z]_{\mathfrak{h}})\\\
-\nu(Y,\theta(X,Z))-({\mathrm{ad}}_{Y}^{\mathfrak{h}})^{*}\theta_{2}(X,Z)=0.\end{cases}$

We get the equation (5) in (12). We have

$\displaystyle\prec K,T\succ$
$\displaystyle=\Omega(X)([Y,Z]_{\mathfrak{h}},T)-\langle\theta(X,T),\theta(Y,Z)\rangle_{A}-\Omega(Y)(Z,[X,T]_{\mathfrak{h}})-\Omega([X,Y]_{\mathfrak{h}})(Z,T)$
$\displaystyle+\langle\omega(Z,T),\theta(X,Y)\rangle_{A}-\Omega(Y)([X,Z]_{\mathfrak{h}},T)+\langle\theta(Y,T),\theta(X,Z)\rangle_{A}+\Omega(X)(Z,[Y,T]_{\mathfrak{h}})$
$\displaystyle=\Delta(\Omega)(X,Y)(Z,T)-\langle\theta(X,T),\theta(Y,Z)\rangle_{A}+\langle\omega(Z,T),\theta(X,Y)\rangle_{A}+\langle\theta(Y,T),\theta(X,Z)\rangle_{A}.$

So $K=0$ if and only if the last equation in (12) holds.

$\bullet$ If the metric is ${\mathrm{R}}$-quadratic, $Q(X,Y,Z)=0$ is
equivalent to

$\begin{cases}F(X)\theta(Y,Z)+F(Y)\theta(Z,X)+F(Z)\theta(X,Y)=0,\\\
K:=\nu(X,\theta(Y,Z))+({\mathrm{ad}}_{X}^{\mathfrak{h}})^{*}\theta_{2}(Y,Z)-\rho(\theta(X,Y),Z)-\nu(Y,\theta(X,Z))-({\mathrm{ad}}_{Y}^{\mathfrak{h}})^{*}\theta_{2}(X,Z)=0.\end{cases}$

We get the equation (3) in (16). We have

$\displaystyle\prec K,T\succ$
$\displaystyle=\langle\omega(X,T),\theta(Y,Z)\rangle_{A}-\Omega(Z)(Y,[X,T]_{\mathfrak{h}})-\langle\theta(Z,T),\theta(X,Y)\rangle_{A}-\langle\omega(Y,T),\theta(X,Z)\rangle_{A}+\Omega(Z)(X,[Y,T]_{\mathfrak{h}})$
$\displaystyle=({\mathrm{ad}}_{T}^{\mathfrak{h}})^{*}\Omega(Z)(X,Y)+\langle\omega(X,T),\theta(Y,Z)\rangle_{A}-\langle\theta(Z,T),\theta(X,Y)\rangle_{A}-\langle\omega(Y,T),\theta(X,Z)\rangle_{A}.$

So $K=0$ if and only if the last equation in (16) holds.

  2. 2.

We have

$\displaystyle X\bullet(Y\bullet\alpha)$
$\displaystyle=X\bullet(\mathrm{ad}^{{\mathfrak{h}}}_{Y})^{*}\alpha$
$\displaystyle=(\mathrm{ad}^{{\mathfrak{h}}}_{X})^{*}(\mathrm{ad}^{{\mathfrak{h}}}_{Y})^{*}\alpha$
$\displaystyle(X\bullet Y)\bullet\alpha$
$\displaystyle=(\epsilon[X,Y]_{{\mathfrak{h}}}+\theta(X,Y)+\theta_{2}(X,Y))\bullet\alpha$
$\displaystyle=(\mathrm{ad}^{{\mathfrak{h}}}_{\epsilon[X,Y]})^{*}\alpha$
$\displaystyle Y\bullet(X\bullet\alpha)$
$\displaystyle=(\mathrm{ad}^{{\mathfrak{h}}}_{Y})^{*}(\mathrm{ad}^{{\mathfrak{h}}}_{X})^{*}\alpha).$

$\bullet$ If the metric is ${\mathrm{L}}$-quadratic then $Q(X,Y,\alpha)=0$ if
and only if $({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$ is a Lie algebra.

$\bullet$ If the metric is ${\mathrm{R}}$-quadratic then $Q(X,Y,\alpha)=0$ if
and only if $({\mathfrak{h}},[\;,\;]_{\mathfrak{h}})$ is a Lie algebra and
$[X,[Y,Z]_{\mathfrak{h}}]_{\mathfrak{h}}=0$ for any $X,Y,Z\in{\mathfrak{h}}$.

  3. 3.

We have

$\displaystyle X\bullet(Y\bullet a)$ $\displaystyle=X\bullet(F(Y)a+\nu(Y,a))$
$\displaystyle=F(X)F(Y)a+\nu(X,F(Y)a)+(\mathrm{ad}^{{\mathfrak{h}}}_{X})^{*}\nu(Y,a)$
$\displaystyle(X\bullet Y)\bullet a$
$\displaystyle=(\epsilon[X,Y]_{{\mathfrak{h}}}+\theta(X,Y)+\theta_{2}(X,Y))\bullet
a$
$\displaystyle=F(\epsilon[X,Y]_{{\mathfrak{h}}})a+\nu(\epsilon[X,Y]_{{\mathfrak{h}}},a)+\theta(X,Y)\bullet_{A}a+\mu(\theta(X,Y),a)$
$\displaystyle Y\bullet(X\bullet a)$ $\displaystyle=Y\bullet(F(X)a+\nu(X,a))$
$\displaystyle=F(Y)F(X)a+\nu(Y,F(X)a)+(\mathrm{ad}_{Y})^{*}\nu(X,a).$

$\bullet$ If the metric is ${\mathrm{L}}$-quadratic then $Q(X,Y,a)=0$ if and
only if $\langle Q(X,Y,a),Z\rangle_{n}=-\langle Q(X,Y,Z),a\rangle_{n}=0$ and

${\mathrm{L}}_{\theta(X,Y)}=[F(X,F(Y))]-F([X,Y]_{\mathfrak{h}}).$

The first condition is satisfied in the previous case and the second one is
equivalent to a part of the equation (3) in (12).

$\bullet$ if the metric is ${\mathrm{R}}$-quadratic then $Q(X,Y,a)=0$ if and
only if

$\begin{cases}{\mathrm{ad}}_{\theta(X,Y)}^{A}=[F(X),F(Y)],\\\
K:=\nu(X,F(Y)a)+(\mathrm{ad}^{{\mathfrak{h}}}_{X})^{*}\nu(Y,a)-\mu(\theta(X,Y),a)-\nu(Y,F(X)a)-(\mathrm{ad}_{Y})^{*}\nu(X,a)=0.\end{cases}$

The first relation is equivalent to the equation (2) in (16). We have

$\displaystyle\prec K,Z\succ$
$\displaystyle=\langle\omega(X,Z),F(Y)a\rangle_{A}-\langle\omega(Y,[X,Z]_{\mathfrak{h}}),a\rangle_{A}-\langle
F(Z)\theta(X,Y),a\rangle_{A}-\langle\omega(Y,Z),F(X)a\rangle_{A}+\langle\omega(X,[Y,Z]_{\mathfrak{h}}),a\rangle_{A}.$

So $K=0$ if and only if

$F(Y)\omega(X,Z)+\omega(Y,[X,Z]_{\mathfrak{h}})+F(Z)\theta(X,Y)-F(X)\omega(Y,Z)-\omega(X,[Y,Z]_{\mathfrak{h}})=0.$

This is equivalent to the equation (4) in (16).

  4. 4.

We have

$\displaystyle X\bullet(a\bullet Y)$ $\displaystyle=X\bullet(G(Y)a+\rho(a,Y))$
$\displaystyle=F(X)G(Y)a+\nu(X,G(Y)a)+(\mathrm{ad}^{{\mathfrak{h}}}_{X})^{*}(\rho(a,Y)),$
$\displaystyle(X\bullet a)\bullet Y$ $\displaystyle=(F(X)a+\nu(X,a))\bullet Y$
$\displaystyle=G(Y)F(X)a+\rho(F(X)a,Y),$ $\displaystyle a\bullet(X\bullet Y)$
$\displaystyle=\epsilon
G([X,Y]_{\mathfrak{h}})a+\rho(a,\epsilon[X,Y]_{\mathfrak{h}})+a\bullet_{A}\theta(X,Y)+\mu(a,\theta(X,Y)).$

$\bullet$ If the metric is ${\mathrm{L}}$-quadratic then $Q(X,a,Y)=0$ if and
only if

$\begin{cases}{\mathrm{R}}_{\theta(X,Y)}=[F(X),G(Y)]-G([X,Y]_{\mathfrak{h}})\\\
K:=\nu(X,G(Y)a)+(\mathrm{ad}^{{\mathfrak{h}}}_{X})^{*}(\rho(a,Y))-\rho(F(X)a,Y)-\rho(a,[X,Y]_{\mathfrak{h}})-\mu(a,\theta(X,Y))=0.\end{cases}$

We have

$\displaystyle\prec K,Z\succ$
$\displaystyle=-\langle\theta(X,Z),G(Y)a\rangle_{A}+\langle\omega(Y,[X,Z]_{\mathfrak{h}}),a\rangle_{A}+\langle\omega(Y,Z),F(X)a\rangle_{A}+\omega([X,Y]_{\mathfrak{h}},Z),a\rangle_{A}+\langle
G(Z)a,\theta(X,Y)\rangle_{A}.$

So $Q(X,a,Y)=0$ if and only if the equation (3) and equation (6) in (12) hold.

$\bullet$ If the metric is ${\mathrm{R}}$-invariant $Q(X,a,Y)=0$ if and only
if

$\begin{cases}{\mathrm{ad}}_{\theta(X,Y)}^{A}=[F(X),F(Y)]\\\
K:=-\nu(X,F(Y)a)+(\mathrm{ad}^{{\mathfrak{h}}}_{X})^{*}(\rho(a,Y))-\rho(F(X)a,Y)-\mu(a,\theta(X,Y))=0.\end{cases}$

We have

$\displaystyle\langle K,Z\succ$
$\displaystyle=-\langle\omega(X,Z),F(Y)a\rangle_{A}-\langle\theta(Y,[X,Z]_{\mathfrak{h}}),a\rangle_{A}-\langle\omega(Y,Z),F(X)a\rangle_{A}-\langle
F(Z)a,\theta(X,Y)\rangle.$

So $K=0$ if and only if

$F(Y)\omega(X,Z)-\theta(Y,[X,Z]_{\mathfrak{h}})+F(X)\theta(Y,Z)+F(Z)\theta(X,Y)=0.$

So $Q(X,a,Y)=0$ if and only if the equations (2) and (5) in (16) hold.

  5. 5.

We have

$\displaystyle X\bullet(a\bullet b)$
$\displaystyle=X\bullet(a\bullet_{A}b+\mu(a,b))$
$\displaystyle=F(X)(a\bullet_{A}b)+\nu(X,a\bullet_{A}b)+(\mathrm{ad}^{{\mathfrak{h}}}_{X})^{*}(\mu(a,b))$
$\displaystyle(X\bullet a)\bullet b$ $\displaystyle=(F(X)a+\nu(X,a))\bullet b$
$\displaystyle=F(X)a\bullet_{A}b+\mu(F(X)a,b)$ $\displaystyle
a\bullet(X\bullet b)$ $\displaystyle=a\bullet(F(X)b+\nu(X,b))$
$\displaystyle=a\bullet_{A}F(X)b+\mu(a,F(X)b).$

$\bullet$ If the metric is ${\mathrm{L}}$-quadratic then $Q(X,a,b)=0$ if and
only if $\langle Q(X,a,b),Y\rangle_{n}=-\langle Q(X,a,Y),b\rangle_{n}=0$ and
$F(X)$ is a derivation for any $X$. We have already get the condition for
$\langle Q(X,a,Y),b\rangle_{n}=0$.

$\bullet$ If the metric is ${\mathrm{R}}$-quadratic then $Q(X,a,b)=0$ if and
only if $F(X)$ is a derivation for any $X$ and

$K:=\nu(X,a\bullet_{A}b)+(\mathrm{ad}^{{\mathfrak{h}}}_{X})^{*}(\mu(a,b))-\mu(F(X)a,b)-\mu(a,F(X)b)=0.$

We have

$\displaystyle\prec K,Y\succ$
$\displaystyle=\langle\omega(X,Y),[a,b]_{A}\rangle_{A}-\langle
F([X,Y]_{\mathfrak{h}})a,b\rangle-\langle
F(Y)F(X)a,b\rangle-\langle_{A}F(Y)a,F(X)b\rangle_{A}.$

So $K=0$ is equivalent to
${\mathrm{ad}}^{A}_{\omega(X,Y)}=[F(X),F(y)]-F([X,Y]_{\mathfrak{h}})$ which is
equivalent the equation (1) in (16).

  6. 6.

We have

$\displaystyle a\bullet(b\bullet X)$ $\displaystyle=a\bullet(G(X)b+\rho(b,X))$
$\displaystyle=a\bullet_{A}G(X)b+\mu(a,G(X)b))$ $\displaystyle(a\bullet
b)\bullet X$ $\displaystyle=(a\bullet_{A}b+\mu(a,b))\bullet X$
$\displaystyle=G(X)(a\bullet_{A}b)+\rho(a\bullet_{A}b,X)$ $\displaystyle
b\bullet(a\bullet X)$ $\displaystyle=b\bullet(G(X)a+\rho(a,X))$
$\displaystyle=b\bullet_{A}G(X)a+\mu(b,G(X)a).$

$\bullet$ If the metric is ${\mathrm{L}}$-quadratic then $Q(a,b,X)=0$ if and
only if

$\begin{cases}G(X)(a\bullet_{A}b)=a\bullet_{A}G(X)b-b\bullet_{A}G(X)a,\\\
K:=\mu(a,G(X)b))-\rho(a\bullet_{A}b,X)-\mu(b,G(X)a)=0.\end{cases}$

We have

$\displaystyle\prec K,Y\succ$ $\displaystyle=-\langle
G(Y)a,G(X)b\rangle_{A}+\langle\omega(X,Y),a\bullet_{A}b\rangle_{A}+\langle
G(Y)b,G(X)a\rangle_{A}.$

So $K=0$ is equivalent to

${\mathrm{L}}_{\omega(X,Y)}^{A}=G(X)^{*}G(Y)-G(Y)^{*}G(X).$

So $Q(a,b,X)=0$ if and only if the second part of the first equations and the
second equation in (12) hold..

$\bullet$ If the metric is ${\mathrm{R}}$-quadratic gives that $Q(a,b,X)=0$ if
and only if $F$ is a derivation an $K=0$ where

$K=-\mu(a,F(X)b))-\rho([a,b]_{A},X)+\mu(b,F(X)a)=0$ $\displaystyle\prec
K,Y\succ$ $\displaystyle=-\langle
F(Y)a,F(X)b\rangle_{A}-\langle\theta(X,Y),[a,b]_{A}\rangle_{A}+\langle
F(Y)b,F(X)a\rangle_{A}.$

So $K=0$ if and only if the second equation in (16) holds.

  7. 7.

We have

$\displaystyle a\bullet(b\bullet c)$
$\displaystyle=a\bullet(b\bullet_{A}c+\mu(b,c))$
$\displaystyle=a\bullet_{A}(b\bullet_{A}c)+\mu(a,b\bullet_{A}c)$
$\displaystyle(a\bullet b)\bullet c$
$\displaystyle=(a\bullet_{A}b+\mu(a,b))\bullet c$
$\displaystyle=(a\bullet_{A}b)\bullet_{A}c+\mu(a\bullet_{A}b,c)$
$\displaystyle b\bullet(a\bullet c)$
$\displaystyle=b\bullet(a\bullet_{A}c+\mu(a,c))$
$\displaystyle=b\bullet_{A}(a\bullet_{A}c)+\mu(b,a\bullet_{A}c).$

$\bullet$ If the metric is ${\mathrm{L}}$-quadratic then $Q(a,b,c)=0$ is
equivalent to $\langle Q(a,b,c),X\rangle_{n}=-\langle Q(a,b,X),c\rangle_{n}=0$
and $(A,\bullet_{A})$ is a left Leibniz algebra.

$\bullet$ If the metric is ${\mathrm{R}}$-quadratic then $Q(a,b,c)=0$ is
equivalent to $(A,[\;,\;]_{A})$ is a Lie algebra and

$K=\mu(a,[b,c]_{A})-\mu([a,b]_{A},c)-\mu(b,[ac]_{A})=0.$

We have

$\displaystyle\prec K,X\succ$ $\displaystyle=\langle
F(X)a,[b,c]_{A}\rangle_{A}-\langle F(X)[a,b]_{A},c\rangle_{A}-\langle
F(X)b,[a,c]_{A}\rangle_{A}$

which is equivalent to $F(X)$ is a derivation.

  8. 8.

We have

$\displaystyle a\bullet(X\bullet Y)$
$\displaystyle=a\bullet(\epsilon[X,Y]_{{\mathfrak{h}}}+\theta(X,Y)+\theta_{2}(X,Y))$
$\displaystyle=G(\epsilon[X,Y]_{{\mathfrak{h}}})a+\rho(a,\epsilon[X,Y]_{{\mathfrak{h}}})+a\bullet_{A}\theta(X,Y)+\mu(a,\theta(X,Y))$
$\displaystyle(a\bullet X)\bullet Y$ $\displaystyle=(G(X)a+\rho(a,X))\bullet
Y$ $\displaystyle=G(Y)G(X)a+\rho(G(X)a,Y),$ $\displaystyle X\bullet(a\bullet
Y)$ $\displaystyle=X\bullet(G(Y)a+\rho(a,Y))$
$\displaystyle=F(X)G(Y)a+\nu(X,G(Y)a)+(\mathrm{ad}^{{\mathfrak{h}}}_{X})^{*}(\rho(a,Y)).$

$\bullet$ if the metric is ${\mathrm{L}}$-quadratic then $Q(a,X,Y)=0$ if and
only if

$\begin{cases}{\mathrm{R}}^{A}_{\theta(X,Y)}=F(X)G(Y)+G(Y)G(X)-G([X,Y]_{{\mathfrak{h}}}),\\\
K:=\rho(a,[X,Y]_{{\mathfrak{h}}})+\mu(a,\theta(X,Y))-\rho(G(X)a,Y)-\nu(X,G(Y)a)-(\mathrm{ad}^{{\mathfrak{h}}}_{X})^{*}(\rho(a,Y))=0.\end{cases}$

We have

$\displaystyle\prec K,Z\succ$
$\displaystyle=-\langle\omega([X,Y]_{\mathfrak{h}},Z),a\rangle_{A}-\langle
G(Z)a,\theta(X,Y)\rangle_{A}+\langle\omega(Y,Z),G(X)a\rangle_{A}+\langle\theta(X,Z),G(Y)a\rangle_{A}-\langle\omega(Y,[X,Z]_{\mathfrak{h}}),a\rangle_{A}.$

So $K=0$ if and only if

$G(Z)^{*}\theta(X,Y)-G(Y)^{*}\theta(X,Z)-G(X)^{*}\omega(Y,Z)+\omega([X,Y]_{\mathfrak{h}},Z)+\omega(Y,[X,Z]_{\mathfrak{h}})=0.$

This equation resembles to equation (6) and by taking the difference, we have
$Q(a,X,Y)=0$ if and only if the second part of equation (3) and the equation
(7) in (12) hold.

$\bullet$ If the metric is ${\mathrm{R}}$-quadratic then $Q(a,X,Y)=0$ if and
only if

$\begin{cases}{\mathrm{ad}}^{A}_{\theta(X,Y)}=[F(X),F(Y)],\\\
K:=\mu(a,\theta(X,Y))+\rho(F(X)a,Y)+\nu(X,F(Y)a)-(\mathrm{ad}^{{\mathfrak{h}}}_{X})^{*}(\rho(a,Y))=0.\end{cases}$

We have

$\displaystyle\prec K,Z\succ$ $\displaystyle=\langle
F(Z)a,\theta(X,Y)\rangle_{A}+\langle\theta(Y,Z),F(X)a\rangle_{A}+\langle\omega(X,Z),F(Y)a\rangle_{A}+\langle\theta(Y,[X,Z]_{\mathfrak{h}}),a\rangle_{A}.$

So $K=0$ if and only if

$F(Z)\theta(X,Y)+F(X)\theta(Y,Z)-F(Y)\omega(Z,X)-\theta(Y,[X,Z]_{\mathfrak{h}})=0.$

Finally, $Q(a,X,Y)=0$ if and only if the equations (2) and (5) in (16) hold.

  9. 9.

We have

$\displaystyle a\bullet(X\bullet b)$ $\displaystyle=a\bullet(F(X)b+\nu(X,a))$
$\displaystyle=a\bullet_{A}F(X)b+\mu(a,F(X)b)$ $\displaystyle(a\bullet
X)\bullet b$ $\displaystyle=(G(X)a+\rho(a,X))\bullet b$
$\displaystyle=G(X)a\bullet_{A}b+\mu(G(X)a,b)$ $\displaystyle
X\bullet(a\bullet b)$ $\displaystyle=X\bullet(a\bullet_{A}b+\mu(a,b))$
$\displaystyle=F(X)(a\bullet_{A}b)+\nu(X,a\bullet_{A}b)+(\mathrm{ad}^{{\mathfrak{h}}}_{X})^{*}(\mu(a,b)).$

$\bullet$ If the metric is ${\mathrm{L}}$-quadratic then $Q(a,X,b)=0$ if and
only if $\langle Q(a,X,b),Y\rangle_{n}=-\langle Q(a,X,Y),b\rangle_{n}=0$ and

$F(X)(a\bullet_{A}b)=a\bullet_{A}F(X)b-G(X)a\bullet_{A}b$

which is equivalent to the first equation in (12).

$\bullet$ If the metric is ${\mathrm{R}}$-quadratic then $Q(a,X,b)=0$ if and
only if $F(X)$ is a derivation and

$K:=\mu(a,F(X)b)+\mu(F(X)a,b)-\nu(X,[a,b]_{A})-(\mathrm{ad}^{{\mathfrak{h}}}_{X})^{*}(\mu(a,b))=0.$

We have

$\displaystyle\prec K,Y\succ$ $\displaystyle=\langle
F(Y)a,F(X)b\rangle_{A}+\langle
F(Y)F(X)a,b\rangle_{A}-\langle\omega(X,Y),[a,b]_{A}+\langle
F([X,Y]_{\mathfrak{h}})a,b\rangle_{A}.$

So $K=0$ is equivalent to

$-F(X)F(Y)+F(Y)F(X)-{\mathrm{ad}}^{A}_{\omega(X,Y)}+F([X,Y]_{\mathfrak{h}})=0$

which is equivalent to the first equation in (16).

  10. 10.

$Q(X,a,\alpha)=Q(a,X,\alpha)=Q(a,b,\alpha)=0$ are satisfied.

##  8  Low dimensional metrised, ${\mathrm{L}}$-quadratic or
${\mathrm{R}}$-quadratic left Leibniz algebras

We illustrate all our results by giving the list of metrised symmetric Leibniz
algebras of dimension 4,5 and 6 obtained from indecomposable non abelian
quadratic Lie algebras, the list of ${\mathrm{L}}$-quadratic non Lie left
Leibniz algebras of $\leq$ 4 and the list of ${\mathrm{R}}$-quadratic non Lie
Leibniz algebras of dimension $\leq$ 5.

Algebra | The non vanishing bracket | The metric
---|---|---
$\mathrm{os}(4,\lambda)$ | $[e_{2},e_{3}]=e_{1},[e_{4},e_{2}]=\lambda e_{3},[e_{4},e_{3}]=-\lambda e_{2},e_{4}\circ e_{4}=\mu e_{1}$ | $\\{(2,2)=(3,3)=\frac{1}{\lambda},(1,4)=1\\}$
$\mathfrak{g}_{1,4}$ | $[e_{4},e_{2}]=e_{2},\;[e_{4},e_{3}]=-e_{3},\;[e_{2},e_{3}]=e_{1},e_{4}\circ e_{4}=\mu e_{1}$ | $\\{(1,4)=(2,3)=1\\}$
$\mathfrak{g}_{1,5}$ | $[e_{2},e_{3}]=e_{1},[e_{3},e_{4}]=-e_{1},[e_{5},e_{2}]=e_{3},$ | $\\{(1,5)=-(2,2)=(3,3)=(4,4)=1\\}$
| $[e_{5},e_{3}]=e_{2}-e_{4},[e_{5},e_{4}]=e_{3},e_{5}\circ e_{5}=\mu e_{1}$ |
$\mathrm{osc}(6,(\lambda_{1},\lambda_{2}))$ | $[e_{2},e_{4}]=[e_{3},e_{5}]=e_{1},[e_{6},e_{2}]=\lambda_{1}e_{4}$ | $\\{(2,2)=(3,3)=(4,4)=(5,5)=\frac{1}{\lambda}$
|  $[e_{6},e_{4}]=-\lambda_{1}e_{2},[e_{6},e_{3}]=\lambda_{2}e_{5},[e_{6},e_{5}]=-\lambda_{2}e_{3}$,$e_{6}\circ e_{6}=\mu e_{1}$ | $(1,6)=1\\}$
$\mathfrak{n}_{1}(2,2)$ | $[e_{6},e_{3}]=e_{2},[e_{6},e_{5}]=e_{4},[e_{3},e_{5}]=e_{1}.$ | $\\{(1,6)=(2,5)=-(3,4)=1\\}$
$\mathfrak{n}_{2}(2,2)$ | $[e_{6},e_{2}]=e_{2}+te_{3},[e_{6},e_{3}]=-te_{2}+e_{3},[e_{6},e_{4}]=-e_{4}+te_{5}$ | $\\{(1,6)=(2,5)=-(3,4)=1\\}$
| $[e_{6},e_{5}]=-te_{4}-e_{3},[e_{2},e_{4}]=-te_{1},[e_{2},e_{5}]=e_{1},$ |
| $[e_{3},e_{4}]=-e_{1},[e_{3},e_{5}]=-te_{1},\;e_{6}\circ e_{6}=\mu e_{1}\quad t>0$ |
$\mathfrak{n}_{3}(2,2)$ | $[e_{6},e_{2}]=e_{3},[e_{6},e_{3}]=-e_{2},[e_{6},e_{4}]=\epsilon e_{2}+e_{5},\epsilon^{2}=1,e_{6}\circ e_{6}=\mu e_{1}$ | $\\{(1,6)=(2,5)=-(3,4)=1\\}$
| $[e_{6},e_{5}]=\epsilon e_{3}-e_{4},[e_{2},e_{4}]=-e_{1},[e_{3},e_{5}]=-e_{1},[e_{4},e_{5}]=\epsilon e_{1}.$ |
$\mathfrak{n}_{4}(2,2)$ | $[e_{6},e_{2}]=e_{3},[e_{6},e_{3}]=-e_{2},[e_{6},e_{4}]=te_{5},e_{6}\circ e_{6}=\mu e_{1}$ | $\\{-(2,2)=-(3,3)=(4,4)=(5,5)=1$
| $[e_{6},e_{5}]=-te_{4},[e_{2},e_{3}]=-e_{1},[e_{4},e_{5}]=te_{1}.\quad(t>0)$ | $(1,6)=1\\}$
$\mathfrak{n}_{5}(2,2)$ | $[e_{6},e_{2}]=e_{2},[e_{6},e_{3}]=e_{2}+e_{3},[e_{6},e_{4}]=-e_{4},[e_{6},e_{5}]=e_{4}-e_{5},$ | $\\{(1,6)=(2,5)=-(3,4)=1\\}$
| $[e_{2},e_{5}]=e_{1},[e_{3},e_{4}]=-e_{1},[e_{3},e_{5}]=e_{1},e_{6}\circ e_{6}=\mu e_{1}.$ |
$\mathfrak{n}_{6}(2,2)$ | $[e_{6},e_{2}]=e_{2},[e_{6},e_{3}]=-e_{3},[e_{6},e_{4}]=te_{4},e_{6}\circ e_{6}=\mu e_{1}$ | $\\{(1,6)=(2,3)=(4,5)=1\\}$
| $[e_{6},e_{5}]=-te_{5},[e_{2},e_{3}]=e_{1},[e_{4},e_{5}]=te_{1},t\geq 1$ |

Table 1: Metrised symmetric Leibniz algebras of dimension 4,5 and 6 obtained
from indecomposable non abelian quadratic Lie algebras $(\mu\not=0)$.

Algebra | The non vanishing bracket | The metric
---|---|---
$\mathfrak{g}_{3,1}$ |  $e_{3}\bullet e_{1}=\lambda e_{2},\;e_{3}\bullet e_{2}=-\lambda e_{1}$, $\lambda\neq 0$ | $\\{(1,1)=(2,2)=1,(3,3)=\pm 1\\}$
$\mathfrak{g}_{3,2}$ |  $e_{3}\bullet e_{1}=\lambda e_{2},\;e_{3}\bullet e_{2}=\lambda e_{1}$, $\lambda\neq 0$ | $\\{(1,1)=-1,(2,2)=(3,3)=1\\}$
$\mathfrak{g}_{3,3}$ | $e_{2}\bullet e_{2}=-\mu e_{3},\;e_{2}\bullet e_{3}=\mu e_{1},\mu\neq 0$ | $\\{(1,2)=(2,1)=(3,3)=1\\}$
$\mathfrak{g}_{4,1}$ | $e_{1}\bullet e_{3}=e_{4},\;e_{1}\bullet e_{4}=-e_{3},\;e_{2}\bullet e_{3}=e_{4},$ | $\\{(1,1)=b,(2,2)=(3,3)=1,(4,4)=a\\},$
| $e_{2}\bullet e_{4}=-e_{3}$ | $a^{2}=b^{2}=1$
$\mathfrak{g}_{4,2}$ | $e_{1}\bullet e_{1}=ae_{4},\;e_{1}\bullet e_{2}=-ae_{3},\;e_{2}\bullet e_{1}=be_{4},\;e_{2}\bullet e_{2}=-be_{3}$ | $\\{(1,3)=(3,1)=(2,4)=(4,2)=1\\}$
$\mathfrak{g}_{4,3}$ | $e_{1}\bullet e_{2}=e_{1},\;e_{2}\bullet e_{1}=-e_{1}+ae_{4},\;e_{2}\bullet e_{2}=-ae_{3},$ | $\\{(1,3)=(3,1)=(2,4)=(4,2)=1\\}$
| $e_{1}\bullet e_{3}=-e_{4},\;e_{2}\bullet e_{3}=e_{3}$ |
$\mathfrak{g}_{4,4}$ | $e_{4}\bullet e_{4}=-\alpha e_{2},\;e_{4}\bullet e_{2}=\lambda e_{3}+\alpha e_{1},\;e_{4}\bullet e_{3}=-\lambda e_{2}$ | $\\{(1,6)=(2,5)=-(3,4)=1\\}$
$\mathfrak{g}_{4,5}$ | $e_{2}\bullet e_{3}=\lambda e_{1},\;e_{3}\bullet e_{2}=-\lambda e_{1},\;e_{4}\bullet e_{2}=\lambda e_{3},\;e_{4}\bullet e_{3}=-\lambda e_{2},$ | $\\{(1,6)=(2,5)=-(3,4)=1\\}$
| $e_{2}\bullet e_{4}=-\lambda e_{3},\;e_{3}\bullet e_{4}=\lambda e_{2}$ |
$\mathfrak{g}_{4,6}$ | $e_{4}\bullet e_{4}=-\alpha e_{2}-\beta e_{3},\;e_{3}\bullet e_{2}=\gamma e_{1},e_{4}\bullet e_{2}=\alpha e_{1},\;$ | $\\{(1,6)=(2,5)=-(3,4)=1\\}$
| $e_{4}\bullet e_{3}=\beta e_{1},\;e_{3}\bullet e_{4}=-\gamma e_{2},\quad\beta\gamma=0$ |

Table 2: ${\mathrm{L}}$-quadratic non Lie left Leibniz algebras of dimension
$\leq$ 4.

Algebra | The non vanishing products | The metric
---|---|---
${\mathrm{L}}_{1,4}$ | $e_{1}\bullet e_{1}=-\lambda_{1}e_{4},e_{1}\bullet e_{2}=-\lambda_{2}e_{4},e_{2}\bullet e_{1}=\lambda_{1}e_{3},e_{2}\bullet e_{1}=\lambda_{2}e_{3}$ | $\\{(1,3)=(2,4)=1\\}$
${\mathrm{L}}_{1,5}$ | $e_{1}\bullet e_{1}=-\lambda_{1}e_{5},\;e_{1}\bullet e_{2}=-\lambda_{2}e_{5}+\rho e_{3},e_{2}\bullet e_{1}=\lambda_{1}e_{4}-\rho e_{3}$ | $\\{(1,4)=(2,5)=(3,3)=1\\}$
| $e_{2}\bullet e_{2}=\lambda_{2}e_{4},e_{1}\bullet e_{3}=\mu e_{5},e_{2}\bullet e_{3}=-\mu e_{4},e_{3}\bullet e_{1}=\rho e_{5},e_{3}\bullet e_{2}=-\rho e_{4}$ |  $\rho=0$ or $\rho=-\mu$

Table 3: ${\mathrm{R}}$-quadratic non Lie Leibniz algebras of dimension $\leq$
5\. .

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